for
ABOUT THE AUTHOR
Vinay Kumar (VKR) graduated from IIT Delhi in
Mechanical Engineering.
Presently, he trains IIT aspirants at VKR Classes,
Kota, Rajasthan.
for
Second Edition
Vinay Kumar
B.Tech., IIT Delhi
McGraw Hill Education (India) Private Limited
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Integral Calculus for JEE Main & Advanced, 2/e
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PREFACE
T
his book is meant for students who aspire to join the Indian Institute of Technologies (IITs) and
various other engineering institutes through the JEE Main and Advanced examinations. The content
has been devised to cover the syllabi of JEE and other engineering entrance examinations on the topic
Integral Calculus. The book will serve as a text book as well as practice problem book for these competitive examinations.
As a tutor with more than thirteen years of teaching this topic in the coaching institutes of Kota,
I have realised the need for a comprehensive textbook in this subject.
I am grateful to McGraw-Hill Education for providing me an opportunity to translate my years of
teaching experience into a comprehensive textbook on this subject.
This book will help to develop a deep understanding of Integral Calculus through concise theory
and problem solving. The detailed table of contents will enable teachers and students to easily access
their topics of interest.
Each chapter is divided into several segments. Each segment contains theory with illustrative examples. It is followed by Concept Problems and Practice Problems, which will help students assess the
basic concepts. At the end of the theory portion, a collection of Target Problems have been given to
develop mastery over the chapter.
The problems for JEE Advanced have been clearly indicated in each chapter.
The collection of objective type questions will help in a thorough revision of the chapter. The
Review Exercises contain problems of a moderate level while the Target Exercises will assess the students’
ability to solve tougher problems. For teachers, this book could be quite helpful as it provides numerous
problems graded by difficulty level which can be given to students as assignments.
I am thankful to all teachers who have motivated me and have given their valuable recommendations. I thank my family for their whole-hearted support in writing this book. I specially thank Mr. Devendra
Kumar and Mr. S. Suman for their co-operation in bringing this book.
Suggestions for improvement are always welcomed and shall be gratefully acknowledged.
Vinay Kumar
CONTENT
About the Author
ii
Preface
v
CHAPTER 1 INDEFINITE INTEGRATION
1.1 – 1.184
1.1
Introduction
1.1
1.2
Elementary Integrals
1.4
1.3
Integration by Transformation
1.10
1.4
Integration by Substitution
1.16
1.5
Integrals Involving Sine and Cosine
1.27
1.6
Rationalization by Trigonometric Substitution
1.36
1.7
Integrals of the Form
1.40
1.8
Integrals of the Form
1.45
1.9
Integrals of the Form
1.50
1.10
Integration of Trigonometric Functions
1.55
1.11
Integration by Parts
1.65
1.12
Special Integrals
1.73
1.13
Multiple Integration by Parts
1.76
1.14
Integration by Reduction Formulae
1.81
1.15
Integration of Rational Functions Using Partial Fractions
1.88
1.16
Special Methods For Integration of Rational Functions
1.101
1.17
Integration of Irrational Functions
1.106
1.18
Integrals of the Type
1.19
Integration of a Binomial Differential
1.118
1.20
Euler’s Substitution
1.120
1.21
Method of Undetermined Coefficients
1.124
dx
P Q
1.112
viii | Content
1.22
Non-elementary Integrals
1.127
Target Problems for JEE Advanced
1.130
Things to Remember
1.142
Objective Exercises
1.147
Review Exercises for JEE Advanced
1.160
Target Exercises for JEE Advanced
1.162
Previous Year’s Questions (JEE Advanced)
1.164
Answers
1.166
CHAPTER 2 DEFINITE INTEGRATION
2.1 – 2.204
2.1
Introduction
2.1
2.2
Definite Integral as a Limit of Sum
2.5
2.3
Rules of Definite Integration
2.12
2.4
First Fundamental Theorem of Calculus
2.19
2.5
Second Fundamental Theorem of Calculus
2.27
2.6
Integrability
2.41
2.7
Improper Integral
2.52
2.8
Substitution in Definite Integrals
2.63
2.9
Integration by parts for Definite Integrals
2.72
2.10
Reduction Formula
2.78
2.11
Evaluation of Limit of sum using Newton-leibnitz Formula
2.83
2.12
Leibnitz Rule for Differentiation of Integrals
2.91
2.13
Properties of Definite Integral
2.95
2.14
Additional Properties
2.122
2.15
Estimation of Definite Integrals
2.124
2.16
Determination of Function
2.134
2.17
Wallis’ Formula
2.138
2.18
Limit under the sign of Integral
2.143
2.19
Differentiation under the sign of Integral
2.144
2.20
Integration of Infinite Series
2.148
2.21
Approximation of Definite Integrals
2.151
Target Problems for JEE Advanced
2.154
Things to Remember
2.166
Objective Exercises
2.169
Review Exercises for JEE Advanced
2.180
Target Exercises for JEE Advanced
2.184
Content | ix
Previous Year’s Questions (JEE Advanced)
2.188
Answers
2.194
CHAPTER 3 AREA UNDER THE CURVE
3.1 – 3.86
3.1
Curve Sketching
3.1
3.2
Area of a Curvilinear Trapezoid
3.7
3.3
Area Bounded by a Function which Changes Sign
3.10
3.4
Area of a Region Between two Non-intersecting Graphs
3.13
3.5
Area of a Region Between two intersecting Graphs
3.17
3.6
Area by Horizontal Strips
3.21
3.7
Area of a Region Between Several Graphs
3.26
3.8
Determination of Parameters
3.30
3.9
Shifting of Origin
3.35
3.10
Area Bounded by a Closed Curve
3.37
3.11
Areas of Curves given by Parametric Equations
3.42
3.12
Areas of Curves given by Polar Equations
3.44
3.13
Areas of Regions given by Inequalities
3.46
Target Problems for JEE Advanced
3.51
Things to Remember
3.62
Objective Exercises
3.64
Review Exercises for JEE Advanced
3.74
Target Exercises for JEE Advanced
3.76
Previous Year’s Questions (JEE Advanced)
3.78
Answers
3.80
CHAPTER 4 DIFFERENTIAL EQUATIONS
4.1 – 4.99
4.1
Introduction
4.1
4.2
Formation of a Differential Equation
4.3
4.3
Solution of a Differential Equation
4.7
4.4
First Order and First Degree Differential Equations
4.11
4.5
Reducible to Variable Separable
4.17
4.6
Homogeneous Differential Equations
4.23
4.7
Linear Differential Equations
4.30
4.8
Solution by Inspection
4.38
4.9
First Order Higher Degree Differential Equation
4.42
x
| Content
4.10
Higher Order Differential Equation
4.46
4.11
Integral Equation
4.50
4.12
Problems in Trajectories
4.53
4.13
Applications of Differential Equation
4.55
Target Problems for JEE Advanced
4.61
Things to Remember
4.72
Objective Exercises
4.73
Review Exercises for JEE Advanced
4.84
Target Exercises for JEE Advanced
4.86
Previous Year’s Questions (JEE Advanced)
4.89
Answers
4.92
C H A P T E R
1
INDEFINITE
INTEGRATION
1.1 INTRODUCTION
Integral calculus is to find a function of a single variable
when its derivative f(x) and one of its values are
known. The process of determining the function
has two steps. The first is to find a formula that
gives us all the functions that could possibly have
f as a derivative. These functions are the so-called
antiderivatives of f, and the formula that gives them
all is called the indefinite integral of f. The second
step is to use the known function value to select
the particular antiderivative we want from the
indefinite integral.
A physicist who knows the velocity of a particle might
wish to know its position at a given time. Suppose an
engineer who can measure the variable rate at which
water is leaking from a tank wants to know the amount
leaked over a certain time period. In each case, the
problem is to find a function F whose derivative is a
known function f. If such a function F exists, it is called
an antiderivative of f.
There are two distinct ways in which we may approach
the problem of integration. In the first way we regard
integration as the reverse of differentiation; this is the
approach via the indefinite integral. In the second way
we regard integration as the limit of an algebraic
summation; this is the approach via the definite
integral. For the moment we shall consider only the
first and we begin with the formal definition.
Indefinite integration is the process which is the
inverse of differentiation, and the objective can be
stated as follows : given a function y = f(x) of a single
real variable x, there is a definite process whereby we
can find (if it exists) the function F(x) such that
dF(x)
f(x)
dx
Definition
A function F(x) is called the antiderivative (primitive)
of the function f(x) on the interval [a, b] if at all points
of the interval F'(x) = f(x).
Find the antiderivative of the function f(x) = x2.
From the definition of an antiderivative it follows that
the function f(x)=
x3
3 = x2.
x3
is an antiderivative, since
3
It is easy to see that if for the given function f(x) there
exists an antiderivative, then this antiderivative is not
the only one. In the foregoing example, we could take
the following functions as antiderivatives:
x3
x3
– 7 or,
1, F(x) =
3
3
x3
generally, F(x) =
+C
3
(where C is an arbitrary constant), since
F(x) =
1.2
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
x3
2
3 C x
On the other hand, it may be proved that functions of
x3
C exhaust all antiderivatives of the
the form
3
function x2. This is a consequence of the following
theorem. The example shows that a function has
infinitely many antiderivatives. We are going to show
how to find all antiderivatives of a given function,
knowing one of them.
Constant Difference Theorem
If F1(x) and F2(x) are two antiderivatives of a function
f(x) on an interval [a, b], then the difference between
them is a constant.
Proof By virtue of the definition of an anti-
F1 (x) f(x)
derivative we have
F2 (x) f(x)
...(1)
for any value of x on the interval [a, b].
Let us put F1(x) – F2(x) =g(x)
...(2)
Then by (1) we have
F1 ( x ) – F2 ( x ) = f(x) – f(x) = 0
or g(x) = [F1(x) – F2(x)]' = 0
for any value of x on the interval [a, b].
But from g(x) = 0 it follows that g(x) is a constant.
Indeed, let us apply the Lagrange's theorem to the
function g(x), which, obviously, is continuous and
differentiable on the interval [a, b]. No matter what the
point x on the interval [a, b], we have, by virtue of the
Lagrange's theorem, g(x) – g(a) = (x – a) g(c) where
a < c < x.
Since g(c) = 0,
g(x) – g(a) = 0
or, g(x) = g(a)
...(3)
Thus, the function g(x) at any point x of the interval
[a, b] retains the value g(a), and this means that the
function g(x) is constant on [a, b]. Denoting the
constant g(a) by C, we get from (2) and (3),
F1(x) – F2(x) = C.
Thus, if for the function F1 and F2 there exists an interval
[a, b] such that F1'(x) = F2'(x) in [a, b], then there exists a
number C such that F1(x) = F2(x) + C in [a, b].
From this theorem it follows that if for a given function
f(x) some one antiderivative F(x) is found, then any
other antiderivative of f(x) has the form F(x) + C, where
C is an arbitrary constant.
For example, if
F '(x) = 6x2 + 2x, then
F(x) = 2x3 + x2 + C,
for some number C. There are no other antiderivatives
of 6x2 + 2x.
If F '(x) = 4x – 3 and F(1) = 3,
then
F(x) = 2x2 – 3x + C, for some number C.
Since, F(1) = 2 – 3 + C = 3, we have C = 4.
Thus F(x) = 2x2 – 3x + 4,
and there is just one function F satisfying the given
conditions.
The equation F '(x) = 4x – 3 is an example of a differential
equation, and the condition F(1) = 3 is called a
boundary condition of F.
Given F ' and a boundary condition on F, there is a
unique antiderivative F of F ' satisfying the given
boundary condition. This function F is called the
solution of the given differential equation.
Definition
If the function F(x) is an antiderivative of f(x), then the
expression F(x) + C is the indefinite integral of the
function f(x) and is denoted by the symbol f(x) dx. It
is the set of all antiderivatives of f(x). Thus, by
definition
f(x) dx = F(x) + C,
if F(x) = f(x).
Here, the function f(x) is called the integrand, f(x) dx
is the element of integration (the expression under
the integral sign), the variable x the variable of
integration, and is the integral sign.
Thus, an indefinite integral is a family of
functions y = F(x) + C
(one antiderivative for each value of the constant C).
The symbol for the integral was introduced by
Leibnitz. This elongated S stood for a "sum" in his
notation.
The graph of an antiderivative of a function f(x) is
called an integral curve of the function y = f(x). It is
obvious that any integral curve can be obtained by a
translation (parallel displacement) of any other integral
curve in the vertical direction.
From the geometrical point of view, an indefinite integral
is a collection (family) of curves, each of which is obtained
by translating one of the curves parallel to itself upwards
or downwards (that is, along the y-axis).
INDEFINITE INTEGRATION
Existence of Antiderivative
A natural question arises : do antiderivatives (and, hence,
indefinite integrals) exist for every function f(x) ?
The answer is no.
Let us find an antiderivative of a continuous function
x 1
f(x) =
2
3 x
, 0 x 1
, on the interval [0, 2].
, 1 x 2
On integrating both the formulae we get
x2
0 x 1
2 x C1 ,
F(x) =
x3
C2 ,
3x
1 x 2
3
To ensure that F '(1) = f(1) = 2, we first make F(x)
continuous :
F(1–) = F(1+)
1
1
+ 1 + C1 = 3 – + C2
3
2
8
3
C1 +
= + C2
3
2
7
C 1 = + C2
6
7
x2
0 x 1
2 x 6 C2 ,
Now, F(x) =
x3
3x
1 x 2
C2 ,
3
We further observe that F (1+) = F (1–).
Thus, we obtain the antiderivative F(x) of the function f(x).
Let us note, however, without proof, that if a function
f(x) is continuous on an interval [a, b], then this function
has an antiderivative (and, hence, there is also an
indefinite integral).
Now, let us find an antiderivative of a discontinuous
1
function f(x) = 2 .
x
1
1
F(x) = – is an antiderivative of f(x)= 2 on (– , 0)
x
x
and on (0, ). However, it is not an an antiderivative
on [–1, 1] since the interval includes 0 where F(x) does
not exist.
We adopt the convention that when a formula for a
general indefinite integral is given, it is valid only on
an interval. Thus, we write
1
1.3
1
x dx x + C
2
with the understanding that it is valid on the interval
(0, ) or on the interval (– , 0). This is true despite
the fact that the general antiderivative of the function
f(x) = 1/x2, x 0, is
1 C if x 0
1
x
.
F(x) =
1 C if x 0
2
x
Example 1. Prove that y = sgn x does not have an
antiderivative on any interval which contains 0.
1 , x 0
0 , x0
Solution y = sgn (x) =
1 , x0
Here we present two antiderivatives :
x c1 ,
f(x) = x c ,
2
x0
x0
x c , x 0
g(x) =
xc , x0
f(x) is discontinuous at x = 0, if c1 c2. g(x) is continuous
at x = 0. None of these functions are differentiable at
x = 0 i.e. we are unable to ensure that f '(0) or g'(0) is 0.
Hence, y = sgn x does not have an antiderivative on
any interval which contains 0. However, the function
has an antiderivative (either f or g) on any interval
which does not contain 0.
Indefinite Integration
The finding of an antiderivative of a given function
f(x) is called indefinite integration of the function f(x).
Thus, the problem of indefinite integration is to find
the function F(x) whose derivative is the given function
f(x) i.e. given the equation
dF(x)
f (x) , we have to
dx
find the function F(x).
The process of integration is not so simple. Although
rules may be given which cover this operation with
various types of simple functions, indefinite
integration is a tentative process, and indefinite
integrals are found by trial.
This chapter is devoted to working out methods by
means of which we can find antiderivatives (and
indefinite integrals) for certain classes of elementary
functions.
1.4
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Note: Though the derivative of an elementary
function is always an elementary function, the
antiderivative of the elementary function may not
prove to be representable by a finite number of
elementary functions. We shall return to this question
at the end of the chapter.
1.2
ELEMENTARY
INTEGRALS
(ii)
d
d
k f (x) dx k
f (x) dx = kf (x)
dx
dx
A constant moves past the derivative symbol.
(b) We show that the derivative of f(x) + g(x) dx
is f(x) + g(x) :
d
f (x) dx g(x) dx
dx
d
d
f (x) dx
g(x) dx = f (x) + g (x).
dx
dx
(c) The proof of property (c) is similar to that of
property (b).
The last two parts of theorem extend to any finite
number of functions. For instance,
=
d
dx
(a)
Assume that f and g are functions with antiderivatives
f(x) and g(x) dx. Then the following hold :
(a) kf(x) dx = k f(x) dx for any constant k.
(b) (f(x) + g(x)) dx = f(x) dx + g(x) dx,
(c) (f(x) – g(x)) dx = f(x) dx + g(x) dx.
Proof
(a) It is only required to show that the derivative
of k f(x)dx is cf(x). The differentiation shows :
2 x
(f (x) – g(x) h(x)) dx
= f (x) dx– g(x) dx + h(x) dx.
Note:
2 x
x
x
= x2ex – 2xex + 2ex – 2C = (x2 – 2x + 2) ex + C
where C = 2C
Wh en a rbit rary consta nts are algebra ical ly
combined with other numbers, the final algebraic
expression is just as arbitrary.
(iii) Note the following representations :
Rules of integration
x e dx = x e – 2(xe – e + C)
f (x)dx f (x)
(b) d f (x)dx f (x ) dx
(c)
f '(x)dx f(x) C
(d)
df (x) f (x) C
Elementary formulae
We begin by listing a number of standard forms, that
is to say formulae for integrals which we shall be free
to quote once we have listed them. Each formula is of
the the type
f (x)dx F(x)
and its validity can be etablished by showing that
d F(x ) f (x )
dx
These formulae should be known and quoted, without
proof, whenever needed.
(i)
dx = kx + C, where k is a constant
(ii)
n
x dx =
(iii)
x dx = ln |x| + C
x n 1
+ C, where n ¹ 1
n 1
1
ax
C , where a > 0
n a
(i) We have (1 x)dx 1dx xdx = (x + C1) +
(iv) a x dx =
2
x2
C2 = x + x + C + C
1
2
2
2
Here, we have two arbitrary constants when one
will suffice. This kind of problem is caused by
introducing constants of integration too soon and
can be avoided by inserting the constant of
integration in the final result, rather than in
intermediate computations.
e dx = e + C
(vi) sin x dx = – cos x + C
(vii) cos x dx = sin x + C
(viii) sec x dx = tan x + C
(v)
x
x
2
1.5
INDEFINITE INTEGRATION
cosec x dx = – cot x + C
(x) sec x tan x dx = sec x + C
(xi) cosec x cot x dx = – cosec x + C
2
(ix)
(xi)
1
1 x dx = sin x + C
1
–1
We have some additional results which will be
established later :
1
(xiv) tan x dx = ln sec |x| + C
a
1
ln sin|x|+ C
a
(xvi) sec x dx = ln |sec x + tan x| + C
x
+C
4 2
= – ln |sec x – tan x| + C
(xvii) cosec x dx = n |cosec x cot x| + c
x
+c
2
= n |cosec x + cot x|
Here, we must analyse carefully the formula
1
x dx n x C
We have two cases :
(i) Let x > 0, then |x| = x and the formula attains the
1
.
x
(ii) Let x < 0, then |x| = –x and the formula has the form
dx
x = n(–x) + C
1
x dx n x C .
In the formula and examples where integrals of this
type occurs, i.e., where the value of an integral involves
the logarithm of a function and the function may
become negative for some values of the variable of
the function, the absolute value sign enclosing the
function should be given, but it has generally been
omitted, though it is always understood to be present
and it should be supplied by the students.
Direct integration is such a method of computing
integrals in which they are reduced to the elementary
formulae by applying to them the principal properties
of indefinite integrals.
For example :
dx
x 3 ln | x 3 | C
x 2x
x
x
3 .4 dx (3.16) dx 48 dx
4
2
48x
C
n48
4x dx = 6 x + C = 3 x + C.
5
6
6
Evaluate
dx
nx C
x
Differentiating, we get (nx + C)' =
d
1 1
.
n(–x) =
dx
x x
Direct integration
= ln tan
form
1
1
dx ln( x) C .
x
Hence, both these results will be included if we write
2
= n tan
1
d
(n x) = ,
dx
x
Therefore when x < 0,
| x | x 1 dx = sec x + C
(xv) cot x dx =
1
1
x
x
x dx ln x C is defined for x > 0.
When x < 0, i.e. – x > 0,
1
dx = tan–1 x + C
(xii)
1 x2
(xiii)
Since ln x is real when x > 0 and
so
–1
2
Differentiating, we have (n(–x) + C)' =
7
2
x 5x 4 x x dx
3
2
7
2
x 5x 4 x x dx
3
2
= x 3 dx + 5x 2 dx – 4dx +
2
7
dx
dx +
x
x
1.6
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
2
= x 3 dx + 5 . x dx – 4 . 1 . dx
+ 7.
=
1
dx + 2. x 1/ 2 dx
x
Extension of elementary formulae
x1/ 2
x4
x3
=
+5
– 4x + 7 ln | x | + 2 1/ 2 + C
4
3
5
x4
=
+ x3 – 4x + 7 ln | x | + 4 x + C
3
4
Evaluate
2 x 3x
2 x 3x
5x
If
f (x)dx F(x) and a, b are constants, then
1
f (ax b)dx a F(ax b)
1 d
1 d
dy 1
F(y)
F(y). = f(y). a = f(ax + b).
a dx
a dy
dx a
For example:
(ax b) n 1
+ C, n 1
a (n 1)
2
2
3
3
= x x dx = 5 5 dx
5
5
(i)
(2 / 5) x (3 / 5) x
=
+C
ln 2 / 5 ln 3 / 5
(ii)
sin(5x 2)dx 5 cos(5x 2) + C
(iii)
sec (3x 5)dx 3 tan(3x 5) C
x
x
x
x
Evaluate
(i)
5 e dx
(ii)
log x
2 4 dx
log x
(ax + b)n dx =
1
2
1
(iv) sec (ax + b) . tan (ax + b) dx
=
(i)
log x
log 5
5 e dx = x e dx =
(ii)
2 4 dx = 2 2
log 2 x
log x
x loge 5 1
+C
log e 5 1
dx = 2
1/ 2 log 2 x
= 2log2 x dx = x dx =
dx
x 3/ 2
+C
3/ 2
2 3/ 2
x C
3
Evaluate e x ln a ea ln x ea ln a dx
2log 4 x dx
e
...(1)
We prove (1) by observing that, when y = ax + b,
dx
dx
5x
ax
x a 1
+
+ aa. x + C.
ln a a 1
x ln a
e
a ln x
e
a ln a
dx
1
sec (ax + b) + C
a
(v) cos 7x + sin (2x – 6)) dx
1
= 1 sin 7x – cos (2x – 6) + C
2
7
dx
1
ln 2x 1 C
(vi)
2x 1 2
x
1
2x
1
1
dx
dx 1
(vii)
dx
2x 1
2 2x 1
2
2x 1
1
1
x ln 2x 1 + C
=
2
2
In this example, we break the function up into parts
= (a x x a a a ) dx
1
whose integrals we know from the
2x 1
list of elementary integrals.
= a x dx + x a dx + a a dx
(viii)
x
a
a
= eln a eln x eln a dx
like 1 and
dx
dx
1
dx
2
25
4 5 2
25 4x
4 x2
x2
4
2
1.7
INDEFINITE INTEGRATION
From the standard result we obtain,
1 sec –1 x C
,xa
a
a
1
x x 2 a 2 dx = 1 –1 x
– sec a C , x a
a
1 1
x
1
2x
C
. tan 1 C
tan 1
5
5
4
10
5
=
2
2
dx
dx
1
(ix)
a b x a (bx) ab tan
(x)
1 9x
2
2
2
2
dx
2
=
(i)
bx
a +C
1
2
dx
1
9 x2
9
dx
3
1
x2
9
2 / 3
1
dx
1
x
1
sin 1 C sin 1 3x C
1
3 1 2
3
3
x2
3
3
1
dx
x 2 is a 'convenient form' for x 2 dx and such
symbols are commonly used. Strictly, the first
symbol has no meaning save as a shorthand for
the second symbol; as the definition shows, there
can be no question here of 'dividing dx by x2'.
(ii) In the chapter of indefinite integration, the
simplification of square root functions are done
without much consideration to the sign of the
expressions. However, this must be taken
seriously in the chapter of definite integration.
For example,
= (sin x cos x)2 dx
= (sin x + cos x) dx
= – cos x + sin x + C
A more elegant way of handling the situation is
illustrated below :
1 sin 2x dx = (sin x cos x) dx
= sin x cos x dx
2
= sgn (sin x + cos x) · (sinx + cos x) dx
= sgn (sin x + cos x) · {– cos x + sin x} + C.
(iii) We notice that the integral of
|x| x a
1
x x a2
2
.
2 / 3
2
5 2
=
6
3 6
This is a wrong result since the integral of a negative
function must be negative. This happened because
the antiderivative used in the calculation is wrong.
2 / 3
1
dx = – sec–1 x
x x2 1
2
2 / 3
2
5 2
.
6
3
6
=
Alternatively, we have
dx
1
x x a a sec
2
2
1
x
C
a
Thus,
2
dx
2 / 3
x x 1
2
sec 1 | x |
2
2 / 3
.
4 6 12
Determination of function
Let f be a polynomial function such
that for all real x, f(x2 + 1) = x4 + 5x2 + 3 then find the
primitive of f (x) w.r.t. x.
f (x2 + 1) = (x2 + 1)2 + 3x2 + 2
= (x2 + 1)2 + 3(x2 + 1) – 1
We replace x2 + 1 by x
f (x) = x2 + 3x – 1
Now we integrate f(x) w.r.t. x :
2
2
1
x
sec 1 + C. But we would
a
a
like to know the integral of
dx = sec–1 x
f (x)dx (x 3x 1) dx
1
2
1
x x2 1
2
1 sin 2x dx
where a > 0 is
,
x 3 3x 2
xC
3
2
3
Given f ''(x) = cos x, f' = e and
2
f(0) = 1, then find f (x).
f''(x) = cos x
Integrating w.r.t. x :
=
1.8
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
f(x) = sin x + C1
3
Now, f ' = e
2
e=–1+C
C1 = e + 1
f ' (x) = sin x + e + 1
Integrating again w.r.t. x :
f (x) = – cos x + (e + 1)x + C2
f (0) = 1 1 = – 1 + C2 C2 = 2
f (x) = (e + 1)x – cos x + 2
Example 7. A curve y = f (x) such that f ''(x) = 4x
at each point (x, y) on it and crosses the x-axis at
(–2, 0) at an angle of 45°. Find the value of f (1).
Solution f ''(x) = 4x f'(x) = 2x2 + c
f ' (–2) = tan 45° = 1 = 8 + c c = – 7
Now
f ' (x) = 2x2 – 7
2
f (x) = x 7x d
3
16
Since, f (–2) = 0 –
+ 14 + d = 0
3
26
d=–
3
1
f (x) = [2x3 – 21x – 26]
3
45
= – 15.
And, thus
f (1) = –
3
Example 8. Find the antiderivatives of the
function y = x + 2 which touch the curve y = x2.
Solution Since the function y = x + 2 is a derivative
of any of its antiderivatives, it follows, that the
equation for finding the abscissa of the point of
tangency has the form 2x = x + 2.
The root of this equation is x = 2. The value of the
function y = x2 at the point x = 2 is equal to 4.
Consequently, among all the antiderivatives of the
functions y = x + 2,
1
i.e. the function F(x) = x2 + 2x + C,
2
we must find that whose graph passes through the
point P(2, 4). The constant C can be found from the
condition F(2) = 4 :
1
. 4 + 2.2 + C = 4 C = – 2.
2
1
F(x) = x2 + 2x – 2.
2
Example 9. Deduce the expansion for tan–1x from
1
= 1 – x2 + x4 – x5 + ... when x < 1,
1 x2
Solution We have
1
= 1 – x2 + x4 – x5 + .....
1 x2
Integrating both sides w.r.t. x, we have
the formula
dx
x3 x5 x7
...
x
–
–
3
5
7
1 x2
No constant is added since tan–1x vanishes with x.
tan–1x =
A
1.
Find an antiderivative of the function :
(i) f(x) = 1 – 4x + 9x2
(ii) f(x) = x x + x – 5
3.
x,
F(x) = x,
(iii) f(x) x 1
(iv) f(x) = (x/2 – 7)3.
4
2.
1
3
(3x + 4)2 and G(x)= x2 + 4x
6
2
differ by a constant by showing that they
are antiderivatives of the same function.
(b) Find the constant C such that F(x) – G(x) = C by
evaluating F(x) and G(x) at a particular value
of x.
(c) Check your answer in part (b) by simplifying
the expression F(x) – G(x) algebraically.
(a) Show that F(x) =
Let F and G be defined piecewise as
4.
5.
x 0
x 2, x 0
and G(x) = 3
x 0
x , x0
(a) Sh ow t h a t F a n d G h a ve t h e sa m e
derivative.
(b) Show that G(x) F(x) + C for any constant C.
(c) Do parts (a) and (b) violate the constant
difference theorem ? Explain.
(a) Graph some representative integral curves
of the function f(x) = ex/2.
(b) Find an equation for the integral curve that
passes through the point (0, 1).
Prove that the following functions donot have an
antiderivative on any interval which contains 0.
x 1, x 0
x0
(i) y = x,
x 2,
(ii) y = 1,
x0
x0
1.9
INDEFINITE INTEGRATION
6.
7.
8.
9.
Find the function satisfying the given equation
and the boundary conditions.
(i) F'(x) = 3(x + 2)3, F(0) = 0
(ii) s"(t) = 8, s'(0) = 7, s(–1) = –3
(iii) f '(x) = x2 + 5, f(0) = –1.
If f ''(x) = 10 and f ' (1) = 6 and f (1) = 4 then find
f (–1).
Evaluate the following integrals :
e2x 1
(i) 2x . ex dx
(ii)
dx
ex
ax b
px q
dx
dx
(iii) e
(iv) a
Evaluate the following integrals :
e dx
1
(iii) n e dx
n x
(i)
(ii)
e
n x 2
dx
(iv) e m ln x dx
10. Evaluate the following integrals :
x
a .b dx
e e
(iii)
dx
e e
mx
(i)
nx
3x
5x
x
x
(2x + 3x)2 dx
n 2 n x
(iv) e
dx
12. Evaluate the following integrals :
x dx
2x 1
dx
(ii)
(i)
a bx
x 2 dx
2
x
x4
dx
(iv)
dx
(iii)
1 x2
1 x2
13. Evaluate the following integrals :
cos xº dx
(iii) cot x dx
11. Evaluate the following integrals :
dx
n
(ii) (ax + b) dx
(i)
2· x
dx
(1+ x)3
(iii)
(iv)
dx
3 2x
x
2
(ii)
2
14. Evaluate the following integrals :
dx
dx
(i)
(ii)
2
(x 1) x 2 2x
x 25x 2
dx
(iii)
(2x 1) (2x 1)2 4
15. Evaluate the following integrals :
(ii)
sec (ax b) dx
1 cos x
(iv) 1 cos x dx
(i)
(i)
(iii)
x
cos 1
2
dx
(ii)
x
2
sin 1
2
1
x2
3 4
dx
1
3 (2 3x) dx
2
A
16. Integrate the following functions :
1
2 5x 2
3 2x
4
1
(i) 3 sin 3 5 cos 5 1 2x
(3x 1
(ii) (7–4x)3 +
7
2
4cosec2 (4x 3)
(3 7x)
16 9x2
17. Show that
I=
dx
1
x
4sin x 5cos x 41 n | tan 2 2 | C
where = tan–1
5
.
4
18. Prove that
1 x 2m
1
1
1
dx x x 3 x 5 ...
x 2m 1 .
2
3
5
2m
1
1 x
19. Given the continuous periodic function f(x),
x R. Can we assert that the antiderivative of that
function is a periodic function ?
xx
+C
2
21. Find all the antiderivatives of the function
f(x) = 3/x whose graphs touch the curve y = x3.
22. In each case, find a function f satisfying the
given conditions.
(a) f(x2) = 1/x for x > 0, f(1) = 1.
(b) f(sin2 x) = cos2x for all x, f(1) = 1.
(c) f(sin x) = cos2x for all x, f(1) = 1.
20. Show that | x |dx =
1 for 0 x 1,
(d) f(ln x) =
, f(0) = 0.
x 1,
x for
1.10
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
23. Use the following information to graph the
function f over the closed interval [– 2, 5].
(a) The graph of f is made of closed line segments
joined end to end.
(b) The graph starts at the point (– 2, 1).
(c) The derivative of f is shown below:
Y
y= f (x)
1
–2
0
1
3
5
X
–2
24. Is there a function f such that f(0) = – 2, f(1) = 1, and
f(x) = 0 for all x? If so, how many such functions
are there?
25. Find all functions f(x) such that f(x) = 2 sin 3x.
26. A function g, defined for all positive real
numbers, satisfies the following two conditions:
g(1) = 1 and g(x2) = x3 for all x > 0.
Compute g(4).
27. Find a polynomial P of degree 5 with P(0) = 1,
P(1) = 2, P(0) = P(0) = P(1) = P(1) = 0.
28. Find a function f such that f(x) = x + cos x and
such that f(0) = 1 and f (0) = 2.
1.3
INTEGRATION BY
TRANSFORMATION
Standard methods of integration
The different methods of integration all aim at reducing
a given integral to one of the fundamental or known
integrals. There are several methods of integration :
(i) Method of Transformation, i.e., it is useful to
properly transform the integrand and then take
advantage of the basic table of integration
formulae.
(ii) Method of Substitution, i.e., a change of the
independent variable helps in computing a large
number of indefinite integrals.
(iii) Integration by parts
In some cases, when the integrand is a rational
fraction it may be broken into Partial Fractions
by the rules of algebra, and then each part may be
integrated by one of the above methods.
In some cases of irrational functions, the method of
Integration by Rationalization is adopted, which is a
special case of (ii) above.
In some cases, integration by the method of Successive
Reduction is resorted to, which really falls under case
(iv) It may be noted that the classes of integrals which
are reducible to one or other of the fundamental forms
by the above processes are very limited, and that the
large majority of the expressions, under proper
restrictions, can only be integrated by the aid of infinite
series, and in some cases when the integrand involves
expressions under a radical sign containing powers of
x beyond the second, the investigation of such
integrals has necessitated the introduction of higher
classes of transcedental function such as elliptic
functions, etc.
In fact, integration is, on the whole, a more difficult
operation than differentiation. We know that
elementary functions are differentiated according to
definite rules and formulae but integration involves,
so to say, an "individual" approach to every function.
Differential calculus gives general rules for
differentiation, but integral calculus gives no such
corresponding general rules for performing the inverse
operation. In fact, so simple an integral in appearance
as
sin x
x cos xdx, or x dx
cannot be worked out that is, there is no elementary
function whose derivative is x cos x, or (sin x)/x,
though the integrals exist. There is quite a large number
of integrals of these types.
Method of Transformation
In the method of transformation, we find the integrals
by manipulation i.e. by simplifying and converting the
given integrand into standard integrands. It requires
experience to find an appropriate transformation of the
integrand, thus reducing the given integral to a
standard form.
The student must not, however, take for granted that
whenever one or other of the transformations is
applicable, it furnishes the simplest method of
integration. The most suitable transformation in each
case can only be arrived at after considerable practice
and familiarity with the results introduced by such
transformations.
INDEFINITE INTEGRATION
5(x 3) 2
Example 1.
Evaluate I =
Solution I =
5x 2 30x 45
x x
= 5 x dx 30
1/ 2
x x
dx
x
dx
45 x
3 / 2
dx
2
5. x 3 / 2 30.2 x 45( 2x 1/ 2 ) C
3
10 3 / 2
x 60 x 90 x 1 / 2 C .
3
Example 2.
x4
dx
x2 1
x4
x4 1 1
dx
=
x2 1
x 2 1 dx
Solution
=
Evaluate
1
x4 1
+ 2
dx
2
x 1
x 1
= (x 2 1) dx +
1
dx
x 1
2
x3
=
– x + tan –1 x + C
3
Example 3.
Solution
Evaluate
x2 3
dx
x 6 (x 2 1)
(x 2 1) 2
x 6 (x 2 1) dx
(x 1) x
1
= 6 dx + 2 6 2
dx
x (x 1)
x
6
=
1
1
x4 x2 1
dx
+
2
dx – 2 2
dx
6
6
x
x 1
x
=
1
1
1
1
1
dx + 2 2 4 6 dx– 2
dx
6
x
x
x
x
1 x2
=
1
1 1 1
–1
5 +2
x 3x 3 5x 5 – 2 tan x + C.
5x
1
1 x4
Let I = ( 3 / 4
+ sec x tan x dx
1 x
x
I = x -3/4 dx + (1 + x + x 2 + x 3 ) dx + sec x tan x dx ,
[ (1 – xn) / (1 – x) = 1 + x + x2 + x3 + .... + xn – 1]
I = 4x1/4 + x + (x2/2) + (x3/3) + (x4/4) + sec x + C.
1 sin x
dx.
Example 5. Evaluate I =
1 cos x
1 sin x
Solution Here I = 1 cos x dx
1
1
1 2sin x cos x
2
2
dx
=
2 1
2sin x
2
1
1
1
cosec2 x dx + cos x dx
2
2
2
1
1
= –cot x + 2 ln (sin x) + C.
2
2
=
Example 6.
Evaluate I =
Solution
We have
I=
dx
3 cos x sin x
dx
3 cos x sin x
dx
3
1
2
cos x sin x
2
2
1
1
cosec x dx
1
3
2
sin x
3
1
1
= ln tan x ( / 6) + C.
2
2
1 sin
d
Example 7. Evaluate
cos
Solution We break the integrand into two
summands
sin
1 sin
1
cos d = cos cos d
=
2
1
dx
dx + 6 2
6
x
(x
1)
x
6
1 1 x4
(
Evaluate
Example 4.
x3/ 4 1 x + sec x tan x dx.
Solution
dx
1.11
1
2
dx
= (sec – tan ) d
= ln |sec + tan | + ln |cos | + C.
Since ln A + ln B = ln AB, the answer can be simplified
to ln (|sec + tan | |cos |) + C.
But sec cos = 1 and tan cos = sin . The answer
becomes even simpler :
1.12
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1 sin
cos d = ln (1 + sin ) + C.
dx
Evaluate
sec x cos ecx
dx
1 2sin x cos x
sec x cos ecx = 2 sin x cos x dx
=
1 (sin x cos x) 2 1
dx
2
sin x cos x
1
= 1 ( sin x + cos x) dx –
2
2 2
1
sin x
4
=
dx
1
x
1
ln tan + C
( – cos x + sin x) –
2 4
2
2 2
sin 2x
Evaluate sin 5x sin 3x dx
sin 2x
sin(5x 3x)
sin 5x sin 3x dx = sin 5x sin 3x dx
sin 5x cos 3x cos 5x sin 3x
dx
=
sin 5x sin 3x
= (cot 3x – cot 5x) dx
= 1/3 ln |sin 3x| – 1/5 ln |sin 5x| + C.
sin x
Evaluate
dx
cos 3x
2 sin x cos x
sin x
cos3 x dx = 2 cos x cos3 x dx
sin(3x x)
= 2 cos x cos 3x dx
sin 3x cos x cos 3x sin x
=
dx
2 cos x cos 3x
1
= 2 (tan 3x – tan x) dx
=
1
1
ln sec 3x ln sec x + C.
6
2
Integrals of the form
cos ax cos bx dx , sin ax sin bx dx
sin ax cos bx dx , in which a b.
We use the addition formulae to change products to
sums or differences, which can be integrated easily.
For example :
1
cos x cos 2x dx = 2 2 cos x cos 2x dx
1
= (cos 3x cos x) dx
2
1 sin 3x sin x
+C
=
1
2 3
sin mx sin nxdx
1
= cos(m n)xdx cos(m n)xdx
2
sin(m n)x sin(m n)x
2(m n)
2(m n)
=
x
sin
2nx
2
4n
if m 2 n 2
if
m n
1
sinmxcosnxdx 2 sin(m n)xdx sin(m n)xdx
cos(m n)x cos(m n)x
2(m n) 2(m n)
=
cos 2nx
4n
if m 2 n 2
if
m n
1
cos mx cos nxdx 2 cos(m n)xdx cos(m n)xdx
sin(m n)x sin(m n)x
2(m n) 2(m n)
=
x sin 2nx
2
4n
if m 2 n 2
if
m2 n 2
Evaluate sin 8x sin 3x dx
We have sin 8x sin 3x =
and so sin 8x sin 3x dx
=
1
(cos 5x cos 11x) dx
2
=
1
1
sin 5x –
sin 11x + C.
10
22
1
(cos 5x – cos 11x),
2
INDEFINITE INTEGRATION
If a question has one of the functions
like sin2x, cos2x, sin3x or cos3x, then we replace them
by
1 cos x 1 cos 2x 3sin x sin 3x 3cos x cos 3x
,
,
,
2
2
4
4
respectively. The idea is to first express the function in
terms of multiple angles as above and then integrate it.
Also tan2x and cot2x should be replaced by sec2x – 1
and cosec2x – 1 respectively.
For example :
therefore sin6 x =
=
x 1
= sin 2x + C,
2 4
1
sin2x dx = 2 (1 – cos 2x) dx
x 1
= sin 2x + C.
2 4
1
sin3xdx = 4 (3sin x–sin x)dx
3
1
= cos x +
cos3x + C.
4
12
cos3 2x =
(1) becomes
sin 6x =
=
1
(cos 6x + 3 cos 2x),
4
3
1
[1 – 3 cos 2x + (1 + cos 4x)
8
2
1
15
3
1
[10 x – sin 2 x + sin 4 x sin 6 x] C ...(3)
32
2
2
6
If the exponents are not too big, this method works
well.
Evaluate I = sin 3 x cos3 x dx
I=
1
(1 + cos 2x)}2
2
1
3
1
+ cos 2x + cos 4x
8
8
2
3 1
1
+ cos 2x + cos 4x) dx
8 2
8
1
sin 3 2x dx
8
=
1 3sin 2x sin 6x
dx
8
4
=
1
32
=
1
1 3
cos 2 x cos 6 x + C.
6
32 2
(3 sin 2x sin 6x) dx
Evaluate
Evaluate sin6x dx .
Since sin2 x =
1
(1 – cos 2x),
2
1
(2sin x cos x)3 dx
8
=
3
1
1
x + sin 2x +
sin 4x + C.
8
4
32
To evaluate sin6x dx we should first
expressed sin6 x as a sum of sines/cosines of multiples
of x and then integrate each term of the sum to obtain
the result.
...(2)
From (2) we have sin 6 xdx
1
{1 + 2 cos 2x + cos2 2x}
4
1
1
= [1 + 2 cos 2x + (1 + cos 4x)]
4
2
cos4 x dx = (
1
(1 + cos 4x),
2
1
(cos 6x + 3 cos 2x)},
4
1
=
[10 – 15 cos 2x + 6 cos 4x – cos 6x]
32
=
=
...(1)
–
Evaluate cos4x dx .
cos4x = {
1
(1 – cos 2x)3,
8
1
[1 – 3 cos 2x + 3 cos2 2x – cos 3 2x]
8
Writing cos2 2x =
cos2x dx = (1 + cos 2x) dx
1.13
I
dx
sin(x )sin(x )
1
sin[(x ) (x )]
dx
sin( ) sin(x ) sin(x )
=
1
sin(x) cos(x) cos(x)sin(x)
dx
sin()
sin(x )sin(x )
1
[cot(x ) cot(x )]dx
sin( )
1.14
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
(ln sin(x ) ln sin(x ) ] C .
sin( )
1
sin(x )
ln
C .
sin( ) sin(x )
Example 16. Evaluate tan(x –)·tan (x + ) ·tan 2xdx
Solution
tan 2x = tan ( x x )
tan(x ) tan(x – )
1 tan(x )·(x – )
or, tan 2x – tan 2x · tan (x + ) tan (x – )
= tan (x + ) + tan (x + )
tan (x + ) tan (x + ) tan 2x
= tan 2x – tan (x + ) – tan (x – )
=
I = [tan 2x – tan (x + ) – tan (x – )] dx
1
ln |sec 2x| – ln sec |(x + )|
2
– ln sec |(x – )| + C
= ln | sec 2 x · cos (x + ) · cos(x – )| + C.
=
Example 17.
Evaluate
dx
tan x cot x sec x cosec x
Solution
dx
I = tan x cot x sec x cosec x
sin x cos x dx
1 sin x cos x
sin xdx
=
sec x tan x 1
Multiplying and dividing by (1+ tan x – sec x), we get
=
sin x(1 tan x sec x)
= (1 tan x)2 sec2 x dk
=
sin x(1 tan x sec x)
dk
2 tan x
1
cos x (1 + tan x – sec x) dx
=
2
1
(cos x sin x 1) dx
=
2
1
= sin x cos x x + C.
2
Example 18.
Integrate the function
3
5cos x 2sin 3 x
+
2sin 2 x cos2 x
(1 sin 2 x)
1 2 sin x 1 cos 2x
w. r. t. x.
1 cos 2x
cos 2 x
Solution The given function may be written as
+
5cos3 x 2 sin 3 x
(cos2 x sin 2 x 2sin x cos x)
2sin 2 x cos2 x
+
1
2 sin x 2sin 2 x
cos2 x cos2 x 2 cos2 x
5
cosec x cot x + sec x tan x + cos x + sin x
2
+ sec2 x + 2 sec x tan x + 2 (sec2 x – 1)
5
= cosec x cot x + 3 sec x tan x + cos x + sin x
2
+ 3 sec2 x – 2.
Now integrating, we get
5
I = cosec x cot x dx + 3 sec x tan x dx
2
=
+ cos xdx + sin xdx + 3 sec2 x dx – 2 dx
5
= – cosec x + 3 sec x + sin x – cos x
2
+ 3 tan x – 2x + C.
cos 5x cos 4 x
dx .
Example 19. Evaluate
1 2 cos 3x
Solution
cos 5x cos 4 x
1 2 cos 3x dx
sin 3x(cos 5x cos 4x)
sin 3x 2 cos 3x sin 3x dx
3x
3x
9x
x
cos . 2 cos cos
2
2
2
2 dx
sin 3x sin 6 x
3x
3x
9x
x
sin cos cos cos
2
2
2
2 dx
4
9x
3x
2 cos sin
2
2
3x
x
2 cos cos dx (cos 2 x cos x) dx
2
2
sin
2x
sin x C .
2
2 sin
INDEFINITE INTEGRATION
1.15
B
1.
Evaluate the following integrals :
(i)
(1 x)
(iii)
3
dx
x 4 x 5x 2
x 2 2x 1
dx
6.
81 x 41x
cos x dx
(ii)
sin4x dx
(iii)
sin 2 x . cos2 x dx
3
sin 6 x cos 6 x
a sin3 x b cos3 x
8.
sin 2 x cos2 x
Evaluate the following integrals :
cos 2x cos 3x dx
(ii) cos x sin 4 x dx .
(i)
2
(iii)
sin 2x . cos2 x dx
(iv) 4 cos
4.
x
21
· cos x · sin x dx
2
2
Evaluate the following integrals :
(i)
(ii)
1 sin x
cos 4 x sin 4 x
1 cos 4x
2
2
2
2
2
(iv) (cot x cos x ) dx
Evaluate the following integrals :
dx
dx
(ii)
(i)
1 cos x
1 sin x
dx
cos x cos 2x
(iii)
(iv)
dx
1 sin 3x
1 cos x
Evaluate the following integrals :
1 tan 2 x
1 cos 2 x
dx
(ii)
dx
(i)
1 tan 2 x
1 cos 2x
1 tan 2 x
cos 2 x
dx
(iv)
dx
(iii)
1 cot 2 x
cos 2 x sin 2 x
Evaluate the following integrals :
cos 5x cos 4 x
dx
(i)
1 2 cos 3x
3
3
cos x sin x
dx
(ii)
cos x sin x
cos 2x 2 sin 2 x
dx
(iii)
cos 2 x
cos ecx tan 2 x sin 2 x
dx
(iv)
sin x
Evaluate the following integrals :
dx
(iii) sin x sin 2 x sin 3x dx
(iv) sin x cos x cos 2x cos 4x dx
2
7.
(i)
2
2
2
dx
2x
Evaluate the following integrals :
(iv)
3.
x x 1
(iv)
2.
sec x cosec x dx
(ii) cot x cos x dx
(iii) tan x sin x dx
2
2 1 x2
Evaluate the following integrals :
(i)
x 1 x 2 dx
4
(ii)
5.
2
9.
(i)
(3 sin x cos x – sin x) dx
(ii)
sin sin (x ) sin 2 dx
(iii)
sin 8 4 sin 8 4 dx
2
3
2 9
x
2x
2 7
x
(iv) cot 3 2 x cos 4 x dx.
4
B
10. Evaluate the following integrals :
dx
(i)
25 4 x 2
(ii)
dx
x1 x
1.16
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(iii)
x 4 5x 3 15x 9 9
6
4
x4
(iii) 3 x 3x 2
dx
(x 4 x 3x 12) / x 5
x sin x sec x dx
2
2
2
1 x2
sin x
(1 – 3 sin3x) dx
cos 2 x
11. Evaluate the following integrals :
(iv)
(i)
(ii)
x 4 x 4 2
x
3
(iv)
2x 3 2x 3
(iv)
1 x
2
x
2
x
12. Evaluate the following integrals :
(i)
2 x3 3x2 4 x 5
dx
2x 1
(ii)
4 2x 1 x 2 . 4 4x 1 x 2 1 2 x
(iii)
x 6 64
x2
2
x
1 x 1
2
2
x x 1
4x 2 (2x 1)
dx
x 1
dx
x 1
2
1 x 1
dx.
1 x 1/ 1 x
13. Evaluate the following integrals :
(iv)
(i)
(ii)
1.4
dx
6
3
x1/2 x1/2 x3/2 x1/2 x1/2 dx
3
6
dx
x x x x
2
1- x 2 - x2
(ii) (cos x sin x) dx
x
(iii) sin x cos dx
2
x 1 x x dx
6
5cos3 x 3sin 3 x
dx
sin 2 x cos2 x
(i)
2
(iii)
Ú
x + 2 - x2
1- x2
14. Evaluate the following integrals :
dx
dx
3
2
( x 1)(x x )
dx.
x x x x
1 x 2 1
1 x
1 x
dx
x
1 x2 x1 1 x 1 x
INTEGRATION BY
SUBSTITUTION
The method of substitution is one of the basic methods
for calculating indefinite integrals. Even when we
integrate by some other method, we often resort to
substitution in the intermediate states of calculation.
(iv)
dx
3 cos x sin x
15. Evaluate the following integrals :
sin 3 x cos 3 x
(i)
sin 2 x cos2 x dx
(ii)
(iii)
cos x sin x (2 2 sin 2x)dx
sin 2x sin 5x sin 3x
dx
cos x 1 2 sin 2 2x
cos x sin x
cot 2 2x 1
cos8x cot 4x dx
(iv)
2 cot 2x
16. Evaluate the following integrals :
cos x
dx
dx
(ii)
(i)
1 cos x
sin xsin(x )
(iii) {1 cot(x ) cot(x } dx
17. Let f(0) = 0 and f '(x) =
1
for – 1 < x < 1,
(1 x 2 )
show that f(x) + f(a) = f{x (1 a 2 ) a (1 x 2 ) }.
The success of integration depends largely on how
appropriate the substitution is for simplifying the given
integral. Essentially, the study of methods of
integration reduces to finding out what kind of
substitution has to be performed for a given element
of integration.
1.17
INDEFINITE INTEGRATION
Substitution – change of variable
Let I = f(x) dx, and let x = (u).
Then, by definition,
Now,
dI
dx
= f(x) and
= '(u).
dx
du
dI
dI dx
= f '(x) '(u) = f{(u)} '(u).
du dx du
I = f{(u)} '(u) du.
Thus, in the integral f(x)dx, we put x = (u) in the
expression f(x) and also we replace dx by '(u) du and
then we proceed with the integration with u as the
new variable. After evaluating the integral we need to
replace u by the equivalent expression in x.
Note that though from x = (u) we can write
dx
= '(u) in making our substitution in the given
du
integral, we generally use it in the differential form
dx = '(u) du. It really means that when x and u are
connected by the relation x = (u), I being the function
of x whose differential coefficient with respect to x is
f(x), it is, when expressed in terms of u, identical with
the function whose differential coefficient with respect
to u is f{(u)} '(u) which later, by a proper choice of
(u), may possibly be of a standard form, and therefore
easy to find out.
Sometimes it is found convenient to make the
substitution in the form g(x) = u where corresponding
differential form will be g'(x) dx = du.
Let g be a function whose range is an interval I, and
let f be a function that is continuous on I. If g is
differentiable on its domain and F is an antiderivative
of f on I, then
f(g(x))g(x) dx = F(g(x)) + C.
If
(iii) Rewrite the integral in terms of the variable u.
(iv) Evaluate the resulting integral in terms of u.
(v) Replace u by g(x) to obtain an antiderivative in
terms of x.
The main challenge in using substitution is to think
of an appropriate substitution. One should try to
choose u to be some function in the integrand whose
differential also occurs (except for a constant factor).
If that is not possible, try choosing u to be some
complicated part of the integrand. Finding the right
substitution is a bit of an art. It is not unusual to
guess a wrong substitution; if the first guess does
not work, try another substitution.
Find x3 cos (x4 + 2) dx.
We make the substitution u = x4 + 2
because its differential is du = 4x3 dx, which, apart
from the constant factor 4, occurs in the integral. Thus,
using x3 dx = du/4, we have
x3 cos (x4 + 2) dx
1
cos u . 4 du = 4 cos u du
=
1
1
sin u + C = sin (x4 + 2) + C
4
4
Evaluate
(n x)2
dx
x
1
dx = dt
x
I =
dx
t3
t . x = t dt = 3 + C =
2
2
(n x)3
+ C.
3
2
Evaluate (1 sin x) cos x dx
f(u)du = F(u) + C.
(i) Choose a substitution u = g(x). Usually, it is best
to choose the inner part of a composite function,
such as a quantity raised to a power.
(ii) Compute du = g(x) dx.
(n x)2
dx
x
Put nx = t
u = g(x), then du = g(x) and
Steps for substitution
1
=
Put sinx = t cosx dx = dt
(1 t 2 ) dt = t +
= sin x +
t3
+C
3
sin3 x
+C
3
1.18
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
dx
x 4x 5
Evaluate
2
dx
x 4x 5 u
2
2
where a =
Evaluate
du
2
u u a
=
1 1 u
sec
+C
a
a
2
=
1
5
sec 1
dx du / 2 ,
I=
2| x|
+C
5
x dx
x
4 x dx = 4 x
3
3
x dx
2
dt
2 1 x 3/2
I
C.
= sin
3
3
4 t2
2
10x 9 10 x ln10
10x x10 dx
Put 10x + x10 = t
(10 ln10 + 10 x9) dx = dt
x
t dt ln t = ln |10 + x | + C.
x
Evaluate
Put xx = t
x2(1 + ln x) dx = dt
dt
t 1 = ln (t + 1)
= ln(xx + 1) + C.
d(x 2 1)
d(x 2 1)
x 2 = (x 1) 1
2
2
(x 2 1) 1 + C
[considering x2 + 1 as the variable of integration]
=2
x 2 2 + C.
x 2 dx
3
1 d ( 4 3x )
4 3x 3 9 4 3x 3
=
1 n 4 3x 3 C
9
Deduction 1
2
dt
3
1
= 2t + C
=2
Put x3/2 = t
I=
2 t dt
t
I=
x
dx .
4 x3
Here integral of
2
I = 2 x 2 2 + C.
Alternative :
2
x is x 3/2 and 4 – x3 = 4 – (x3/2)2
3
I=
2 xdx
I=
Evaluate
2
Put x2 + 2 = t2 2x dx = 2t dt
2
Evaluate
x 2
x 2
I=
du
=
2
We know d(x2 + 1) = 2x dx
5
Put u 2x, x u / 2,
x 2
d(x 2 1)
I=
u2 a 2
d(x 2 1)
10
x x (1 (ln x)dx
.
( x x 1)
The General Poser Rule for integration
[ f ( x )]n 1
+ C , n –1.
n 1
In other words it means that if we are to integrate any
function of x raised to the power n and multiplied by
the derivative of that function, then we shall apply the
above power formula on that function.
Proof
n
[f (x)] f ' (x)dx
I = [ f ( x)]n f ' ( x )dx
Put f(x) = t f(x) dx = dt
n 1
[f (x )]n 1
+C
n 1
5
2
I = t dt t
n 1
n
For example, tan x sec xdx
tan 6 x
+C
6
Evaluate (x 2 + 1) 2 (2x) dx.
Letting g(x) = x2 + 1, you obtain
INDEFINITE INTEGRATION
g(x) = 2x and f(g(x)) = [g(x)]2.
From this, we can recognize that the integrand follows
the f(g(x)) g(x) pattern. Thus, we write
[g( x )]2
1 2
g(x)
3
(x 2 1)2 (2 x) dx = 3 (x + 1) + C.
cos x
Evaluate I = 7 dx
sin x
Here power formula is applicable on sin x
as its derivative i.e. cosx is present in the numerator.
But you should note the form of power formula which
1
will be 1n dx
,
x
(n 1)x n 1
i.e. when power of x is in the denominator, then
decrease the power by one and multiply the
denominator by the decreased power with sign
changed.
1
+C
Here , I =
6 sin 6 x
4x
dx
Evaluate
(1 2x 2 )2
f '(x) dx = dz.
dz
ln | z | ln | f (x) | C
I
z
Hence, if the integrand be a fraction such that its
numerator is the differential coefficient of the
denominator, then the integral is equal to
ln |(denominator)| + C.
x 2 dx
.
For example, consider I =
7 x3
Here the derivative of the denominator (7 + x3) is 3x2.
1 3x 2 dx
1
= n7 + x3+ C
I=
3 7 x3
3
Also,
2ax b
1
u ( 1)
(1 2x 2 )1 C =
Find
1
2
2x 1
+ C.
x 2 dx
(x 2) .
3
5
Let u be the value in the parenthesis,
3
that is, let u = x – 2. Then du = 3x2 dx. so by substitution
:
x 2 dx
du / 3
(x 2) u
3
=
5
5
1 5
u du
3
1 3
1 u 4
+C=–
(x – 2)–4 + C.
12
3 4
Deduction 2
I=
f '(x)
f (x) dx ln | f (x) | C
We prove this as follows :
Put f(x) = z.
2
2
Find
dx
1 e
x
We write the integrand as follows :
1
ex 1
ex
1 ex ex 1 ex ex 1
–x
–x
u = e + 1, du = –e dx
2
cos x sin x
sin x cos x dx ln | (sin x cos x) | C
ax bx c dx ln | (ax bx c) | C
u
du
4x
2 2
dx (1 2x ) ( 4x)dx
(1 2x 2 )2
1.19
dx
e x dx
du
x
x
u
1 e
e 1
= –ln|u| + C = –ln (e–x + 1) + C
Remember, e–x + 1 > 0 for all x, so
ln |e–x + 1| = ln(e–x + 1).
Integrate 2 sin x dx .
5 3 cos x
I = 2 3 sin x dx
3 5 3 cos x
Now, since the numerator of the integrand is the
differential coefficient of the denominator,
2
I = – ln|(5 + 3 cos x)| + C.
3
With the help of the deduction 2, we can prove some
of the standard results mentioned earlier.
(i) tan x dx = ln|sec x| + C
Proof :
Put cos x = z , then – sin x dx = dz.
I
sin x
dx
dx
ln z+ C
cos x
z
= – ln cos x + C = ln|sec x| + C.
1.20
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Alternative :
sec x tan x
tan x dx
dx ln | sec x | C
sec x
(ii) cot x dx = ln |sin x| + C
Proof:
By substituting sin x = z, this result follows.
Alternative :
cos x
cot x dx
dx ln | sin x | C
sin x
x
(iii) cos ec x dx ln tan + C
2
Proof:
dx
dx
cos ec x dx
1
1
sin x
2 sin x cos x
2
2
1
We multiply the numerator and denominator by sec2 x
2
to get numerator as the differential coefficient of the
denominator.
1 21
sec x
x
2 dx = ln tan + C.
= 2
2
1
tan x
2
Alternative 1 :
cosec x(cos ecx cot x)
cos ec x dx cosec x cot x dx
= ln|(cosec x – cotx)| + C
Alternative 2 :
dx
sin x
cos ec x dx
dx
sin x
sin 2 x
d(cos x)
dz
, where z = cos x
2
1 cos x
1 z2
1 1 z 1 1 cos x
ln
ln
+C
2 1 z 2 1 cos x
It should be noted that the different forms in which
the integral of cosec x is obtained by different methods
can be easily shown to be identical by elementary
trigonometry.
We have,
1
2 sin 2 x
1 1 cos x 1
1
1
2
ln
ln
ln tan 2 x
1
2 1 cos x 2 2 cos 2 x
2
2
2
1
ln tan x .
2
x
(iv) sec x dx = ln tan + C
4
2
= ln |(sec x + tan x)| + C.
Proof:
dx
dx
sec x dx
1
cos x
sin x
2
dx
=
x
x
2sin cos
4 2
4 2
1 2 x
sec dx
2
4 2
=
x
tan
4 2
x
= ln tan + C
4 2
Alternative 1 :
secx(secx tan x)
secxdx secx tan x dx = ln|(sec x + tan x)| + C
since the numerator is the derivative of the
denominator.
While this is the shortest method, it does seem artificial.
The next method may seem a little less contrived.
Alternative 2 :
cos x
cos x
sec x dx
dx =
dx
2
1 sin 2 x
cos x
The substitution z = sin x and dz = cos x dx transforms
this last integral into the integral of a rational function :
dz
1 1
1
1 z 2 1 z 1 z dz
2
=
1
[ln 1 + z – ln1 – z] + C
2
=
1 1 z
ln
+ C.
2 1 z
1 1 z
1 1 sin x
ln
= ln
.
2 1 z
2 1 sin x
The reader may check that this equals ln |sec x + tan x|
Since z = sin x,
by showing that
1 sin x
= (sec x + tan x)2.
1 sin x
INDEFINITE INTEGRATION
Neither method gives, ln |tan(x/2 + /4)|. However, we
can to show that tan(x/2 + /4) = sec x + tan x.
1 x
dx
Evaluate
1 x2
1 x
1
1 2x
1 x 2 dx 1 x 2 dx 2 1 x 2 dx
1
2
= tan–1x + ln(1 x ) + C
2
=
1
ln3 + 2 lnsinx + C
2
x 9
x 3 9x dx
Integrate
x 9 x2 x2
dx
I=
x ( x 2 9)
dx
dx
x
2
dx
= 2
x
x 9
x 9
=
(i)
(ii)
Find
1
x cos (1 ln x) dx
I = dt/cos 2 = sec2 t dt
= tan t = tan (1 + ln x) + C.
1
x(1 ln x) dx .
m
Putting 1 + ln x = t, so that (1/x) dx = dt, we have
I=
x
e x (1 x)
dz
dx
sec 2 z dz
2
x
cos (xe )
cos2 z
= tan z + c = tan(xex) + C.
Now
dx
(ax b) (ax b)
dx
(ax b) (ax b) 1
I
Now, d
dx
(ax b) 1 2 (axa b)
I=
2
a
a
2 (ax b)
[ (ax b) 1]
dx
2
ln (ax b) 1 + C
a
Rationalizing Substitutions
2
Putting 1 + ln x = t, so that (1/x) dx = dt, we have
(ii) Here I =
2
Here integrand contains expression of
the form cos, where = xex is a function of x, therefore,
put z = xex.
Let z = xex, then dz = (ex + xex) dx = ex(1 + x)dx
(i) Here I =
e x (1 x)
cos (xe ) dx .
Evaluate I
1 1 x
1
ln| x | ln(x 2 9) C .
tan
3
3
2
Integrate
1
dx
x cos2 (1 ln x)
1
dx , m 1.
x(1 ln x)m
1.21
dt
t m 1
tm m 1
Some irrational functions can be changed into rational
functions by means of appropriate substitutions. Very
often in integration processes emphasis is mainly laid
upon finding a change of variable which reduces the
given integral to an integral of a rational function;
then the further course of integration becomes clear.
We say that such a change rationalizes the integral.
In particular, when an integrand contains an
expression of the form n g(x) , then the substitution
u = n g(x) may be effective.
(1 ln x) m 1
1
(1 ln x)1m + C.
(1 m)
(1 m)
Evaluate
cot x
cot x
3 2ln sin x dx
1
2cot x
3 2ln sin x dx 2 (3 2ln sin x)dx
Evaluate
dx
.
2 x (x 1)
Put x = t2 dx = 2t dt
dx
2 t dt
I=
= 2 t(t 2 1)
2 x (x 1)
dt
1
x C
= tan–1 t + C tan
1 t2
1.22
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
x4
dx.
x
Let u x 4 . Then u2 = x + 4,
2
so x = u – 4 and dx = 2u du.
(u 8 6u 6 8u 4 )du
Evaluate
u
x4
2u du
dx = 2
u 4
x
Therefore,
= 2
4
u2
du 2 1 2
du
u 4
u 4
2
= 2 x 4 + 2 ln
Simplifying integrals step by step
( 4 sin 2 x)dx = dt
Put sin x = t so that cos x dx = dt.
Then the given integral
(4 t 2 ) dt = (22 t 2 ) dt
1
22
2
2
t (2 t ) sin–1 (t/2) + C
2
2
(4 sin
Let
If we do not know what substitution to make, we
try reducing the integral step by step, using a trial
substitution to simplify the integral a bit and then
another to simplify it some more. You will see what
we mean if you try the sequences of substitutions
in the following integrals.
(i)
x) + 2 sin–1 (1/2 sin x) + C.
Integrate 2 4 x dx
2 4 x dx = u, 2 4 x = u
4 x = u2 – 2
4 x = (u2 – 2)2
x ( u 2 2) 2 4
x = [(u2 – 2)2 – 4]2
dx = 2[(u2 – 2)2 – 4]2 · 2(u2 – 2) · 2u du
= 8u(u2 – 2) (u4 – 4u2) du
= 8(u3 – 2u) (u4 – 4u2) du
= 8(u7 – 4u5 – 2u5 – 8u3 ] du
I = 8 u(u 7 6u 5 8u 3 )du
1
1 cos x dx
2
1
1
x + = t, so that
2
4
1
1
= 2 cos x + C.
4
2
x4 2
+ C.
x4 2
2
We have I = (1 sin x) dx
Now putting
Evaluate
1
= sin x .
2
2 4 x .
1
1
1
1
2 sin 2 x dx 2 sin x dx.
4
2
2
4
u2
1
= 2u + 8 .
2 . 2 ln u 2 + C
=
where u =
Evaluate I = (1 sin x) dx
du
2 du 8 2
u 4
cos x
u 9 6 u 7 8u 5
= 8
C,
7
5
9
(ii)
2
18 tan 2 x sec 2 x
dx
(2 tan 3 x)2
(a) u = tan x, followed by v = u3,
then by w = 2 + v
(b) u = tan3 x, followed by v = 2 + u
(c) u = 2 + tan3 x.
1 sin (x 1) sin (x – 1) cos (x – 1) dx
2
(a) u = x – 1, followed by v = sin u,
then by w = 1 + v2
(b) u = sin(x – 1), followed by v = 1 + u2
(c) u = 1 + sin2(x – 1)
sec x dx
cos(2 x ) cos
sec x dx
I =
2 cos(x ) cos x
1
sec x dx
=
2
(cos x cos sin x sin ) cos x
INDEFINITE INTEGRATION
=
2
cos tan x sin
1
=
f (x) = ex + sin x + x and g (x) = – x
f (x) + g (x) = ex + sin x.
sec2 x dx
1
Evaluate
sec2 x dx
2 sin
cot tan x
Put cot – tan x = t2
– sec2x dx = 2t dt
1
I=
2 sin
2
=–
sin
=C
2 t dt
t
2
2 sin
dt
cot tan x C
2(cot tan x)
sin
sin 3 x dx
(cos4 x 3cos2 x 1) tan 1 (sec x cos x)
3
sin x dx
(cos 4 x 3 cos 2 x 1) tan 1 (sec x cos x)
Put tan–1 (sec x + cos x) = t
1
1 (sec x cos x)
2 (sec x tan x – sin x) dx = dt
sin3 x dx
cos4 x 3cos2 x 1
I =
= dt
dt
= ln | t | + C
t
= ln |tan–1 (sec x + cos x)| + C.
Special forms
cos x sin x 1 x
dx
e x sin x x
= ln |f(x)| + g(x) + C where C is the constant of
integration and f (x) is positive, then find f (x) + g (x).
If
(e x cos x 1) (e x sin x x)
dx
e x sin x x
= ln |ex + sin x + x| – x + C
I=
x
x e (sin x cos x) sin x cos x
dx
(x 2 2e x sin x cos2 x)2
1 2x 2e x (sin x cos x) 2sin x cos x
dx
I=
2
(x 2 2e x sin x cos2 x)2
Put x2 + 2ex sin x – cos2x = t
[2x + 2(ex cos x + ex sin x) + 2 sin x cos x]dx = dt
[2x + 2ex(cos x + sin x) + 2 sin x cos x] dx = dt
I=
Evaluate
I=
1.23
1 dt
1 1
=C– ·
2 t2
2 t
=C–
1
2
x
2(x 2e sin x cos2 x)
.
Evaluate
e2x ex 1
(e x sin x cos x )(e x cos x sin x ) dx
f (x) = ex sin x + cos x
f(x) = ex cos x + sin x ex – sin x
g (x) = ex cos x – sin x
g(x) = cos x · ex – ex sin x – cos x
Now f (x) · g(x)
= (ex sin x + cos x)(cos x · ex – ex sin x – cos x)
= e2x sin x cos x – e2x sin2x – ex sin x cos x
+ ex cos2x – ex sin x cos x – cos2x
...(1)
and g(x) · f(x)
= (ex cos x – sin x)(ex cos x + sin x ex – sin x)
= e2x cos2x + e2x sin x cos x – ex sin x cos x
– ex sin x cos x – ex sin2x + sin2x
...(2)
f (x) · g(x) – g (x) · f(x) = e2x – ex + 1
I=
g(x)
f (x)g '(x) g(x)f '(x)
dx = ln
+C
f (x)g(x)
f(x)
e x cos x sin x
= ln x
+C.
e sin x cos x
1.24
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Taking xn common
Many integrals can be evaluated by taking x
common from some bracketed expression and then
using substitution. Some of the suggested forms
are given below :
dx
n N
(i)
x(x n 1)
Take xn common and put 1 + xn = t.
dx
(ii)
n N
(n 1)
n
x 2 x n 1
Take xn common and put 1+xn = t n.
dx
(iii)
1/n
x n 1 x n
Take xn common and put 1 + xn = t .
dx
Example 31. Evaluate
x(x 4 1)
dx
dx
=
Solution Let I =
5
x(x 4 1)
x 1 14
x
Put 1 14 = t
x
4
dt
dx = dt 15 dx =
4
x5
x
1
1 dt
1
I = = ln |t| + C = 1 ln 1 4 + C.
4 t
4
x
4
dx
Example 32. Evaluate 4 3
x (x 1)2
n
Solution
Let I =
dx
dx
=
2
3
2
1
x (x 1)
x10 1
4
x
3
Put 1 + 13 = t
x
1
– 34 dx = dt 6 = (t – 1)2 14 dx = – dt
x
3
x
x
2
1 t 1 2 t
( t 1)2
dt
dt =
I = 1
2
3
t2
3
t
1 1
t 2 ln | t | C
= 1 1 12 2 dt =
3
t
3
t
t
1
where t = 1 + 3 .
x
Example 33.
Let I =
Solution
=
=
5
2
5x 4 4 x 5
dx
( x 5 x 1)2
1
1
x10 1 4 5
x
x
5 / x6 4 / x5
1
1
1
x4 x5
2
2
dx
5
1
1
4
= t 5 6 dx = dt
x4 x5
x
x
dt 1
I= 2 +C =
t
t
=
5x 4 4 x 5
(x x 1) dx.
x 4 (5 4 x)dx
Put 1 +
Evaluate
1
+C
1
1
1 4 5
x
x
x5
+ C.
x5 x 1
Example 34.
Solution
I=
Evaluate
dx
x 2 (x 1 x 2 )
Let
1
dx
=
3
x 2 (x 1 x 2 )
x 1 12 1
x
2
Put 1 + 12 = t2 3 dx 2t dt
x
x
I=–
1
tdt
t 11
=–
dt = – 1 t 1 dt
1 t
1 t
= – t + ln | t + 1| + C, where t 1
Example 35.
Solution I =
Evaluate
(x 4 1)dx
x 2 (x 4 x 2 1)
4
(x 1)dx
1
x .x x 2 1
x
2
2
1
.
x2
a
x 3 dx
x
1
x2 2 1
x
INDEFINITE INTEGRATION
1
2
Put x 12 t 2 x 3 dx dt
x
x
(x 4 x 2 1
dt (t 1)
1
C =
+ C.
2 (t 1)
x
Example 36. Given that f(0) = f (0) = 0 and
f(x) = sec4 x + sec2 x tan2 x + 4, find f(x).
Solution f(x) = sec2 x + sec2 x tan2 x + 4
Integrating we get
I=
f (x) = tan x +
tan 3 x
+ 4x + C
3
But,f (0) = 0
0 =0+0+0+C
C=0
Then, f (x) = tan x +
f (x) = tan x +
1
2
tan2 x + log |sec x| + 2x2 + D
6
3
But,f(0) = 0
0=0+0+0+D
D= 0
2
1
Then, f(x) = ln |sec x| + tan2 x + 2x2.
3
6
f(x) =
3x 2
Example 37. Let F(x) be the primitive of
x 9
w.r.t. x. If F(10) = 60 then find the value of F(13).
3x 2
Solution F(x) = x 9 dx
Let x – 9 = t2 dx = 2t dt
3( t 2 9) 2
3
tan x
+ 4x
3
1
(sec2 x 1) tan x + 4x
3
1
2
tan x sec2 x + tan x + 4x
3
3
Integrating both sides w.r.t. x we get
f (x) =
1.25
F(x) =
t
· 2 t dt
= 2 (29 3t ) dt = 2 [29t + t3]
2
F(x) = 2 [29 x 9 (x 9)3/2 ] + C
Given F(10) = 60 = 2 [29 + 1] + C C = 0
F(x) = 2 [29 x 9 (x 9)3 2 ]
F(13) = 2 [29 × 2 + 4 × 2]
= 4 × 33 = 132.
C
1.
(a) Evaluate (5x 1) dx by two methods :
2.
first square and integrate, then let u = 5x – 1.
(b) Explain why the two apparently different
answers obtained in part (a) are really
equivalent.
Avni (using the substitution u = cos ) claims that
(ii)
2
6.
2 cos sin d = – cos2, while Meet (using the
3.
4.
5.
substitution u = sin ) claims that the answer is
sin2. Who is right?
2
Integrate 2x 2 by the substitutions
(x 1)
(a) x = tan ,
(b) u = x2 + 1,
and verify the argeement of the results.
dx
dx
Evaluate (i)
(ii)
x ln x
x ln x ln ln x
Evaluate the following integrals :
tan x sec 2 x
dx
(i)
x
1
2 x tan x .sec x dx
(iii) cosx cos(sinx)dx
(iv) (x + 1) cos(x + 3x + 2)dx
2
4
2
3
Evaluate the following integrals :
(i)
(ii)
(iii)
x2
dx
1 x
x
dx
1 x3
x 2 3x 1
dx
(x 1)2
9x
12x 1/3
(iv) (27e e ) dx
7.
Evaluate the following integrals :
ex 1
(ii) x
(i) dx
dx
e 1
ex 1
log x 2
(iii) 1 12 e x 1 / x dx (iv) x e ·log ex (x)dx
x
1.26
8.
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Evaluate the following integrals :
(i)
sin x cos x dx
3 sin x cos x
x
x
x
(ii) e tan(e )sec(e ) dx
(iii)
( 3 sin x cos x – sin x) dx
(iv)
log(tan x) dx
2
3
sec x cosec x
C
9.
Evaluate the following integrals :
(i)
(ii)
(iii)
(3x 2) 2x 1 dx
x2 1
3
( x 3 3x 6 )
1
dx
x (4 3 x ) 2
dx
x
5x
(iv) 55 55 5 x dx
10. Evaluate the following integrals :
cos 2x
(i)
(cos x sin x)2
dx
(ii)
2
xsin (1 log x)
dx
(iii)
2
(1 x ) (tan 1 x 3)
(x 2 1)dx
(iv) (x 4 3x 2 1) tan 1 x 1 / x
11. Evaluate the following integrals :
tan x dx
(i)
a b tan 2 x
cot x
(ii) ln sin x dx
sec x dx
(iii)
ln(sec x tan x)
a cos x b sin x
dx
(iv)
a sin x b cos x c
12. Evaluate the following integrals :
(i)
(tan x – x tan x)dx
3
2
(iii)
(iv)
(1 x2 ) sin(tan 1 x)
e3x
dx
(1 e x )3
(i)
(ii)
(tan x)
sec 4 x dx
2xsin 1 x 2
dx
(1 x 4 )
(iv) sin 2x cos
(iii)
11/2
x(1 cos 5/2 x)1/2 dx
14. Evaluate the following integrals :
(i)
(ii)
(x 1)(x log x)2
dx
2x
x dx
(2x 1)3
cos2 x
dx
2 sin x
x x e x
(iv) ln x dx
e x
15. Evaluate the following integrals :
(iii)
(x x2 )
dx
x3
(i)
(ii)
x (1 x )
(iii)
x (x 1 x )
tan x sec x
(a 2 b 2 tan 2 x)2 dx
cos(tan 1 x)dx
13. Evaluate the following integrals :
2
(ii)
tan x dx
logsec x
(iv)
dx
2
4 3/ 4
dx
2
x
2
(x 4 1)
2
x4 x2 1
dx
1.27
INDEFINITE INTEGRATION
16. Evaluate the following integrals :
dx
x (1 x )
(i)
2
(ii)
5 4/5
x2 1
x (1 x ) dx
4
17. Show that the integral
(( x 1)( x 1))
2
2 / 3
dx
1.5 INTEGRALS INVOLVING
SINE AND COSINE
Positive integral powers of sine and cosine
1. Odd positive index
Any odd positive power of sines and cosines can be
integrated immediately by substituting cos x = z and
sin x = z respectively as shown below.
Example 1.
Solution
Evaluate sin 3 x dx
I = sin 2 x sin x dx
= – (1 – z2) dz, putting z = cos x
1
1
z z 3 C cos x cos3 x C
3
3
5
Example 2. Evaluate cos x dx
Solution
I = cos 4 x cos x dx
= (1 – z2)2 dx, putting z = sin x
= (1 – 2z2 + z4)dz = z –
2 3 1 5
z + z +C
3
5
2 3
1
sin x + sin5x + C.
3
5
3
Example 3. Integrate cos 4x dx.
= sin x –
Solution
=
cos 4x dx. = dx cos 4x cos 4x dx
3
2
(1 – sin24x)cos 4x dx
= cos 4x dx – sin 2 4x cos 4x dx
1
1
= sin 4x –
sin 3 4x + C.
4
12
can be evaluated with any of the following
substitutions.
(a) u = 1/(x + 1)
(b) u = ((x – 1)/(x + 1))k
for k = 1, 1/2. 1/3, –1/3, –2/3, and –1
(c) u = tan–1 x
(d) u = tan–1 x
(e) u = cos–1 x
angles by means of trigonometry and then integrate it.
The simplest examples are the two integrals
cos x dx and sin x dx,
2
which can be integrated by means of the identities
1
cos2x = (1 + cos 2x),
2
1
sin2x = (1 – cos 2x)
2
1
We get cos 2 x dx = (1 + cos 2x) dx
2
x 1
= sin 2x + C,
2 4
1
2
sin x dx = 2 (1 – cos 2x) dx
x 1
= sin 2x + C.
2 4
Going on to the higher powers, consider the integral
cos x dx,
2n
where n is an arbitrary positive integer. We write
n
1
cos2n x = (cos2x)n = (1 cos 2x)
2
We expand using binomial theorem and integrate the
terms using previous methods.
Similarly,
n
sin2n x = (sin2x)n =
Example 4.
Solution
2. Even positive index
In order to integrate any even positive power of sine
and cosine, we should first express it in terms of multiple
2
=
1
4
1
2 (1 cos 2x) .
Evaluate sin 4 x dx
sin4x dx = 12 (1 – cos 2x)2 dx
2
(1 – 2 cos 2x + cos22x) dx
1.28
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
=
1 x – sin 2 x 1 (1 cos 4 x )dx
4
2
=
1 3 x – sin 2 x sin 4 x C
.
4 2
8
Note: It should be noted that when the index
is large, it would be more convenient to express the
powers of sines or cosines of angles in terms of
multiple angles by the use of De Moivre's theorem, as
shown below.
Let z = cos x + i sin x
...(1)
1
= cos x – i sin x
...(2)
z
From (1) and (2),
1
2 cos x = z +
...(3)
z
1
...(4)
2 i sin x = z –
z
From De Moivre's Theorem
n = cos nx + i sin nx
...(5)
1 = cos nx – i sin nx
...(6)
zn
From (5) and (6),
...(7)
zn + 1n = 2 cos nx
z
...(8)
zn – 1n = 2 i sin nx
z
z
Example 5.
Solution
1 1
4
[ sin8x – sin 6x+7sin4x – 28sin 2x+35x] + C.
3
27 8
Example 6. Evaluate cos8 x dx.
=
Solution Using (3), we have
28 cos8 x = z8 + 8C1 z6 + 8C2 z4 + 8C3 z2 + 8C4
1
1
1
1
+ 8C5 2 + 8C6 4 + 8C7 6 8
z
z
z
z
8 1 6 1
4 1
2 1
= z 8 8 z 6 28 z 4 56 z 2 +70
z
z
z
z
= cos 8x + 8.2 cos 6x + 28.2 cos 4x + 56.2 cos 2x + 70
[ using (7) ]
1
cos8 x dx =
(cos 8x + 8 cos 6x
128
+ 28 cos 4x + 56 cos 2x + 35) dx
1 sin8x sin 6x
sin 4x
sin 2x
=
8
28.
56.
35x +C
128 9
6
4
2
Integral powers of tangent and cotangent
Any integral power of tangent and cotangent can be
readily integrated. Thus,
(i)
(ii)
Using (4), we have
8
sin x dx
= 2–7 (cos 8x–8cos 6x + 28cos 4x–56 cos 2x + 35)dx
=
1 2
tan x – n |secx| + C.
2
cot4x dx = cot2x (cosec2x – 1)dx
= cot2xcosec2xdx – cot2x dx
= – cot2xd(cotx) – (cosec2x – 1)dx
Positive integral powers of secant and
cosecant
1. Even positive index
Even positive powers of secant or cosecant admit of
immediate integration in terms of tan x or cot x. Thus,
(i) sec4x dx = (1 + tan2x)sec2xdx
8
1 sin8x 8sin 6x
sin4x
sin 2x
+28
+ 35x + C
56
7 8
6
4
2
2
2
1
= cot3x + cotx + x + C.
3
28 i8 sin8 x z
– 56.2 cos 2x + 70 [using (7)]
sin8x = 2–7(cos 8x – 8 cos 6x + 28 cos 4x
– 56 cos 2x + 35)
3
= tanx d(tanx) – tanxdx =
Evaluate sin 8 x dx.
1
z
1
1
z8 8 8 z 6 6
z
z
1
1
28 z 4 4 56 z 2 2 70
z
z
tan x dx tanx . tan x dx
= tanx(sec2x–1)dx
= sec2 x dx + tan2x d(tan x)
= tan x +
(ii)
1 3
tan x + C.
3
cosec6 x dx = cosec4 x. cosec2 x dx
INDEFINITE INTEGRATION
= (1 + cot2x)2 cosec2 x dx
= – cosec x cot x – cosec x cot2 x dx
= – (1 + 2 cot2 x + cot4 x)d (cot x)
= – cosec x cot x – cosec x (cosec2x – 1) dx
= – cot x –
= – cosec x cot x + cosec x dx – cosec3 x dx
2
1
cot3x – cot5x + C
3
5
Transposing cosec3x dx and writing the value of
2. Odd positive index
Odd positive powers of secant and cosecant are to be
integrated by the application of the rule of integration
by parts (to be dealt later).
(iii)
1.29
sec3 x dx = sec x . sec2x dx
1
cosec x dx cosec3 x dx = – 2 cosec x cot x +
1
x
ln tan + C.
2
2
Example 7.
= sec x tan x – sec x tan2 x dx
Find tan6x dx.
= sec x tan x – sec x(sec2x – 1)dx
Solution Here power of tanx is even positive
integer therefore change tan2x into sec2x – 1 and then
put z = tanx.
= sec x tan x + sec x dx – sec3 x dx.
Now tan6x dx = (tan2x)3 dx = (sec2x – 1)2dx
Transposing sec3 x dx to the left side, writing
= (sec6x – 3sec4x + 3sec2x – 1)dx
the value of sec x dx and dividing by 2, we get
= sec6x dx – 3 sec4x dx + 3 sec2x dx – dx
1
1
x
= sec6x dx – 3 sec4x dx + 3tanx – x
sec3 x dx = 2 sec x tan x + 2 ln tan 4 2 + C.
Let z = tanx then dz = sec2x dx
(iv) sec5 x dx = sec3 x sec2 x dx
Now sec6x dx – 3 sec4x dx
= sec3 x tan x – 3 sec3 x tan2 x dx
= sec3 x tan x – 3 sec3 x (sec2 x – 1)dx
= sec3 x tan x + 3 sec3c x dx – sec5 x dx
Now, transposing 3
sec5 x dx and writing the value of
sec3 x dx we get ultimately
sec x dx
5
3
tan x sec x
4
3 tan x sec x 3 1
ln tan + C.
4
2
42
4 2
(v)
cosec3 x dx =
cosec x cosec2 x dx
...(1)
= sec4x sec2x dx – 3 sec2x sec2x dx
= (1 + tan2x)2 sec2x dx – 3 (1 + tan2x)sec2x dx
= (1 + z2)2 dz – 3 (1 + z2)dz
= (1 + 2z2 + z4)dz – 3 (1 + z2)dz
z2.
z8 z 5
z3
3z 3.
3
5
3
z5 z3
tan 5 x tan 3 x
2z
2 tan x
5
3
5
3
Putting in (1) we get
tan6x dx=
tan 5 x tan3 x
– 2tanx + 3tanx – x+ C
5
3
tan 5 x tan 3 x
tan x x C .
5
3
1.30
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Integrals Involving Sine and Cosine
Together
1.
sinm x cosn xdx
If the power of the sine is odd and positive,
save one sine factor and convert the remaining
factors to cosine. Then expand and integrate.
Convert tocos ine
Save for du
Odd
2 k 1
n
sin
x cos xdx = (sin 2 x)k cosn x sin xdx
= (1 cos x) cosnx sin x dx
2
k
Here, we put cos x = t.
sin 3 x
dx
cos2 x
Put cos x = t – sin x dx = dt
Example 8.
Solution
3
Evaluate 3
2
3
4
(1 t )
sin x
dx = – 2 / 3 dt = cos2 x dx
2
sin x
t
cos x
=
(1 sin 2 x )2
dx = (cosec2x – 2 + sin2x) dx
sin 2 x
1 cos2x
= cosec 2 x 2
dx .
2
= – cotx –(3/2)x – (1/4)sin 2x + C.
If the power of the cosine is odd and positive,
sa ve on e cosin e factor a nd convert t he
remaining factors to sine. Then, expand and
integrate.
Odd
m
2 k 1
sin
x
cos
xdx
2.
Convert to sin e
for du
Save
= sin m x (cos 2 x) k cos xdx
= sin x(1 – sin2x)k cos x dx
m
Here, we put sin x = t.
Example 9.
Solution
Evaluate
cos 3 x
sin 4 x dx
(1 – sin 2 x ) cos xdx
=
sin 4 x
3
cos x
dx
sin 4 x
cos 2 x cos xdx
sin 4 x
Denoting sin x = t, cos x dx = dt, we get
(1 – t 2 )dt
t4
dt
dt
1 1
= 4 2 – 3 C
t
t
t
3t
= – 13 1 C .
3 sin x sin x
3. If the powers of both the sine and cosine are
even and nonnegative, make repeated use of
1 cos 2 x
t h e i den ti t i es si n 2 x =
and
2
1 cos 2 x
cos2 x =
.
2
Putting them into the integral we get
cos 3 x
sin 4 x dx
sin2p x cos2q x dx
p
q
= 1 – 1 cos 2 x 1 1 cos 2 x dx
2 2
2 2
Powering and opening brackets, we get terms
containing cos 2x to odd and even powers. The terms
with odd powers are integrated as indicated in Case 1.
We again reduce the even exponents by formulae given
above. Continuing in this manner we arrive at terms of
the form cos kx dx, which can easily be integrated.
4. If both exponents are even, and atleast one of
them is negative, then the preceding technique
does not give the desired result. If in the expression
sinmx cosnx, m + n is a negative even integer, then
one should make the substitution tan x = t (or
cot x = t).
1 , sin =
x ,
We have cos =
1 x2
1 x2
dx ,
and d=
1 x2
The integraltransforms into
x mdx
mn
1
(1 x 2 ) 2
Hence, if m + n = – 2r, this becomes
xm(1 + x2)r–1dx,
a form which is immediately integrable.
Take, for example,
sin 2 d .
cos 6
1.31
INDEFINITE INTEGRATION
Let x = tan , and we get
tan 3 tan 5
C .
x 2 (1 x 2 )dx
3
5
d
Next, to find
sin cos5
making the same substitution, we obtain
(1 x 2 ) 2 dx
.
x
Hence, the value of the proposed integral is
tan 4
tan 2 ln tan C
4
d
Again, to find 3 / 2
sin cos 5 / 2
I
3
2
2
tan 2 –
+ C.
3
tan1/2
In many cases it is more convenient to assume
x = cot .
For example, to find d4 .
sin
d
, if cot = x, the transformed
Since d(cot ) = –
sin 2
3
cot
+ C.
3
tan x
dx
Example 10. Evaluate
sin 2x
Solution Here m = –1/2, n = –3/2,
so that m + n = – 2, a negative integer.
Hence, we put tan x = t, sec2 x dx = dt
tan x
sin 2x dx = 2 sin x cos x cos2 x dx
cos 2 x
2
t
1
dt =
dt = t + C = tan x + C.
2 t
2t
If in the expression sinmx cosnx, m + n is a negative
odd integer, then one should multiply the
integrand by suitable power of (sin2x +cos2x) and
expand it into simpler integrals.
2
cos2 x
sin 2 x
2
2
sin x cos x dx sin x cos x dx
ln tan
sin x
cos x dx
x
dz
2
2
z
2
x
cosec x dx ln tan 2
[z = cosx]
x
1
x
+ + c = ln tan + sec x + c.
2
z
2
Consider some more illustrations.
= ln tan
Example 12.
Evaluate sin4x cos2x dx.
Solution Here power of neither cosx nor sin x is
odd and positive and sum of their powers is even and
positive.
Let z = cos x + i sin x, then cos x
and sin x
Also zn +
1
1
z
2
z
1
1
z
2i
z
1
1
n
n = 2cos nx and z – n = 2i sin nx
z
z
4
2
4
1
1
1 1
Now, sin4x cos2x = z z
z
z
2i 2
2
2
2
1
1
1
1
z z z
64
z
z
z
1
1
1
z z2 2
64
z
z
1 2
1
1
z 2 2 z4 2 4
64
z
z
=
5.
cos2 x sin 2 x
dx
sin x cos x sin x cos x dx
tan x
Solution Here sum of powers of sin x and cos x is
odd and negative, therefore multiply by suitable power
of (cos2x + sin2x).
cos ecx dx
(1 x 2 )dx
Here the transformed expression is
, and
x 3/2
accordingly the value of the proposed integral is
integral is – (1 + x2) dx = – cot –
dx
Evaluate sin x cos 2 x
Example 11.
2
2
2
2
1.32
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1 6
1
2
4
z 2z 2 2z
64
z
2
2 1
4 4 z2 2 6
z
z z
1 6
1 4 1
z 6 2z 4
64
z
z
1
1
z2 2 z2 4
2
z
z2
1
=
(2cos 6x – 2cos 4x – 2cos 2x + 4)
64
1
=
(cos 6x – cos 4x – cos 2x + 2)
32
sin x cos xdx
4
2
Example 15.
Integrate
sin
1/ 2
dx
.
x cos 7 / 2 x
1 7
= – 4.
2 2
Put tan x = z, then sec2 x dx = dz.
Solution
Here, p + q = –
Now, I
sec 4 x dx
1 z2
dz
tan1 / 2 x
z1 / 2
= (z–1/2 + z3/2)dz = 2 z1/2 +
= 2 tan1/2 x +
Example 16.
2 5/2
z +C
5
2
tan5/2 x + C
5
Evaluate sin 5 x cos4x dx.
It is the exponent of the sine which is
an odd positive integer. Hence
Solution
1 sin 6x sin 4x sin 2x
2x C .
32 6
4
2
= z2(1 – z2)2 dz, Putting z = sin x
sin5x cos4 x dx = (sin2x)2 cos4x sin x dx
= (1 – cos2x)2 cos4x sin x dx
= (1 – 2 cos2x + cos4x) cos4x sin x dx
= cos4x sin x dx – 2 cos6x sin x dx
+ cos8 x sin x dx
= (z2 – 2z4 + z6) dz
=
Example 13.
Solution
Integrate sin2x cos5x dx.
I = sin2x cos4x cos x dx
= sin2x(1 – sin2x)2d (sin x)
1
2
1
cos5x + cos7x – cos9x + C.
5
7
9
1 3 2 1 7
z z z +C
3
5
7
Example 17.
1 3
2
1
sin x sin x sin 7 x + C.
3
5
7
Solution
Example 14.
Integrate
sin 2 x
dx .
cos 6 x
Here p + q = 2 – 6 = – 4.
Put tan x = z, then sec2x dx = dz.
Solution
Now, I = tan2x . sec4 x dx
= z2(1 + z2) dz =
=
1 3 1 5
z + z +C
3
5
1
1
tan3x + tan5x + C.
3
5
Evaluate
sin 2 x
cos6 x dx
sin 2 x
dx
cos6 x
sin 2 x(sin2 x cos2 x)2
dx
cos6 x
= tan2x (1 + tan2x)2 dx
Put tan x = t, then x = tan–1t, dx =
sin 2 x
2
2 2
dt
and we get
1 t2
dt
cos 6 x dx t (1 t ) 1 t 2
3
5
= t2 (1 + t2) dt = t t C
3 5
=
tan 3 x tan 5 x C
3
5
1.33
INDEFINITE INTEGRATION
Evaluate
Example 18.
Let I =
Solution
cos5 x
sin x dx.
cos5 x
cos4 x
2
2
sin x dx = sin x cos x dx
(1 sin 2 x)2
cos x dx.
sin 2 x
Put sin x = t so that cos x dx = dt.
=
Then I =
=
2 2
I = sin3x cos22x dx
2
4
(1 t )
1 2t t
dt =
dt
2
t
t2
1
2
t2
1
t 2 t dt = – t 2t 3
2
=–
1
(7sin x – 5 sin 3x + 3 sin 5x – sin 7x) dx
16
1
= – 7 cosx + 5 cos3x – 3 cos5x +
cos7x+ C
112
80
18
16
The integral may of course be obtained in different
forms by different methods. For example
I=
2
1
sin3 x
– sin x +
sin x
3
1 3
sin x + C.
3
dx
.
Example 19. Evaluate
3
sin x cos5 x
Solution Here the integrand is sin–3 x cos –5 x. It
is of the type sinm x cosn x, where
m + n = –3 – 5 = –8
i.e. a negative even integer.
= (4 cos4x – 4 cos2x + 1) (1 – cos2x) sin x dx,
which reduces, on making the substitution cos x = t, to
(4t6 – 8t4 + 5t2 – 1) dt
8
5
4
cos7x – cos5x + cos3x – cos x.
7
5
3
It may be verified that this expression and that
obtained above differ only be a constant.
=
= cosec x – 2 sin x +
I
dx
(sin x / cos x)cos x.cos x
3
3
=
sec xdx
tan x
3
3
3
6
5
2
sec x.sec xdx
tan 3 x
(1 tan 2 x)3 sec 2 xdx
=
tan 3 x
Now put tan x = t so that sec2 x dx = dt.
I=
(1 t 2 )3 dt
1 3
3 3t t 2 dt
t
t3
t
= – {1/2t2)} + 3 ln|t| +
3
1
tan2 x + tan4 x + C.
4
2
Example 20. Evaluate I = sin3x cos22x.
Solution Here we use the formulae
sin3x = 1/4 (3 sin x – sin3x),
cos22x = 1/2 (1 + cos 4x),
Multiplying these two expressions and replacing
sin x cos 4x, for example by 1/2 (sin 5x – sin 3x), we
obtain
Evaluate
Example 21.
Solution
cos 3 x
dx.
sin11 x
I = cos3/2 x · sin–11/2x dx.
Here 3 11 – 4 negative even integer..
2
2
So, we put tan x = z, then sec2x dx = dz
I = cos3/2 x · sin–11/2x dz2
sec x
= cos7/2 x · sin–11/2x dz
=
cos 7 / 2 x
· cosec2x dz
sin 7 / 2 x
= cot7/2 x · cosec2x dz
–7
–7
– 11
= z 2 · 1 12 dz z 2 z 2 dz
z
–
5
–
9
z 2 z 2
C.
=
5
9
–
–
2
2
=
2
2
cot5/2 x +
cot9/2 x + C.
5
9
Example 22.
Solution
Evaluate
dx
(cos x sin x) .
3
5
Here the integrand is of the type cosm x
sinn x. We have m = –3/2, n = –5/2, m + n = – 4 i.e., and
1.34
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
even negative integer.
dx
3/2
3
5
cos xsin 5/2 x
(cos xsin x)
=
dx
cos x(sin
3/2
=
5/2
dx
x / cos5/2 x).cos 5/2 x
Integrals Involving Secants and Tangents
sec x tan xdx
m
n
If the power of the secant is even and positive,
save a secant-squared factor and convert the
remaining factors to tangents. Then, expand
and integrate.
sec
2k
Convert to tan gents
x tan xdx =
n
Save for du
2
k 1
n
(sec x)
tan xsec 2 xdx
2
k 1
n
2
= (1 tan x) tan x sec x dx
Here, we put tan x = t.
2. If the power of the tangent is odd and positive,
save a secant-tangent factor and convert the
remaining factors to secants. Then, expand and
integrate.
Odd
m
2k 1
sec
x
tan
xdx
Convert to secants
Save for du
= sec m 1 x (tan 2 x)k sec x tan x dx
= sec
m 1
= tan
n 2
x(sec2 x – 1)dx
m
If the integral is of the form sec xdx , where
m is odd and positive, use integration by parts.
5. If none of the first four cases applies, try
converting to sines and cosines.
A similar strategy is adopted for
4.
Even
dx
sec x
dx
5/2
cos x tan x
tan 5/2 x
4
(1 tan 2 x)
(1 t 2 )
2 x dx =
sec
dt,
tan 5/2 x
t 5/2
putting tan x = t and sec2 x dx = dt
2
=
(t–5/2 + t–1/2) dt = – t–3/2 + 2t1/2 + C
3
2
= – (tan x)–3/2 + (tan x)1/2 + C
3
2
= 2 (tan x) – (tan x)–3/2 + C.
3
1.
Convert to secants
n
tan xdx = tan n2 x (tan2 x) dx
4
sec 2 x
sec2 x dx
tan 5/2 x
=
Here, we put sec x = t.
3. If there are no secant factors and the power of
the tangent is even and positive, convert a
tangent-squared factor to a secant-squared
factor; then expand and repeat if necessary.
x(sec2 x – 1)k sec x tan x dx
cosec x cot xdx .
m
n
Example 23.
Find tan sec4d.
Here power of sec is an even positive
integer, therefore, we put z = tan
Let z = tan, then dz = sec2d
Solution
Now tan sec4d= tan sec2d
= tan (1 + tan2) sec2d
= z(1 + z2) dz
= (z + z5) dz =
z2 z4
C.
2
4
Example 24. Evaluate tan3xdx.
Solution Here power of tanx is an odd positive
integer therefore, we put z = sec x.
Let z = sec x, then dz = sec x tan x dx
tan 2 x sec x tan x dx
Now tan3xdx =
sec x
2
2
sec x 1
=
sec x tan x dx = z 1 dz
sec x
z
2
1
z
= z dz =
– ln | z | + C
z
2
sec 2 x
– ln |secx| + C.
=
2
Example 25. Find tan32xsec2xdx.
Here power of sec2x is not an even
positive integer and power of tan2x is an odd positive
integer therefore, we put z = sec2x
Let z = sec2x, then dz = 2sec2xtan2xdx
Solution
INDEFINITE INTEGRATION
Now tan32xsec2xdx = tan22xsec2xtan2xdx
Again sec4xdx = tan x +
[from previous example]
(sec 2 2 x – 1)2 sec 2 x tan 2 x
dx
2
(z 2 1
dz 1 (z2 – 1) dz
=
2
2
3
sec3 2x sec2 x
1 z
–
C.
= 2 3 z + c =
6
2
=
tan 3 x
3
From (1), tan4xdx
tan 3 x
= tan x +
– 2 tanx + x + C
3
3
tan x
– tanx + x + C.
=
3
Example 26. Find tan6xdx.
Solution Here power of sec x is not an even
positive integer and power of tan x is an even positive
integer therefore, we change tan2x into sec2x–1 and
then put
z = tan x.
Find cot2x cosec4x dx.
Example 27.
Solution Here power of cosecx is even positive
integer, therefore, put z = cotx.
Let z = cotx, then dz = – cosec2x dx
Now, cot2x cosec4x dx = cot2x cosec2x.cosec2x dx
= cot2x(1 + cot2x)cosec2x dx
Now tan4xdx = (tan2x)2dx = (sec2x – 1)2dx
= z2(1 + z2) (–dz)
= (sec4x – 2 sec2x + 1) dx
z3 z 6
= – (z2 + z4) dz = – 3 5 C
cot 3 x cot 5 x
=
+ C.
3
5
= sec4xdx – 2 (sec2xdx + dx
= sec4xdx – 2 tanx + x
1.35
...(1)
D
1.
Evaluate the following integrals :
2.
sin 3 d
(ii)
(i) cos2x sin3x dx
cos
dx
d
(iii)
(iv)
3
3
sin cos
sin x cos5 x
Evaluate the following integrals :
4.
5.
3.
d
1/2
sin cos7/2
(tan x)
dx
(ii)
(i)
sin x cos x
5
sin 7 x dx
(iii) cos x dx
(iv)
Evaluate the following integrals :
4
sec x
tan x dx
(ii)
dx
sin 6 x
(iii)
sin 3 2 x
dx
sin 5 x
(iv)
cos x cos x dx
6.
3
(i)
cos x cosec x dx
(ii)
sin x dx
5
2
cos x
5
2
(i)
Integrate the following functions :
(i) sin5x sec6x
(ii) tan2x sec4x
6
(iii) sec x
(iv) cosec5x.
Evaluate the following integrals :
sec x
(iii)
sin 2x dx
(iv)
dx
x
x
cos3
2
2
Evaluate the following integrals :
sin
(i)
cot x cosec x dx
(ii)
sec x
dx
tan x
(iii)
3
3
4
sin 3 x
cos
2/5
x
dx
1.36
7.
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
have inverses only for selected values of .
dx
4
sin x cos2 x
Evaluate the following integrals :
(iv)
(i)
cos 2 x dx
(iii)
tan
1.6
6
1
3/2
(ii)
x
For reversibility, x = a tan requires = tan –1
a
tan 3x sec 3x dx
3
x sec4 x dx (iv) tan x sec4 x dx
4
RATIONALIZATION BY
TRIGONOMETRIC
SUBSTITUTION
Consider the integrand a 2 x 2 .
If we change the variable from x to by the
substitution x = a sin , then the identity
1 – sin2= cos2
allows us to get rid of the square root sign because
a 2 x 2 a 2 a 2 sin 2
a 2 (1 sin 2 ) a 2 cos2 = a|cos |
Notice the difference between the substitution u = a2 – x2
(in which the new variable is a function of the old
one) and the substitution x = a sin (the old variable
is a function of the new one).
In general we can make a substitution of the form x = g(t).
To make our calculations simpler, we assume that g
has an inverse function; that is, g is one-one. We
want any substitution we use in an integration to
be reversible so that we can change back to the
original variable afterward.
Hence, we make the substitution x = a sin restricting
to lie in the interval , so that it defines a one 2 2
one function. Thus, x = a sin leads to = sin–1
with –
x
a
.
2
2
To sim plify the integration of x 2 a 2 , we
substitute x = a tan Further, we want to be able to
set = tan –1(x/a) after the integration takes place.
If x = a sec is substituted to simplify the
integration of x 2 a 2 we want to be able to set
= sec–1(x/a) when we are done.
As we know, the function in these substitutions
with –
<<
2
2
x
x = a sec requires = sec–1 with
a
0 if x 1
2
a
x
if
1
2
a
To simplify calculations with the substitution
x = a sec , we will restrict its use to integrals in
which x/a 1. This will place in [0, /2) and make
tan 0. We will then have
x 2 a 2 = a 2 tan 2 = |a tan | = a tan ,
free of absolute values, provided a > 0.
Any algebraic expression in x which contains only
one surd of a quadratic form, is capable of being
rationalized by such trigonometric substitutions.
For example :
dx
(i)
Ú (a - x )
2
2 3/2
Let x = a sin , and we get
d tan C
1
x
C
a 3 cos2
a2
a2 a2 x2
(ii)
dx
1
x (1 x 2 ) 2
Let x = tan, and the integral becomes
2
cos d
d(sin )
1
1 x2
–
C –
C
2
2
sin
sin
sin
x
(iii)
dx
x 3 (x 2 1)1 / 2
Let x = sec , and the integral becomes
cos d
2
sin cos
C
2
2
x 2 1 1 sec –1
x+C
2
2x 2
Suppose R denotes a rational function of the entities
involved. The integral of the form
=
1.37
INDEFINITE INTEGRATION
1.
R(x, b a x )dx is simplified by the
2
2 2
substitution x
2.
b
sin
a
2
is simplified by the substitution x
3.
1 (16 9x 2 )5/2
+C
80
x5
b
tan
a
Solution
dx
2
(a x 2 )3
We use the substitution x = a sin , with
–
then dx = a cos d
2
2
dx
2
(a x 2 )3
2
2
2
1
sin
1 sin
C.
C = 2
2
a cos
a 1 sin 2
The restriction on serves two purposes – it
enables us to replace |cos | by cos to simplify
the calculations, and it also ensures that the
substitutions can be rewritten as = sin –1(x/a), if
needed.
=
Example 2. Find
Solution
(16 9x )
x6
dx.
4
4
sin . Then dx = cos d
3
3
16 9 x 2 = 4 cos . Hence
and
(16 9x 2 )3/2
dx
x6
243
=
16
4
(64 cos3 ) cos d
3
4096 6
sin
729
243
cos cosec2 d = –
cot 5 + C
80
4
2x - x2
Then dx = cos dand
2 x x 2 = cos . Hence,
(1 sin )2
cos d
cos
1
3
2
= (1 sin ) d = 2 sin cos 2 d
2
2
I=
3
2 3/2
x–1
a cos d
(a a sin )
a cos d 1
d
1
tan C
=
a 3 cos3 a 2 cos 2 a 2
Let x =
1 (x 1)2
1
Let x – 1 = sin .
Example 1. Compute the integral
=
x 2 dx
2
b
substitution x sec
a
Solution
2x x 2
We have I
R(x, a x b )dx is simplified by th e
2 2
x 2 dx
Find
Example 3.
R(x, a x b )dx
2 2
=
=
3
1
– 2 cos – sin 2 + C
2
4
3
sin –1(x – 1) – 2 2 x x 2
2
1
– (x – 1) 2 x x 2 + C
2
=
=
3 –1
1
sin (x – 1) – (x + 3) 2 x x 2 + C.
2
2
Evaluate
Example 4.
Solution
dx
4x .
2
We set
<<
2
2
4 + x2 = 4 + 4 tan 2 = 4(1 + tan 2 ) = 4 sec2 .
x = 2 tan , dx = 2 sec2 d, –
Then,
2 sec2 d
dx
2
d
4 x = 4 sec = sec
| sec |
2
2
sec 2 = |sec | sec > 0 for
= sec d
= ln |sec + tan | + C
2
2
1.38
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
From this and the fact that the substitution can be
expressed as = sec–1 (x/5), we obtain
4 x2 x
+C
2
2
= ln
= ln | 4 x 2 + C. [Taking C = C – ln 2]
Note: A trigonometric substitution can
som et im es h elp us to eval ua te a n i nt egra l
containing an integer power of a quadratic binomial,
as in the next example.
=
Put x = 2 tan dx = 2 sec2 d,
2 sec 2 d
(4sec 2 )2
2sec 2 d
=
16sec 4
1
16
2 cos d
Evaluate
Example 6.
that x – 5.
x 2 25
dx, assuming
x
dx
(4x 24x 27)
2
dx
= 5 sec tan or, dx = 5 sec tan d
d
x 2 25
dx
x
25 sec 2 25
(5 sec tan )d
=
5 sec
5 | tan |
(5 sec tan ) d
=
5 sec
2
= – 5 tan d
= – 5 (sec 1)d = – 5 tan + 5 + C
2
where tan = –
x 2 25
5
3/2
dx
(4(x 3) 9)
2
3/2
3
sec .
2
3
Then dx = sec tan d and
2
Let x – 3 =
4 x 2 24 x 27 = 3 tan .
3
dx
(4x 24x 27)
So,
2
= 2
3/ 2
sec tan d
27 tan 3
=
1
cose c cot d
18
=
1
x3
1
cosec + C = – 9
+C
2
18
4x 24 x 27
Integrate
Example 8.
The integrand involves a radical of
the form x 2 a 2 with a = 5, so we make the
substitution
<
x = 5 sec ,
2
Thus,
Solution
2
1
1 sin 2 C
(1 cos2) d =
=
2
16
16
1
1
x
C.
=
tan –1
+
8(4 x 2 )
2
16
Solution
dx
x
Also, = tan –1 .
2
I=
Evaluate I = (x 2 4)2
Example 5.
Solution
x
x 2 25
dx = – x 2 25 + 5 sec–1 + C.
5
x
dx
Example 7. Find (4x 2 24x 27)3/2
Here
Solution
dx
x x 1 .
4
x 4 1 ( x 2 )2 1 which is of
the form x 4 a 2 . Hence, the substitution x2 = sec
should be tried.
Now
dx
dx
4
2 2
x x 1 x (x ) 1
...(1)
Let x2 = sec then 2x dx = sec tan d
sec tan
sec tan
d
d
dx
2x
2 sec
Now from (1),
sec . tan
dx
1
d
2
x x4 1
sec sec 1 2 sec
sec tan
1
d
sec .tan 2 sec
1
1
1
d C sec 1 (x 2 ) C .
2
2
2
INDEFINITE INTEGRATION
Note: When integrand involves expressions of the form :
ax
(i)
a x put x = a cos 2
(ii)
x
2
a x put x = a sin
(iii)
x
2
a x put x = a tan
(v)
Example 9.
Evaluate I =
1 x
dx
(x )( x)
...(1)
2x
I = cos–1
+ C.
Note that using (x – ) = () sin2 the answer can
Example 10.
Solution
Evaluate
dx
(x a)(x b)
Put x = a sec2 – b tan2
...(1)
dx = [a.2 sec2 tan – 2b tan sec2 ] d
= 2 (a – b) sec2 tan d
Thus, x – a = a (sec2 – 1) – b tan2
= (a – b) tan2
...(2)
and x – b = a sec2 – b (1 + tan2
= (a – b) sec2
...(3)
2
Solution
2( )sin cos
d = 2 + C = cos–1 (cos 2) + C
=2
But x = cos2 + sin2 ;
2x = (1 + cos 2) + (1 – cos 2)
i.e. ( – ) cos 2= ( – 2x)
or cos 2 = ( – 2x) / ().
From (1), we get
1
Example 11.
( )cos sin d
also be written as 2 sin–1
1 x dx cos x 1 x C .
Put x = cos sin2
so that dx = 2 () sin cos d.
Also (x – ) = () sin2 ,
and ( – x) = () cos2 .
Making these substitutions, the given integral
2( )sin cos d
I =
( ) cos 2 .( )sin 2
=
2 sin 2
2 sin cos d
2 cos2
sin
= –2
2sin cos d = – 2 2sin2 d
cos
sin 2
C
= –2 (1 – cos2)d= –2
2
= –2 + sin2 + C
...(1)
cos2 = x, 2 = cos–1x and sin2
2
= 1 cos 2 2 1 x 2
Hence from (1),
2
Solution
1 cos 2
Now 1 x dx 1 cos 2 ( 2 sin 2)d
x a
or (x a)(b x)
b x
put x = a cos2 b sin2
x a
or (x a)(x b)
x b
put x = a sec2– b tan2
(iv)
1x
1.39
x
+ C.
1 x
Integrate 1 x dx .
Let x = cos 2 then dx = –2sin2 d
dx
(x a)(x b)
2(a b)sec 2 tan
d
(a b) tan sec
= 2 sec d= 2 ln |sec + tan | + C.
From (2) and (3),
xb
and tan =
ab
The given integral
sec =
xa
.
ab
(x b) (x a)
+C
= 2 ln
(a b)
=
2
l n
(a b) (x a) + C ,
1
omitting the constant term – 2 ln (a b) which may
be added to the constant of integration C.
dx
Example 12. Evaluate I = (1 x ) x x 2
Solution
We make the substitution
1.40
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
x = sin2t dx = 2sin t cos t dt
Hence,
2 sin t cos tdt
2dt
I=
1 sin t
(1 sin t ) sin 2 t sin 4 t
1 sin t dt 2 tan t 2 C
cos t
cos 2 t
2 x 2 C 2( x 1) C
=
.
1 x
1 x
1 x
= 2
E
Evaluate the following integrals :
2
x dx
(4 x )
1.
2.
2 /2
x 3dx
a 8 – x8
5.
dx
(9 x 2 )2
dx
7.
(1 x2 ) (1 x2 )
3.
1.7
4.
6.
8.
dx
y 2 49
dy, y< –7
y
x
(a 3 x 3 )
dx
dx
2
2
Some Standard Integrals
(i)
1
x
x dxa = a tan a + C
2
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
1
2
dx
1
xa
=
ln
+C
x2 a 2
2a
xa
dx
1
ax
=
ln
+C
a 2 x2
2a
ax
dx
2
2
= ln x x a + C
x2 a 2
dx
2
2
= ln x x a + C
2
2
x a
dx
x
= sin 1
+C
a 2 x2
a
x
x
a2
a 2 x2 +
sin1 + C
2
a
2
x
2
2
2
2
x a dx =
x a
2
a 2 x 2 dx =
(4 x )
13.
(4x x )
15.
(x ) (x )( x)
x 2ax – a 2
ax bx c , ax bx c , ax bx c dx
dx
11.
2 3/2
INTEGRALS OF THE
FORM
2
(4x x )
dx
x2
dx
(a x 6 )
9.
10.
6
x 2 dx
2 /2
dx
2 3/2
dx
x (x 4 1)
1 x x2
12.
(1 x )
14.
(9 x )
2 3/2
dx
dx
2 2
dx
a2
2
2
ln x x a + C
2
x
2
2
(ix)
x a dx = 2 x 2 a 2
a2
2
2
ln x x a + C
2
Let us evaluate some of these standard integrals.
dx
1
x
tan 1 C
(a) I =
a
x2 a2 a
Proof : Put x = a tan, then dx = a sec2 d
a sec 2 d 1
1
1
x
d tan 1 + C
I
2
2
a
a
a
a
a sec
+
Note: Putting x = a cot , the above integral
takes up the form –(1/a) cot–1(x/a), which evidently
differs from the previous form by a constant.
Sometimes,
dx
a 2 x 2 is written in the form – 1 cot–1 x + C
a
a
It is an established convention of the calculus that the
integral of a real function shall be presented as a real
function and not as a function involving complex numbers.
For example (anticipating formulae from the chapter of
differentiation, to illustrate the point)
dx
1 1 x
x 2 a 2 a tan a (a, a > 0)
INDEFINITE INTEGRATION
uses only real numbers and is the standard form for
this integral.
The calculation
1 1
1
1
x 2 a 2 dx = 2ai x ai x ai dx
1
x ai
n
2ai x ai
though it can be fitted to the theory, is rejected because
it does not give the integral in terms of real numbers.
(b)
dx
x a
1
x a 2a ln x a C
2
2
Proof :
dx 1 1 1 dx
2a x a x a
x a2
2
1 dx
dx
2a x a
x a
=
1
{ln|(x – a)| – ln|(x + a)|} + C.
2a
xa
1 ln
2a x a + C
Note: The above is an example of integration
by breaking up the integrand into partial fractions.
dx
1
ax
C
ln
2
2
2a a x
a x
The proof is as before, noticing that
1
1 1 1
2
2
2
a
a
x
a
x
a x
(c)
(d)
dx
x a ln (x x a C
2
2
Proof : Put
x 2 a 2 = t – x,
or, t = x +
x2 a2
2
2
2x
t
dx
dt 1
dx
2
2
2
2
2
x
a
x
a
dx
dx
2 2 t ln |t | C
(x a )
= ln (x + x 2 a 2 C
The first of the integrals (d) can also be evaluated by
putting x = a tan, so that dx = a sec2 d. Thus,
a sec 2 d
dx
x a a sec
2
2
1.41
sec d
sec (sec tan )
d
sec tan
= lnsec + tan
=
ln | (1 tan 2 ) tan | C'
ln 1
x (x 2 a 2 )
x2
x
C' ln
C'
2
a
a
a
Similarly, put x = a sec, in the other integral.
dx
x
sin 1 C, (| x | | a |)
(e)
2
2
a
a x
Put x = a sin , then dx = a cos d.
a cos d
d sin 1 x + C
I
a
a cos
Note: Putting x = a cos, the integral
x
dx
2 2 can be put in the form – cos–1 a + C ' instead
(a x )
x
of sin–1 + C.
a
x x2 a 2
a2
x
(f)
x 2 a 2 dx
sin 1 + C
2
2
a
Proof : Putting x = a sin, so that dx = a cos d, we get
a x dx a cos d
1
= a2 . (1 + cos 2) d
2
2
2
2
2
1 2
1
a [ cos 2 d +
d]
2
2
1
1
= a2 [ sin 2 + ] + C
2
2
1
1
= a2 . sin cos + a2 + C
2
2
=
1 2x
x2
1
x
a
1 2 a 2 sin 1 + C
2 a
2
a
a
x x2 a2
a2
x
sin 1 C
2
2
a
Note: For the evaluation of the integrals
x 2 a 2 dx and x 2 a 2 dx, refer the section
of Integration by Parts.
1.42
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 1.
Evaluate
1
4 9x dx
2
1
4 9x 2
1
1
1
1
=
dx =
dx
2
4
(2
/
3)
x2
9
9
x2
9
1
1
x
+C
tan –1
= .
(2
/
3)
2 /3
9
Solution
We have
1
3x
tan –1 + C.
2
6
Example 2. Evaluate the following integrals :
=
(a)
16 x dx,
(c)
x 16 dx,
(b) x 2 16 dx,
2
2
Solution
Thus, in each case write
ax2 + bx + c = a x
2
b 4ac b 2
2a
4a
b
u and use the standard formulae.
and put x
2a
1
Example 3. Evaluate 4 (2 3x)2 dx.
Here we shall apply the formula
1
1 1 x
a 2 x 2 dx a tan a
But in place of x, we have (2 – 3x) and hence we shall
divide by the derivative of 2 – 3x which is –3.
1 1
2 3x
1 1 2 3x
. tan 1
tan
C
C.
I=
2
6
2
3 2
dx
Example 4. Evaluate (2x 2 x 1) .
Solution
We have
Solution
(a) The integrand is of the form a 2 x 2 , where a=4.
dx
1
dx
1
1
x 2 x 2
16 x 2 dx = 1/2 x 16 x 2 + 8 sin–1 (x + 4) + C
(2x x 1) 2
(b) The integrand is of the form x 2 a 2 , where a = 4.
1
=2
x 2 16 dx = 1/2 x x 2 16 –8ln x 2 16 + x+ C.
(c) The integrand is of the form x 2 a 2 , where a = 4.
x 16 dx = 1/2 x x 2 16
2
+ 8 ln x 2 16 + x + C.
The integrals of the form
dx
dx
ax bx c ,
2
ax bx c , and ax bx c dx can be
2
2
evaluated by first expressing ax2 + bx + c in the form
of a perfect square and then applying the standard
results.
For example x2 + 2x + 2 = (x + 1)2 + 1.
and hence the substitution u = x + 1, du = dx yields
1
1
x 2 2 x 2 dx = u 2 1 du
= tan –1 u + C = tan –1(x + 1) + C.
In general, the objective is to convert ax2 + bx + c
into either a sum or difference of two squares–either
u2 ± a 2 or a 2 – u2.
2
dx
2
2
1 1 1
x 4 2 16
1 3
1
dx
1 1 4 x 4 4
. . ln
2
2
2 2 3 x 1 3 +C
1
9
x
4 4
4 16
1
2x 1
1
2x 1 1
C ln
ln 2 + C
= ln
2
2(x 1)
3
x 1
3
1
= ln |(2x – 1)/(x – 1)| + C1.
3
1
dx
Example 5. Evaluate
2
x x 1
1
dx
Solution
x2 x 1
1
2
= x x 1 1 1 dx
4 4
=
1
(x 1 / 2) 3 / 4 dx
2
INDEFINITE INTEGRATION
1
=
(x 1 / 2)
2
3 / 2
dx
2
1
2
1
2
=
1
2
{(41 / 16) (x 3 / 4) }
=
1
x 1 / 2
+C
tan –1
3/2
3/2
=
2
2x 1
+ C.
=
tan –1
3
3
Evaluate
Example 6.
dx
x 4x 14 .
2
Solution x2 – 4x + 14 = (x2 – 4x + 4) + 10 = (x – 2)2 + 10
I=
dx
(x 2) 10
=
dt
I=
2
= ln t t 10 + C
t 2 10
2
= ln x 2 x 4x 14 + C
Example 7. Evaluate
Solution
=
dx
2 3
2 x x
2
dx
9 2 3
9
2 x x
2
16
16
dx
2
dx
{( 41 / 4)2 (x 3 / 4)2}
2
Put x – 2 = t dx = dt
=
1
= 2
1.43
1
2
x 8x 9
dx
1
dx
1
(x 4) 5 dx
2
=
x 2x 1 4 dx
=
1
(x + 1)
2
=
2
= ln x 4 x 8x 9 C .
Evaluate
Solution
dx
dx
(4 3x 2x ) .
1
2
2
(4 3x 2x )
(x 1)2 2 2
1
. (2)2 ln ((x + 1) + (x 1)2 2 2 ) + C
2
1
(x + 1)
2
x 2 2x 5
+ 2 ln ((x + 1) +
x 2 2x 5 ) + C
Evaluate
2x 3x 4 dx
2
Example 10.
dx
Solution
{2(3 / 2)x x }
2x 3x 4 dx
2
=
2
2
+
2
2
= ln x 4 ( x 4) 5 C
x 2x 5 dx
2
2
Example 8.
Evaluate
x 2x 5
Solution
2
x 2 8x 16 25
4x 3
sin 1
C.
2
41
1
Example 9.
x 8x 9 dx
1
3
x
4 C
sin 1
2
( 41 / 4 )
1
2 (x 2
2
3
x 2) dx
2
2
1.44
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
2
2
2
3 23
x
dx
4 4
2
=
3
1
2 x 2x 2 3x 4
4
2
=
1 23
3
ln x 2 2x 2 3x 4 C
2 16
4
1
4 2
(4x 3) 2x 2 3x 4
23
3
ln x 2x 2 3x 4 + C
4
16 2
Example 11.
Solution
=
=
=
Evaluate (x 1)(2 x) dx.
We have (x 1)(2 x) dx
(x 3x 2) dx
2
2
3 9
2 x dx
2 4
2
1
3
x
dx
2
4
1
3
= x
2
2
Solution
We evaluate
x2
x a
2
2
dx
x
2
tan sec d
=a
2
(sec 1)sec d
= a2
2
3
sec d sec d
3
1
1
sec tan sec d sec d
= a2 2
2
=
a2
(sectan ln|sec tan|) C
2
x 2
a2
2
x
a
ln
= 2
2
Solution
2
x a dx
2
2
x2
x a dx
2
a 2 tan 2
a sec 2 d
a sec
=a
x2 a 2 x
C
a
a
Evaluate cos sin d
5 cos 2
Example 13.
2
1
3
x
2
4
Evaluate
x 2 a 2 a 2 (1 tan 2 ) = a|sec | = a sec ,
Then,
and
< x< .
2
2
x 2
a2
x a 2 ln | x 2 a 2 x | C ,
2
2
where C = C + (a2 ln a)/2.
1
1
2
(2x – 3) (3x x 2) sin–1 (2x – 3) + C.
4
8
Example 12.
dx = a sec2 d, –
=
1 1
3
. sin 1 x / (1 / 2) + C
2 4
2
=
by letting x = a tan ,
2
3 23
dx
x
4 4
2
2
cos sin
5 cos 2 d
cos
sin
=
5 cos 2 d + 5 cos 2 d
=
6 2sin d + 4 2 cos2 d
cos
Put sin = u
sin
2
Put cos = v
du
dv
.
–
6 2u 2
4 2v 2
Now, the above integrals can be evaluated easily.
=
1.45
INDEFINITE INTEGRATION
F
Evaluate the following integrals :
dx
9x 2 4
2.
dx
1
dx
2x 2 x 1
dx
x2 x 1
dx
1.
3.
x 2 9x 20
4.
5.
x x 1
6.
2ax x
7.
2x 3x2 dx
8.
9.
5x 8x 4 dx
10. 3x 2 6x 10 dx
11.
2ax x dx
12.
dx
2
1
2
2
2
1.8
2
4 3x 2 x 2 dx
dx
(x2 2x cos 1)
INTEGRALS OF THE
FORM
px q
px q
ax bx c dx , ax bx c dx , and
2
2
(px q) ax bx c dx
px q
ax bx c dx . (a 0, p 0)
2
1.
2
Here, noting that the differential of
ax2 + bx + c = (2ax + b)dx,
the given integral can be written as
2aq
2aq
b
2ax
(2ax b)
p
p
p
p
dx
dx
2a ax 2 bx c
2a
ax 2 bx c
p (2ax b)
2aq pb
dx
dx
2
2
2a ax bx c
p
ax bx c
The first integal is equal to lnax2 + bx + c, since the
numerator of the integral is equal to the differential
coefficient of the denominator. The second integral is
evaluated as in the previous section.
cos x
4 sin x dx
cos x dx
14. sin x6sin x 12 dx
13.
Note:
(i) To express px + q as the sum of two terms we
might also proceed thus :
Let px + q (derivative of the denominator)
+ m, where the constants , m are to be determined
2
2
sin x dx
15.
cos2 x4 cos x1
16.
x[(log x) 2 log x 3]
dx
2
17. (a) Evaluate the integral
1
dx in three
2x x2
ways: using the substitution u = x , using
the substitution u = 2 x , and completing
the square.
(b) Show that the answers in part (a) are
equivalent.
by comparing the coefficients.
(ii) If ax2 + bx + c breaks up into two real linear factors,
then instead of proceeding as above, we may
break up the integrand into the sum of partial
fractions, and then integrate each separately.
px q
2.
dx . (a 0, p 0)
ax 2 bx c
px q
Let I =
dx
ax 2 bx c
Observe that the derivative of ax2 + bx + c is 2ax + b.
We therefore write (px + q) (2ax + b) + m.
Comparing coefficients of x on both sides of the above
identity, we have p = 2a
...(1)
Also, comparing the constant terms on both sides, we
have q = b + m
...(2)
From (1) and (2), = p/(2a), m = q – bp/(2a)
...(3)
With these values of and m, we can write
I=
(2ax b) m
ax 2 bx c
2ax b
=
2
dx,
dx m
dx
ax bx c
2
ax bx c
= 2 ax 2 bx c + mJ,
dx
.
where J =
2
ax bx c
The method of integrating J has been discussed earlier,
and therefore we can evaluate I.
1.46
3.
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(px q) ax bx c dx . (a 0)
2
To integrate this, transform
p
bp
px + q =
(2ax + b) + q
2a
2a
Then the integral reduces to the sum of two integrals,
(px + q) ax2 bx c dx
where J = ax bx c dx can be integrated as
3x 1
dx
2
2x 2x 3
1
3
5 4
=
1
3
=
–
1 5 4
2
xx
6 9 3
+
4
9
2
3
3
(4x 2) 1
4
2 dx
(2x 2 2x 3)
=
3
4x 2
5
1
dx
dx
4 2x 2 2x 3
2 2x 2 2x 3
=
5
dx
3
ln (2x2 – 2x + 3) +
2
2.2 x x (3 / 2)
4
=
=
=
=
2
(x 2)dx
2
xx
9
3
1
4
2
2x 2
2
3
3 dx
5
4
x x2
9 3
1/2
2x 4
dx
3
dx
5 2 4
x x
9
3
1/2
=
dx
3
5
ln (2x2 –2x+ 3)+
2
1 3 1
4
4
x 2 2 4
dx
3
5
ln (2x2 – 2x + 3) +
2
4
4
1
2
x ( 5 / 2)
2
(x 2)dx
(5 12x 9x )
=
3x 1
2x 2x 3 dx .
Here (d/dx) (2x2 – 2x + 3) = 4x – 2.
Solution
I=
Evaluate
2
We have
before. Recall that = p/(2a), m = q – bp/(2a).
Example 1.
(x 2)dx
(5 12x 9x )
Solution
= (2ax + b) ax 2 bx c dx + mJ,
2
=
(ax2 + bx + c)3/2 + mJ,
3
2
Evaluate
Example 2.
5 4 x x2
1 9 3
1
6
2
4
dx
2
9
5
2 4
x
3 9
9
=
– 1 5 4 x x 2 4
3 9 3
9
x 1
1
5
1
2
tan
+
+C
4 ( 5 / 2)
( 5 / 2)
=
x 2
1
4 1
2
3
(5 12x 9x ) sin
+C
9
9
1
3
2x 1
5
+ C.
ln(2x2 – 2x + 3) +
tan–1
5
4
2
=
–
3
ln (2x2 – 2x + 3)
4
dx
2
2
1 x
3
1
4
3x 2
(5 12x 9x 2 ) sin 1
+ C.
9
9
3
INDEFINITE INTEGRATION
Evaluate
Example 3.
3x 1
x 4x 1 dx .
=–
2
The linear expression in the numerator
can be expressed as
Solution
d 2
(x + 4x + 1) + m
dx
3x + 1 = l
3x 1
3 / 2 (2x 4) 5
x 4x 1
x 4x 1
3
2x 4
dx
=
5
2
2
2
x 4x 1
x 4x 1
3
2x 4
3 dt
Let I1
2
t
x 2 4x 1 2
2
2
=
2
=
We have (2x – 5) = –(3 – 2x) – 2.
1
(2x 1)
2
2
x2 x dx – 2 x x dx
2
=
2
17
3
x
dx
2
4
3
2
1
2 3/2
= – (2 + 3x – x ) – 2 2 x 2
3
2
3
17
x
2
4
x 3 / 2
1 17
. sin 1
+ C
2 4
(17 / 4)
11
2
2
x 1 1
dx
2 2
2
2
1 t 3/2
11 1 x 1 x 1 1
–
2
2 2
2 3/2
2 2
1
1 1
1 1 2
– ln x 2 x 2 2 + C
2 2
2
2
(2x – 5) (2 3x x ) dx
= – (3–2x) (2 3x x ) dx– 2 (2 3x x ) dx
(2 3x x 2 )3/2
2
=–
3/2
2
11
1
11
t dt –
2
2
2
where t = x + x
(2x–5) (2 3x x ) dx.
2
11
2
x 2 x dx – 2 x x dx
=
= 3 x 2 4x 1 5 ln | x 2 x 2 4x 1 | C
1
2 (2x 1) 2 x x dx
1
= (2x 1)
2
=
= 5 ln | x 2 (x 2) 2 3 | C
I = I 1 – I2
Solution
1
11
and = –
2
2
2
2
Evaluate
d 2
(x + x) + . Then,
dx
(x 5) x x dx
dx
2
Let (x – 5) = .
1 = 2 and + = – 5 =
dx
x 4x 1 5 (x 2) 3
Example 4.
2
Evaluate (x 5) x x dx
x – 5 = (2x + 1) + .
Comparing coefficients of like powers of x, we get
= 3 t C 3 x 2 4x 1 C
Let I2 = 5
17 1 2x 3
sin
+ C.
4
17
Solution
(where t = x2 + 4x + 1)
dx
2
1
(2 + 3x – x2)3/2 – (2x – 3) ( 2 x 3x x 2 )
3
2
Example 5.
3x + 1 = l (2x + 4) + m
Comparing the coefficients of x and constants both
sides.
3 = 2l & 1 = 4l + m
l = 3/2 & m = – 5
I
–
1.47
2
1 3/2 11
t –
3
2
2x 1 2
1
1
x x n x x 2 x +C
4
8
2
=
1 2
(x + x)3/2
3
–
1
1
11 2 x 1 2
x x n x x 2 x + C
8
2
2 4
1.48
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 6.
The integral becomes
Solution
2
2
x – 2 x cos 1
2
sin dx
(x – cos )2 sin 2
–
Hence,
(x cos 1)dx
cos ln x2 – 2x cos + 1
2
x 2 2 x cos 1
– sin tan–1
Example 7.
I=
Solution
x cos
+ C.
sin
Evaluate
2 sin 2 x cos x
6 cos x 4 sin x dx
2
2 sin 2 x cos x
6 cos x 4 sin x dx
2
( 4 sin x 1) cos x
2
( 4t 1)dt
(t 4t 5)
2( 2 t 4) 7
t 4t 5 dt
[using (1) and (2)]
2
2( 2 t 4) 7
dt + 7
t 2 4t 5
= 2ln (t2 – 4t + 5) + 7
dt
t 2 4t 5
dt
t 4t 4 4 5
= 2 ln (t2 – 4t + 5) + 7
Evaluate I =
2 x 2 3x 4
dx
x 2 6 x 10
l (x 2 6 x 10) m(2 x 6) n
dx
x 2 6 x 10
2x2 + 3x + 4 = l(x2 + 6x + 10) + m(2x + 6) + n
On comparing the coefficients of x2, x and constant
term, we have
l = 2, 6l + 2m = –3, 10l + 6m + n = 4
I=
Solution
= 2x –
....(1)
2
2
9 and n = 11.
2
9
(2 x 6 )
dx
I= 2dx 2 x 2 6 x 10 dx 11 (x 3)2 1
Now, let (4t – 1) = (2t – 4) +
Comparing coefficients of like powers of t, we get
2 = 4, – 4 + = – 1
= 2, = 7
...(2)
=2
ax bx c dx , and
In this case express the numerator as
Nr = l (Dr) + m (derivative of Dr) + n
i.e. put ax2 + bx + c
= l (px2 + qx + r) + m ( 2ax + b) + n
where l, m and n are to be so chosen by comparing the
coefficients of x2, x and constant term, that it becomes
equal to the given numerator.
l = 2, m =
(4 sin x 1) cos x
=
dx
sin 2 x 4 sin x 5
Put sin x = t, so that cos x dx = dt.
I=
px 2 qx r
2
Example 8.
6 (1 sin x) 4 sin x dx
I=
px 2 qx r
dx ,
ax 2 bx c
(ax bx c) (ex fx g) dx
cos (x cos ) dx
=
Integrals of the form
(x cos 1) dx
x 2 – 2x cos 1
2
dt
(t 2) (1)
2
2
= 2ln (t2 – 4t + 5) + 7tan–1 (t – 2) + C
= 2 ln (sin2 x – 4 sin x + 5) + 7tan–1(sin x – 2) + C.
9
nx2 + 6x + 10 + 11tan–1(x + 3) + C
2
Example 9.
Evaluate
x 2 3x 1
1 x2
dx
Solution
Let x2 – 3x + 1 = l (1 – x2) + m
d
(1 x 2 ) n
dx
Comparing the coefficients like powers of x
l = –1 ,m = 3/2, n = 2
3
(1 x 2 ) ( 2x) 2
2
dx
dx
1 x2
1 x2
3x
1
2
dx 2
dx
= – 1 x dx
2
1 x
1 x2
x 1 x 2 1 1
sin x
= –
2
2
x 2 3x 1
INDEFINITE INTEGRATION
1
1 2x
3
dx 2
dx
2
2 1 x
1 x2
Each of the three integrals on the right hand side of
the last equality can be evaluated using formulae.
x
3
1 x 2 sin 1 x 3 1 x 2 C
2
2
Evaluate
( x 4 x 7)
2
x2 x 1
.
Solution
Let x2 + 4x + 7 = A(x2 + x + 1) + B(2x + 1) + C
Comparing the coefficients of x2, x and constant
term,we get A = 1, A + 2B = 4, A + B + C = 7
3
9
A = 1, B = 2 , C = 2
3 (2x 1)dx
2
So I = x x 1 dx 2 2
x x 1
9
dx
2
x2 x 1
1 3
Now x x 1 = x
2 2
2
x 2 3x 1 dx
1 4x 2x 2
1
( 2 u 1 2 u 2 sin 1 2 u)
=
4 2
6x
3
1 x 2 sin 1 x C .
=
2
2
Example 10.
2
2
1 1 2 u 2 1 sin 1 2 u C
2
2
2
1
3
1
2u C
= 4 (2 u) 1 2 u 4 2 sin
On returning to the original variable x we arrive at the
x 2 3x 1 dx
1 4x 2x 2
final result :
2
1
3
1
2 (x 1) C
= 4 (3 x ) 1 4 x 2 x 4 2 sin
Form
p 0 x n p1x n 1 p 2 x n 2 ....p n
dx
ax 2 bx c
In this case divide the Nr by Dr and express it as
px q
, where quotient will consist
ax bx c
of certain terms which we shall integrate by power
Quotient +
2
px q
1
x 2
3
1
2
2
x x 1 n x x x 1
2
8
2
formula and
9
1
2
3 x 2 x 1 2 n x 2 x x 1 C .
Example 12.
I=
Example 11.
Evaluate
x 2 3x 1 dx
.
1 4x 2x 2
First of all, the quadratic trinomial under
the radical –1 + 4x – 2x2 is brought to the form :
–1 + 4x – 2x2 = –[1 + 2(x2 – 2x)] = 1 – 2(x – 1)2
Now we change of the variable of integration.
Put u = x–1 in the integral, which leads to
Solution
=
2
x 2 3x 1 dx (u 1) 3(u 1) 1du
1 4x 2x 2
1 2u 2
2
1.49
u
u
du
du du
2
2
1 2u
1 2u
1 2u 2
2
ax bx c
explained before.
Solution
will be integrated as
x 4 x 3 2 x 1 dx
x2 x 1
2
( 3x 2 )
dx
I = (x 1) 2
( x x 1)
Evaluate I =
(2x 1)
2
1
dx
= x 1 3 2
2
2
2 x x 1
3
1
2 x
2 2
=
x 3 x 3 n( x 2 x 1) 1 tan 1 2 x 1
+C
2
3
3
3
Example 13.
(x 3 3)dx
Evaluate .
(x 1)
2
1.50
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
We have
Solution
3
(x 3)dx
–
2
(x 1)
x(x 1) x 3
x(x 2 1)
x dx
x dx
2
2
2
2
(x 2 1)
dx
(x 1) dx – (x 1) + 3 (x 1)
=
1
(2x) (x 2 1) dx
2
2x dx
dx
2
2
(x 1) 3 (x 1)
1 2 2
1
(x 1)3/2 [2 (x 2 1)]
2 3
2
=
=
1
2
x2 1 ) + C
+ 3 ln (x +
1 2 2/3
(x + 1) – (x 2 1) + 3 ln (x + x 2 1 ) + C.
3
=
G
Evaluate the following integrals :
x dx
1.
x 2x 1
3.
x
x 2 5x 6 dx
2
3x 5
2
x 2x 3
5.
7.
3x 5x 1 dx
6x 5
2
2.
x 2 3x 10 dx
2x 3
12.
4.
14.
6.
x 1
dx
2
x x3
2x 1
3 2x x
8.
(x 1) 1 x x dx
10. (3x 5) x 2x 3 dx
2
dx
(3x 5)dx
x 4x 3
2
2
9.
1.9
(x 2) 2 x 2 x 1 dx
ax2 + b
x + px + q
4
2
Evaluate I =
1 x
2
1 x2 x 4 dx
15.
(2 x 2 3x) dx
(x x 1) dx
2
x2 2x 3
3x 3 8 x 5
x 4x 7
2
2 x2 2 x2
4 x4
Solution
2
x 2 dx
1 2x x .
2
x 3 dx
(x 2 x 2)
(x 1) x 2
21.
dx
x 2 dx
x 16
17.
19.
2
2
x2
x3 1
x x dx
2
dx.
We divide Nr and Dr by x².
1 1 dx
1
x2
Put x +
=t
1
x
2
x 2 1
x
I=
–
t 2 1 = – 2 n t 1 + C
dx , q > 0
Here we have a very interesting method for evaluating
the integral of a rational function in which the
denominator is of degree 4 and the numerator is either
of degree 2 or is constant. Moreover the odd powers
of x occur neither in the numerator nor in the
denominator.
ax 2 b
dx
Let us consider first the integral 4
x px 2 1
1
We divide Nr and Dr by x² and then put x ±
= t.
x
Example 1.
x 2 2x 1
18.
22.
INTEGRALS OF THE
FORM
2x 2 3x 1 dx
13.
2
x 2x 5
2
11.
x 1 4 ln x ln x
16.
20.
2
ln x dx
dt
1
t 1
1
x 1
1
x
+C
= – n
1
2
x 1
x
Example 2.
Evaluate
x 2 1 dx
.
x4 1
x 2 1 dx
x4 1
Divide above and below by x2.
Solution
I=
dx
INDEFINITE INTEGRATION
1 12
I = x dx
x 2 12
x
H
e r e
For I1, we write x –
as a perfect square of x –
1
by subtracting and adding 2.
x
1
Put x – 1 = t 1 2 dx dt
x
x
dt
1
t
C
tan 1
I= 2
2
2
2
t 2
1
x x
1
+C
1
tan
2
2
1 tan 1 x 2 1
=
+ C.
2
2x
Evaluate
1
dx
4
x 1 5x 2
We express the numerator in terms of
(x + 1) and (x2 – 1).
1 = (x2 + 1) + m (x2 – 1)
+ m = 0, – m = 1
= 1/2, m = – 1/2
1
1
Hence, 1 = (x2 + 1) – (x2 – 1)
2
2
2
1
2
dx
4
2 x 1 5x 2
1
1 x2
1
x2 1
dx
dx
= 4
2 x 1 5x 2
2 x 4 1 5x 2
I
1
1 (1/ x 2 )
1
1 (1/ x 2 )
dx
dx
2 x 2 (1/ x 2 ) 5
2 x 2 (1/ x 2 ) 5
(I1 I 2 ) / 2
1
tan 1
7
dv
I2
1
u
7
x 1/ x
7
For I2, we write x +
1
1
v 1 2 dx dv
x
x
1
tan 1
v
x 1/ x
tan 1
3
3
1
v ( 3)
3
3
Combining the two results, we get I = (I1 – I2)/2
1
x 1/ x
1
x 1/ x
tan 1
tan 1
C
=
2 7
7
2 3
3
x2
dx
Example 4. Evaluate I =
x 4 x2 1
x2
dx
Solution Let I = 4 2
x x 1
We express the numerator as
1
1
x2 = (x2 + 1) + (x2 – 1)
2
2
2
I=
Solution
=
1
du
1 12 dx
1 12 dx
x2
I= x
2
1
x 2 22
x 1 2 2
x
x
Example 3.
1
1
u 1 2 dx du
x
x
I1 u 2 ( 7 )2 7 tan
1 12 is d x 1 hence we express the Dr
dx
x
x
1.51
=
1 (x 2 1) (x 2 1)
dx
2
x4 x2 1
1 (1 1 / x 2 ) (1 1 / x 2 )
dx
2
x 2 1 (1 / x 2 )
1 (1 1 / x 2 )dx 1 (1 1 / x 2 )dx
2 (x 1 / x)2 3 2 (x 1 / x)2 1
1
In the first integral put x –
= t so that
x
1 1
dx = dt, and in the second integral
x2
1
1
put x + = z so that 1 2 dx = dz.
x
x
1
dt
1
dz
I=
2
2
2
2 t ( 3)
2 z 1
t
1
1
z 1
1
ln
=
tan–1
+C
3 2 2 1 z 1
2 3
=
=
(x 1 / x) 1 (x 1 / x) 1
ln
tan–1
+C
3 4 (x 1 / x) 1
2 3
1
1.52
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
2
2
x 1 1 x x 1
ln 2
=
tan–1
+ C.
2 3
( 3)x 4 x x 1
1
Evaluate
Example 5.
Solution
2
2
dx
I= 1 x 3
– 1 4 x – 23
dx
6 x 8x 9
6 x 4 8x 2 9
We have,
1
dx =
4
x 1
=
1
dx
x 1
4
1
x2
x2
2
1
x2
dx
1
1
x2
x2
x2
1
1
2
1 1 x2
1 1 x2
dx
–
dx
=
2 x2 1
2 x2 1
x2
1
2
1
1 2
1
x
dx –
2
2
x 1 2
x
x2
2
= 1 1 3/x
dx – 1 1 3 / x
dx
6 x4 3 / x2 8
6 x 4 3 / x2 8
1 2
2
1
1 1 x2
x
= x
dx =
dx
2 x2 1
2 x2 1 x2 1
=
1 = (x2 + 3) + m (x2 – 3)
+ m = 0, 3 – 3m = 1
= 1/6, m = – 1/6
Hence 1 = 1/6 (x2 + 3) – 1/6 (x2 – 3)
1
1 2
x
dx
2
x 1 2
x
2
2
= 1 1 3/ x
– 1 1 3 / x2
dx
dx
6 (x 3 / x) 2
6 (x – 3 / x)2 14
Put x – 3/x = u
Put x + 3/x = v
(1 – 3/x2) dx = dv
(1 + 3/x2) dx = du
du
I= 1 2
– 1 2dv
6 u 14
6 v 2
Now, the above integrals can be evaluated easily.
x
dx
x x2 1
Evaluate
Example 7.
4
We have,
Solution
x
x
dx =
dx
(x 2 )2 x 2 1
x4 x2 1
Let x2 = t, then d (x2) = dt
=
dt
2x
1
1
= u in 1st integral and x +
= v in
Putting x –
x
x
2x dx = dt dx =
2nd integral, we get
=
t t 1 . 2x = 2 t t 1 dt
=
1
2
=
=
1
1
du
dv
–
2
2 u2 2
2 v2 2 2
v 2
u 1 1
–
tan –1
ln
+C
v
2
2 2 2 2
2 2
1
x 1/ x 2
x 1/ x 1
tan
ln
+C
=
–
2 2
2 4 2 x 1/ x 2
1
=
x
–1
x2 1
1
x2 2 x 1
tan–1
ln 2
+C
–
x x 2 1
2x 4 2
2 2
1
dx
x 4 8x 2 9
Solution We express the numerator in terms of
(x2 + 3) and (x2 – 3). Note that 3 is the square root of the
constant term 9 in the denominator.
Example 6.
Evaluate I =
dt
1
1
2
2
1
2
t 1 3
2 2
2
dt
1
t
1
1
2
tan –1
+C
= .
2
3
3
2
2
=
2x 2 1
1
1
2t 1
–1
tan–1
+
C
=
tan
+ C.
3
3
3
3
x4 1
dx
x6 1
(x 2 1)2 2 x 2
dx
I=
x6 1
Example 8.
Solution
Evaluate I =
1.53
INDEFINITE INTEGRATION
cos x
1 cos 2 x
t 1 1
cos x 1
1
dt
I2 = 2
= ln
= ln
t 1 2
cos x 1
t 1 2
2
( x 2 1)
dx
dx – 6x
4
2
(x x 1)
x 1
In the second integral, put x3 = t 3x2 dx = dt
=
=
1 12
I = x dx – 2 2dt
3 t 1
2 1
x 2 1
x
In the first integral, put x – 1/x = v 1 + 1/x2 dx = dv
1
1 2
x
dx – 2 tan–1t
I=
2
3
1
2
x 1
x
dv
= 2 2 – 2 tan–1x3
v 1
3
1 1 cos x
ln
2 1 cos x
=
I=
1 1 1 cos x cos x
ln
+C
2 2 1 cos x sin 2 x
Example 10.
Solution
tan–1 x 1 – 2 tan–1 x3 + C
x 3
Example 9. Using substitution only to evaluate
=
cosec x dx .
3
Solution
Put cos x = t
sin x
sin x
dx
dx =
I=
4
(1 cos 2 x ) 2
sin x
dt
dt
I=–
2 2 =– 4
(1 t )
t 2t 2 1
1 ( t 2 1) ( t 2 1)
dt
2 t 4 2t 2 1
1
t2 1
1
t 2 1
dt
dt
I=– 4
2
2
2
2
t
2t
1
t4
2t
1
=–
Put x –
–
=–
I1
Let, I =
(1 x ) 1 x x
2
2
(1 x 2 )dx
(1 x ) 1 x x
2
2
4
1
x 2 1 2 dx
x
1
1
x x x 2 1 x 2
x
x
(1 1 / x 2 )dx
(1 1 / x) (x 1 / x) 3
2
1
= t,
x
dt
t t 3
2
Again put t2 + 3 = s2 2t dt = 2s ds.
s ds
ds
2
=–
2
s(s 3)
s ( 3)2
...(1)
I2
1
Hence, I = (I2 – I1)
2
1
1
1 2
1 2
t dt
t
dt =
Consider, I1 =
2
1
1
t2 2 2
t
t
t
1
Put t – = y
t
dy
t
1
1
=– 2
Hence, I1 = 2 = – = –
y
1
t 1
y
t
t
Evaluate
(1 x 2 )dx
1
2 3
1
=–
ln
s 3
s 3
+C
t2 3 3
=– 2 3
ln
1
(x 1 / x) 2 3 3
2 3
=–
ln
1
2 3
t2 3 3
+C
(x 1 / x) 2 3 3
1
1 3
x2
+C
1
x2 2 1 3
x
x2
ln
+C
4
1.54
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Evaluate I =
Example 11.
I=
Solution
x 1
x 1
dx
x 1 x x x
3
2
dx
1
x 1
x
x(x 1)
Form
2
(x 1)
dx
1
x 1
x
1 12 dx
x
=
x 1 2 x 1 1
x
x
Put x + 1/x + 1 = t2 (1 – 1/x2) dx = 2t dt
2 t dt
= 2 dt = 2 tan–1 t + C
I = 2
( t 1) t
t2 1
= x(x 1)
2
= 2 tan–1
x 1/ x 1 + C
Example 12.
Evaluate
1
dx
x x 3x 2 1
4
1
dx
x x 3x 2 1
1
=
dx
2
2
x x 1 / x2 3
Now, 1 = (x2 + 1) + m (x2 – 1)
+m=0,–m=1
= 1 + m , 1 + 2m = 0
m = – 1/2, = 1/2
Solution
=
4
1
{ln(u u 2 5} – 1 {ln (v +
2
2
f ( x)
(ax 2 bx cx 2 bx a) dx
4
3
2
1
1
a x 2 2b x c
x
x
and hence the substitution is
1
1
x z or, x z ,
x
x
according as f(x) is expressible in the form
1
1
1
1
x x g x x or, x x g x x .
1
1
2
2
If b = 0, the substitution x 2 z or x 2 z
x
x
is sometimes useful.
1 x2
dx
.
Example 13. Integrate
2
1 x (1 x 2 x 4 )
x
I
1
= 2
11/ x
2
=
2
2
1 12 dx du
x
1 12 dx dv
x
dv
du
I= 1
– 1
2
2
2
u 2 ( 5 )2
v 12
(1 1 / x ) dx
dx – 1
2
1
(x 1 / x)2 1
x x 5
Put x – 1/x = u Put x + 1/x = v
2
where f(x) is a rational function of x.
The denominator can be written as
Solution
1 / 2(x 2 1) 1 / 2(x 2 1)
dx
I=
x 2 x 2 12 3
x
v 2 1 )} + C
1
x 2 1 2 dx
x
1
1
x x x 2 x 2 2 1
x
x
1
1 2 dx
x
2
1
1
x x 1
x
x
dz
z (z 1)
2
cos ec cot
cos ec cot d
putting x +
1
=z
x
putting z = cosec
1
= d C cos ec z C
x2 1
x
1
cos ec 1
C.
C sin
2
1 x
x
Example 14. Evaluate
tan x
4
dx.
2
3
cos x tan x tan 2 x tan x
1.55
INDEFINITE INTEGRATION
Let I =
Solution
=
tan x
4
2
3
cos x tan x tan 2 x tan x
(1 tan 2 x)dx
(1 tan x) cos x tan x tan x tan x
I=
2
2
3
2
1 2
1
sec xdx
tan 2 x
tan x 2 1 tan x 1 1
tan x
tan x
1
tan x
2
sec 2 x
dx
2ydy = sec x
tan 2 x
2ydy
I=
= –2tan–1y + C
(y 2 1)y
Put y2 = tan x 1
= – 2tan–1 tan x 1
1
+C
tan x
H
Evaluate the following integrals :
1.
3.
5.
x2 1
dx
4
x 7x 2 1
dx
x4 x2 1
dx
x2 1
x 2 1 . x4 1
dx
x4 x2 1
2.
4.
x 2 dx
x4 x2 1
7.
2
(1 x )dx
2
13.
4
1.10 INTEGRATION OF
TRIGONOMETRIC
FUNCTIONS
Integrals of the form
(i)
dx
a cos x 2bsin x cos x bsin x
2
2
dx
a b sin 2 x
Here, we shall find the integral of the fractions whose
Nr is a constant and Dr contains cos2x, sin2x and sin x
cos x or any of them and a constant.
In such cases we divide the Nr and Dr by cos2x or sin2x
and then Nr becomes either sec2x or cosec2x and Dr
becomes something in terms of sec2x and tan2x or
something in terms of cosec2x and cot2x.
Then in the new Dr we shall replace sec2x by 1 + tan2x
and cosec2x by 1 + cot2x. Hence the Nr shall be either
sec2x or cosec2x and Dr in terms of tan x or cot x and
(ii)
dx
a cos2 x b
(x 2 1)dx
x x 3x 1
4
2
(x 2 a 2 )dx
4
2 2
4
2
(1 x ) (1 3x x ) .
2
8.
10.
x a x a
11. tan x dx
(x 1)dx
12.
x (x ax 1)(x bx 1)
9.
2
6.
dx
x 4 18 x 2 81
x 2 dx
x4 a4
(iii)
2
2
(1 x )
dx
2
2
(1 x ) 1 x x
4
we shall either put tan x = t or cot x = t as the case may
be. The above shall be more clear by the following
examples:
dx
Example 1. Evaluate
2 sin 2 x
Solution Put tan x = t
dt
dx
2
2 sin 2 x =
t
2
2
(1 t )
1 t2
dt
1
tan–1 t C
2
2t
2
2
= 1 tan–1 tan x C
2
2
Example 2.
Solution
Evaluate
dx
1 3cos x
2
2
dx
sec x dx
1 3cos x = tan x 4
2
2
1.56
=
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
tan x
+ C.
tan –1
2
2
dx
3sin x 4 cos 2 x
dx
sec 2 x
dx
Solution
3sin 2 x 4 cos 2 x 3 tan 2 x 4
dt
2
where t = tan x
3t 4
1
dt
1
t
=
tan 1
2
C
2
3 t 2/ 3
2 3
2/ 3
Evaluate
Example 3.
2
3
tan
tan x C .
=
2 3
2
1
1
dx
1 3 sin 2 x
r
Here divide the N and Dr by sin2x
Evaluate I
Example 4.
Solution
cos ec 2 xdx cos ec 2 xdx
1 cot 2 x 3
cos ec 2 x 3
Put cotx = t
–cosec2x dx = dt.
dt
1
t
1
1
tan 1 C tan 1 cot x C
I=
2
2
2
2
4t
2
I=
Note: We could have divided the Nr and Dr
by cos2x as well.
dx
sec 2 xdx
2
2
1 3 sin x
sec x 3 tan 2 x
sec 2 xdx
=
.
1 tan 2 x 3 tan 2 x
Now put tan x = t
sec2x dx = dt
dt
1
1
tan 1 (2t) C tan 1 (2tan x) C .
I=
2
2
2
1 4t
The two answers obtained in different forms can be
easily shown to be differing only by a constant.
dx
Example 5. Evaluate (2sin x 3cos x)2 .
sec2 xdx
dx
=
Solution
(2 tan x 3)2
(2sin x 3cos x)2
Put tan x = t, so that sec2 x dx = dt
I=
=
dt
(2t 3) = – 2(2t1 3) C,
2
1
C.
2(2 tan x 3)
Example 6. Evaluate
=–
d
5cos 4cos sin 2sin .
2
2
Divide the Nr and Dr by cos2
Solution
sec 2 d
.
5 4 tan 2 tan 2
Now, put tan = t
sec2 d = dt.
dt
dt
1
2
I = 5 4t 2t 2 2
7
(t 1) 2
2
7
(t 1)
1 1
I= .
ln | 2
| +C
2
7
7
2
(t 1)
2
2
7
tan 1
1
ln 2
+C
=
2 14
7
tan 1
2
Example 7. Evaluate
I=
I=
dx
sin x cos x sin x cos x .
Solution
4
4
I=
2
2
sec 4 x dx
tan 4 x 1 tan 2 x
Put tan x = t, sec2x dx = dt
1 12 dt
(1 t 2 ) dt
t
I= 4 2
=
2
1
t t 1
t 2 1
t
1 12 dt
t
=
2
t 1 3
t
Now, put t 1 = v 1 12 dt = dv
t
t
dv
= 1 tan–1 v + C
I=
2
2
3
3
v 3
1
tan x
tan x C
1
–1
tan
.
=
3
3
INDEFINITE INTEGRATION
Example 8.
Solution
=
dx
Evaluate I = (cos2 x 4 sin 2 x)2
Integrals of the form
dx
cos 4 x 16 sin 4 x 8 cos 2 x sin 2 x
(i)
a + b sinx
(iii)
a + b sin x + c cos x
I=
sec 4 x dx
1 16 tan 4 x 8 tan 2 x
sec 4 x dx
=
1 2 tan x 2 2 tan x
4
2
Now, put 2 tan x = tan 2 sec2x dx = sec2 d
1 tan 2
2
sec d
4
1
=
2
sec 4
1
= (4 cos2 + sin2) d
8
Now, the above integral can be evaluated easily.
3 tan
5
+C
I = tan–1(2x) +
21 4 tan 2 x
2
dx
Example 9. Evaluate I =
(sin x 2 sec x)2
dx
Solution I = 1
2
4 sin x
sec x
2
2
sec
xdx
1
=
2
4 2
tan x 1
tan x
2
2
sec
xdx
1
=
2
4
2
tan x 1 15
4 16
15
1
tan
Put tan x +
4
4
15
sec 2 d
sec2x dx =
4
15
sec 2
1
4
2
= 4 15
d
4
16 sec
2
1 15 16
.
.
cos2 d
4 4 15
Now, the above integral can be evaluated easily.
8
4tanx 1
4tanx 1
tan1
+C
3/2
2
15
15 15(2tan x tanx 2)
=
dx
(ii)
1.57
dx
a + b cosx
dx
In general, the integrals of the form
F (cos x, sin x) dx,
where F( cosx, sin x) is a rational function of cos x and
sin x, can often be evaluated by the substitution
x
tan = t
2
x
1
We then have sec2 dx = dt,
2
2
i.e. dx = 2 (1 + t2)–1 dt,
cos x = (1 – t2)/(1 + t2), sin x = 2t/(1 + t2), and
consequently, F (cos x, sin x) dx
1 t 2 2t 2dt
F
,
2
2
2
1 t 1 t 1 t
This substitution enables us to integrate any function
of the form F(cos x, sin x). For this reason it is sometimes
called a "universal trigonometric substitution".
However, in particular it frequently leads to extremely
complex rational functions. It is therefore convenient
to know some other substitutions (in addition to the
"universal" one) that sometimes lead more quickly to
the desired end.
1
dx
Example 10. Evaluate
1 sin x cos x
1
Solution Let = 1 sin x cos x dx
1
dx
=
2 tan x / 2
1 tan 2 x / 2
1
1 tan 2 x / 2 1 tan 2 x / 2
1 tan 2 x / 2
dx
=
2
1 tan x / 2 2 tan x / 2 1 tan 2 x / 2
sec 2 x / 2
=
dx
2 2 tan x / 2
x
1
x
Putting tan = t and sec2
dx = dt, we get
2
2
2
1
x
dt = ln | t + 1| + C = ln tan 1 + C
=
t 1
2
=
1.58
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Integrate I =
Example 11.
dx
4 5cos x
Example 13.
Solution
I=
dx
4(cos2 x / 2 sin 2 x / 2) 5(cos2 x / 2 – sin 2 x / 2)
=
dx
.
9 cos (x / 2) – sin 2 (x / 2)
=
2
sec 2 ( x / 2) dx
...(1)
9 – tan 2 (x / 2)
Putting tan x/2 = t, 1/2 sec2 (x/2) dx = dt in (1), we have
I=
dt
= 1 ln 3 t C,
9 t2
3
3t
Solution I
Integrate
Solution Putting 3 = r cos , 4 = r sin ,
so that r = 5, = tan–1 (4/3). We have
dx
3cos x 4sin x 5
dx
=
5 cos( x – ) 5
= 1 sec 2 1 (x – ) dx,
2
16
1
=
[2tan(x – )/2] + C,
10
1
= tan[(x – tan–1 (4/3))/2] + C.
5
dx
Example 14. Integrate 13 3 cos x 4 sin x
Solution
dx
5 13 sin x
dx
x
x
2x
2 x
5 sin cos 13.2sin cos
2
2
2
2
x
Multiplying the numerator and denominator by sec2 ,
I=
dx
x
x
x
x
x
x
13 sin2 cos2 3 cos2 sin2 4.2sin cos
2
2
2
2
2
2
Multiplying the numerator and denominator by
x
2
sec2 , then
2
we get
x
dx
2
x
x
5 tan 2 1 26 tan
2
2
2 dz
x
, Putting tan = z
5z 2 26z 5
2
2
dz
2
2
5
13
12
z
5
5
2
du
13
12
5 u 2 a 2 , where u = z – 5 and a = 5
2 1
ua
1
z5
ln
C
ln
C
5 2a u a
12 z 1 / 5
sec 2
dx
3cos x 4sin x 5
= 1 ln 3 tan x / 2 C.
3
3 tan x / 2
Example 12.
Evaluate
x
1 5tan 2 25
ln
+ C, on restoring the value of z.
12 5tan x 1
2
I=
sec 2
x
dx
2
x
x
8 tan 16
2
2
2 dx
x
. Putting z = tan
10z 2 8z 16
2
1
dz
1
du
2
2
5
5 u2 a2
2
6
z
5
5
2
6
where u = z + , a =
5
5
11
1
5z 2
1 u
C tan 1
C
tan
5a
a
6
6
1
5 tan x 2
1
1
2
tan
+C
6
6
a bsin x
dx
Example 15. Evaluate
(b a sin x)2
10 tan 2
Solution
Here I =
a bsin x
(b a sin x) dx
2
INDEFINITE INTEGRATION
a2
b (b a sin x)
b b
=
dx
a
(b a sin x)2
I=
a 2 b2
dx
b
dx
a
(b a sin x)2 a (b a sin x)
a2
a sin x b b
dA
b
b
dx
a (b a sin x)2
b
dx
(a 2 b 2 )
dx
a b a sin x
a
(b a sin x)2
a 2 b2
dx
b
dx
– A ...(2)
a
a b a sin x
(b a sin x)2
From (1) and (2),
b
dx
b
dx
A
a (b a sin x)
a (b a sin x)
I = –A+ C
I=–
cos x
I=–
+ C.
b a sin x
Evaluate
Let A =
Solution
dx
(16 9 sin x)
2
cos x
16 9 sin x
dA (16 9 sin x)( sin x) cos x(9 cos x)
dx
(16 9 sin x)2
dA 16 sin x 9
dx (16 9 sin x)2
dA
dx
256
16
(9 sin x 16)
9
9
9
(16 9 sin x)2
16
(1 tan 2 x / 2)dx
9 16 16 tan 2 x / 2 18 tan x / 2
175
dx
16
2dt
2 =A+
2
9 (16 9 sin x)
9 16 t 18t 16
where tan x/2 = t
175
dx
2
dt
=A+
9 (16 9 sin x)2
9 t2 9 t 1
8
2
dt
=A+
2
2
9
9 175
t
16 16
2
16
16 t 9
+ C1
= A+
tan–1
175
9
175
dx
9
cos x
2
175 (16 9 sin x)2
(16 9 sin x)
2
16 tan x / 2 9
+C
+
tan–1
(175)3 / 2
175
Integrals of the form
Example 16.
...(2)
Integrating both sides of (2) w.r.t. 'x', we get
16
dx
175
dx
A=–
9 16 9 sin x
9 (16 9 sin x)2
175
dx
9 (16 9 sin x)2
=A+
dA
b
1
a 2 b2
2 ,
dx
a b a sin x b(b a sin x)
Integrating both sides w.r.t. ‘x’, we get
A=
dA
16
1
175
dx
9 (16 9 sin x) 9(16 9 sin x)2
cos x
dA bsin x a
b a sin x
dx (b a sin x)2
Now let, A =
...(1)
1.59
...(1)
acos x bsin x c
dcos x esin x f dx
Since cos x, sin x, and 1 are linearly independent, therefore
we can find real numbers p, q, and r such that
a cos x + b sin x + c p (d cos x + e sin x + f)
+ q(–d sin x + e cos x) + r
...(1)
Equating the coefficients of cos x and sin x on both
sides, and also equating the constant terms on both
sides, we have
a = pd + de, b = pe – qd, c = pf + r
....(2)
Solving the equations (2) we get the values of p, q and
r. Substituting the expression the expression for a cos x
+ b sin x + c from (1) in the integrand, we have
a cos x b sin x c dx
I=
d cos x e sin x f
1.60
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
r dx
= p + q d sin x e cos x dx +
d cos x e sin x f
d cos x e sin x f
= p + q lnd cos x + e sin x + f + rI1
...(3)
dx
.
where I1 =
d cos x e sin x f
To evaluate I1 we use the substitution tanx/2 = t.
Substituting the value of I1 in (3), we get the expression
for I from (3).
Note: We could have used the substitution
tanx/2 = t from the very beginning.
Example 17.
Evaluate
2 cos x 3 sin x 4
dx
3 cos x 4 sin x 5
Let
2 cos x + 3 sin x + 4 (3 cos x + 4 sin x + 5)
+ m (– 3 sin x + 4 cos x) + n
...(1)
Equating the coefficients of cos x and sin x on both
sides of (1), and also equating the constant terms on
both sides, we have
2 = 3 + 4m, 3 = 4 – 3m, 4 = 5 + n
...(2)
Solving the system of equations (2), we have
= 18 , m = – 1 , n = 2 .
...(3)
5
25
25
Substituting the expression for 2 cos x + 3 sin x + 4
from (1), the given integrand and the values of , m, n
from (3), we have
2 cos x 3 sin x 4
3 cos x 4 sin x 5 dx
18 x – 1
=
ln3 cos x + 4 sin x + 5
25
25
dx
2
+
5 3 cos x 4 sin x 5
18 x – 1
=
ln3 cos x + 4 sin x + 5
25
25
2
+
tan[(x – (tan–14/3))/2] + C.
25
2 sin x 3 cos x
dx
Example 18. Evaluate
3 sin x 4 cos x
Solution Let 2 sin x + 3 cos x
= (denominator) + m(derivative of denominator)
= (3 sin x + 4 cos x) + m(3 cos x – 4 sin x)
= (3 – 4m) sin x + (4 + 3m) cos x
Now comparing the coefficients of sin x and cos x of
both sides, we get
3 – 4m = 2 and 4 + 3m = 3,
Solution
18
1
,m
.
25
25
2 sin x + 3 cos x
18
1
=
(3 sin x + 4 cos x) +
(3 cos x – 4 sin x)
25
25
18
1 3 cos x 3 sin x
I
dx
dx
25
25 3 sin x 4 cos x
18
1
x
ln (3 sin x 4 cos x) + C
25
25
3cos x 2
dx.
Example 19. Evaluate
sin x 2 cos x 3
3cos x 2
Solution Let = sin x 2 cos x 3 dx.
Let 3 cos x + 2 = (sin x + 2 cos x + 3)
+ (cos x – 2 sin x) + .
Comparing the coefficients of sin x, cos x and constant term on both sides, we get
– 2 = 0, 2 + = 3, 3 + = 2
6
3
8
, and = –
5
5
5
(sin x 2 cos x 3) (cos x 2 sin x)
=
dx
sin x 2 cos x 3
cos x 2sin x
dx
= dx
sin x 2 cos x 3
1
+
dx
sin x 2 cos x 3
= x + log | sin x + 2 cos x + 3 | + 1, where
1
1 =
dx
sin x 2 cos x 3
=
Putting sin x =
we get 1 =
2 tan x / 2
1 tan 2 x / 2
,
cos
x
=
1 tan 2 x / 2
1 tan 2 x / 2
1
dx
2(1 tan 2 x / 2)
3
1 tan 2 x / 2
1 tan 2 x / 2
2 tan x / 2
1 tan2 x / 2
=
2tan x / 2 2 2 tan x / 2 3(1 tan x / 2) dx
=
tan x / 2 2 tan x / 2 5 dx
2
Putting tan
sec2 x / 2
2
x
1
x
= t and sec2
= dt
2
2
2
2
INDEFINITE INTEGRATION
where A and B are constants.
AQ(x) BQ(x)
dx
Then I =
Q(x)
Q(x)
dx
= A dx B
Q(x)
= Ax + B ln |Q(x) | + C
From (1), by comparing coefficients of same type of
terms, one gets constants A and B.
x
dx = 2 dt, we get
2
2dt
1 = 2
t 2t 5
dt
2
t 1
=2
tan –1
2
2 =
(t 1) 2
2
2
or sec2
tan x 1
2
= tan –1
2
Hence, = x + log | sin x + 2 cos x + 3 |
tan x 1
2
+ tan–1
+ C,
2
6
3
8
where = , = and = – .
5
5
5
Example 20.
Evaluate
(5 sin x 6)dx
.
sin x 2 cos x 3
Let 5 sin x + 6 = A(sin x + 2 cos x + 3)
+ B(cos x – 2 sin x) + C
Equating the coefficients of sin x, cos x and constant
term, we get
A 2B 5
2A B 0 A = 1, B = –2, C = 3
3A C 6
(cosx 2sinx)dx
dx
x 2n | sin x 2 cos x 3 | 3I1
x
x
t sec 2 dx 2dt
2
2
2dt
2dt
So, I1 2
=
( t 1) 2 4
t 2t 5
In I1, put tan
1 tan x
2 C
.
2
Example 21.
Solution
Evaluate
To evaluate I =
assume P(x) = AQ(x) +BQ(x)
I=
4e x 6e x
dx.
9e x 4e x
P(x)
Q(x) dx ,
...(1)
x
x
A(9e x 4e x ) B(9e x 4e x )
dx
9e x 4e x
9e x 4e x
dx
9e x 4e x
= Ax + B ln | 9ex – 4e–x | + C
19
35
ln | 9e x 4e x | C .
= x
36
36
dx B
Note:
ae x be x c
dx , we can find real
pe x qe x r
numbers l, m, and n such that a ex + b e–x + c
l (a ex + b e–x + c) + m(a ex – b e–x ) + n
and integrate the three integrals.
cos x sin x
dx
Integrals of the forms
f (sin 2 x)
d
We have
(sinx ± cos x) = cos x sin x
dx
and (sin x ± cos x)2 = 1 ± sin 2x.
Example 22. Evaluate
To evaluate
t 1
1
C = tan
= tan
2
4e x 6e x
9e 4e
Denominator Q(x) = 9ex – 4e–x
Numerator P(x) = 4ex + 6e–x
As Q(x) = 9ex + 4e–x, we take
4ex + 6e–x = A(9e x – 4e–x) + B (9ex + 4e–x)
By comparing the coefficients of ex and e–x, we get
4 = 9A + 9B
6 = –4A + 4B
19
35
A= , B
36
36
=A
sinx 2cosx 3 3 sinx cosx 3
1
In the present problem, I =
Solution
I = dx 2
1.61
(cos x sin x ) (3 4 sin 2x )dx .
Solution I = (cos x sin x ) (3 4 sin 2 x )dx
1.62
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Here derivative of sin x + cos x is cos x – sin x
and 3 + 4 sin 2x = 3 + 4((sin x + cos x)2–1)
Put sin x + cos x = t
(cos x – sin x )dx = dt
2
So I = (3 4( t 1)dt
=
t 2
[4t 3] C
3
sin x cos x
2
[ 4(sin x cos x ) 3] + C
=
3
sin x cos x
=
(1 4sin 2x) C .
3
Evaluate I = cos x – sin x dx
sin 2 x
Solution Put sin x + cos x = u
(cosx – sin x) dx = du
Also, u2 = 1 + sin 2x
Hence, I = du
u2 1
Example 23.
2
= ln u u 1 + C, where u = sin x + cos x.
Example 24.
Evaluate I =
sin x 2 cos x
dx
9 16 sin 2x
Solution Let sin x + 2 cos x
= (cos x – sin x) + m (sin x + cos x)
– + m = 1 and + m = 2
= 1/2 and m = 3/2
1 cos x sin x dx 3 sin x cos x dx
+
2 9 16 sin 2x
2 9 16 sin 2x
Put sinx + cos x = u and sin x – cos x = v
(cos x – sin x) dx = du (cos x + sin x) dx = dv
du
3
dv
1
I = 2 9 16 (u 2 – 1) + 2 9 16 (1 v 2 )
1
du
3
dv
=
+
2
2 16u – 7 2 25 v 2
Now, the above integrals can be evaluated easily.
I=
Example 25.
Solution
Evaluate tan x dx.
I = tan x dx.
= 1 [ ( tan x cot x ) dx + ( tan x cot x ) dx] .
2
sin x cos x
Now, I1 = ( tan x cot x ) dx = sin x cos x dx
=
2d(sin x cos x)
d(sin x cos x)
= 2
sin 2x
1 (1 sin 2x)
= 2
d(sin x cos x)
1 (sin x cos x)2
dz ,
where z = sin x – cos x
1 z2
= 2 sin–1z + c1 = 2 sin–1(sinx – cos x) + c1.
= 2
I2 = ( tan x – cos x ) dx =
=
sin x cos x dx
sin x cos x
– d(sin x cos x )
d(sin x cos x)
= – 2
sin x2 x
(1 sin 2x) – 1
= – 2
= – 2
d(sin x cos x )
(sin x cos x ) 2 – 1
dz
, where z = sin x + cos x
z 2 1
= – 2 ln(z +
z 2 1 ) + c2
= – 2 ln (sin x + cos x +
sin 2 x ) + c2
1
1
I + I
2 1 2 2
= 1 sin–1 (sin x – cos x)
2
I=
– 1 ln (sin x + cos x + sin 2 x ) + C.
2
dx
.
cos x cos ecx
sin x
dx
dx
Solution I =
sin x·cos x 1
cos x 1
sin x
(sin x cos x ) (sin x cos x )
2
sin
x
dx
dx
=
2 sin 2 x
2 sin 2 x
= sin x cos x dx + sin x cos x
2 sin 2x
2 sin 2 x
d (sin x cos x)
d (sin x cos x)
dx
dx
=
dx –
dx
3 (1 sin 2x)
1 (1 sin 2x)
d (sin x cos x)
d (sin x cos x)
dx
dx
dx –
dx
=
3 (sin x cos x )2
1 (sin x cos x )2
= I1 – I2.
For I1, put sin x – cos x = z. Then
dz
I1 = dz 2 =
3z
( 3)2 z 2
1
3 z
c1
ln
=
23 3 z
Example 26.
Evaluate
INDEFINITE INTEGRATION
=
1
ln 3 sin x cos x c1
2 3 3 sin x cos x
For I2, put sinx + cos x = t. Then
dt
= tan–1 t + c2
I2 =
1 t2
= tan–1 (sin x + cos x) + c2
I = I 1 – I2
1
ln 3 sin x cos x – tan–1 (sinx + cosx) + C.
=
2 3 3 sin x cos x
dx
Example 27. Evaluate
2 sin x sec x
dx
Solution Let I = 2 sin x sec x
cos xdx
1 2 cos xdx
=
=
sin 2 x 1 2 1 sin 2 x
(cos x sin x) (cos x sin x)
=
dx
(sin 2 cos2 x 2 sin x cos x)
1 cos x sin x
(cos x sin x)
=
dx +
dx
2 (sin x cos x)2
(sin x cos x)2
1 dv
1
dx
+
, where v = sin x + cos x
=
2 sin x cos x 2 v 2
1
dx
1
=
2 2 1 sin x 1 cos x 2 v
2
2
1
dx
1
=
2 2 sin x 2(sin x cos x)
4
1
=
ln cosec x – cot x
4
4
2 2
1
–
+C
2(sin x cos x)
Integrals of the forms
sec x a dx , cosec x a dx ,
tan x a dx , cot x a dx
2
2
2
2
Method :
(i) Write sec 2 x a
2
sec x
2
sec x a
a
sec 2 a
sec 2 x a
cos x
2
1 a cos x
1.63
In the first part, put u = tanx and in the second part, put
v = sinx.
(ii) In case of cosec2 x a proceed as in (i) and put
put u = cot x in the first part, and v = sinx in the
second part.
(iii) In case of tan 2 x a or cot 2 x a , change
tan2x into sec2x – 1 and cot2x into cosec2x – 1 and
then proceed as in (i) and (ii).
Evaluate tan 2 x 3 dx
Example 28.
2
2
tan x 3 dx = sec x 4 dx
Solution
2
= sec x 4 dx
sec 2 x 4
sec 2 x dx
tan 2 x 3
=
–
4
dx
sec 2 x 4
sec 2 x dx
cos x
–4
dx
tan 2 x 3
1 4 cos 2 x
Put tan x = u
Put sinx = v
cosx dx = dv
sec2x dx = du
Now, the above integrals can be evaluated easily.
=
Evaluate
Example 29.
2 2 cos 2 x
3 cos 2 x dx
=
dx
cos x
cos 2 x
Solution
sec x 1 dx =
=
2
=
2
3 cos 2 x dx
cos x
2
2
sec 2 x 1 dx
sec 2 x 1
sec 2 x
dx
2
2
tan x 2 dx sec x 1
2
cos xdx
sec x dx
2
2
1 cos2 x
tan x 2
Put tan x = u
Put sinx = v
cosx dx = dv
sec2x dx = du
Now, the above integrals can be evaluated easily.
=
Example 30.
Solution
I
Evaluate
cos ec x 2 dx .
2
cos ec 2 x 2
cos ec x 2 dx
2
1.64
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
cos ec 2 x
cos ec x 2
2
dx 2
dx
cos ec x 2
cos ec 2 x
2
sin x
cot x 1 dx 2 1 2 sin x dx
Let I1
Let I2 2
2
2
...(1)
Now I2
cos ec 2 x
2
then dz = –cosec3x dx
dz
z2 1
ln(cot x
2
1 2sin x
dx 2
sin x
2 cos x 1 dx
2
Put z 2 cos x, then dz 2 sin x dx
cot x 1 dx , put z = cotx,
Now I1
sin x
ln(z z2 1)
2
dz
z 1 2 ln(z z 1)
2
2
2 ln( 2 cos x 2 cos 2 x 1)
2 ln ( 2 cos x cos2x )
From (1), I = I1 + I2
= –ln(cot x + cot 2 x 1 )
cot 2 x 1)
+ 2 ln( 2 cos x cos2x ) C .
I
1.
Evaluate the following integrals :
dx
dx
(i)
(ii)
2
2
4 5 sin 2 x
cos x 3sin x
dx
(iii)
(3 sin x 4 cos x)2
1
dx
(iv)
(sin x 2 cos x) (2 sin x cos x)
Evaluate the following integrals :
dx
sin x
dx
(i)
(ii)
2
sin 3x
3 cos x
dx
cos x
dx
(iii)
(iv)
(sin x cos x)2
cos 3x
Evaluate the following integrals :
1
dx
dx
(i)
(ii)
5 4 cos x
13 12 cos x
dx
dx
(iii)
(iv)
5 4 sin x
1 2sin x
Evaluate the following integrals :
dx
(i)
3 2 sin x cos x
dx
(ii)
2 sin x 3 cos x 5
dx
(iii)
2 sin x cos x 3
dx
(iv)
5 4 sin x 3 cos x
Evaluate the following integrals :
d
(ii) (1 sin ) d
(i)
2 3 cos 2
2 cos 3
6.
7.
4.
8.
5.
(iv)
3.
cos x dx
5 3 cos x
2.
cos x
dx
cos 2 x
Evaluate the following integrals :
d
dx
(ii)
(i)
1 sin 4
sin 4 x cos 4 x
dx
(iii)
1 cos 4 x
d
(iv)
(a cos 2 b sin 2 ) 2
Evaluate the following integrals :
(iii)
9.
3sin x2 cos x
(i)
x dx
3cos x2sin x dx (ii) 2 sincos
x 3 cos x
(iii)
5sin x 4 cos x dx
(iv)
3 4sin x 5cos x dx
4sin x 5cos x
6 3sin x 14 cos x
Evaluate the following integrals :
cos x sin x
dx
9sin 2x 6
cos x
4 dx
2 sin 2x
(i)
(iii)
cos x sin x (2 + 2 sin 2x) dx
(iv)
sin x tan x
(ii)
cos x sin x
dx
Evaluate the following integrals :
cos x
1
dx (ii)
(i)
a b cot x
cos x sin x
INDEFINITE INTEGRATION
1 sin 2x
1 cos2x dx
(iii)
10. Evaluate the following integrals :
2 tan x dx (ii) 1 sec x dx
(iii) 1 cose c x dx
(iv) cose c x cos2x dx
2
(i)
11. Evaluate the following integrals :
1.11 INTEGRATION BY PARTS
Integration by parts is a method to compute integrals
of the form
u(x).v(x) dx in which u can be
1.65
1
(i)
(cos x 2sin x) dx
(ii)
(sin x 2 cos x) dx
(iii)
(5 4 cos )2
(iv)
sin x cos x
2
dx
2
2
2
cos d
dx
6
6
and certain integrals containing inverse trigonometric
functions are evaluated by means of integration by parts.
Note:
differentiated easily and g can be integrated easily.
While using integration by parts, we choose u
and v such that
Formula for integration by parts
(a)
vdx is simple and
(b)
d x . v d x dx is simple to integrate.
du
u.v dx = u v dx d x . v d x dx
The formula comes from the product rule of
differentiation
d
du
dw
(u.w) .w u.
dx
dx
dx
Integrating both sides w.r.t. x
du
dw
u.w .w dx u.
dx
dx
dx
dw
du
u. dx dx u.w dx .w dx
dw
Let
= v, then w = v dx
dx
du
Hence, u.v dx = u v dx d x . v d x dx
It is most frequently used in the integration of
expressions that may be represented in the form of a
product of two factors u and v in such a way that the
finding of the function v from its derivative, and the
evaluation of the integral vdu should, taken together,
be a simpler problem.
The rule for integration by parts is widely used for
integrals of the form.
xk sin ax dx, xk cos ax dx
xk eax dx, xk ln x dx
du
This is generally obtained, by keeping the order of u
and v as per the order of the letters in ILATE, where
I Inverse function, L Logarithmic function,
A Algebraic function, T Trigonometric function
and E Exponential function. When the integrand
of an integration by parts problem consists of the
product of two different types of functions, we should
let u designate the function that appears first in the
acronym ILATE, and let v denote the rest.
However, to become skilled at breaking up a given
integrand into the factors u and v, one has to solve
problems; we shall show how this is done in a number
of cases.
For example, since the integrand of x cos x dx is
the product of the algebraic function x with the
trigonometric function cos x, we should let u = x
and v = cos x.
d( x)
x cos xdx = x cos xdx dx cos xdx dx
= x sin x – 1. sin x dx
= x sin x – (–cos x) + C
= x sin x + cos x + C
1.66
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 1.
Solution
Integrate xe x dx
I x e x dx dx e x dx dx
dx
xe x 1.e x dx = xex – ex + C
Note:
In the above integral, instead of taking x as the first
function and ex as the second, if we take ex as the first
function and x as the second, then applying the rule
for integration by parts we get
x
x 1 2
x 1 2
(e . x)dx e . 2 x e . 2 x dx .
The integral 1 e x x 2 dx on the right side is more
2
complicated than the one we started with, for it involves
x2 instead of x.
Thus, while applying the rule for integration by parts
to the product of two functions, care should be taken
to choose properly the first function, i.e., the function
to be differentiated.
A little practice and experience will enable the student
to make the right choice.
Note:
1.
When determining the function v from its
derivative, we can take any arbitrary constant,
since it does not enter into the final result. This
can be seen by adding the arbitrary constant C in
the formula of integration by parts. It is however
convenient to consider this constant equal to zero.
u.v dx
du
= u vdx C1 d x . vdx C1 dx
du
. v d x dx – u C
= u vdx + u C1
1
d
x
du
= u vdx d x . v d x dx
We see that the constant C1 cancels out, thus giving
the same solution obtained by omitting C1 , thereby
ju st i fyin g t he om issi on of t he const a n t of
integration when calculating v in integration by
parts. Also note that v dx should be taken as same
in both terms.
2. Sometimes, a constant can be used to simplify
the calculations. For example,
1
ln(1 x) dx ln(x 1).(x 1) x 1 .(x 1)dx
= (x + 1). ln(x+1) – x + C.
Thus, we have simplified the computation of the second
term by introducing a constant of integration C1 = 1.
3. The method sometimes fails because it leads
back to the original integral. For example, let
us try to integrate x dx by parts.
1
1
1
x dx x . x – x x dx
1
2
x dx = – 1 + x dx + C.
1
1
...(1)
and we are back where we started.
This example is often used to illustrate the
importance of paying attention to the arbitr ary
constant C. If (1) is written without C, it leads to
the equation x dx = –1 + x dx, which is false.
1
Example 2.
Solution
1
Evaluate
x tan x dx
1
x tan x dx
1
1
x2
x2
–
dx
2 .
1 x
2
2
1 x2 1 1
x2
tan –1 x –
dx
=
2
2
x2 1
1
1
x2
1 2
tan –1 x –
dx
=
2
x 1
2
1
x2
tan–1 x – [x – tan –1 x] + C.
=
2
2
= (tan –1 x)
Example 3.
Solution
Evaluate x ln(1 x) dx
x ln(1 x) dx
1
x2
x2
–
.
dx
x 1 2
2
1
x2
x2
=
ln (x + 1) –
dx
2 x 1
2
1 x2 1 1
x2
=
ln (x + 1) –
dx
2
2
x 1
= ln (x + 1) .
1.67
INDEFINITE INTEGRATION
1 x2 1
1
x2
ln (x + 1) –
+
dx
2 x 1
x 1
2
1
1
x2
dx
=
ln (x + 1) – (x 1)
x 1
2
2
2
1 x
x2
=
ln (x + 1) – 2 x ln | x 1| + C
2
2
=
the integral of ln x, we get
ln xdx = ye dy
y
y ln x,x e x , dx e y dy
= yey – ey + C
= x ln x – x + C.
For the integral of cos–1 x we get
cos x dx = y cos y – cos y dy
Note:
1
Very often an integral involving a single logarithmic
function or a single inverse circular function can be
evaluated by the application of the rule for integration
by parts, by considering the integral as the product of
the given function and unity; and taking the given
function as the first function and unity as the second.
The principle is illustrated below.
Example 4. Evaluate ln x dx
Solution
ln x . 1 dx
Example 5.
Solution
= x cos–1 x – cos y dy
= x cos–1 x – sin y + C
= x cos–1 x – sin (cos–1 x) + C.
Generally, we can integrate f–1(x), the inverse function
of f(x), if we can integrate f(x).
Evaluate ln( x x 2 a 2 )dx
Example 6.
d
ln x dx (ln x) . dx dx
dx
1
= ln x . x – . x dx
x
= x ln x – dx
= x ln x – x + C.
where y = cos–1 x, x = cos y
Integrating Inverses of Functions
Integration by parts leads to a rule for integrating
inverses that usually gives good results :
f (x)dx yf (y)dy
1
y = f–1(x), x = f(y), dx = f(y)dy
= yf(y) – f ( y)dy = xf–1(x) – f ( y)dy
The idea is to take the most complicated part of the
integral, in this case f–1(x), and simplify it first. For
d
–
ln x x 2 a 2
dx
. 1dx dx
ln(x x 2 a 2 .x
x a .x dx
x ln(x x 2 a 2 )
tan 1 x dx d tan 1 x dx dx
dx
1
–1
= tan x . x –
. x dx
1 x2
= x tan–1x – 1 2 x dx
2 1 x2
1
= x tan–1x – ln(1 + x2) + C.
2
x 2 a 2 ).1dx
ln(x x 2 a 2 ) 1dx
Evaluate tan 1 x dx
I tan 1 x . 1 dx
I ln(x
Solution
To evaluate
1
2
2
x dx
(x 2 a 2 )
x dx
(x a ) , put x2 + a2 = z2, so that
2
2
x dx = z dz.
I x ln( x x 2 a 2 ) x 2 a 2 + C.
Example 7.
Solution
Evaluate
We have
xn ln x.
xn ln x dx = (ln x). xn dx
x n 1
1 x n 1
.
dx,
n 1
x n 1
n
[Integrating by parts taking x as the second function]
= (ln x) .
= (ln x).
x n 1
x
dx
n 1
n 1
n 1
= (ln x) .
x
x n 1
+ C.
n 1 (n 1)2
1.68
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Evaluate
Example 8.
Solution
ln(sec 1 x)dx
x (x 2 1)
Put sec–1 x = t so that
.
1
x (x 2 1)
dx = dt.
Then the given integral
=
ln t dt = (ln t). 1 dt
= (ln t). t –
= t ln t –
1
t dt
t
dt = t ln t – t +C
sec 1 x
t
–1
= t ln + C = (sec x) ln
+ C.
e
e
1 x
Evaluate tan 1
dx.
1 x
Solution Put x = cos so that dx = – sin d.
The given integral
Example 9.
1 1 cos
(– sin ) d
= tan
1 cos
{tan–1 (tan 1/2 ) sin d
1
1
= – sin d= – sin d
2
2
=–
1
= – [. (– cos ) – (– cos ) d]
2
1
= [cos – cos d]
2
1
= [cos – sin ] + C
2
1
= [x cos–1 x – (1 x 2 ) + C.
2
Solution
Evaluate
We have
x2 tan–1 x dx
x2 tan–1 x dx
x3
x3
1
.
tan–1 x –
3
3 1 x2
Integrating by parts taking x2 as the second function
=
3
=
2
x
1 x(x 1) x
tan–1 x –
dx
3
3
1 x2
Note: The operation of integration by parts
can be used successively, i.e. we can repeat it several
times.
Example 11.
= t (ln t – 1) + C = t. (ln t – ln e) + C
Example 10.
1
1 1
2x
x3
tan–1 x –
x dx + .
dx
3
3 2 1 x2
3
1
1
x3
tan–1 x – x2 + ln (1 + x2) + C.
=
6
6
3
=
Find x2e–x dx.
x2e–x dx = x2(–e–x) – (–e–x) 2x dx)
= –x2e–x + 2 xe–x dx
= – x2 e–x + 2 [x(–e–x) – (–e–x) dx]
2 –x
–x
–x
Solution
= x e – 2xe – 2e + C
= –e–x (x2 + 2x + 2) + C.
Example 12.
3 2x
Evaluate I = x e dx
e 2x
e 2x
3x 2 . dx
2
2
2x
2x
e
1 3 2x 3 2 e
= 2 x e 2 x 2 2 x 2 dx
1 x 3e 2 x 3 x 2 e 2 x 3 xe 2 x dx
=
2
4
2
1
3
3 e2x
e 2y
= x 3e2x x 2e2x x.
dx
1.
2
4
2 2
2
I = x3e 2x dx x3
Solution
1 x 3e 2 x 3 x 2 e 2 x 3 xe 2 x 3 e 2 x
dx
2
4
4
4
1 3 2x 3 2 2x 3 2x 3 2x
= x e x e xe e + C
2
4
4
4
1 e 2 x 4x 3 6x 2 6x 3
+C
=
8
=
Example 13.
Solution
Evaluate ( p 6p) sin p dp
3
I = (p 6p)sin p dp
3
= – (p3 + 6p) cos p + (3p 6) cos p dp
2
= – (p3 + 6p) cos p + (3p2 + 6) sin p – 6p sin p dp
= – (p3 + 6p) cos p + (3p2 + 6) sin p
[ p cos p cos p dp]
= – (p3 + 6p) cos p + (3p2 + 6) sin p + 6p cos p
– 6 sin p + C
= 3p2 sin p – (p3 + 6p) cos p + 6p cos p + C
1.69
INDEFINITE INTEGRATION
= 3p2 sin p – p3 cos p + C
= (3 sin p – p cos p) p2 + C.
Example 16.
Evaluate sec3x dx
Example 14.
Solution
Here both functions are trigonometric and integral of
sec2x is tanx which is simpler than the integral of secx
which is ln(secx + tanx). Therefore, we take sec2x as y.
= x x2 a2 –
=x
= secx tanx – secx(sec2x – 1)dx
2I = secx tanx + ln(secx + tanx)
sec x tan x 1
ln(sec x tan x) C .
I
2
2
Note: While performing integration by parts,
sometimes the original integral appears on the right
hand side, which we name as I and transfer on the left
hand side.
Evaluate I =
Example 15.
2
2
Integrating by parts, taking the first
Solution
function
a x dx
2
a x
2
a x .1dx
=
1 (–2x)
x dx
a2 x2 x – · 2
2
2 a x
2
2
Solution
2
2
2
Evaluate I =
x a dx.
2
2
Integrating by parts, taking the first
function x 2 a 2 , and second function 1, we have
2x
x dx
I = x 2 a 2 .x – 1
2 x2 a2
= x x2 a2 –
Note:
dx
a2 x2
= x a 2 x 2 – I + a2 sin–1(x/a) + C1
2I = x a 2 x 2 + a2 sin–1(x/a) + C1
1
or I = 1 x a 2 x 2 + a2sin–1 (x/a) + C,
2
2
where we have written C for 1/2 C1.
or
Example 17.
(x 2 a 2 ) – a 2
x2 a 2
dx
2I = x x 2 a 2 + a2 ln x 2 a 2 + x + C1
1
1 2
or I = x x 2 a 2 +
a ln x 2 a 2 + x + C,
2
2
where we have written C for 1/2 C1.
dx
= x a x – a x dx + a2
2
2
or
2
a2 x2
2
2
2
= x x 2 a 2 – I + a2 ln x a + x + C1
x 2
(a 2 x 2 ) – a 2
x a dx – a2 x2dx a 2 ,
dx
= x x 2 a 2 – x 2 a 2 dx + a2 2 2
x a
a x dx
=x a x –
2
dx,
2I = x x 2 a 2 – a2 ln x 2 a 2 + x + C1
1
1
or I = x x 2 a 2 – a2 ln x 2 a 2 + x + C
2
2
where we have written C for 1/2 C1.
2
x2 a2
or
2
2
2
= x a x –
x2 a2 –
(x 2 a 2 ) a 2
= x x 2 a 2 – I – a2 ln x 2 a 2 + x,
and second function 1, we have
I=
2
function x a 2 , and second function 1, we have
1·
2x
x dx
I = x2 a2 . x – 2
2
x a2
= secx tanx – secx tan2x dx
or
2
2
I = secx tanx – (secx tanx) tanx dx
= secx tanx – I + secx dx [ I = sec3x dx]
x a dx.
Integrating by parts, taking the first
Solution
I = sec3x dx = secx . sec2x dx
Evaluate I =
e ax
(i)
e . sin bx dx = 2 2 (a sin bx b cos bx) + C
a b
e ax
(ii)
eax .cos bx dx = 2 2 (a cos bx + b sin bx) + C
a b
Proof:
Let us evaluate I = eax sin bx dx.
Integrating by parts taking sin x as the second
function, we get
ax
1.70
I=
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
e ax cos bx
cos bx
ae ax
dx
b
b
e ax cos bx a ax
e cos bx dx.
b
b
Again integrating by parts taking cos bx as the second
function, we get
=–
I=–
or
e ax cos bx a e ax sin bx
sin bx
ae ax
dx
b
b
b
b
(a sin bx – b cos bx).
Thus, e2x sin x dx =
a
and Q = eax sin bx dx.
P + iQ = eax (cos bx + i sin bx) dx
= eax eibx dx = e(a+ib)x dx
a2
e ax
1 2 I = 2 (a sin bx – b cos bx)
b
b
1 2 2
1
(a + b ) I = 2 eax (a sin bx – b cos bx).
b2
b
eax
a
Let P = eax cos bx dx
e ax
a2
(a
sin
bx
–
b
cos
bx)
–
I.
b2
b2
b
the integrals eax cos bx dx and Q = eax sin bx dx.
a2
Transposing the term – 2 I to the left hand side, we get
b
I=
b
f (x) dx u(x) dx i v(x) dx
Now this provides an alternative method to evaluate
or
b
a
e ax cos bx a ax
2 e sin bx
I=–
b
b
I=
f(x) dx = u(x) dx + i v(x) dx = F(x) + C
The definite integral of a complex function of a real
variable, f(x) = u(x) + iv(x), is defined as follows :
a2
– 2 eax sin bx dx
b
or
is an antiderivative of the function f(x), then any
antiderivative of f(x) is of the form F(x) + C, where C is
an arbitrary complex constant. We will call the
expression F(x) + C the indefinite integral of a complex
function of a real variable and we will write
1 2x
e (2 sin x – cos x) + C.
5
We can proceed similarly to evaluate .
eax
cos bxdx.
Integration of a complex function of a real
variable
We can define a complex function as
f(x) = u(x) + iv(x)
of a real variable x and also its derivative
f (x) = u(x) + iv(x).
A function F(x) = U(x) + iV(x) is called an antiderivative
of a complex function of a real variable f(x) if
F(x) = f(x)
...(1)
that is, if U(x) + iV(x) = u(x) + iv(x)
...(2)
From (2) it follows that U(x) = u(x), V(x) = v(x), that is,
U(x) is an antiderivative of u(x) and V(x) is an
antiderivative of v(x).
It follows, from this definition, that if
F(x) = U(x) + iV(x)
=
e(a ib)x
a ib
C iD 2
eax eibx + C + iD
a ib
a b2
=
e ax
(a – ib) (cos bx + i sin bx) + C + iD.
a b2
2
(aeax cosbx beax sinbx) i(aeax sinbx beax cosbx)
a2 b2
+ C + iD.
Equating real and imaginary parts we get the values of
P and Q as before.
eax (a cos bx b sin bx)
P
a 2 b2
=
Q=
e ax (a sin bx b cos bx)
.
a 2 b2
Example 18.
Solution
Integrate
cos3 x
dx .
e3x
I = e–3x cos3 x dx
1
e–3x(cos 3x + 3 cos x)dx
4
1
= [ e–3x cos 3x dx + 3 e–3x cos x dx]
4
=
3x
1 e
e3x
(3cos3x 3sin3x) 3
(3cosx sinx) +C
4 18
10
INDEFINITE INTEGRATION
e 3 x 1
3
(sin 3x cos 3x) (sin x 3 cos x) + C.
8 3
5
Example 19.
Solution
=
Evaluate I =
1
I=
2
e2x sin 3x cos x dx.
x ln x
1
[
2
1 e 2x
e x
sin(4x tan 1 2)
sin(2x tan 1 1) + C
2 20
8
e 2x 1
1
sin(4x tan 1 2)
sin 2x + C.
2 20
4
8
e2x sin 4x dx + e2x sin 2x dx]
Example 20. Evaluate xex cos x dx
Solution xex cos x dx
=x
ex
(cos x + sin x)
2
ex
(cos x + sin x) dx
2
= xe (1 i ) x dx xe
1 i
I1 =
x ln x
1 x dx
2
Put x = sin, then dx = cos d
x ln x
dx sin ln sin d
1 x2
= – cos ln sin – – cos . cot d
= – cos ln sin +
= – cos ln sin +
cos2
d
sin
1 sin 2
d
sin
= –cos ln sin + (cosec – sin) d
= –cos ln sin + ln(cosec – cot) + cos
ex
ex
– 1
(cos x + sin x) +
(sin x – cos x) + C.
2
2 2
Alternative :
Let I2 = xex cos x dx and I2 = xex sin x dx
Consider I1 + i I2 = xex (cosx + i sin x) dx
= xex · eix dx
(1 i ) x
...(1)
2
Now in order to evaluate
=
1.
x
2
[Taking sin–1x as the first function]
1 2x
e (sin 4x + sin 2x) dx
2
ex
(cos x + sin x) –
2
I = (x ln x – x)sin–1x
1 x dx 1 x dx
–
e2x . 2 sin 3x cos x dx
=x
Evaluate I = ln x . sin–1x dx.
Example 21.
Solution
1.71
(1 i ) x
– e
+C+iD
(1 i)2
x ex cos x dx
(1 i)x
e(1 i)x
e
–
C iD
= real part of x
2i
1 i
xe x (cos x i sin x)
= real part of
(1 – i)
2
–
e x (cos x i sin x )
i
2
=
xe x
ex
(cos x + sin x) –
sin x + C.
2
2
+ C + iD
1 1 x2
2
1 x 2 ln x ln
1 x
x
From (1),
1 1 x2
+C.
I = (x lnx – x)sin–1x + 1 x 2 ln x – ln
x
Example 22.
Evaluate
x3 sin 1 x
(1 x )
2 3/2
x3 sin 1 x
Solution Let I = (1 x 2 )3/2 dx.
Put x = sin or sin–1 x = .
Then dx = cos d
I=
sin 3
cos cos d=
3
sin (1 cos2 )
d.
cos2
[sec tan – sin ] d
= (sec + cos ) – 1. (sec + cos ) d
=
(Integrating by parts taking sec tan
– sin as the second function)
1.72
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
= (sec + cos ) – ln|sec + cos | – sin + C
we have
–
2
sin 2 d
I=
–
2
[ sin–1 x + cos–1 x = /2]
= 2 2 sin 2 d
2
4
= sin 2 d – sin 2 d
= 4 cos 2 cos 2 d cos 2
2
2
2
4 cos 2 sin 2 cos 2 C
=
2
4
2
2
2
= [ (2 sin2 – 1) + sin cos ] + 1 2 sin C
2
2
1
–1
x x 1 x ] + 2x C
= [(2x – 1) sin
2
2
–1
x x 1 x ] – x + C.
= [(2x – 1) sin
1
sin
1 cos
– ln
– sin + C
=
cos cos
cos
=
=
=
=
(1 cos2 )
– ln
cos
1 sin
(1 sin 2 )
(1 1 sin 2 )
– sin + C
1/2
1 sin
– ln
– sin + C
2
(1 sin )
1 sin
2 sin 2
2
(1 sin )
2 x2
(1 x 2 )
1
1 sin
ln
– sin + C
2
1 sin
sin–1 x – x –
1 x
1
ln
+ C.
1 x
2
Example 23. Evaluate I =
sin–1 x – cos–1 x dx
.
sin–1 x cos–1 x
Putting x = sin2 and dx = 2 sin cos d,
Solution
D
1.
Evaluate ln(2 x 3)dx using integration by
parts. Similify the computation of v du by
2.
3
introducing a constant of integration C1 = when
2
going from dv to v.
1
Evaluate x tan x dx using integration by parts.
Simplify the computation of
3.
vdu by
1
introducing a constant of integration C1 = when
2
going from dv to v.
What equation results if integration by parts is
applied to the integral
1
x ln x dx with the choices
1
1
and dv =
dx ?
ln x
x
In what sense is this equation true ? In what
sense is it false ?
Evaluate the following integrals :
u=
4.
(i)
x ln xdx
3
2
1
dx
(ii) lnx ·
( x 1) 2
5.
6.
x dx
1 sin x
Evaluate the following integrals :
(iii)
ln (1 x ) dx
(iv)
(i)
sec x
(ii) x sin–1x
(iii)
sin
1 x
–1
–1
(iv) sin
x
1
2x
dx
1 x2
Prove that
(ln|u|)2 du = u(ln|u|)2 – 2u ln|u| + 2u + C.
3
7.
x
Prove that e dx
8.
Prove that
ex
3
3x
2
3
2 ex
dx .
3 x3
f(x) F(x) dx = f(x) F(x) – f(x) F(x) + f(x) F(x) dx
and generally f(n)(x) F(x) dx = f(n–1)(x) F(x)
– f(n–2)(x) F(x) + ... + (– 1)n f(x) F(n)(x) dx.
1.73
INDEFINITE INTEGRATION
J
9.
Evaluate the following integrals :
ln cos x
(i)
dx
cos 2 x
(ii) sin x . ln tan x dx
ln(1 + 2x + x ) dx
(iii)
2
4
12. Evaluate the following integrals :
2 sin x dx
(iii) e sinx sin2x dx
(i)
2
11. Evaluate the following integrals :
ln (x x a )dx
(ii) ln ( x 1 x ) dx
1 x
dx
(iii) x ln
1 x
(i)
2
2
(iv)
(iii)
sin
–1
ex{f(x) + f '(x)}dx
Integrating by parts ex f(x) dx, we have
ex f(x)dx = f(x)ex dx
= f(x)ex – f '(x) ex dx
Transposing the second integral to the left hand side
ex{f(x) + f '(x)}dx = exf(x) + C.
Alternatively, we may integrate by parts exf '(x) dx,
and derive the same result.
(ii)
sin 1 x
(1 x )
2 3/ 2
dx
2
2
16. Evaluate the following integrals :
3 x
(ii) x 3 cos x dx
(i) x e dx
(iii)
(i)
1
x sin x cos x dx (ii) x sec x tan x d x
(iii) x cos x cos 2 x dx
3
Proof:
x
x
dx (iv) x 3 tan 1 xdx
ax
1
(i)
ln x
x
e sin x sin 3x dx
cos 1 x
1 1 x
dx
dx
(iv) tan
3
1 x
x
15. Evaluate the following integrals :
(x 1) dx
e [f(x) + f (x)] dx = e . f(x) + C
(ii) [f(x) + xf (x)] dx = x f(x) + C
(ii)
cos x dx
(iii)
2
Using integration by parts we can establish the
following integrals :
x
sin (l n x) d x
(i)
1.12 SPECIAL INTEGRALS
(i)
(i)
14. Evaluate the following integrals :
2
2
2
13. Evaluate the following integrals :
x
cosec x ln sec x dx.
(ii) cos x ln(cosecx + cot x) dx
(iii) sin x . l n (sec x + tan x)d x
(iv) secx . ln(secx + tanx)dx
(i)
x
2x
x
(iv) e (1 x ) ln ( xe ) dx
10. Evaluate the following integrals :
x
3 cos 3x dx
(iv) e sin x dx
(ii)
x
x l n x dx
3
2
Note:
ex (x)dx, when (x) can be broken up as the sum of
two functions of x such that one is the differential
coefficient of the other, can be easily integrated as above.
[f(x) + xf (x)] dx
Integrating by parts f(x). 1 dx, we have
f(x) dx = f(x). x – f '(x). x dx
(ii)
Transposing the second integral to the left hand side
[f(x) + xf (x)] dx = x f(x) + C.
xe
Example 1.
Evaluate
dx.
(x 1)
x
2
xe
x
Solution We have (x 1)2 dx = (xex)
1
dx.
(x 1)2
1
as the second
(x 1)2
x
function and xe as the first function, we have
Integrating by parts taking
1.74
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
xe x
dx
(x 1)2
1
= (xex)
–
x 1
t
I=– e
1
(ex + xex) x 1 dx.
1
1
is –
]
(x 1)2
x 1
[Note that the integral of
=
xe x
(x 1)2
x
sin t
2
1
et
e cos x
=C–
=C–
.
1 cos t
1 x
x
Solution
x 1 x
e
x
C
+ C = ex
+ C.
= ex 1
x 1
x 1
x 1
Alternative :
1
Example 4.
xe
xe
+ ex dx = –
+ ex + C
x 1
x 1
We have
t
e 1 cos t (1 cos t) dt
1
e (x + 1) x 1 dx
x
=
=–
(1 cos t) sin t
dt
(1 cos t)2
x4 2
x
dx
Evaluate e
2 52
(1 x )
2 2
2
x (1 x ) 1 2x
I= e
dx
(1 x 2 )5 2
1
x
x
= e
2 32
2
(1 x )
1 x
xe x
(x 1) 1
dx e x
dx
2
(x 1)
(x 1)2
x
1 2x 2
dx
2 32
(1 x 2 )5 2
(1 x )
1
x 1
= e
dx
x
1
(x
1)2
1
x
x
=e
2 3 2 + C.
2
(1
x
)
1 x
ex [f(x) + f(x)] dx, where
Example 5.
1
1
f(x) =
= ex
+ C.
x 1
x 1
x 2 sin 2x
dx
Example 2. Evaluate e
1 cos 2x
x 2 sin 2x
Solution We have e 1 cos 2x dx
Solution
Evaluate sin 4x·e tan x dx
sin 4x·e
= 2 cos 2x·2
=
2
tan 2 x
dx
2
sin x
cos2 x e tan x dx
cos x
2 cos 2x cos x·2 tan x·e
2
tan 2 x
dx
2
sin 2x
x
dx
= e
1 cos 2x 1 cos 2x
= 2 cos 4x cos x·2 tan x sec x·e
2
2sin x cos x
x
dx
= e
2
2 cos2 x
2 cos x
Put tan2x = t 2 tan t sec2t dx= dt
4
= 2
=
ex [sec2 x + tan x] dx
ex[f(x) + f(x)] dx, where f(x) = tan x = ex tan x + C.
1
cos x
·
Example 3. Evaluate e
Solution
(x 1) 1 x2
Let cos–1x = t –
(x 1)2 1 x2
dx
1 x2
= dt
dx .
2
tan 2 x
dx
t
(1 t) e t dt
= 2 e (1 t)dt
2
(1 t) (1 t)
(1 t)3
= – 2
e t [t 1 2]dt
(1 t)3
1
2
t
dt
=– 2 e
2
(t
1)
(t
1)3
2
1
C = C – 2 e tan x cos4 x .
= – 2 e t
2
(t 1)
INDEFINITE INTEGRATION
Example 9.
Note:
ex (f + g + f + g) dx = ex (f + g) + C
Example 6. Evaluate I = ex (2 sec2x – 1) tan x dx
Solution I = ex (2 sec2x tan x – tan x) dx
= ex (2 sec2x tan x + sec2 x – sec2x – tan x) dx
= ex (sec2 x – tan x) + C.
esin x (xcos3 x sinx)
dx
cos2 x
Solution I = esin x (x cos x – tan x secx) dx
Put sin x = t cos x dx = dt
dt
dt
=
dx =
cos x
1 t2
t –1
ett
I = e sin t (1 t 2 )3/2 dt
t
t
–1
dt
= e sin t –
2 3/2
(1
t
)
1
1
1
t
–1
–
–
dt
= e sin t
2 3/2
2
2
(1
t
)
1 t
1 t
We have
Example 7. Evaluate I =
ex (f + g + f + g) dx = ex (f + g) + C
1
t
–1
= e sin t –
C
1 t2
Example 8. Evaluate
I= e
tan x
= – etan x · cosx
+ e
tan x
2
sec x cos x dx e
= – etan x · cos x + C.
tan x
sec x dx
Note: We have eg (fg + f) dx = eg f + C,
because
eg (fg + f) dx = eg gf dx + egfdx
Integrating the first integral eg gf dx by parts,
= eg f – eg fdx + egfdx
= egf + C.
x 2 cos2 x x sin x cos x
dx
I = e(x sin x + cos x)
x2
2
– x sin x – cos x dx
= e(x sin x + cos x) cos x
x2
cos x cos x dx
= e(x sin x + cos x) x cos x
x x
= eg (gf + f) dx = eg f + C
= e(x sin x + cos x) · cos x + C.
x
Evaluate I = x sin x dx
1 cos x
1
sin x dx
Solution I = x
1 cos x 1 cos x
sec 2 x / 2
tan x dx
= x
2
2
x
= x tan + C.
2
Alternative :
x sin x
x
sin x
dx
dx
dx
I
1 cos x
1 cos x
1 cos x
Example 10.
x
x
2sin cos
2
2
dx
dx
x
x
2 cos2
2 cos2
2
2
1
x
x
x . sec 2 dx tan dx
2
2
2
x
x
tan
tan
1
2
2 dx ] tan x dx
[ x.
1.
2
1/ 2
1/ 2
2
x
x
x
x
x tan tan dx tan dx x tan C
2
2
2
2
Example 11. Evaluate I = ln(ln x ) 1 2 dx
(ln x )
Solution I = ln(ln x) x – 1 x 2 dx
x ln x ln x x(ln x)
x
tan x
tan x
I = e sin x dx e sec x dx
Solution
Evaluate
x 2 cos2 x x sin x cos x
dx
I = e(x sin x + cos x)
x2
Solution We have
sin x sec x dx
1.75
1.76
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
= x ln (ln x) – x + C.
ln x
1
1
dx .
Integrate
ln
x
(ln
x)2
Solution Let z = ln x
1
dx
dz dx z
x
e
or dx = ezdz
1
1 1
1
Now I
dx 2 e z dz
2
ln
x
(ln
x)
z z
1 1
e z 2 dz
z z
1
ez f (z) f '(z) dz, where f (z)
z
1
x
z
z
C.
= e f(z) + C = e . C
z
ln x
1.
4.
Example 12.
K
Evaluate the following integrals :
ex(sin x – cosx) dx
(ii) ex (tanx – ln cosx) dx
(iii) ex sec x. (1 + tanx) dx.
(i)
2.
(i)
x 2 sin 2x
e 1 cos 2x dx
(iii)
5.
e 2 x (sin 4 x 2)
dx
1 cos 4 x
e x (1 x x3 )
dx
(1 x 2 )3/2
Evaluate the following integrals :
(iv)
3.
(i)
2
1 x
e x
2
dx
1 x
e
6.
3
x 2)
dx
(x 2 1) 2
x (x
2
e ( x 1)
1 x
dx (iv) e x
dx
( x 1)3
1 x
x
(iii)
(ii)
ex
(x 2 3x 3)
dx
(x 2) 2
e x ( x 2 1)
dx
(ii)
( x 1) 2
Evaluate the following integrals :
1 sin x
ex
dx
(i)
1 cos x
(ii)
Evaluate the following integrals :
1.13 MULTIPLE
INTEGRATION BY PARTS
In calculating a number of integrals we had to use the
method of integration by parts several times in
succession. The result could be obtained more rapidly
and in a more concise form by using the so-called
generalized formula for integration by parts (or the
(1 x)2
dx
(1 x 2 )2
x2 ex
(iv)
dx
( x 2) 2
Evaluate the following integrals :
(iii)
(i)
ex [ l n(sec x+tan x) + sec x ] d x
(ii)
e log x x dx
ex
x
1
2
Evaluate the following integrals :
2x sin 2x
dx
(i)
1 cos 2x
(ii)
(iii)
(tan(ln x) + sec2(ln x)}dx
x (1 x 2 ) sin 1 x
(1 x 2 )
dx
formula for multiple integration by parts) :
Let u and v be two differentiable functions of x
(differentiable n times), and let us denote
u' by u' and v dx by v1,
(u')' by u" and v1 dx by v2, etc.
...(1)
Now, uv dx = uv1 – u'v1 dx
(by integrating by parts)
...(2)
Again, u'v1 dx = u'v2 – u"v2 dx,
INDEFINITE INTEGRATION
u"v2 dx = u"v3 – u"'v3 dx,
u"'v3 dx = u"'v4 dx – u(4)v4 dx,
...(3)
...(4)
where u(4) denotes fourth derivative of u.
Combining (1), (2), (3) and (4) we get
And generally,
u(4)v4 dx
uv dx = uv1 – u'v2 + u"v3 – u"'v4 + .....
+ (–1)n–1 u(n–1) vn + (–1)n un vn dx
2x 6
(2x 6)(70x3 45x 2 396x 897) C .
5.7.9
Now, consider some more illustrations on integration
by parts.
...(5)
2x 2
1
Example 2. Evaluate sin
dx
2
4x 8x 13
2x 2
Solution sin 1
dx
2
4x 8x 13
2x 2
1
= sin
dx
2
2
(2x 2) 3
Put, 2x + 2 = 3 tan 2 dx = 3 sec2 d
1 3tan 3
sec2 d
= sin
3sec 2
...(6)
where u(n) denotes nth order derivative of u.
The use of the generalized formula for integration by
parts is especially advantageous when calculating the
integral Pn (x)(x)dx where Pn(x) is a polynomial of
degree n, and the factor (x) is such that it can be
integrated successively n + 1 times.
For example,
I = x4 cos x dx.
= x4 sin x – 4x3 (– cos x) + 12x2(–sin x)
– 24x(cos x) + 24 sin x + C
= x4 sin x + 4x3 cos x – 12x2 sin x – 24x cosx
+ 24 sin x + C
Example 1. Applying the generalized formula for
integration by parts, find the following integrals :
3
2
(i)
(x 2x 3x 1) cos 2xdx
(ii)
(2x 3x 8x 1) 2x 6dx
3
2
Solution
(i)
3
2
(x 2x 3x 1) cos 2xdx
cos 2 x
= (x3 –2x2 + 3x–1) sin 2x –(3x2 –4x+ 3)
4
2
sin 2 x
cos 2x
–6
+ (6x – 4)
+C
16
8
sin 2x 3
cos 2x 2
=
(2x – 4x2 + 3x) +
(6x –8x + 3) + C
4
8
(ii) (2 x 3 3x 2 8x 1) 2 x 6 dx
(2x 6)3/2
= (2x3 + 3x2 – 8x + 1)
3
5/2
(
2
x
6
)
– (6x2 + 6x – 8)
3 .5
(2 x 6) 7 / 2
(2 x 6) 9 / 2
– 12
+C
3 .5 .7
3 .5 .7 .9
=
uv dx = uv1 – u'v2 + u"v3 – u"'v4
+ (–1)4
+ (12x + 6)
1.77
3
3
sec2 d { tan –
2
2
3
= { tan – ln |sec |} + C
2
=
=
tan d}
2
3 2x 2
2x 2
1 2x 2
tan
ln
1
3
3 + C
2 3
3 2
2 3
2
3
1
2
= (x 1) tan (x 1) ln 4x 8x 13 +C
2
3
= (x + 1) tan–1 (x 1) ln (4x2 + 8x + 13) + C.
3
4
Example 3.
Evaluate
x 2 (x sec2 x tan x)dx
(x tan x 1)2
xsec 2 x tan x
dx
(x tan x 1)2
1
1
= x2
– 2x .
dx ...(1)
(x tan x 1)
(x tan x 1)
Solution
2
I= x .
using,
x sec 2 x tan x
dt
(x tan x 1) dx t
2
2
1
1
t
(x tan x 1)
x2
I=–
x tan x 1
2x(cos x)dx
xsin x cos x
1.78
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Put, x sin x + cos x = u
(x cos x + sin x – cos x) dx = du
I =–
x2
du
x2
+2
+2 ln |u| + C
(x tan x 1)
u
x tan x 1
2
=–
x
+ 2 ln|x sin x + cos x| + C.
x tan x 1
Evaluate
Example 4.
Solution
sec x(2 sec x)
dx
(1 2sec x)2
Let
sec x(2 sec x)
=
cos x(cos x 2) sin 2 x
dx
(2 cos x)2
2
2
sin x
sin 2 x
sin 2 x
dx
dx
2
2 cos x
(2 cos x)
(2 cos x)2
I=
sin x
+ C.
2 cos x
Example 5.
Solution
Evaluate
Let, I =
x 2 dx
(xsin x cos x)2
x 2 dx
dx
(xsin x cos x)2
We know,
d
1
{sin x x cos x sin x}
dx x sin x cos x
(x sin x cos x)2
x cos x
(x sin x cos x)2
x cos x
1
(xsin x cos x) dx (xsin x cos x)
I=
x
+ tanx + C.
cos x(xsin x cos x)
I=
2 cos x 1
(1 2sec x) dx (2 cos x) dx
cos xdx
sin 2 x
dx
2 cos x
(2 cos x)2
In the first integral integrating by parts taking cos x as
the second function and keeping the second integral
unchanged, we have
I=
2
2
Evaluate 3x tan 1 – x sec 1 dx
x
x
Example 6.
I=
1
x
vu = cos x . (x sin x cos x)
cos x x sin x
1
.
dx
2
x
sin
x
cos x
cos x
x
+ sec2 x dx
=
cos x(xsin x cos x)
2
x . cos x
x
(xsin x cos x) . cos x dx
2
Solution
2
2 1
1
3x tan x – x sec x dx
1
= 3x2 tan 1 dx – x sec2 dx
x
x
1 3
= tan 1 . x3 – sec2 1
x dx
x
x x 2
– x sec2
Example 7.
Solution
1
1
dx = x3 tan
+ C.
x
x
Evaluate cot–1 (x2 + x + 1) dx
Let I = cot–1 (x2 + x + 1) dx
1
dx
= tan–1 2
x x 1
(x 1) (x)
= tan–1
dx
1 x(x 1)
= {tan–1 (x + 1) – tan–1 (x)} dx
= tan–1 (x +1) dx – tan–1 (x) dx
applying integration by parts, we get
1
dx
= {tan–1 (x + 1)} x – x .
1 (x 1)2
1
1
dx
– (tan x)x x.
1 x2
x
= x {tan–1 (x + 1) – tan–1 x} +
dx
1 x2
x
dx
–
1 (1 x 2 )
1
I = x {tan–1 (x + 1) – tan–1 (x)} + ln (1 + x2) – I1 ...(1)
2
1.79
INDEFINITE INTEGRATION
Let, I1 =
x
1 (1 x) dx
put 1 + x = t, dx = dt
t 1
t
1
dt
dt
=
dt
2
2
1 t
1 t
1 t2
=
1
ln|1 + (1 + x)2| – tan–1 (1 + x)
2
From (1) and (2),
I = x{tan–1 (x + 1) – tan–1 (x)} +
2
...(2)
1
ln|1 – x2|
2
n (1 x 2 )
dx
Solution I =
1 x
1 x = t 2 dx = 2 t d t
= 2 l n (1 + (1 t 2)2) d t
2
4
2
= 2 t n ( t 2 t 2)
= 2 t l n (t 2 t + 2) + 8
4
2
(4 t 3 4 t ) t d t
4
2
4
t t
dt
t4 2 t2 2
2
dt
t4 2 t2 2
I = 2 t l n (t 4 2 t 2 + 2) + 8t + 8I1
+ 8
Now I1 =
t2
dt
4
t 2 t2 2
dt
2
t 2 22
t
I2 (say)
2
t2
dt
t 2 2 22
t
I3(say)
1 t2 1 t2 d t
t 2 2
t
2
2
2
2
1 t2 d t
t 2 2 2 1
t
2
2
1 2 dt
t
2
t t 2 2 1
1
1 t2 d t
=
2
t t2 2 2 1
2
...(1)
2 dt
=
t4 2 t2 2
1
2
2
1 t2 d t
t t2 2 2 1
1 2 dt
2 1
t
I =I I =
2
2
t t 2 2 1
( t 2 t 2) ( t 2)
2
2
2
2
= 2 t l n (t 4 2 t 2 + 2)
4
2
2
2
t 2t 2
2
2
= 2 l n (t 2 t + 2).1 d t
4
1 t2 d t
t t2 2 2 1
1
=
2
I
2
1
I3 =
2
n (1 x 2 )
dx
1 x
Evaluate
2
2
1
+
2
1
– ln |1 + (1 +x)2} + tan–1 (1 + x) + C.
2
Example 8.
2
2
2
1
ln |1 + t2| – tan–1 (t)
2
I1 =
1 t2 d t
t 2 2 2 2
t
1 t2 d t
1
+
2 t 2 2 2 2
t
1 t2 d t
1
I =
2
t t2 2 2 1
1
I2 =
2
2
2
2
2
2
1
+
2
3
2 1
2
2
1 t2 d t
t t2 2 2 1
2
2
Now we can compute I from (1).
2
2
1.80
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 9.
Evaluate
x 2 n(n 1)
dx
(xsin x n cos x)2
x 2 n(n 1)
dx
Solution
(xsin x n cos x)2
Multiplying and dividing by x2n – 2, we get
Here I =
2
(x sin x nx
dx
2
Also,
=
2
x
2
2a2
2
dx
2
2 2
dx
2
2
2
2
2 2
dx
x
2
2 2
2
dx
1
2
and I1 =
Here,
=
2 2
x
2
cos sin
d
cos 2 . ln
cos sin
on applying integration by parts.
2
sin 2
cos sin . sin 2
.
d
= ln
cos
sin
2
cos2
2
=
2
Evaluate
Let I =
1
1 x
2
dx
(x a )
2
2 3
dx
(x a )
2
...(1)
2 3
1
(x a ) dx
2
...(2)
2 2
1
(x a ) . 1 dx
2
uv =
cos sin
d
= (2 cos2 – 1) ln
cos sin
1 x
2
(x a ) 2a x a a tan a + C.
Solution
cos sin
cos sin
2 cos ln
ln
d
cos
sin
cos sin
1
(x a ) x a a tan a
Example 12.
cos2
ln
1 sin 2 d
2
If cos > sin > 0, then evaluate
2
2
(xsec x)
+ tan x + C.
(xsin x n cos x)
2x
1
2
1 1 x
x
1
x
dx
2 tan 1 2a 2
tan
a
a x2 a 2
a
a
(x 2 a 2 )2
1 sin 2
I = ln 1 sin 2
I=
1
2
From (1) and (2),
(xsec x)
sec2 x dx
(xsin x n cos x)
cos2
...(1)
x a . 1 dx = x a x (x a ) x dx
dx
Solution
1 x
x
x2 a 2 a 2
2
dx
x2 a 2
(x 2 a 2 )2
2
(x n sec x tan x nx n 1 .sec x)
(x n sin x nx n 1 cos x)dx
Example 10.
1
2
x a x a 2 x a 2a (x a ) ...(2)
1
= xn sec x . n
n 1
(x sin x nx cos x)
=–
2 2
x a a tan a
(x 2 n(n 1)).x n 2 .cos x n
. x . sec x dx
I=
(x n sin x nx n 1 cos x)2
vu and applying integration by parts, we get
=–
2
Here, we know
Solution
2n 2
dx
(x a )
Evaluate
Example 11.
(x n(n 1))x
dx
n
n 1
cos x)2
We put xn sin x + n xn – 1 cos x = t
(n xn – 1 sin x) + (xn cos x) + n (n – 1) xn – 2 cos x
– (n xn – 1 sin x) dx = dt
xn – 2 cos x. (x2 + n (n – 1)) dx = dt
Keeping this in mind we put,
I=
sin 2 cos sin 1
ln |cos 2 | + C.
ln
2
cos sin 2
=
2 2
1
2(2x)
x dx
.x
2 2
(x a )
(x 2 a 2 )3
2
x
x2 a 2 a 2
4
dx
(x 2 a 2 )2
(x 2 a 2 )3
dx
x
1
4
dx – 4a2
2
2 2
2
2 2
(x
a 2 )3
(x a )
(x a )
x
I1 = 2
+ 4I1 – 4a2 . I [using (1) and (2)]
(x a 2 )2
=
2
INDEFINITE INTEGRATION
x
3
I
...(3)
(x 2 a 2 )2 4a 2 1
{Using, previous example
dx
x
1
x
I1 = 2 2 2 2 2 2 3 tan 1 + c}
(x a )
2a (x a ) 2a
a
4a2 I =
1.81
x
4a 2 (x 2 a 2 )2
3
x
1
x
2 2 2
3 tan 1 + C.
2
4a 2a (x a ) 2a
a
I=
L
Evaluate the following integrals using multiple
integration by parts :
1.
x2 e3x d x
2.
( x 3x 1)e
3.
x3 cos 2x dx
x
2
e cos xdx
3
2
3x
(x 2x 5)e dx
2
(x 2x 3)nx dx
4.
5.
6.
3
7.
dx
1.14 INTEGRATION BY
REDUCTION FORMULAE
In some cases of integration, we take recourse to the
method of successive reduction of the integrand which
mostly depends on the repeated application of
integration by parts. This is specially the case when
the integrands are complicated in nature and depend
on certain parameter or parameters. These parameters
may be positive, negative, or fractional indices, as for
example, xn eax, tannx, (x2 + a2)n/2, sinm x cosn x, etc.
To obtain a complete integral of these trigonometric
or algebraic functions, we first of all define these
integrals by the letters I, J, U, etc., introducing the
parameter or parameters as suffixes, and connect them
with certain similar other integral or integrals whose
suffixes are lower than that of the original integral.
Then by repeatedly changing the value of the suffixes,
the original integral can be made to rest on much
simpler integrals. This last integral can be easily
evaluated and knowing the value of this last integral,
by the process of repeated substitution, the value of
the original integral can be found out.
The formula in which a certain integral involving some
parameters is connected with some integrals of lower
order is called a Reduction Formula. In most of the
cases the reduction formula is obtained by the process
of integration by parts. In some cases the method of
differentiation or other special devices are adopted.
2
x 2 7x 1
dx
3
2x 1
Evaluate the following integrals using integration by
parts :
cos 2x ln(1 tan x)dx
9.
8.
3x
2
(3x x 2) sin (3x 1)dx
10.
11.
1
x2 1
ln 1 2 dx
4
x
x
e x (1 nx n 1 x 2n )
dx .
(1 x n ) 1 x 2n
Find the reduction formulae for
Example 1.
tan xdx and cot xdx .
Solution In = tan xdx tan x tan xdx
tan x(sec x 1)dx
= tan x sec xdx tan xdx
tan x
tan xdx
– tan n 2 xdx
n 1
n
n
n 2
n
n 2
2
n 2
2
2
n 2
n 1
n
tan n 1 x
– In–2.
n 1
n
n 2
2
Similarly, In = cot xdx cot x cot xdx
In
= cot
= cot
n 2
x(cos ec x 1)dx
n 2
x cos ec 2 x dx cot n 2 x dx
2
cot n 1 x
cot n 2 xdx
cot n xdx
n 1
cot n 1 x
In = –
– In–2.
n 1
Above are the required reduction formulae which
reduce the power of tanx or cotx by 2. By successive
application of these formulae we shall ultimately
depend on
tan xdx or cot xdx
1.82
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
when n is odd and which are lnsec x or lnsinx
respectively or will depend on
tan xdx or cot xdx
i.e. (sec x 1)dx or (cos ec x 1)dx
2
2
2
I 7 sec7 x dx
which are (tanx – x) or (– cotx – x) respectively.
For example,
tan6 x tan 4 x tan2 x
– lnsecxC.
6
4
2
tan xdx
6
Example 2.
tan 5 x tan 3 x tan x
x + C.
5
3
1
Obtain the reduction formula for
I n sec x dx .
Solution
I n sec n x dx sec n 2 x . sec 2 x dx
Integrating by parts,
In = secn–2 x . tan x
– (n – 2) secn–3 x sec x . tanx . tan x dx
In order to put this equation into the desired form, let
us replace tan2 x by sec2 x – 1 in the new integral to
yield
= secn–2 x tan x – (n – 2) secn–2 x (sec2x – 1)dx
= secn–2 x tan x – (n–2) [ secn x dx – secn–2 x dx]
= secn–2 x tan x – (n – 2) In + (n – 2) In–2
Transposing and simplifying,
sec n 2 x tan x n 2
I n 2
n 1
n 1
This is the required reduction formula.
With the aid of this formula, using it several times, we
can integrate any positive integral power of the secant.
For example, assuming, n = 3, we get
In
sec x dx 2 sec x tan x sec x dx
3
1
1
(sec x tan x + ln |sec x + tan x|) + C ;
2
Assuming n = 4, we get
=
sec x dx 3 sec x tan x 2 sec x dx
4
=
1
2
2
1
(sec2 x tan x + 2 tan x) + C, and so on.
3
sec 5 x tan x 5
I5 .
6
6
I5
sec3 x tan x 3
I3 .
4
4
I3
sec x tan x 1
I1
2
2
I1 sec x dx ln| sec x tan x | .
n
7
Using the above reduction formula,
Solution
2
tan7 xdx
Evaluate sec x dx .
Example 3.
I7
sec5 x tan x 5 sec3 x tan x 3.5 sec x tan x
6
6
4
4.6
2
1.3.5
ln | sec x tan x | C .
2.4.6
Example 4. Find the reduction formula for
In = x sin xdx .
n
In = x sin n xdx x sin n 1 x sin xdx
Integrating by parts.
In = (x sinn–1x)(–cosx)
Solution
+ cos x{sin
n 1
x x(n 1)sin n 2 x cos x}dx
In = –xcosx sinn–1x + sin
+ (n 1) x sin
n 2
n 1
x cos xdx
x(1 sin 2 x)dx
sin n x
+ (n – 1)(In – 2 – In)
n
sin n x
(1 + (n – 1))In = –xcosx sinn – 1x +
+ (n – 1) In – 2
n
x cos xsin n 1 x sin n x
2
xsin n xdx
or,
n
n
n 1
x sin n 2 xdx
n
This is the required reduction formula.
Example 5. Find a reduction formula for the integral
In = –x cosxsinn–1x +
sin nx
sin x dx .
Solution
Let In =
sin nx
dx
sin x
1.83
INDEFINITE INTEGRATION
We havesin nx – sin(n – 2) x = 2 cos(n – 1) x sin x
sin nx
sin(n 2)x
= 2 cos(n – 1)x +
sin x
sin x
Integrating both sides we get
sin nx
2sin(n 1)x
sin(n 2)x
dx
dx
sin x
(n 1)
sin x
integrate 1/(x2 + k)n – 1 by parts, taking unity as the
second function.
Thus, In–1 =
2sin(n 1)x
+ In–2
(n 1)
Above is the required reduction formula.
In
If In = x n a 2 x 2 dx, prove that
Example 6.
n 1
2
2
= x .{x a x } dx
Applying integration by parts we get
(a
= xn – 1 .
2
x 2 )3/2
3
2
2 3/2
(a x )
(n 1)x n 2 .
dx
3
n 1 2
2 3/2
x (a x )
=–
3
(n 1) n – 2 2 2
x . (a – x ) a 2 x 2 dx
3
x n 1 (a 2 x 2 )3/2 (n 1)a 2
(n 1)
I n 2
In
In = –
3
3
3
(n 1)
In
In +
3
x n 1 (a 2 x 2 )3/2 (n 1)a 2
I n 2
3
3
n 1 2
2 3/2
2
n 2 I x (a x ) (n 1)a I
n
n 2
3
3
3
n 1
2
2 3/2
2
(a x )
(n 1)a
I n 2 .
(n 2)
(n 2)
Example 7. Find a reduction formula for the integral
dx
(x k) .
2
Solution
n
To obtain a reduction formula, we
n 1
1 . dx
x
(n 1)
. 2x dx
x. 2
n 1
(x k)
(x k)n
x
(x 2 k) k
dx
n 1 + 2(n – 1)
(x k)
(x3 k) n
x
In – 1 = 2
(x k)n 1
dx
dx
k
+ 2 (n – 1)
2
n 1
2
(x k)n
(x k)
x
In – 1 = 2
+ 2 (n – 1) In – 1 – 2k (n – 1) In
(x k)n 1
x
2k (n – 1) In = 2
+ {2 (n – 1) – 1} In – 1
(x k)n 1
x
2k (n – 1) In = 2
+ (2n – 3) In – 1
(x k)n 1
2
In = x n a 2 x 2 dx
Solution
x
2
2
In – 1 =
x n 1 (a 2 x 2 )3/2 (n 1) 2
a In – 2.
(n 2)
(n 2)
In = –
In =
=
1
(x k)
Hence,
dx
(x k)
2
n 1
x
(2n 3)
dx
.
2
n 1
2
2k(n 1) (x k)n 1
2k(n 1)(x k)
This is the required reduction formula. By repeated
application of this formula the integral shall reduce to
1
1
x
.
which is
tan–1
(x k)
k
k
dx
Example 8. Evaluate
.
(x 2 3)3
Solution By the above reduction formula, we get
that of
2
dx
x
3
dx
(x 3) 12(x 3) 12 (x 3)
2
3
2
2
2
2
[Putting n = 3 and k = 3 in the formula]
x
1
x
1
dx
12(x 2 3)2 4 6(x 2 3) 6 (x 2 3)
[On applying the same reduction formula by
putting n = 2 and k = 3]
=
=
x
x
1
x
tan–1
+ C.
2
2
12(x 3)
24(x 3) 24 3
3
2
1.84
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 9.
Here (d/dx) (2x2 + 4x +3) =4x + 4.
Solution
(x 2)dx
(2x 4x 3)
2
=
=
2
1
(4x 4) 2 1
4
dx
2(x 2 4x 3)2
1
(4 4)dx
(2 1)dx
2
2
2
4 (2x 4x 3)
3
2
x 2x
2
1
(2x2 + 4x + 3)–2 (4x + 4) dx
4
1
+
4
=–
3
2
x 2x
2
Solution Here, since two parameters m, n are
involved, we shall denote the integral by the symbol
dx
2 .
(x 1)2
1
4(2x 2 4x 3)
1 (x 1)
1
2 tan { 2(x 1)} + C.
4 (x 1)2 1
2
Example 10. If Im = (sin x + cos x)m dx, then
show that m Im = (sin x + cos x)m – 1 . (sin x – cos x) + 2
(m – 1) Im –2.
=
Im = (sin x + cos x)m – 1 (sin x –cos x)
+ (m – 1) Im – 2 – (m – 1) Im
(m – 1) Im + Im
= (sin x + cos x)m – 1 (sin x – cos x) + 2 (m – 1) Im – 2
m I m = (sin x + cos x) m – 1 (sin x –cos x)
+ 2 (m – 1) Im – 2.
Example 11. Obtain a reduction formula for
xm(ln x)n dx where m, n are positive integers).
2
1
1
4(2x 2 4x 3) 4
Solution
2(sin x + cos x)m – 2 dx
–(m – 1) (sin x + cos x)m dx
+ (m – 1)
dx
1
2
Now put x + 1 = t and then applying the reduction
formula, we get
I=
+ (m – 1) [(sin x + cos x)m – 2
{2 – (sin x + cos x)2 ] dx
= (sin x + cos x)m – 1 (sin x – cos x)
(x 2)dx
Evaluate
.
(2x 2 4x 3)2
Im =
(sin x + cos x)m dx
(sin x + cosx)m – 1 . (sin x + cos x) dx,
(applying integration by parts.)
= (sin x + cos x)m – 1 (cos x + sin x) – [(m –1) sinx
+ cos x)m – 2. (cos x – sin x) . (sin x – cos x) ]dx
= (sin x + cos x)m – 1 (sin x – cos x) + (m – 1)
(sin x – cos x)m – 2 (sin x + cos x)2 dx
Since, (sin x + cos x)2 + (sin x – cos x)2 = 2,
Im = (sin x + cos x)m – 1 (sin x – cos x)
Im , n =
xm(ln x)n dx. Integrating by parts,
xm1
1
1
(ln x)n
n(ln x)n1 .xm1dx
m 1
m 1
x
x m 1
n
(ln x) n
x m (ln x) n 1 dx
m 1
m 1
x m 1
n
(ln x) n
I m,n 1 ,
m 1
m 1
x m 1
n
(ln x) n
I m ,n 1 .
I m ,n
m 1
m 1
Im, n
Note:
1.
Here we have connected Im, n with Im, n – 1 and by
successive change the power of ln x can be
reduced to zero, i.e., after n operations we shall
get a term Im, o, i.e., xm dx, which is easily
integrable. Thus, by step by step reduction, Im, n
can be evaluated. It may be noted that when two
parameters are involved this is the usual practice.
2. Students must be cautious in denoting these
integrals. For example, Im, n In, m in general.
Example 12. Obtain reduction formulae for
xm(1 – x)n dx.
Solution
=
Let Im, n =
xm(1 – x)n dx
x m 1
n
. (1 x)n
x m 1 . (1 x)n 1 dx
m 1
m 1
1.85
INDEFINITE INTEGRATION
x m 1 (1 x) n
n
x m (1 x)n 1 {1 (1 x)}dx
m 1
m 1
x m 1 (1 x) n
n
I m, n 1 I m, n .
m 1
m 1
Transposing and simplifying,
x m 1 (1 x) n
n
I m, n
I m , n 1 .
m n 1
m n 1
Find the reduction formula for
Example 13.
m
Im,n = cos x sin nxdx
Solution
Integrating by parts, Im,n
cos nx cos nx .m. cosm 1 x( sin x)dx
+
= cosmx
n n
cos nx m cosm 1 x(cos nx sin x)dx
–
...(1)
n
n
Now sin(nx–x) = sinnx cosx – cosnx sinx.
cosnx sinx = sinnx cosx – sin(n – 1)x.
Subtituting in (1)
= –cosmx
cosm x cos nx
cos x sin nxdx
n
m cosm 1 x[sin nx cos x sin(n 1)x]dx
n
sinm+2 cosn–2 d
...(1)
In like manner, if the integral be written in the form
– sinm–1 cosn d (cos ),
we obtain
m 1
sinm cosn d = n 1 sinm–2 cosn+2 d
sin m–1 cos n 1
...(2)
n 1
It may be observed that formula (2) can be derived
from (1) by substituting /2 – for , and
interchanging the letters m and n in it.
–
Case o f one P ositive and o ne Negative
Index
The results in (1) and (2) hold whether m or n be
positive or negative. Accordingly, let one of them be
negative (n suppose), and on changing n into – n,
formula (2) becomes
sin m
n
m
cosm x cos nx m cosm x sin nxdx
n
n
m cosm 1 x sin(n 1)xdx
n
Transposing the middle term to the left, we get
=
cosm x cos nx
mn
cosm xsin nxdx
m
cosm 1 x sin(n 1)xdx
mn
Reduction formulae for sinm cosn d
sinm cosn d = cosn–1sinmd (sin)
Consequently, if we assume
sin m 1
,
m 1
The formula for integration by parts gives
u = cosn–1, v =
sinm cosn d =
cos n–1 sin m 1 n 1
+
m 1
m 1
sin m–1
cos d (n 1) cos
n 1
–
m 1 sin m 2
d
...(3)
m 1 cos n–2
in which m and n are supposed to have positive signs.
sin m
is made to
cos n
depend on another in which the indices of sin and
cos are each diminished by two. The same method is
applicable to the new integral and so on.
If m be an odd integer, the expression is integrable
immediately.
If m be even, and n even and greater than m, the
substitution of tan = x is applicable.
If m = n, the expression becomes tanmd, which can
be treated as before.
By this formula the integral of
If n < m, the integral reduces to that of sinm–n d.
Again, if n be odd, and n > m, the integral reduces to
d
sin m–n 1 d
.
, and if n < m reduces to
n m
cos
cos
The mode of finding these latter integrals will be
considered subsequently.
Again, if the index of sin be negative, we get, by
changing the sign of m in (1),
1.86
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
cosn
cosn 1
or, transposing and multiplying by
sin m d (m 1)sinm–1
n–2
– n 1 cosm 2 d
...(4)
m 1 sin
We shall next consider the case where the indices are
both positive.
Case of both indices being positive
If sinm (1 – cos2) be written instead of sinm+2 in
formula (1), it becomes
sinm cosn d =
n 1
+
m 1
cos n 1 sin m 1
m 1
cos n 1 sin m 1 n 1
+
sinm cosn–2 d
m 1
m 1
n 1
sinm cosn d.
–
m 1
Now, transposing the latter integral to the other side,
and dividing by m n , we get,
m 1
n 1
sinm cosn d = cos msin1
n 1
sinm cosn–2 d
m 1
In like manner, from (2), we get
n 1
sinm cosn d =
m 1
m 1
+
...(5)
It remains to consider the case where the indices of
sin and cos are both negative.
Writing – m and – n instead of m and n, in formula (1),
it becomes
d
–1
m
n
sin cos (m n)cos n 1 sin m–1
d
sin m cos n 2
1
(n 1)sin
m 1
cos n 1
...(7)
Again, if we substitute n for n + 2 in this, it becomes
1
d
sin m cosn = (n 1)sin m1 cosn 1
mn–2
d
...(8)
n 1
sin m cos n
Making a like transformation in formula (8), it becomes
–1
d
sin m cosn = (m 1) sin m 1 cosn 1
mn–2
d
.
...(9)
m 2
m 1
sin cos n
In each of these, one of the indices is reduced by two
degrees, and consequently, by successive
applications of the formulae, the integrals are reducible
ultimately to those of one or other of the forms
+
d
The formulae of reduction for
are given as follows :
sinm–2 cosn d
Case of both indices being negative
n 1
mn
=
mn
d
n 1 sin m cosn
d
sin m 1 cos n 1 .
...(6)
mn
m
n
By aid of these formulae the integral of sin cos is
made to depend on another in which the index of either
sin, or of cos, is reduced by two. By successive
application of these formulae the complete integral
can always be found when the indices are integers.
+
+
n 2
cos and sin and these can be easily integrated.
–
m
+
sinm (cosn–2 – cosn) d
=
d
sin cos
mn
,
n 1
d
and
sin n
d
cos
n
d
sin
n2
d
...(10)
d
– cos
n2
d
...(11)
cosn (n 1) cosn –1 + n 1 cos n – 2
sin n (n 1)sin n–1 + n 1 sin n–2
It may be here observed that, since sin2 + cos2 = 1,
we have immediately.
d
d
d
sin m cosn = sinm2 cosn sinm cosn2 ...(12)
and a similar process is applicable to the latter integrals.
This method is often useful in elementary cases.
Application of Method of Differentiation
The formulae of reduction given above can also be
readily arrived at by direct differentiation. Thus, for
example, we have
d sin m msin m–1 n sin m 1
;
d cos n
cos n–1
cos n 1
and, consequently,
INDEFINITE INTEGRATION
sin m 1
1 sin m m sin m 1
d
–
n 1
n cos n n cos n 1
cos
Again,
d
(sinm cosn) = m sinm–1 cosn+1 – n sinm+1 cosn–1.
d
If we substitute for cosn+1 its equivalent
cosn–1 (1 – sin2), we get
d
(sinm cosn) = m sinm–1 cosn–1 – (m + n) sinm+1
d
cosn–1, hence we get sin m 1 cos n 1 d
m
n
= – sin cos + m sin m–1 cosn 1 d .
mn
mn
We can simplify the following forms using
trigonometric substitutions :
x m dx
(i) 2
(a x 2 ) n/2
We put x = a tan ,
x m dx
the integral
transforms into
(a 2 x 2 )n/2
1.87
sin m cos n–m–2d (neglecting a constant
multiplier).
x m dx
(ii)
(a 2 x 2 )n/2
In a like manner, the substitution x = a sin transforms
x m dx
a m–n 1 sin m d
2
2 n/2 into
(a x )
cos n–1
x m dx
and, if x = a sec , the integral 2 2 n/2
(x a )
n–m–2
cos
d
transforms into
sin n–1
(neglecting the constant multiplier).
A similar substitution may be applied in other cases.
x n dx
,
For example, to find the integral
(2ax x 2 )1/2
2
let x = 2a sin , then dx = 4a sin cos d, and the
transformed integral is 2n+1an
be solved easily.
sin2n d which can
M
1.
2.
3.
4.
Derive the reduction formula
1
n 1
n
n 1
n 2
cos x dx n cos xsin x n cos x dx .
Obtain a reduction formula for the following
integrals
(ii) (ln x) n dx(n 1)
(i)
x e dx (n > 1)
Evaluate the following integrals :
d
tan4d
(ii)
(i)
tan 5
d
(iii)
(iv) cos6d
sin 3
Evaluate the following integrals :
d
(i)
sin cos2
8.
x 1 x
3
2
(ii)
x dx
(a x )
2
2 2
cot 6 x cot 4 x cot 2 x
6
4
2
+ lnsinx + C
cot 7 x cot 5 x cot 3 x
7
5
3
+ cot x + x + C.
Derive the formula
sin n 1 x
1 sin n x n sin n 1 x
dx.
m 1 dx = m
cosm x m cosm 1 x
cos x
If Im, n = x cos nx dx (n 0), then show that ,
m
x m sin nx mx m 1 cos nx m(m 1)
+
–
Im–2, n.
n2
n
n2
10. If In =
4
cot 8 x
x3dx
(a x 2 )3/2
2
8
Show that cot xdx
Im, n =
Evaluate the following integrals :
dx
Show that
9.
(iv)
2 7/2
9
2
x 2 dx
(a cx )
cot xdx 8
7.
(ii) cos2sin4d
d
(iii)
sin cos
(i)
6.
n x
3
5.
(iii)
cos nx cosec x dx, then show that ,
In – In–2 =
2 cos(n 1)x
n 1
1.88
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1 cos nx
dx, then show that
1 cos x
2sin(n 1)x
In + In–2 = 2In–1 +
.
n 1
n
x
12. If In =
dx (n 2), then show that
2
x a2
a 2 (n 1)
x n 1 x 2 a 2
–
In–2.
In =
n
n
11. If In =
1.15 INTEGRATION OF
RATIONAL FUNCTIONS
USING PARTIAL
FRACTIONS
A rational function (a ratio of polynomials) is found
by combining two or more rational expressions into
one rational expression. Here, the reverse process is
considered : given one rational expression, we
express it as the sum of two or more rational
expressions. A special type of sum of simpler fractions
is called the partial fraction decomposition ; each term
in the sum is a partial fraction. In this section we
show how to integrate any rational function by
expressing it as a sum of partial fractions, that we
already know how to integrate.
To illustrate the method, observe that by taking the
fractions 2/(x – 1) and 1/(x + 2) to a common
denominator we obtain
2
1
2(x 2) (x 1)
x5
2
x 1 x 2
(x 1)(x 2)
x x2
If we now reverse the procedure, we see how to
integrate the function on the right side of this
equation :
x5
2
1
dx
dx
x2 x 2
x 1 x 2
= – 2 ln |x – 1| – ln |x + 2| + C.
There are four types of partial fractions :
A
(i)
x–a
A
(ii)
(k a positive integer 2),
(x – a) k
Ax B
(iii) x 2 px q (the roots of the denominator are
imaginary, that is, p2 – 4q < 0,
13. Prove that
x
n
ax b
dx
=–
(n 1)bx n 1
ax b
(2n 3)a
dx
+ C (n 1).
(2n 2)b x n 1 ax b
14. Deduce the reduction formula for
dx
dx
.
n =
4 n and hence evaluate 2 =
(1 x )
(1 x 4 )2
–
Ax B
(iv) (x 2 px q) k (k a positive integer 2; the roots
of the denominator are imaginary).
These are called partial fractions of the type (i), (ii),
(iii) and (iv).
It is known that every proper rational fraction may be
represented as a sum of partial fractions. We shall
therefore first consider integrals of partial fractions.
The integration of partial fractions of type (i), (ii) and
(iii) does not present any particular difficulty.
A
dx A ln | x – a | C
(i)
xa
A
(x a)– k 1
C
dx A (x a)– k dx A
(ii)
k
–k 1
(x a)
A
C.
=
(1 – k)(x – a) k–1
A
Ap
(2x p) B –
Ax
B
2
2 dx
(iii)
x2 px q
x 2 px q
2x p
dx
dx B – Ap
= A 2
2 x px q
2 x 2 px q
= A ln |x2 + px + q|
2
Ap
dx
+ B–
2
2
2
x p q p
2
4
2x p
2B Ap
C
= A ln |x2 + px + q| +
tan–1
2
2
4q
p
2
4q p
The integration of partial fractions of type (iv) requires
more involved computations. Suppose we have an
integral of this type :
Ax B
dx
(iv)
(x 2 px q) k
We perform the transformations :
INDEFINITE INTEGRATION
A
Ap
(2x p) B –
Ax B
2
2
dx
dx
2
k
(x px q)
(x 2 px q) k
2x p
Ap
dx
A
dx B –
2
=
2
k
2 (x px q)k
2 (x px q)
The first integral is considered via the substitution,
x2 + px + q = t, (2x + p) dx = dt :
dt
t – k 1
2x p
–k
C
t
dt
=
=
dx
k
t
1 k
(x 2 px q) k
1
= (1 k)(x 2 px q) k–1 C
We write the second integral (let us denote it by Ik) in
the form
dx
dx
k
Ik = (x 2 px q) k
2
p
p2
x q –
2
4
dt
=
2
(t m 2 ) k
p
p2
= t, dx = dt, q –
= m2
2
4
We then do as follows :
(t 2 m 2 ) – t 2
dt
1
dt
Ik =
(t 2 m 2 ) k
(t 2 m 2 ) k m 2
assuming x +
=
1
t2
dt
1
– 2 2
dt
2 k 1
2 2
(t m 2 ) k
(t m )
m
m
Integrating by parts we have
t 2 dt
(t 2 m 2 )k
1
dt
1
2 k
t 2
(t 2 m 2 ) k–1
2(k – 1) (t m )
Putting this expression into (1), we have
=–
Ik =
+
=
dt
1
dt
(t m ) m (t m )
2
2 k
2
2
2 k–1
1
1
1
dt
2
2 k–1
m 2 2(k – 1) (t 2 m 2 ) k–1
(t m )
t
2k – 3
–
2m 2 (k – 1)(t 2 m 2 )k–1 2m 2 (k – 1)
dt
(t m )
2
2 k–1
...(1)
1.89
On the right side is an integral of the same type as Ik, but the
exponent of the denominator of the integrand is less by
unity (k – 1); we have thus expressed Ik in terms of Ik–1.
Continuing in the same manner we will arrive at the
familiar integral.
dt
1
t
tan 1 C
I1 = 2
2
m
m
t m
Then substituting everywhere in place of t and m their
values, we get the expression of integral (iv) in terms
of x and the given number A, B, p, q.
To see how the method of partial fractions works in
general, let’s consider a rational function
P(x)
f(x) =
Q(x)
where P and Q are polynomials. It is possible to
express f as a sum of simpler fractions provided that
the degree of P is less than the degree of Q. Such a
rational function is called proper.
To form a partial fraction decomposition of a rational
expression, we use the following steps:
Step 1 : If f is improper, that is, deg(P) deg(Q), then
we must take the preliminary step of dividing Q into
P (by division) until a remainder R(x) is obtained such
that deg(R) < deg(Q). The division statement is
P(x)
R(x)
f(x) =
= S(x) +
Q(x)
Q(x)
where S and R are also polynomials. For example,
x 4 3x3 x 2 5x
14x 6
x 2 3x 2 2
.
x3 3
x 3
Then, apply the following steps to the remainder,
which is a proper fraction.
Step 2 : Factor Q(x) completely into factors of the
form (ax + b)m or (cx2 + dx + e)n, where cx2 + dx +
e is irreducible and m and n are integers.
It can be shown that any polynomial Q can be factored
as a product of linear factors (of the form ax + b) and
irreducible quadratic factors (of the form ax2 + bx + c,
where b2 – 4ac < 0). For instance, if Q(x) = x4 – 16, we
could factor it as
Q(x) = (x2 – 4) (x2 + 4) = (x – 2)(x + 2)(x2 + 4)
Note:
If the equation Q(x) = 0 cannot be solved algebraically,
then the method of partial fractions naturally fails and
recourse must be had to other methods.
1.90
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Step 3 :
(I) For each distinct linear factor (ax + b), the
decomposition must include the term A .
ax b
(II) For each repeated linear factor (ax + b)m, the
decomposition must include the terms
A1
A2
Am
....
ax b (ax b)2
(ax b)m
(III) For each distinct quadratic factor (cx2 + dx + e),
the decomposition must include the term
Bx C
.
2
cx dx e
(IV) For each repeated quadratic factor (cx2 + dx +
e)n, the decomposition must include the terms
B1x C1
B2 x C 2
2
cx dx e (cx 2 dx e)2
Bn x C n
(cx 2 dx e) n
Step 4 : Use techniques to solve for the constants
in the numerators of the decomposition.
We explain the details for the four cases that occur.
.....
Case I : The denominator Q(x) is a product
of distinct linear factors
This means that we can write
Q(x) = (a1x + b1) (a2x + b2) .... (akx +bk)
where no factor is repeated. In this case the partial
fraction theorem states that there exist constants
A1, A2, ....., Ak such that
A1
A2
Ak
R(x)
.....
Q(x) a1x b1 a 2 x b 2
a k x bk
These constants can be determined as in the following
example.
x 2 2x 1
dx.
2x3 3x 2 2x
Solution Since the degree of the numerator is
less than the degree of the denominator, we don’t
need to divide. We factor the denominator as
2x3 + 3x2 – 2x = x(2x2 + 3x – 2) = x(2x – 1) (x + 2)
Since the denominator has three distinct linear factors,
the partial fraction decomposition of the integrand
has the form
Example 1.
Evaluate
x 2 2x 1
A
B
C
x(2x 1)(x 2) x 2x 1 x 2
...(1)
Method of Comparision of Coefficients
To determinator the values of A, B, and C, we multiply
both sides of this equation by the product of the
denominators, x(2x – 1) (x + 2), obtaining
x2 + 2x – 1 = A(2x – 1) (x + 2) + Bx(x + 2)
+ Cx(2x – 1)
...(2)
Expanding the right side of Equation 2 and writing it in
the standard form for polynomials, we get
x2 + 2x – 1 = (2A + B + 2C) x2
+ (3A + 2B – C) x – 2A
...(3)
The polyomials in Equation 3 are identical, so their
coefficients must be equal. The coefficient of x2 on
the right side, 2A + B +2C, must equal the coefficient
of x2 on the left side – namely, 1. Likewise, the
coefficients of x are equal and the constant terms are
equal. Ths gives the following system of equation for
A, B and C.
2A + B + 2C = 1
3A + 2B – C = 2
–2A = –1
1
1
1
Solving, we get A = , B = , and C = –
, and so
2
5
10
1 1
x 2 2x 1
1 1 1 1
2x3 3x2 2x dx = 2 x 5 2x 1 10 x 2 dx
=
1
1
1
ln |x| +
ln|2x – 1| –
ln |x + 2| + k
2
10
10
Method of Particular Values
Example 2.
Solution
Let
or
I=
Evaluate
dx
(x 1)(3 2x)
2
dx
(x 1)(3 2x)
2
dx
(x 1)(x 1)(2x 3)
1
B
C
A
+
+
(x 1)(x 1)(2x 3) x 1 x 1 2x 3
1 A (x – 1) (2x + 3) + B(x + 1) (2x + 3)
+ C(x + 1)(x – 1)
Putting x = 1, – 1, –
1 = B · 2 · 5,
1 = A(–2) · 1,
3
successively, we get
2
1 5
1 = C · 2 2
INDEFINITE INTEGRATION
B=
1
1
4
,A= – , C = .
10
2
5
B
C
A
I = x 1 x 1 2x 3 dx
=–
1
1
2
ln x + 1 +
ln x – 1 + ln 2x + 3 + k.
2
10
5
Heaviside Cover-up Method for Linear
Factors
When the degree of the polynomial P(x) is less than
the degree of Q(x), and Q(x) is a product of n distinct
linear factors, each raised to the first power, there
is a quick way to express P(x)/Q(x) into partial
fractions.
Example 3. Find A, B, and C in the partial
fraction expansion
x2 1
A
B
C
...(1)
(x 1)(x 2)(x 3) x 1 x 2 x 3
Solution If we multiply both sides of equation (1)
by (x – 1) to get
x2 1
B(x 1) C(x 1)
A
(x 1)(x 2)(x 3)
x2
x3
and set x = 1, the resulting equation gives the value
of A :
(1)2 1
= A + 0 + 0, A = 1.
(1 2)(1 3)
Thus, the value of A is the number we would have
obtained if we had covered the factor (x – 1) in the
denominator of the original fraction
x2 1
(x 1)(x 2)(x 3)
and evaluate the rest at x = 1:
(1)2 1
2
= 1.
(x 1) (1 2)(1 3) ( 1)( 2)
Similarly, we find the value of B in (1) by covering
the factor (x – 2) and evaluating the rest at x = 2 :
A=
B=
(2)2 1
5
= – 5.
(2 1) (x 2) (2 3) (1)( 1)
Shortcut Method :
Consider x – a1 = 0, then x = a1, put this value of x in all
the expressions other than x – a1 and so on , e.g.
1.91
x2 1
0 1
x(x 1)(x 1) x(0 – 1)(0 1)
11
11
.
1(x 1)(1 1) –1(–1 – 1)(x 1)
Case II : Q(x) is a product of linear factors, some of
which are repeated.
Suppose the first linear factor (a1x + b1) is repeated r
times; that is, (a1x + b1)r occurs in the factorization of
Q(x). Then instead of the single term A1/(a1 x + b1), we
would use
A1
A2
Ar
....
a1x b1 (a1x b1 )2
(a1x b1 ) r
For example, we write
x3 x 1 A B
C
D
E
x 2 (x 1)3 x x 2 x 1 (x 1)2 (x 1)3
x 4 2x 2 4x 1
dx
Example 4. Find
x3 x 2 x 1
Solution The first step is to divide. The result of
long division is
+
4x
x 4 2x 2 4x 1
=x+1+ 3
2
3
2
x
x
x 1
x x x 1
The second step is to factor the denominator
Q(x) = x3 – x2 – x + 1. Since Q(1) = 0, we know that
x – 1 is a factor and we obtain
x3 – x2 – x + 1 = (x – 1)2 (x + 1)
Since the linear factor x – 1 occurs twice, the partial
fraction decompostion is
4x
A
B
C
2
2
x 1
(x 1) (x 1) x 1 (x 1)
Multiplying by the least common denominator,
(x – 1)2 (x + 1), we get
4x = A(x – 1) (x + 1) + B(x + 1) + C(x – 1)2
= (A + C)x2 + (B – 2x) x + (–A + B + C)
Now we equate the coefficients :
A+C=0
B – 2C = 4
–A + B + C = 0
Solving, we obtain A = 1, B = 2, and C = –1,
so
x 4 2x 2 4x 1
dx
x3 x 2 x 1
1
2
1
= x 1
dx
2
1
x
1
x
(x
1)
=
2
x2
+ x + ln |x – 1| –
– ln |x +1| + k
x 1
2
1.92
=
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
x 1
2
x2
+x–
+ ln
+ k.
x 1
x 1
2
Note: The Heaviside cover-up method
applies recursively to handle the more general case of
repeated linear factors, as the following example
illustrates.
5x 3
1
5x 3
.
(x 1)(x 2)2 x 2 (x 1)(x 2)
=
1 2
7
x 2 (x 1) x 2
=
2
7
(x 1)(x 2) (x 1)2
=
(2) / (1 2) (2) / (2 1)
7
x 1
x2
(x 2)2
2
2
7
=
.
x 1 x 2 (x 2)2
Application of limit
A
A2
Ar
P(x)
1
....
..... ..(1)
(x a)r Q(x) (x a) (x a)2
(x a)r
Now in order to obtain A1, A2, .... Ar, following steps
are useful.
(i) To obtain the constant of the partial fraction
corresponding to (x – a), multiply both sides of
the identity (1) by (x – a) and then let x .
(ii) To obtain the constant of the partial fraction
containing (x – a)r in the denominator put x = a in
the given fraction except for (x – a)r present in the
P(a)
.
denominator. This gives A r
Q(a)
(iii) To obtain the other constants, we can multiply
both sides of the identity again and again by (x –
a) and then let x , or
put some particular values of x (say 0, 1, –1 etc.)
except the roots of denominator in the identity
thus forming a system of linear equations and
solving it.
Example 5.
x3 2
dx .
Evaluate
(x 1) (x 2)3
Solution
x3 2
A
B
C
D
(x 1)(x 2)3 x 1 x 2 (x 2)2 (x 2)3
13 2
B
C
23 2
3
2
(x 1)(1 2) x 2 (x 2) (2 1)(x 2)3
3
B
C
10
= (x 1) x 2 (x 2)2 (x 2)3
...(1)
To get B we multiply (1) by (x – 2) and let x
So, xlim
x3 2
(x 1)(x 2)3
C
10
3(x 2)
B
= xlim
(x 1)
(x 2) (x 2)2
1 = –3 + B + 0 + 0 B = 4.
To get C let us put x = 0 in (1)
3 B
2
C
10
(1) ( 2)3 1 2 (2)2 (2)3 C = 2
x3 2
(x 1)(x 2) dx
3
3
4
2
10
= x 1 x 2 (x 2)2 (x 2)3 dx
= –3 ln|x – 1| + 4 ln|x – 2| –
2
5
C.
x 2 (x 2)2
x2 x 2
dx
(x 1)(x 1)3
Example 6.
Evaluate
Solution
x2 x 2
dx
(x 1)(x 1)3
C
D
x2 x 2 = A + B +
2 +
(x – 1)3
(x 1)(x – 1)3 (x 1) (x –1) (x 1)
...(1)
x2 – x + 2 = A (x – 1)3 + B(x + 1) (x – 1)2
+ C (x + 1) (x – 1) + D (x + 1)
x = – 1 4 = – 8A A = – 1/2
x = 1 2 = 2D D = 1.
Multiply both sides of (1) by (x – 1) and taking limit,
x2 x 2
x (x 1)(x 1)2
lim
A(x 1)
B C D 2
= xlim
(x
1)
(x
–
1)
(x
1)
1
1
0 = – + B B = .
2
2
Multiply by (x – 1) again
INDEFINITE INTEGRATION
2
lim x x 2
x (x 1)(x 1)
A(x 1)2
B(x 1) C D
= xlim
(x 1)
(x 1)
1 (x 2 1 2x) 1 (x 2 1)
D
2
2
C
x
(x
1)
(x
1)
= lim
x2 1 x 1 x2 1
2 2
2
2 C D
= xlim
(x 1)
(x 1)
x 1
C D
= xlim
x
1
(x
1)
1 = 1 + C + O C = 0
Now, the integral can be evaluated easily.
2x 1
Example 7. Evaluate (x 2 (x 3)2 dx .
Solution
Assume
2x 1
...(1)
A B C
(x 2) (x 3)2 x 2 x 3 (x 3)2
The Heaviside method of determining coefficients
gives
4 1
3
6 1 7
,C
25
3 2 5.
( 2 3)2
Also, multiplying by x, and then making x , we
3
find A + B = 0 or B
.
25
The integral is therefore
A
–
3
3
7
ln(x + 2) +
ln(x – 3) –
+ C.
25
25
5(x 3)
Method of differentiation
Example 8.
Find A, B and C, in the equation
x 1
A
B
C
.
3
x 1 (x 1) (x 1)3
(x 1)
We first clear of fractions :
x – 1 = A(x + 1)2 + B(x + 1) + C.
Substituting x = – 1 shows C = –2.
We then differentiate both sides with respect to x,
obtaining
1 = 2A(x + 1) + B.
Solution
1.93
Substituting x = –1 shows B = 1.
We differentiate again to get 0 =2A, which shows
A = 0. Hence
x 1
1
2
.
(x 1)3 (x 1) (x 1)3
Case III : Q(x) contains irreducible quadratic factors,
none of which is repeated.
If Q(x) has the factor ax2 + bx + c, where b2 – 4ac < 0,
then, in addition to the partial fractions taken earlier,
the expression will have a term of the form
Ax B
ax 2 bx c
where A and B are constants to be detemined. The
term given above can be integrated by completing
the square and using the formula
dx
1
x
tan 1 C
x2 a 2 a
a
For instance, the function given by
x
has a partial fraction
f(x) =
2
(x 2)(x 1)(x 2 4)
decomposition of the form :
x
A
Bx C Dx E
.
(x 2)(x 2 1)(x 2 4) x 2 x 2 1 x 2 4
Example 9.
Evaluate
2x 2 x 4
x 4x
3
dx.
Since x3 + 4x = x(x2 + 4) cannot be
factored further, we write
Solution
2x 2 x 4 A Bx C
2
x x 4
x(x 2 4)
Multiplying by x(x2 + 4), we have
2x2 – x + 4 = A(x2 + 4) + (Bx + C)x
...(1)
= (A + B)x2 + Cx + 4A
Equating coefficients, we obtain
A + B = 2,
C = –1, 4A = 4
Thus A = 1, B = 1 and C = –1.
Alternatively, we can put x = 0 in (1) to obtain A = 1,
and put x = 2i to get
– 4 –2i = – 4 B + 2i C
...(2)
Separating real and imaginary parts in (2),
we get B = 1 and C = – 1.
2x 2 x 4
1 x 1
dx 2
dx
3
x 4x
x x 4
In order to integrate the second term we split it into
two parts :
Now,
1.94
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
x 1
x
1
dx
dx 2
dx
2
2
x 4
x 4
x 4
We make the substitution u = x2 + 4 in the first of
these integrals so that du = 2x dx. We evaluate the
second integral by means of the formula (1) :
2x 2 x 4
1
x
1
dx
dx 2
dx 2
dx
2
x
x(x 4)
x 4
x 4
= ln |x| +
1
1
ln(x2 + 4) – tan–1(x/2) + K.
2
2
Example 10. Evaluate
3x 2 5x 8
dx .
(x 2 2x 5)(x 1)
Solution
3x 2 5x 8
A
B(2x 2) C
2
x
1
(x 2x 5)(x 1)
(x 2x 5)
Note : Here we prefer to write B(2x + 2) + C instead
of the usual expression Bx + C, looking at the needs
of the integration process.
Multiplication of both sides by (x2 + 2x + 5)2 (x – 1)
leads to 3x2 + 5x + 8 = A(x2 + 2x + 5)
+ [B(2x + 2) + C](x – 1) ...(1)
If x = 1, equation (1) reduces to 16 = 8A, or A = 2.
If x = –1, 6 = 4A – 2C C = 1
If x = 0, 8 = 5A – 2B – C B = 1/2.
f(0) = c = –3
f(1) = a + b + c = –3
3f(2) = 3 (4a + 2b + c) = – 3
On solving we get a = 1, b = –1, c = –3
f(x) = x2 –x – 3.
f(x)
x2 x 3
dx
dx
3
x 1
(x 1)(x 2 x 1)
Using partial fractions, we get,
I=
A
Bx C
(x 2 x 3)
=
+ 2
2
(x
1)
(x
x 1)
(x 1)(x x 1)
where, A = –1, B = 2, C = 2
(2x 2)
1
x 1 dx (x x 1) dx
I=
2
= – ln |x – 1| +
2
2
(1 / 2)(2x 2) 1
= x 1 (x 2 2x 5) dx
(2x 2)
1
1
= x 1 dx 2 (x2 2x 5) dx (x2 2x 5) dx
= 2n|x – 1| +
+
1
n (x2 + 2x + 4)
2
x 1
1
tan –1 2 + k.
2
f(x)
dx, where f(x) is a
x3 1
polynomial of degree 2 in x such that
f(0) = f(1) = 3f(2) = –3.
Solution Let f(x) = ax2 + bx + c
Given f(0) = f(1) = 3f(2) = –3
f(0) =f(1) = 3f(2) = –3
Example 11.
Evaluate
1 dx
2
2
= – ln |x – 1| + ln (x2 + x + 1)
+
dx
(x 1 / 2) ( 3 / 2)
2
= ln| x – 1| +ln (x2 + x + 1) +
2
2
2x 1
+ C.
tan–1
3
3
Case IV : Q(x) contains a repeated
irreducible quadratic factor
If Q(x) has the factor (ax2 + bx + c)r, where
b2 – 4ac < 0, then instead of the single partial fraction,
the sum
3x 2 5x 8
dx
2
(x 2x 5)(x 1)
2
(2x 2)
(x x 1) x x 1
A1x B1
A 2 x B2
A r x Br
....
2
2
2
ax bx c (ax bx c)
(ax 2 bx c) r
occurs in the partial fraction decomposition.
2x 2 3
Find
dx.
Example 12.
(x 2 1)2
2x 2 3 Ax B Cx D
2
Let
.
Solution
(x 2 1)2
x 1 (x 2 1)2
Then
2x2 + 3 = (Ax + B)(x2 + 1) + Cx + D
= Ax3 + Bx2 + (A + C)x + (B + D)
Compare coefficients : A = 0, B = 2, A + C = 0,
B + D = 0. Hence, C = 0, D = 1. Thus,
2x 2 3
2
1
dx 2
dx
dx
2
2
2
(x 1)
x 1
(x 1)2
1
dx
= 2 tan–1x +
( x 2 1)2
1.95
INDEFINITE INTEGRATION
In the second integral, let x = tan . Then
1
dx =
2
(x 1)2
sec 2 d
=
sec 4
cos d
2
1
( + sin cos )
2
1
tan
1 1
x
= tan x 2
=
2
2
tan 2 1
x 1
=
Thus,
1 x
5 –1
2x 2 3
tan x +
+C
2
2 dx =
(x 1)
2 x2 1
2
Example 13.
1 x 2x 2 x3
Evaluate
dx.
x(x 2 1)2
The form of the partial fraction
decomposition is
Solution
2
3
1 x 2x x
A Bx C Dx E
2
x x 1 (x 2 1)2
x(x 2 1)2
Multiplying by x(x2 + 1)2, we have –x3 + 2x2 – x + 1
= A(x2 1)2 + (Bx + C) x (x2 + 1) + (Dx + E)x
= A(x4 + 2x2 + 1) + B(x4 + x2) + C(x3 + x) + Dx2 + Ex
= (A + B) x4 + Cx3 + (2A + B + D)x2 + (C + E)x + A
If we equate coefficients, we get the system
A + B = 0 C = –1 2A + B + D = 2
C + E = –1
A=1
The solution is A = 1, B = –1, C = –1, D = 1, and E = 0.
Thus
=
1 x 2x 2 x3
dx
x(x 2 1)2
1 x 1
x
2
x 2
2 dx
x
1
(x
1)
dx
x
dx
xdx
2
dx 2
2
x
x 1
x 1
(x 1)2
1
1
ln(x2 + 1) – tan–1 x –
+ k.
2(x 2 1)
2
x 1
dx
Example 14.
(x 2 2x 3) 2
= ln |x| –
Solution
1
(2x 2) (–1 – 1)
2
dx
(x 2 2x 3)2
x 1
dx
2
(x 2x 3)2
=
1
2x 2
dx
dx – 2
2
2
2
2 (x 2x 3)
(x 2x 3)2
1
dx
1
–2
2
(x 2 2x 3)2
2 (x 2x 3)
We apply the substitution x + 1 = t to the last integral:
=–
dx
dx
(x 2x 3) ((x 1) 2)
2
=
2
2
(t 2 2) – t 2
1
dx
(t 2) 2 (t 2)
1
=
2
2
2
dt
1
–
t2 2 2
2
2
2
dt
t2
dt
(t 2 2)2
1 1
t
t 2 dt
1
tan–1
–
2 2
2 2 (t 2 2)2
Let us consider the last integral
=
t 2 dt
t.tdt
(t 2 2)2
(t 2 2)2
Integrating by parts
=–
1 t
1
dt
2
2
2 t 2 2 t 2
t
1
t
tan 1
2(t 2) 2 2
2
We do not yet write the arbitary constant but will take
it into account in the final result.
Consequently,
=–
2
dx
1
(x 2x 3) 2 2 t an
1 x 1
2
2
1
x 1
1
x 1
– –
tan 1
2 2(x 2 2x 3) 2 2
2
Finally we get,
x 1
x2
(x 2x 3) dx – 2(x 2x 3)
2
–
2
x 1
2
C.
tan–1
2
4
Example 15.
Evaluate
(2x 3)dx
(x 2x 3) .
2
2
Here (d/dx) (x2 + 2x + 3) = 2x + 2.
Solution
I=
2
(2x 3)dx
(2x 2 1)dx
(x 2x 3) (x 2x 3)
2
2
2
2
1.96
=
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(2x 2)dx
dx
(x 2x 3) (x 2x 3)
2
2
2
=
2
1
dx
(x 2 2x 3)
(x 2 2x 3)2
=–
Now let I1 =
Put x + 1 =
I1 =
...(1)
P(x)
2 tan t, so that dx =
2 sec2 t dt.
2 sec 2 t dt
2
2
(2 tan t 2) 4 cos t dt
2
[t + sin t cos t] + C
8
sin t =
cos t =
x 1
2
2
{(x 1) 2}
1
2
x 1
2
(x 2x 3)
, and
2
(x 2 2x 3)
x 1
.
Also t = tan–1
2
x 1
2
tan–1
Hence, I1 =
2
8
+
2
x 1
2
.
.
2
2
8
{(x 2x 3)} (x 2x 3)
=
x 1
x 1 1
2
+
tan–1
.
2
2 4 (x 2x 3)
8
...(1)
where Q1(x) is the greatest common divisor of the
polynomial Q(x) and its derivative Q(x), and
1
X(x) and Y(x) are polynomials with undetermined
coefficients, whose degrees are, respectively, less by
unity than those of Q1(x) and Q2(x).
The undetermined coefficients of the polynomials X(x)
and Y(x) are computed by differentiating the identity (1).
Example 16.
x 1
Y(x)
Q(x)
Q2(x) = Q (x) .
2
Now tan t =
X(x)
Q(x) dx Q (x) Q (x) dx
1
2 1
2
(1 cos 2t)dt =
[ t + sin 2t]
2
4 2
8
=
The Ostrogradsky method
If Q(x) has multiple roots, then
dx
.
[(x 2 1)2 2]2
2
x3
2
x 1
tan 1
+ C.
8
4(x 2x 3)
2
2
1
I=– 2
x 2x 3
Find
dx
(x 1)
3
2
Solution
Ax 2 Bx C
Dx 2 Ex F
dx.
3
x 1
x3 1
Differentiating this identity, we get
1
(x – 1)
3
2
1
(2Ax B)(x3 1) 3x 2 (Ax 2 Bx C)
=
(x3 1)2
(x 3 1)2
Dx 2 Ex F
x3 1
or, 1 = (2Ax + B) (x2 – 1) – 3x2 (Ax2 + Bx + C)
+ (Dx2 + Ex + F) (x3 – 1).
Equating the coefficients of the respective degrees of
x, we will have
D = 0, E – A = 0, F – 2B = 0, D + 3C = 0,
E + 2A = 0, B + F = – 1,
+
whence A = 0, B = –
1
2
, C = 0, D = 0, E = 0, F = –
3
3
x 1
1
x 1
2
tan 1
+ C, from (1)
2
4 x 2 2x 3 8
and, consequently,
dx
1 x
2
dx
– 3
...(1)
3
2 = –
3
(x – 1)
3 x 1 3 x 1
To compute the integral on the right of (1), we
=
x 1 4
2
x 1
tan 1
+C
4(x 2 2x 3) 8
2
decompose the fraction
1
into partial fractions :
x 1
3
INDEFINITE INTEGRATION
1
L
Mx N
2
,
x 1 x 1 x x 1
That is, 1 = L (x2 + x + 1) + Mx (x – 1) + N (x – 1).
3
1
Putting x = 1, we get L = .
3
Equating the coefficients of identical degrees of x on
the right and left, we find
L + M = 0; L – N = 1, or M = –
1
2
;N=– .
3
3
Therefore,
dx
1
dx 1
x2
x3 1 3 x 1 3 x2 x 1 dx
1
1
= ln |x – 1| – ln (x2 + x + 1)
3
6
1
2x 1
C
–
tan–1
3
3
and
dx
1 x2 x 1
1 x
ln
–
=
+
(x3 1)2
(x 1)2
3 x3 1 9
–
2
2x 1
C
tan–1
3 3
3
Example 17.
Find
x3 3x 2 2x 3
dx .
(x 2 1)2
Solution There is a repeated quadratic
polynomial in the denominator. Hence,
x3 3x 2 2x 3 A1x B1 A 2 x B2
2
2
(x 2 1)2
x 1
(x 1)2
for some constants A1, B1, A2 and B2.
An easy way to determine these constants is as
follows. By long division,
x
x3 3x 2 2x 3
=x–3+ 2
,
x 1
x2 1
and therefore
x
x3 3x 2 2x 3
x3
= 2
+ 2
2 .
2
2
(x 1)
x 1 (x 1)
Thus A1 = 1, B1 = –3, A2 = 1, and B2 = 0.
We now have
x3 3x 2 2x 3
dx
(x 2 1)2
x
3
x
2
dx 2
dx
dx
2
x 1
x 1
(x 1)2
=
1.97
1
1
ln (x2 + 1) – 3 tan–1 x –
+ C.
2
2(x 1)
2
Example 18. Show that
4x9 21x6 2x3 3x2 3
(x7 x 1)2
dx is a rational function.
Solution This is of the type A/Q2, where
Q x7 – x + 1 , Q' 7x6 – 1
and Q, Q' have no common factor [Q = 0, Q' = 0 have
no common root].
Hence there are polynomials C and D for which
A (x7 – x + 1)C + (7x6 – 1)D
...(1)
We choose the degree of C and D to be as low as the
general theory permits and take C to be degree 5 [one
less than that of 7x6 – 1], D to be of degree 6[one less
than that of x7 – x + 1].
If we take C c0 + c1x +....+ c5x5,
D d0 + d1x +...+ d6x6,
and equate coefficients in (1), we see that our first
needs are c5 + 7d6 = c4 + 7d5 = c3 + 7d4 = 0
and that there is probably a solution with
c3 = c4 = c5 = d4 = d5 = d6 = 0 ;
a little work then leads to
D(x) x3 + 3
C(x) –3x2,
We therefore consider
A
3x 2
Q'
3
dx
Q2 Q dx (x 3) Q2 dx
3x 2
x3 3
3x 2
dx
dx
=
Q
Q
Q
since (Q'/Q2) dx = – d(Q–1).
Hence the integral is –(x2 + 3)/Q and is rational.
Substitutions leading to integral of rational
functions
(i) If an integral is of the form R(sin x) cos x dx, the
substitution sin x = t, cos x dx = dt reduces this
integral to the form R(t) dt, where R(t) is a rational
function.
(ii) If the integral has the form R(cos x) sin x dx, it is
reduced to an integral of a rational function by
the substitution cos x = t, sin x dx = – dt.
(iii) If the integrand is dependent only on tan x, then
dt
the substitution tan x = t, i.e. x = tan–1t, dx =
1 t2
reduces this integral to an integral of a rational
function :
dt
R(tan x) dx = R (t)
1 t2
1.98
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(iv) If the integrand has the form R (sin x, cosx), but
sin x and cos x are involved only in even powers,
then the same substitution is applied : tan x = t
because sin 2x and cos 2x can be expressed
rationally in terms of tan x :
1
1
cos2x =
1 tan 2 x 1 t 2
tan 2 x
t2
1 tan 2 x 1 t 2
dt
dx =
1 t2
After the substitution we obtain an integral of a rational
function.
sin2x =
3
Example 19.
Solution
Compute the integral
sin x
2 cos x dx .
This integral is readily reduced to the
form R (cos x) sin x dx. Indeed,
sin 3 x
dx =
2 cos x
1 cos2 x
=
sin x dx
2 cos x
We substitute cos x = z. Then sinx dx = – dx
z2 1
dz
z2
3
z2
dz =
= z 2
– 2z + 3 ln (z + 2) + C
z2
2
=
cos2 x
– 2 cos x + 3 ln (cos x + 2) + C
2
sin x
dx.
sin 4x
sin x
sin xdx
dx =
Solution
sin 4x
2 sin 2x cos 2x
sin xdx
1
dx
=
4sin x cos x cos 2x 4 cos x cos 2x
1
cos dx
4 cos 2x cos2 x
1
cos x dx
=
4 (1 sin 2 x)(1 2sin 2 x)
Now put sin x = t so that cos x dx = dt.
Example 20.
Evaluate
=
1
dt
1
dt
4 (t 2 1) 4 t 2 (1 / 2)
1 1 t 1 1
= 4 . 2 ln t 1 4 .
1
t (1 / 2 )
ln
2.(1 / 2 ) t (1 / 2 )
=
1 t 1
1
t 2 1
ln
ln
+C
8 t 1 4 2 t 2 1
=
1 sin x 1
1
ln
ln
8 sin x 1 4 2
2 sin x 1
+ C.
2 sin x 1
dx
sin x sin 2x
dx
dx
Solution
sin x sin 2x
sin x 2 sin x cos x
dx
sin x(1 2 cos x)
sin x dx
sin xdx
=
2
2
sin x(1 2 cos x)
(1 cos x)(1 2 cos x)
Now putting cos x = t, so that – sin x dx = dt, we get
dt
dt
=–
(1 t)(1 t)(1 2t)
(1 t 2 )(1 2t)
1
4
1
=–
dt
6(1 t) 2(1 t) 3(1 2t)
1
1
2
= ln 1 – t + ln 1 + t – ln 1 + 2t + C
6
2
3
1
1
= ln (1 – cos x) + ln (1 + cos x)
6
2
2
– ln 1 + 2 cos x + C.
3
(1 sin x)dx
Example 22. Evaluate sin x(1 cos x)
Example 21.
Evaluate
sin 2 xsin xdx
2 cos x
sin 3 x
1 z2
dx =
(– dz) =
2 cos x
2z
dt
1
dt,
2
2
4 (t 1)(2t 1)
resolving into partial fractions
=
(1 sin x)dx
sin x(1 cos x)
Solution
=
sin x(1 cos x) dx 1 cos x
=
(1 cos x)(1 cos x) 2 sec 2 dx
1
dt
1
dt
I=
4 (1 t 2 )(1 2t 2 ) 4 (t 2 1)(2t 2 1)
We have I =
1
dx
sin xdx
1
2
2 x
1.99
INDEFINITE INTEGRATION
For first integral put cosx = t
dt
x
I=–
tan
2
2
(1 t)(1 t)
on integration by partial fractions, we get
1 1 cos x
1
x
tan + C
= n
4 1 cos x 2(1 cos x)
2
=
1
x 1
x
x
n tan 2 sec 2 tan + C
4
2 4
2
2
Integrate
Example 23.
3 tan x d x
I = (tan x)1/3 d x
Put tan x = t3 sec2 x d x = 3 t2 d t
Solution
3t 3 d t
1 t6
=
Put t2 = y 2t d t = dy
y
3
dy
=
1
y3
2
I1 =
y 11
dy
y3 1
=
dy
=
y 21 3
2
2
2
+
Now
=
y 1
.
2
y y 1
y 1
(2 y 1) 1
1
dy =
y2 y 1
y2 y 1
2
dy
1
1
l n (y2 y + 1) y 2 y 1
2
2
I1 =
+
y 2 (y 2 1)
dy
y3 1
2
1
2y 1
l n (y3 + 1)
tan 1
3
3
3
=
1
2y 1 1
tan 1
l n (y3 + 1)
3
3
3
1
l n (y2 y + 1) + C.
2
I=
dx,
=
3
I where y = (tan x)2/3.
2 1
We have
Solution
1
(e 1) dx.
x
2
1
dx =
(e 1)2
x
ex
e x (e x 1)2
dt
t(t 1) , putting ex = t so that ex dx = dt.
2
1
A
B
C
,
t(t 1) t t 1 (t 1)2
...(1)
1 A(t – 1)2 + Bt (t – 1) + Ct,
To find A, putting t = 0 on both sides of (1),
we get A = 1.
To find C, put t = 1 and we get C = 1.
Thus 1 (t – 1)2 + Bt (t – 1) + t.
Comparing the coefficients of t2 on both sides,
we get 0 = 1 + B or, B = –1.
1
1
1
1
.
t(t 1)2 t t 1 (t 1)2
Now
Hence
dy
dy
y 2 y 1 y3 1
Evaluate
Example 24.
dt
1
dt
dt
t(t 1) t dt t 1 (t 1)
2
2
= ln t – ln t – 1 – {1/(t – 1)} + C
= ln ex – ln ex – 1 – {1/(ex – 1)} + C
= x – ln ex – 1 – {1/(ex – 1)} + C.
Example 25.
I=
Solution
Evaluate
tan
x
1
4
tan 1 x
dx.
x4
dx tan 1 x.
1
dx
x4
1
1
1
.
dx
= (tan–1 x) 2
2
1 x ( 3x3 )
3x
tan 1 x 1
dx
,
3
3
3 x (1 x 2 )
3x
Put 1 + x2 = t
2x dx = dt
=–
=–
tan 1 x 1
dt
3
6 (t 1)2 .t
3x
tan 1 x 1
I1
...(1)
6
3x3
1
B
C
A
dt
dt
where, I1 =
2
2
t
(1 t) .t
t 1 (t 1)
Comparing coefficients we get,
I= –
1.100
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
From (1) and (2), we get
tan 1 x 1
1
ln x 2 2 ln(1 x 2 ) + C
I=
6
3x3
x
A= –1, B =1, C = 1
I1 =
1
1
1
(t 1) (t 1) t dt
2
= – ln |t – 1| –
1
+ ln |t|
(t 1)
...(2)
I=–
tan 1 x 1 x 2 1
1
ln
2 + C.
6
3x3
x2
6x
N
1.
Evaluate the following integrals :
6.
x 1
(i)
4 x3 x d x
2 x 2 41x 91
(x 1) (x 3) (x 4) d x
Evaluate the following integrals :
(ii)
2.
(i)
x3dx
x 4 3x 2 2
7.
(iii)
x3 1
dx
x3 x
8.
x 4 2x 3 3x 2 x 3
dx
(iv)
x 3 2x 2 3x
Evaluate the following integrals :
dx
dx
(ii) x (x 2 1)
(i) x3 1
x2
1 x 2/3
dx (iv)
(iii)
dx
2
2
(2x 4x 3)
1 x
Evaluate the following integrals :
dx
dx
(ii)
(i)
sin x(3 2 cos x)
sin 2 x 2 sin x
5.
(iii)
(iv)
(ii)
(iii)
sin tan d
2
2
cos
dx
x
2
ln x (ln x) 3ln x 10
4.
3
Evaluate the following integrals :
x
(i) (x 1) (x 2 4) dx
(ii)
cos 2 x sin 2x
dx
(2 cos x sin x)2
dx
(iv)
x
e 3 2e x
Applying Ostrogradsky's method, find the
following integrals :
dx
dx
(ii)
(i)
(x 1)2 (x 2 1)2
(x 4 1)2
dx
(iii) (x 2 1) 4
x 4 2x 2 2
dx
(iv)
(x 2 2x 2)2
Evaluate the following integrals :
5x 2 12
dx
(i)
2
x2 6x 13
(iii)
x4
(1 x) dx
6x2 12x 4
(ii)
dx
x2 (x 2)2
3.
Evaluate the following integrals :
dx
(i)
sin x(3 cos2 x)
(ii) sec x. sec 2x dx
3
(iv)
x3 x 1
( x 2 2 )2 d x
x6 x 4 4x 2 2
dx
x3 (x 2 1)2
dx
x x 1
4
9.
3
2
Evaluate the following integrals :
(i)
(iii)
x3 x2 x 3
dx (ii)
(x2 1)(x2 3)
x7 2
dx (iv)
(x 2 x 1)2
dx
x (x 1)
4
3
2
3x 4 4
dx
x 2 (x 2 1)3
1.101
INDEFINITE INTEGRATION
10. Evaluate the following integrals :
2x3 x 2 4
(i)
dx
(x 2 4)2
x3 x 2 5x 15
(ii)
(x 2 5)(x 2 2x 3) dx
1.16 SPECIAL METHODS FOR
INTEGRATION OF
RATIONAL FUNCTIONS
1
2
x 1
x(x 3) dx = 3 ln |x3 + 3x| + C.
pt q
A
B
,
(t a)(t b) (t a) (x b)
then the integral becomes
Suppose that
A
B
dx 2
dx .
2
(x a)
x b
In general, if the numerator and the denominator of a
given fraction contains even powers of x only, we can
first write the fraction in a simpler form by putting t for
x2 and then break it up into partial fractions involving
t, i.e., x2, and then integrate it.
Example 1.
Solution
Evaluate
dx
x (1 x )
2
2
Treating 1/x2 (1 + x2) as a function of x2,
we have
1
1
1
,
x 2 (1 x 2 ) x 2 1 x 2
dx
1
tan 1 x + C.
Hence,
2
2
x
x (1 x )
(iv)
4
3
x 5 x 4 4x3 4x 2 8x 4
dx
(x 2 2)3
Solution
Integrate
x2
dx .
x4 x2 2
Putting x2 = z, we have
x2
z
2
4
2
x x 2 z z2
z
A
B
, say..
(z 2)(z 1) z 2 z 1
z = A(z – 1) + B(z + 2).
Putting z = –2 and 1, we get respectively
A
2
px 2 + q
1. Integrals of the form (x 2 + a)(x 2 + b) dx
Assume x2 as t for finding partial fractions.
(x 2x 10)
Example 2.
Note that sometimes partial fractions can be avoided
when integrating a rational function. For instance,
although the integral
x2 1
dx
x(x 2 3)
could be evaluated by the method of partial fractions,
it is much easier, if we put u = x(x2 + 3) = x3 + 3x, then
du = (3x2 + 3) dx and so
dx
(iii)
2
1
,B
3
3
x2
2 1
1 1
x4 x2 2 3 x2 2 3 x2 1
2
dx
1 dx
I
3 x2 2 3 x 2 1
2 1
x
11 x 1
tan 1
ln
C .
3 2
2 32 x 1
Example 3.
Solution
Evaluate
x3 dx
(x 1) .
2
2
Regarding x2/(x2 + 1)2 as a function of
x2, we find (by inspection)
x2
(x 2 1) 1
1
1
2
2
(x 1)
(x 2 1)2
x 1 (x 2 1)2
2
x3 dx
x dx
x dx
2
2
2
2
(x 1)
x 1
(x 1)2
1
1
ln(x 2 1)
+ C.
2
2
2(x 1)
Hence,
2. Integrals of the form
P(x)
dx w here
Q(x)
Q(x) has a linear factor with high index
Substitute the linear factor as 1/t.
1.102
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 4.
Evaluate
3x 1
(x 1) (x 1) dx
3
Put x – 1 = 1/t x = 1/t + 1
Solution
3t 4t 2
(2t 1) dt
1
1
(2t 1)2 t 1
= 2t
=–
dt
2 2(2t 1)
(2t 1)
Now, the integral can be evaluated easily.
dt
dx = 2 .Thus, the integral
t
3 4 dt
3 4t dt
t
t
=
=–
3
1 1 2t 2
1 1 2 t2
t
t3 t
t t
1 (bz ab' a ' b)2
dx
b'3
x3
1
dz
dz
3
2b(ab' a ' b) dz3 (ab' a ' b)2 3
b' z
z
z
2
2
b
2b(ab' a ' b) 1 (ab' a ' b) 1
= 3 ln | z |
+C
z
b'
b'3
2b'3
z2
b2
2b(ab' a ' b)
(ab' a ' b)2
= 3 ln a ' b' x 3
+C
b'
b' (a ' b' x) 2b'3 (a ' b' x)2
3. Substitution
Note: By the same process we can integrate
(a bx)m
dx , where m is a positive integer, n being
(a ' b' x)n
a rational number.
(i) Integrals of the form
m, n N
xm
(ax + b ) n
dx ,
(ii) Integrals of the form m dx
x (ax b) n
Put
Put ax + b = t.
(t b)m dt
.
a
tn
Expanding by the binomial theorem and integrating
each term separately the required integral can be
immediately obtained.
x2
dx
Example 5. Evaluate I =
(x 2)3
Solution Put x + 2 = t
1
The integral becomes
m 1
t 2 4 4t
(t 2)2
dt
=
dt
t3
t3
1 4 4
4 4
dt = ln |t| – 2 + C.
=
t t3 t2
3t
t
where t = x + 2.
(a bx)2
Example 6. Integrate (a ' b' x)3 dx
z a'
Solution Put a' + b'x = z or x = b'
1
dx = dz.
b'
Now the given integral becomes
I=
a
b
(z a ')
b'
z3
2
dz
b'
I=
Solution
Put
, m, n N
ax b
=t
x
Example 7.
I=
Evaluate I =
dx
x (x 2)
2
3
dx
x2
x 5
x
3
x2
2
2
= t 1 +
= t 2 dx = dt
x
x
x
1 (t 1)3
dt
2
8t 3
Now, the integral can be evaluated easily.
I=–
Example 8.
Solution
Integrate
dx
x (a bx)
3
2
Put a + bx = zx, or
a
+ b = z.
x
a
dx= dz
x2
The given integral
Then –
3
1
dz
1 z b dz
a x.z 2 x 2
a a z2
1
3b2 b3
2 dx
z
3b
z
a4
z
1.103
INDEFINITE INTEGRATION
1 z2
b3
3bz 3b2 ln z C
4
z
a 2
2
1 1 a bx
a bx
4
3b
a 2 x
x
a bx
x
3b2 ln
b3
C .
x
a bx
(iii) Integrals of the form
dx
(x a) m (x b) n
This integral can be easily transformed into a form
which is immediately integrable, by the substitution
xa
= t if m < n
xb
(x b)(1) (x a)(1)
ab
dx= dt
dx = dt
2
(x b)
(x b)2
a bt
Also , x =
1 t
(a b)t
ab
,
x–a=
, x–b=
1 t
1 t
m n–2
dt
The integral transforms into (1 t)
m n–1 m
(a b)
t
Expand the numerator by the Binomial Theorem, and
the integral can be immediately obtained.
For example, take the integral
dx
(x a)2 (x b)2
xa
Put
=t
xb
1
1 3 3 t
dt
=
(a b)4 t 2 t
–1 t 2
1
3t 3ln t C .
=
4
t
(a b) 2
xa
.
where t =
xb
dx
Integrate (x 1)2 (x 2)3
Example 9.
x 1
Solution Put x 2 z
1 2z
x
1 z
dz
dx (1 z)2 .
Hence, the integral transforms into
(1 z)3
1 3z 3z 2 z3
dz
dz
2
z
z2
1
1
3ln| z | 3z z 2 C
z
2
2
x 1
x 2
x 1 1 x 1
3ln x 2 3 x 2 2 x 2 + C
x 1
(iv) Integrals of the form
Here, the substitution of xn =
dx
x(a bx n )
1
gives
t
1
dt
n at b
the value of which is obviously
–
–
1
1 xn
ln
ln | at b | C
C.
na
na a bx n
(v) Integrals of the form
where m and n are integers
Put ax2 + b = t.
x 2 m1
dx ,
(ax 2 b)n
(t b)m dt
2a m 1t n
a form which is immediately integrable by aid of the
Binomial Theorem. It is evident that the expression is
made integrable by the same transformation when n is
either a fractional or a negative index.
The integral becomes
x5
dx
(x 2 1)3
Put x2 + 1 = t 2x dx = dt
To evaluate I =
1 (t 1)2
dt
2
t3
1
1
1
= 2
ln x 2 1 C .
x 1 4(x 2 1) 2
In general, if in a fraction, the numerator contains only
odd powers of x and the denominator only even
powers, then it is found more convenient to change
the variable first by putting x2 = z and then break it up
into partial fractions as usual.
I=
1.104
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
It may be also observed that the more general
2
f(x )x
can be integrated by the same
(a cx 2 )n
transformation, where f(x2) denotes an integral
algebraic function of x2.
I=
expression
Example 10.
Integrate
I
1 1
t2
dt
2 t (t 2 t 1)
1
1 2t 1 3
dt
= lnt –
2
4 (1 t t 2 )
1
z dz.
2
z
z
Now, z 2 3z 2 (z 1) (z 2)
2
A
B
, say..
z 1 z 2
We determine A = –1, B = 2 as usual.
Example 13.
dz
z 1
Solution
1
[2 ln(z + 2) – ln(z + 1)] + C
2
1
= ln(x2 + 2) – ln(x2 + 1) + C
2
dx
Example 11. Evaluate
x(1 x3 )2
Solution
put x2 = t.
2
x
x2 1 1
dx
2
(x 1)
(x 2 1)2
x
dx
dx
= 2
2
2
(x 1)
(x 2 1)
(x 2 1)2
dx
1 x
1
2
.
tan 1 x
(x 2 1)2 2 x 2 1 2
dx
1 x
1
or
.
tan 1 x + C.
(x 2 1)2 2 x 2 1 2
dx
Example 14. Evaluate I = (x 4 1)2
Solution We have d(x4 – 1) = 4x3 dx
=
1 x2
x(1 x x )dx
2
2
In order to evaluate it we integrate
3x2 dx = dt.
dt
1
dt
I=
split into partial fractions
3x3 t 2 3 (t 1)t 2
= 1 1 1 12 dt
3 t 1 t t
1
1
= ln t 1 ln t C
2
t
3
1
x
1
= ln
C .
3 1 x3 1 x3
Evaluate
dx
(x 1) .
1
by parts choosing unity as the other
(x 1)2 1
function.
1
1
(2x)
dx 2
x x 2
dx
2
(x 1)
(x 1)
(x 1)2
Put 1 + x3 = t
Example 12.
Evaluate
2
2
4. Integration by parts
Solution
3
2x 2 1
1
tan 1
+C
= ln x – ln(x4 + x2 + 1) –
4
2
3
3
dt
1
1
lnt – ln(t2 + t + 1) – 4
2
2
2
4
t 1 3
2 2
1
x2
1 2 1
3 1
1
4
2
2
= lnx – ln(x + x + 1) – .
+C
tan
2
4
4 3
3
1
z dx
.
2 x 2 3z 2
1
dz
I 2
2 z2
x3 dx
.
x 3x 2 2
4
=
2x dx = dz. x3 dx =
I=
Put x2 = z.
Solution
1
1 t
dt split into partial fractions.
2 t(1 t t 2 )
4
Multiply above and below by x and
2
1
4x 3
· 4
dx
3
4x (x 1)2
1
1
3
1
·
·
dx
=
–
4x3 (x 4 1)
4x 4 (x 4 1)
I=
INDEFINITE INTEGRATION
1
3 1
1
4 dx
4
–
4
4 x 1 x
4x (x 1)
1
1
dx
dx =
Now 4
(x 2 1)(x 2 1)
x 1
= –
3
We have A =
( 3 1)(3 2)
= 2,
( 3 4)
B=
(4 1)( 4 2)
= –6,
(4 3)
1 1
1
dx
2 x2 1 x2 1
Now, the integral can be evaluated easily.
Here are some more examples on integration of rational
functions.
Example 15. Evaluate
=
2e 5x e 4x 4 e 3x 4 e 2x 2 e x
dx
(e 2x 4)(e 2x 1)2
Solution Put ex = y
2y 4 y3 4y 2 4y 2
dy
(y 2 4)(y 2 1)2
I=
y(y 2 4) (y 4 2y 2 1)
dy
(y 2 4) (y 2 1)2
x
1
1 e
tan
=
2 + C.
2 (e 2x 1)
=
dy
Evaluate y 2 (1 y 2 )3
Example 16.
Solution
Put y = tan
cos6
dy
y (1 y ) sin d
2
=
=
2 3
2
2
4
(1 sin 2 )3 d
sin 2
6
(1 3sin 3sin sin )d
sin 2
(cosec – 3 + 3 sin – sin ) d
2
2
4
1 15
1
tan–1 y – sin (2 tan–1 y)
y 8
2
1
sin (4 tan–1 y) + C.
–
32
(x 2 1)(x 2 2)
Evaluate
dx .
Example 17.
(x 2 3)(x 2 4)
Solution We have
=–
(x 2 1)(x 2 2) (y 1)(y 2)
, where y = x2.
(x3 3)(x 2 4) (y 3)(y 4)
(y 1)(y 2)
A
B
Now let
=1+
,
(y 3)(y 4)
y3 y4
resolving into partial fractions.
1.105
(y 1)(y 2)
1
6
=1+
.
(y 3)(y 4)
y3 y4
The given integral
2
6
2
I = 1 2
dx
x 3 x 4
=
dx
dx
2
2
dx + 2 x 3 –6 x 4
= x + 2.
=x+
1
x
1
x
tan–1
– 6 . tan–1 + C
2
2
3
3
2
x
x
– 3 tan–1 + C.
tan–1
3
2
3
Example 18. Evaluate
Solution
(3x + 1) / {(x – 1)3(x + 1)}dx.
Putting x – 1 = y so that x = 1 + y, we get
3x 1
3(1 y) 1
4 3y
3
3
3
(x 1) (x 1)
y (2 y)
y (2 y)
arranging the Nr. and the Dr. in ascending powers of y
=
1
1
1
1 y3
2 y y2
,
3
2
4
4 2 y
y
by actual division
2
1
1 1
1
= y3 2y 2 4y 4 . (2 y)
2
1
1
1
3
2
4(x 1) 4(x 1)
(x 1) 2(x 1)
Hence the required integral
=
1
1
1
2
dx
=
3
2(x 1)2 4(x 1) 4(x 1)
(x 1)
=
1
1
1
1
ln x – 1 + ln x + 1 + C
2
4
2(x 1) 4
(x 1)
=
1
1
1 x 1
ln
+ C.
2
2(x 1) 4 x 1
(x 1)
1.106
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Evaluate the integral
Example 19.
1
d cos 2 2 d 2 cos 2 d
2
1
1 cos 4
sin 2
d
2
2
2
12
sin 4
sin 2 C
=
2
2
8
sin 4 sin 2
C
2 4
16
2
3 sin 2 sin 4
=
+ C where x = sec2 .
4
2
16
1
x (x 1) dx .
=
3
1 cos2 d 2 (cos2 2 2 cos2)d
=2
2
4
Let x = sec2
dx = 2 sec2 tan d
2 sec 2 tan d
2 cos 4 d
I=
sec 6 tan
4
2
2
I = 2 cos d 2 [(cos 2 ) ]d
Solution
2
O
1.
Evaluate the following integrals :
(3x 2 – 2) dx
(i) 4
x 3x 2 4
x 2 dx
(ii)
2
(x 1) (2x 2 1)
3.
(iii)
(iv)
2.
x 2 dx
(a 2 x 2 )2
dx
x 4x 4 x 4x 5
2
1.17 INTEGRATION OF
IRRATIONAL
FUNCTIONS
Integration by Rationalization
Certain types of integrals of algebraic irrational
expressions can be reduced to integrals of rational
functions by a appropriate change of the variable.
Such transformation of an integral is called its
rationalization.
1. Let n be a positive integer. Any rational function
4.
Evaluate the following integrals :
dx
dx
(i)
(ii) (x 2 9)3
(x 4 1)2
dx
x 2 dx
(iii)
(iv) 2
x (1 x 2 )2
(1 x 2 )3
Evaluate the following integrals :
x9 dx
x7
dx
(i) 12
(ii)
(x 4 1)2
(x 1)
2
Evaluate the following integrals :
dx
x dx
(ii)
(i)
x4 x2 2
x(a bx 2 )2
x
(iii)
2
(x 2) (x 2 1) d x
(1 x 2 )dx
x(1 x 2 x 4 )
Evaluate the following integrals :
dx
dx
(i)
(ii)
2
x (a bx)
x(a bx n )2
dx
(iii)
x 4 4x3 5x 2 4x 4
(iv)
5.
of x and n ax b can be transformed into a rational
function by the substitution tn = ax + b and thus
can be integrated by partial fractions.
x n dx
, where n is a positive
(a bx)1/2
integer, we put a + bx = t 2 , then dx =
For example, in
2tdt
t2 a
, and x
. Making these substitutions,
b
b
the integral becomes
easily integrated.
(t 2 a) n dt
, which can be
b n 1
1.107
INDEFINITE INTEGRATION
Example 1.
Evaluate
x 1 x dx.
7
3
3
4
Let x = t3
Solution
t(1 + t4)1/7 . 3 t2 dt = 3 t3(1 + t4)1/7 dt
I
Integrate
1 x
1 x dx
Rationalizing the numerator, we have
1 x
dx
x dx
(1 x ) dx (1 x ) (1 x )
2
2
2
Put 1 – x2 = z2 in the second integral, so that
– 2x dx = 2z dz
2
The second integral = – dz z (1 x )
I = sin–1x – (1 x 2 ) + C.
ax b
Note: Integrals of the type cx d dx (a
0, c 0) of which the above is a particular case can be
evaluated exactly in the same way.
Example 3.
Solution
Evaluate
x dx
(2x 3) x 1 .
2
2
Put x2 – 1 = t2 x dx = t dt
t dt
So I = (2t 2 5)t =
I
2tdt
nx
n 1
.
1
2
dt
since xn = t2 – 1
x.t n t 2 1
2 1 t 1
1
(1 x n ) 1
| + C.
C = ln |
. ln
n
n 2 t 1
(1 x n ) 1
dx
3. Integrals of the form
(a cx2 )3/ 2
1
Here, we put x = and the integral becomes
t
tdt
1
– (at 2 c)3/2 = a(at 2 c)1/2 C
dx
x
So,
+ C.
(a cx 2 )1/2 a(a cx 2 )1/2
dx
Consider the integral
(a 2bx cx2 )3/2
This can be written in the form
21
(1 x 4/3 )8/7 C .
Therefore, I =
32
Example 2.
=
3
21 6
. 7X 7 dX
X C
4
32
Solution
n
nxn – 1 dx = 2tdt.
Put 1 + t4 = X7
4 t3 dt = 7 X6 dX
I=
dx
x (1 x ) .
Put 1 + xn = t2.
Solution
dx = 3 t2
I=
Evaluate
Example 4.
dt
1
=
2t 2 5
2
dt
t2
5
2
1
1 2
2
= 10 tan 5 x 1 C
2. Integrals of the form
Here we put axn + b = t2.
c3/2 dx
{ac b2 (cx b)2}3/2
whi ch is r educed t o th e precedin g form by
substituting cx + b = z.
Hence, we get
dx
(a 2bx cx )
2 3/ 2
=
b cx
+C
(ac b )(a 2bx cx 2 )1/ 2
4. Integrals of the form
x axn + b
xdx
(a + 2bx + cx 2 ) 3/2
1
We substitute x = . The integral becomes
t
dx
(az2 2bz c)3/2
Hence,
dx
2
xdx
(a 2bx cx )
=
2 3/ 2
a bx
+C
(ac b )(a 2bx cx 2 )1/ 2
2
1.108
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Combining this with the previous result, we get
(p qx) dx
(a 2bx cx )
2 1/2
Example 5.
bp aq (cp bq)x
+ C'
(ac b2 )(a 2bx cx 2 )1/2
a 2 x2
dx
x4
Evaluate I =
Hence, I =
a 2 1 / t 2 = – t a 2 t 2 1dt
dt
(1 / t 4 )t 2
Now make one more substitution a 2 t 2 1 = z. Then
2a2tdt = 2zdz and
1
1
I = – 2 z 2 dz 2 z 3 C
a
3a
Returning to t and then to x, we obtain
2
2 3/2
I = (a x )
3a 2 x3
Example 6.
Solution
C.
Find
x x2
dx.
x4
x x2
dx =
x4
2
Let z – 1 = s .
dz
z 1
, x x2 =
,
z2
z
z 1 dz
z z2
= z z 1 dz
1 / z4
2
Then – z z 1 dz = – (s 1) (s) (2s ds)
s 5 s3
= –2 + C.
5 3
(z 1)5/2 (z 1)3/2
=– 2
+C
5
3
5.
z3 (1 z)dz
1 z2
Consequently the value of the integral is
4
4x3/4 + 2x1/2 – 4x1/4 + 4 tan–1(x1/4) – 2 ln(1 + x1/2) + C.
3
dx
Example 7. Find 1/3 1/2
x x
Solution Because 6 is the L.C.M. of 2 and 3, we
set u = x1/6, so that u6 = x and 6u5 du = dx. We now have
dx
6u 5du
x1/3 x1/2
(u 6 )1/3 (u 6 )1/2
5
6u du
6u 5du
6u 3du
=
1 u
u2 u3
u 2 (u 1)
When the degree of the numerator is greater than or
equal to the degree of the denominator, division is
often helpful. By long division,
6
6u 3
= 6u2 – 6u + 6 +
1 u
1 u
6
6u 3du
2
6u 6u 6
du
1 u
1
u
=2u3 – 3u2 + 6u – 6 ln|1 + u| + C
= 2(x1/6)3 – 3(x1/6)2 + 6x1/6 – 6 ln|1 + x1/6| + C
=2x1/2 – 3x1/3 + 6x1/6 – 6ln(1 + x1/6) + C.
1
Put x =
z
The substitution yields dx = –
and
(1 x1/4 )dx
1 x1/2
Let x = z4, where 4 is L.C.M. of 2 and 4, and the
transformed integral is
1
dt
Put x = , dx = – 2
t
t
Solution
the integral can be rationalized by the substitution
x = tm, where m is the L.C.M. of the denominators
q1, q2,....., qk of the several fractional powers. By this
means the integration of such expressions is reduced
to that of rational functions.
For example, to find
(1 x)5/2 (1 x)3/2
+ C.
=– 2
5/2
3x3/2
5x
If the integrand is a rational function of fractional
p1
pk
q1
qk
powers of x, i.e. the function R x, x , . . . .,x , then
Evaluate I =
x x2 6 x
x(1 3 x )
Example 8.
3
Solution The least common multiple of the
numbers 3 and 6 is 6, therefore we make the
substitution :
x = t6, dx = 6t5 dt,
I=6
(t 6 t 4 t)t 5
t (1 t )
6
2
dt = 6
t 5 t3 1
1 t
2
dt
1.109
INDEFINITE INTEGRATION
=6
dt
3
t3 dt + 6 t 1 2 t4 + 6 tan–1 t + C.
2
2
3 3
x + 6 tan–1 6 x + C.
2
6. Again, any algebraic expression containing integral
powers of x along with fractional powers of an
expression of the form a + bx is immediately reduced to
rational functions, by the substitution a + bx = tm,
where m is the L.C.M. of the denominators of the several
fractional powers.
I=
1
Example 9.
Solution
6
Evaluate I =
(2x 3) 2 dx
1
(2x 3) 3 1
The integrand is a rational function of
2x 3 . Therefore we put 2x – 3 = t6 dx = 3t5 dt
1
1
Also (2x 3) 2 = t3 and (2x 3) 3 = t2.
I=
3t 8
dt
2
t 1
= 3 (t6 – t4 + t2 – 1) dt + 3
=3
2
t7
t5
t3
–3
+ 3 – 3t + 3 tan–1 t + C.
7
5
3
7
5
1
1
1
1
6 (2x 3) 6 (2x 3) 2
(2x
3)
=3 7
5
3
1
1
(2x 3) 6 t an 1 (2x 3) 6
+ C.
Example 10.
Find
dx
.
2x 1 4 2x 1
The substitution 2x – 1 = z4 leads to an
integral of the form
dx
2z3 dz
z 2 dz
2
2
z 1
z z
2x 1 4 2x 1
1
2 z 1
dz = (z+1)2 + 2 ln|z–1| + C
z 1
1 4 2x 1
ln 2x 1 1 C .
2
4
dx
3
dx
(x 1) (x 1)1/2
The least common multiple of 2 and 3 is 6. So
substitute x +1 = t6
dx = 6t5 dt
I
1/3
6t 5dt
t 3dt
I t 2 t3 6 1 t
1
I 6 t t 1 1 t ) dt
2
t3 t2
3 2
On substituting t = (1 + x)1/6, we get
(1 x)1/2 (1 x)1/3
I6
3
2
I 6 t ln| t 1| C .
(1 x)1/6 ln (x 1)1/6 1 C .
(x 1)1/2 (x 1)3/4
dx
Example 12. Evaluate I =
x(x 1)1/4
Solution The common denominator of 1/2 and 3/4
is 4.
Putting x + 1 = t4 and dx = 4t3 dt, we have
t2 t3
· 4t3 dt
I=
(t 3 – 1)t
dt
1 t
dt 4
=4 4
(t – 1)(t 2 – 1)
t 1
1 – t 1
dt
=2
t – 1 t2 1
2t
dt
dt
=2
– 2
dt – 2
t –1
t –1
t –1
= 2ln |t – 1| – ln(t2 – 1) – tan–1 t + C
= 2 ln |(x + 1)1/4 – 1| – ln [(x + 1)1/2 + 1]
– 2 tan–1(x + 1)1/4 + C.
Solution
3
x 1 x 1
Let I
Solution
dx
x 1 x 1
dt
1 t
Evaluate
Example 11.
2
Example 13.
(x 7/6 x 5/6 )dx
x (x x 1) x (x x 1)
1/3
2
1/2
1/2
2
Solution
I=
(x 7/6 x 5/6 )dx
x 7/6 .x1/3(x 2 x 1)1/2 x1/2(x 2 x 1)1/3
1/3
1.110
=
=
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(1 x 2 )
x3/2(x 2 x 1)1/2 x 5/3(x 2 x 1)1/3
1
1 2 dx
x
1/2
1/3
x 1 1 x 1 1
x
x
(x b)1/2 (x a)1/2
dx
ba
1 2
2
(x b)3/2 (x a)3/2
=
b a 3
3
=
=
2 1
[(x + b)3/2 – (x + a)3/2] + C.
3 (b a)
1
1
( putting x t 1 2 dx dt )
x
x
Example 15.
dt
=–
1/2
(t 1) (t 1)1/3
(putting (t + 1) = u6)
Solution
=–
3
=–6
2
u3
du, (putting u – 1 = z)
u 1
z3 3z 2
= – 6
3z ln | z | + C,
2
3
1/6
1
where z = x 1 – 1
x
7. If the integrand is a rational function of x and
fractional powers of a linear fractional function of the
ax b
form
, then rationalization of the integral is
cx d
ax b
effected by the substitution
= tm, where m is
cx d
the L.C.M. of the denominators of the several fractional
powers.
Example 14.
Solution
dx
Evaluate
x 1
z2
x2
2
(z 1)3
z3 3z 2 3z 1
dz = – 6
dz
z
z
1
= 6 z 2 3z 3 dz
z
=
x 1 dx
x 2 x 3 .
2z
2z 2 1
1
2
2
2 and x + 3 =
1 z
1 z
1 z2
2zdz
dx =
(1 z 2 )2
2zdz (1 z 2 )
2z 2 dz
I z.
(2 z 2 )(1 z 2 )
(1 z 2 )2 (2 z 2 )
2 4 dz
=
1 z2 2 z2
= ln 1 z 2 ln 2 z C ,
1 z
2 z
x=
6u 5du
u u
Put
Evaluate
dx
(x a) (x b) .
Rationalizing the denominator, we have
(x b) (x a)
(x a) (x b) (x b) (x a) dx
x 1
where z =
.
x2
(1 sin x)(2 sin x)
Example 16. Evaluate (1 sin x)(2 sin x) dx
Solution
We have
cos x
1 sin x
=
1 sin x
1 sin x
I = cos x · 2 sin x dx
1 sin x 2 sin x
Let 1 + sin x = y
cos x dx = dy
2 (y 1) 3 y
2 sin x
=
=
2 (y 1) 1 y
2 sin x
1 3y
I= ·
dy
y 1 y
3y
Put
= t2 3 – y = t2 + t2y
1 y
1.111
INDEFINITE INTEGRATION
y(1 + t2) = 3 – t2 y =
3 t2
dy
1 t2
3 dt
3
2 + C.
3
2 t
4t
=–
=
2t 6t
(t 2 1)( 2t) (3 t 2 )2t
=
dt
2 2
(1 t 2 )2
(1 t )
Returning to x, we get I =
=
8t
dt
(1 t 2 )2
Example 18.
2
1 t t(8t)
t
·
dt = – 8
dt
I=
2
2 2
(3 t 2 )(1 t 2 )
3 t (1 t )
2
2
t2
dt
=8
(t 2 3)(t 2 1)
t2
A
B
2
Let 2
= 2
2
(t 3)(t 1)
(t 3) (t 1)
t2 = A(t2 + 1) + B(t2 – 3)
1
3
A + B = 1, – 4B = – 1 B = A =
4
4
3
1
t2
2
=
2
(t 3)(t 2 1) 4(t 3) 4(t 1)
3
dt
1
dt 2
2
t 1
4 t 3
1
t 3
= 2 3·
tan 1 (t)
ln
2 3 t 3
=
t 3
3 ln
2 tan 1 (t) + C.
t 3
Solution
x2
; therefore let us introduce
x 1
the substitution
x2
x2
= t
= t4,
x 1
x 1
3
t4 2
x – 1 = 4
x= 4
t 1
t 1
4
the substitution
2x
2x
= t
= t3,
2
x
2x
2 2t 3
4t 3
x=
3 2 – x =
1 t
1 t3
12t 2
dx =
dt.
(1 t 3 )2
3
Hence, I = –
x+2=
3t
12t 3
dx = 4
dt.
(t 1)2
t 1
2(1 t 3 )2 t.12t 2 dt
16t 6 (1 t 3 )2
4
Hence, I = –
(t 4 1)(t 4 1)12t 3dt
3.3t 4 t(t 4 1)2
4 dt 4
+ C.
3 t 2 3t
Returning to x, we obtain I =
Here we put
2x
, therefore let us introduce
2x
5
function of x and 4
The integrand is a rational function of x
and the expression 3
3
4
x2
, the integrand is a rational
x 1
Alternative :
Evaluate I =
dx
(x 1) (x 2)
= (x – 1) (x + 2) 4
2 3 2x
dx.
(2 x)2 2 x
Example 17.
Evaluate I =
3
5
4
Since (x 1) (x 2)
Solution
2
Now, I = 8 ·
3 3 2x
+ C.
4 2x
4 4 x 1
+ C.
3 x2
x 1
t
x2
dx
dt
3
dx dt or (x 2)2 3
2
(x 2)
dx
I=
3
x 1
8
4
(x 2)
x2
dx
dt
1
3/4
x 1
(t)3/4 dt
3/4
(x 2)2
3
3(t)
x2
1/4
4 x 1
1 (t)1/4
C
3x2
3 (1 / 4)
C
1.112
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Note: The integrals of the form
dx
(x a) (x b) , where p + q = 2n are solved more
p
n
q
xa
t , as
xb
comfortably using the substitution
shown above.
Evaluate
Example 19.
Solution
I=
(x 3)
dx
.
(x 4)17/16
15/16
dx
(x 3)15/16 (x 4)17/16
=
x3
dx
15/16
(x 4)2
x4
(x 4) (x 3)
x3
t
Put
dx dt
(x 4)2
x4
dt
dx
=
7
(x 4)2
dt
1
1 15/16
16 1/16
t
dt =
t C
So, I =
15/16 =
t
7
7
7
1/16
16 x 3
=
C.
7 x4
P
1.
Evaluate the following integrals :
x dx
2x 1
dx
(ii)
(i)
2
(a bx)1/2
x
xa
(iii)
dx
xb
2.
Evaluate the following integrals :
x3x
(i)
4 x5 6 x7 dx
x 2/3
dx
(ii) 1/3
2x (x 1)1/3
dx
(iii)
x 1
x2 3
x
Evaluate the following integrals :
xdx
1
(ii)
(i)
dx
(px q)3/2
x (1 x6 )
dx
x5
dx
(iv) 3 2
(iii)
2
x x 1
(1 x
Evaluate the following integrals :
dx
(i)
(1 x)3/2 (1 x)1/2
dx
(ii)
4 5 x 5 x
dx
(iii)
(x 2) 4 (x 2)
3.
4.
(iv)
x 1 2
(x 1) x 1 dx
2
1.18 INTEGRALS OF THE
dx
TYPE P Q
Where P and Q are l inear or q uadratic
expressions
1.
Integrals of the form
dx
(ax + b) (cx + d) , where (a 0, c 0)
To find the integral
dx
P Q , where P and Q are linear
algebraic expressions of x, we put Q = t2.
dx
Thus, to integrate
(ax b) (cx d)
2
Put cx + d = t
c dx = 2t dt.
The integral then reduces to
2
t dt
dt
,
2 2
c t2 d
at (bc ad)
b . t
a
c
which can be easily evaluated.
dx
.
Example 1. Evaluate
(2x 1) (4x 3)
Solution Put 4x + 3 = t2, so that 4dx = 2t dt and
(2x + 1) =
2(t 2 3)
t2 3
t2 1
1
1
4
2
2
INDEFINITE INTEGRATION
1
tdt
2
dt
2
1 2
(t 1)
(t 1)t
2
1 t 1
1
(4x 3) 1
C .
= 2 ln t 1 C 2 ln
(4x 3) 1
dx
(2x 1) (4x 3)
dx
, .
(x )( x)
Example 2. Evaluate
2tdt
dt
2
2
2
2
dt
t mt n , which can be easily evaluated.
4
2
x
t
+ C = 2 sin–1
+C
k
x 2 dx
.
Example 3. Evaluate
(x 1) (x 2)
= 2 sin–1
Put (x + 2) = t2, so that dx = 2t dt.
x
=
x dx
(x 1) (x 2)
2
2
(t 2) .2t dt
(t 2 3).t
t 4 4t 2 4
dt
t2 3
2
=2
[t2 – 1) + t 3 ] dt,
1
2
1
t 3
1
ln
t3 – t +
]+C
2
3
t 3
3
=2 (x 2)
3
2.
3/2
(x 2)
1
2 3
2
ln
1
(x 2) 3
+C.
(x 2) 3
dx
(px qx r) ax b
2
1
1
2
by partial fractions
1
dt
1
dt
1
dt
1
dt
=
2 (t 1)2 2 (t 1) 2 (t 1)2 2 (t 1)
=–
1
1
1
1
+ ln t + 1 –
– ln t – 1 + C
2(t 1) 2
2(t 1) 2
=–
1
1
t 1
1
1
[
+
] + ln
+C
t 1
2 (t 1) (t 1)
2
=–
(x 1) 1
ln
x
2
(x 1) 1
+ C.
(x 1) 1
x2
dx.
(x 3x 3) x 1
x2
Solution Let I = (x 2 3x 3) x 1 dx
Putting x + 1 = t 2, and dx = 2t dt, we get
Evaluate
2
I=
(t 2 1) 2t dt
{(t 1) 3(t 1) 3} t
2
2
2
(t 2 1)
dt = 2
4
t t2 1
I=2
I=2
u
dx
P Q , where P is a
1
1
2
Integrals of the form
We now consider the integrals
2
dx
2t dt
dt
2
2
2
2
(t 1) . t
(t 1) (t 1)2
(x 1)
Example 5.
dividing the numerator by the denominator
=2[
dx
x (x 1) .
2 (t 1) (t 1) (t 1) (t 1) dt,
Also x = t2 – 2.
2
Evaluate
Put (x + 1) = t2, so that dx = 2t dt.
Also x = t2 – 1.
2
where k2 = – ,
Solution
dx
(px qx r) ax b into an integral of the form
Solution
{t ( t )} 2 (k t ) ,
I=
quadratic algebraic expression of x and Q is linear.
To find the integral, we put Q = t2.
The substitution (ax + b) = t2 transforms the integral
Example 4.
Put x – = t2
dx = 2t dt and – x = – – t2
Solution
1.113
du
2
3
2
2
1
1 2
t
dt
1
2
t 2 1
t
where t –
1
= u.
t
1.114
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
2
2
u
t
I=
tan–1
tan–1 t + C
+C=
3
3
3
3
I=
t2 1
2
+C
tan –1
3
t 3
=
2
x
tan –1
+ C.
3
3 (x 1)
Also x = (1/t) – 1.
3. Integrals of the form
dx
(px+q) (ax 2 + bx+ c)
(px q) (ax bx c) ,
2
px + q =
1
.
t
1
p
a
1
(2t – 1)–1/2 . 2 dt = –
2
=–
2
1 x
1 x 1 C 1 x + C.
dt
(At Bt C) , which can be integrated easily..
2
Solution
Evaluate
Put 2x + 3 =
(2t 1) ) + C
dx
(2x 3) (x 2 3x 2)
.
1
1
. 2 dx = – 2 dz
z
z
dx
(1 x) (1 x ) .
2
Put (1 + x) = 1/t, so that dx = –(1/t2) dx.
dz
(1 z )
2
1
= – sin–1z + C = – sin–1
+ C.
2x 3
Alternative :
I
=
bt
2
2
2 (1 qt) (1 qt) ct
p
p
Example 6.
2
1 dz
2 z2
11
1
z
and x 3 ,
2z
2x
3
1
dz
I
2
2
1 1 1
31
z2
3
3 2
2
z 2 z
2z
dt
which when simplified takes up the form
=–
2
2
a 1
b 1
q c
q
2
pt
p t
1
t
dt
[t (1 t) ] (2t 1)
dt
t2
dt
=–
Solution
The given integral then reduces to
1
p
2
dt
1 1
so that p dx 2 and x q
pt
t
2
dx = –
dx
1
i.e.
t
(1 x) (1 x ) (1 / t) [1 {(1 / t) 1} ]
dx
P Q , where Q is a
quadratic algebraic expression of x and P is linear.
we put P =
(1 / t 2 )dt
Example 7. Integrate
Here, we consider the integrals
To evaluate
dx
dx
1
(2x 3) (4x 2 12x 8)
4
dx
1
(2x 3)
2
Put 2x + 3 = z
2dx = dz dx =
I
dz
z (z 2 1)
(2x 3)2 1)
1
dz.
2
= sec–1z + C' = sec–1(2x + 3) + C'
INDEFINITE INTEGRATION
Although apparently the forms of the two results are
different, it can be easily shown (by using the
properties of inverse trigonometric functions) that one
differs from the other by a constant.
dx
Example 8. Evaluate (x a)3/2 (x a)1/2 .
dx
Solution I =
(x a) (x 2 a 2 )
1
1
Put x – a = dx = – 2 dt.
t
t
1
1 at
Also, x = a
t
t
(1 at)2 a 2 t 2 1 2at
x2 – a 2 =
t2
t2
=
I=
=
dt
=
=
=
1
(ax + b) (cx + d)
Here we have integrals of the form
Evaluate
dx
(x 1) (x 1) .
2
2
dz
1
2 2
1
2 2
1
2 2
2
dz
2
z 2
1
ln
z 2 2 2
ln
1 (1 / x 2 ) 2
ln
1 (1 / x 2 ) 2
1 (1 / x 2 ) x 2
ln
1 (1 / x 2 ) x 2
2
(1 t 2 ) 2
(1 t 2 ) 2
+C
+C
+ C.
Show that
dt
2
2
2
a
x 2 b2
– C,
b x2 a 2
a b2 a 2
if b2 > a2, C being the constant of integration.
1
cos–1
dx
P Q , where P
2
Put x = 1/t, so that dx = –(1/t2) dt.
2
d dz
[1 (z 1)]z 2 z z 2
2
Also, x2 + a2 =
and Q are pure quadratic expressions. It means that
they do not contain terms of x, i.e. they are of the type
ax2 + b and cx2 + d. In this case put x = 1/ t and then the
expression under the radical should be put equal to z2.
(x 1) (x 1)
2
Solution Put
2
dx
2
x 2 b2
b2 a 2 t 2
2
2=
t
,
so
that
x
x2 a 2
t2 1
2x dx = – [2 (b2 – a2)t/(t2 – 1)2] dt.
dx
I=
t dt
(1 t ) (1 t )
(x a ) x b
4. Integrals of the form
Solution
2
Example 10.
= 1 1 2a C 1 x a C .
a xa
a xa
Example 9.
2
I=–
(1 2at) 2a 2 (1 2at) + C
2
{(1 / t )} {(1 / t ) 1}
Now put 1 + t 2 = z2 so that t dt = z dz. Then
1
2 dt
t
1 (1 2at)
.
t
t
(1 / t 2 )dt
=–
1.115
b2 a 2 t 2
+ a2 = (b2 – a2)/(t2 – 1),
t2 1
b2 a 2 t 2
+ b2
t2 1
= (b2 – a2) t2/(t2 – 1).
x2 + b2 =
dx
(x a ) x b
2
=–
=
2
2
1
(b a 2 )
2
2
dt
b –a t ,
2
2 2
1
cos–1 (at/b) + C,
a (b2 a 2 )
1.116
=
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
a x 2 b2
+ C.
cos–1
b x2 a 2
a b2 a 2
dx
(px + qx + r) (ax + bx + c) .
2
Here we shall consider two cases only.
Case I: If px2 + qx + r breaks up into two linear factors
of the forms (mx + n) and (m'x + n'), then we resolve
1/{(mx + n) (m'x + n)} into two partial fractions and the
integral then transforms into the sum (or difference)
of two integrals.
Case II: If px2 + qx + r is a perfect square, say, (lx + m)2,
then the substitution is lx + m = 1/t.
Example 11. Evaluate
I=
dx
2x 1 x (2 x) 1 x .
Solution
I=
2x 1 x (2 x) 1 x
2
(1 t ) t t 1
1 / z 2 dz
dz
2
3 3
z 2 2
(t 1)(t 1) t t 1
3
= – ln z z 2 3z 3
2
1
For I2, put (t + 1) =
u
du
I2 = –
2
1
3
u 2 4
where z =
1 1
1
dt
2 t 1 t 1 t 2 t 1
2
1
.
1 x 1
dx
2
r
...(1)
dt
...(3)
(x k) ax bx c , where r N
(t 1) t t 1
1
and u =
1 x 1
6. Integrals of the form
1
1
1
1. dt
. dt
2
2 (t 1) t t 1
2 (t 1) t 2 t 1
where I1 =
...(2)
1 3
2
ln z z 3z 3
2 2
2
1
1
Let, I = I1 – I2
2
2
2
1
1
ln u u 2 u 1 + C,
2
2
dt
2
1 1
1 1
1
z z z
2
1
1 1
1
(t 1)(t 1) 2 t 1 t 1
=
1
I=–
dt
2
I1 =
1
2
= – ln u 2 u u 1
2t dt
2
1
for I1,
z
dx
2(1 t ).t 1 t t
=
Here, I =
2
Put (t – 1) =
5. Integrals of the form
2
dt
(t 1) t t 1
and I2 =
Here, we substitute, x – k =
Example 12.
Solution
Evaluate
Put x – 3 =
dx
1
.
t
2
3
2
1
1
dx = – 2 dt
t
t
(x 3) x 6x 10
3
dx
(x 3) x 6x 10 .
1.117
INDEFINITE INTEGRATION
=
1 / t 2 dt
1 / t (1 / t 3) 6(1 / t 3) 10
3
=–
2
t 2 dt
dt
1 t = 1 t – 1 t dt
2
2
2
t
1
2 – ln |t+ 1 t 2 | + C
= ln |t+ 1 t
1
t
2
2
t
1
= ln |t + 1 t 2 | –
1 t2 + C
2
2
2 |–
=
1 1 x 2 6x 10
x 2 6x 10
ln
+ C.
2
| x 3|
| x 3 |2
Integrate
Example 13.
dx
2
Solution
=
dz
=
ax 2 bx c
(dx e) px qx r dx
2
Here, we write,
ax2 + bx + c = A(dx + e) (2px + q) + B(dx + e) + C
where A, B and C are constants which can be obtained
by comparing the coefficients of like terms on both
sides.
2x 2 5x 9
(x 1) x x 1 dx
Evaluate
2
2
(x 1)
(z 2 2)
2
Let
2x2 + 5x + 9 = A(x + 1) (2x + 1) + B(x + 1) + C
2x2 + 5x + 9 = x2(2A) + x(3A + B) + (A + B + C)
A = 1, B = 2, C = 6
by comparing the coefficients of like terms on both
sides.
Solution
dx
{(x 1)2 2}
, putting z = x – 1.
2 sec 2 d
2 tan . 2 sec , putting x = 2 tan.
2
1
1
cos ec cot d cos ec + C.
2
2
1
(z 2 2)
Since tan =
z, cosec =
2
z
8. Integrals of the form
2
I
1
1
1
1
n
C
2
2
x2
4
(x 2)
Example 15.
(x 2x 1) (x 2x 3) .
z
= 4 n x 2 x 2 4x 8
1 (z 2 2)
1 (x 2 2x 3)
I
C
C.
2
z
2
x 1
2 x 2 5x 9
Thus,
(x 1) x x 1 dx
=
(x 1) x x 1 dx
2
(x 1)(2x 1)
2
+ 2
=
7. Integrals of the form
(ax b)dx
2
(x 1) x2 x 1
dx + 6
dx
(x 1) x 2 x 1
dx
2x 1
x x 1 dx + 2 x x 1
2
2
+6
(cx d) px qx r .
x 1
dx
(x 1) x x 1
2
du
u +2
dx
dt
Here we put (ax + b) = A(cx + d) + B, and find the
values of A and B by comparing the coefficients of x
and constant term.
(4x 7)
.
Example 14. Evaluate
(x 2) x 2 4x 8
Solution Let 4x + 7 = A(x + 2) + B
=
A = 4, B = – 1
= 2 x 2 x 1 + 2ln |(x + 1/2) + x 2 x 1 |
dt
–6
(t 1 / 2)2 3 / 4
So, I = 4
dx
dx
x 4x 8 (x 2) x 4x 8
2
2
2
+6
1 3
x 2 4
1
where u = x2 + x + 1 and = x + 1
t
t t 1
2
1.118
=2
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
2
x 2 x 1 + | 2 ln x 2 x x 1
1
2
– 6ln t 2 t t 1 + C
=2
1
2
x 2 x 1 + 2 ln x 2 x x 1
– 6ln
1 x x2 x 1
+ C.
2(x 1)
Q
1.
Evaluate the following integrals :
dx
(i)
(2x 1) (4x 3)
1
(ii)
dx
(x 3) x 1
Evaluate the following integrals :
dx
(i)
4.
2.
5.
x (9x 4x 1)
2
dx
(ii)
(1 x) (1 x x )
(iii)
(1 x) 1 2x x
2
3.
dx
(iv)
2
2x dx
(1 x ) (x 4 1)
Evaluate the following integrals :
dx
(i)
2
(x 1) x
dx
(ii)
2
(x 5x 6) x 1
dx
(iii)
2
(x 4) x 1
2
6.
1.19 INTEGRATION OF A
BINOMIAL
DIFFERENTIAL
Evaluate the following integrals :
dx
(i)
(3 4x 2 )(4 3x 2 )1/2
dx
1 x 2 dx
(ii)
2
2 (iii)
(2x 1) 1 x
2 x2
Evaluate the following integrals :
dx
(i) 2
(x 4x) 4 x 2
dx
(ii)
2
(4x 4x 1) (4x 2 4x 5)
dx
(iii)
2
(x 2x 2) x 2 2x 4
dx
(iv)
3
(x 1) x 2 3x 2
Evaluate the following integrals :
xdx
(i)
2
(x 3x 2) x 2 4x 3
(x 2 1)dx
(ii) 2
(x 2x 2) (x 1)
(2x 3)dx
(iii)
2
(x 2x 3) x 2 2x 4
2.
p is a positive integer. Then, the integrand is
expanded by the formula of the Newton binomial.
p is a negative integer. Then we put x = tk, where k
is the L.C.M of the denominators of the fractions m
and n.
Expressions of the form xm (a + bxn)p, in which where
m, n, p are rational numbers, are called Binomial
Differentials.
3.
m 1
is an integer. We put a + bxn = t, where
n
is the denominator of the fraction p.
xm (a + bxn)p dx is expressed through
4.
Integrals of the form xm(a + bxn)p dx
The integral
elementary functions only in the following situations :
1.
m 1
+ p is an integer. We put a + bxn = txn,
n
where is the denominator of the fraction p.
1.119
INDEFINITE INTEGRATION
Example 1.
3
2
Evaluate I = x (2 x ) dx.
1
1
m 1 11 1
n
4
I = x 3 (2 x 2 )2 dx.
Solution
=–
Here p = 2, i.e. an integer, hence we have
1
4
1
5
1
m 1
5 1
p = –3 is an integer..
n
2 2
We put 1 + x4 = x4t2. Hence,
I = x 3 (x 4x 2 4) dx = (x 3 4x 6 4x 3 ) dx
7
=
11
but
4
3 3 24 6
x x 3x 3 C .
7
11
Solution
Evaluate
x=
2
2 1
x 3 1 x3
.
Here, p = – 1, is a negative integer and
Example 2.
=
1
3dt
1 t2
Evaluate
2
1 1
1 3 x
3
x2
dx.
I = x 3 (1 x 3 ) 2 dx .
2 1
m 1 3
= 1, i.e. an integer..
1
n
3
Here,
Example 4.
Solution
1 3
t2 dt = 2t3 +C = 2 (1 x 3 ) 2 + C.
Evaluate
Here p = –
1
x 11 (1 x 4 ) 2 dx.
1
2
5
2(t 2 1) 4
We have m = –3, n = 3, p = –1/3.
m 1
integer, but
n
x3(z3 – 1) = 1, x
2
1 3
x dx = 2t dt.
3
t dt
m 1
p 1 , (an integer).
n
We put 1 + x3 = z3x3.
1
Hence, I = 6
1
Solution
Let us make the substitution 1 + x 3 = t2
2
1
1
Here m = – , n = , p =
3
3
2
5
2(t 2 1) 4
1 2
t 5 t3 t
(t 1)2 dt + C.
2
10 3 2
Returning to x, we get
1
I=– 1
(1 x 4 )5 6 (1 x 4 )3
10x10
3x
1
3 1 x 4 + C.
2x
dx
.
Example 5. Evaluate
33
x (1 x3 )
=–
= 3tan (x ) C .
Solution
t dt
dx = –
1
1/3
Example 3.
1
(t 2 1) 4
t2 2
1
2
4
(t
1)
2
I=– 2
t 1
t
2
2 1 2
So I = t (1 t ) 3t dt
1
Substituting these expression into the integral, we
obtain
m and n are rational numbers.
3
Put
dx 3t2 dt
x
5
also a fraction,
2
is a fraction ,
dx
z2
dz .
(z 1)4/3
3
The denominator = x4z =
1
.
(z 1)1/3
3
z
.
(z3 1)4/3
1
1 (1 x 2 )2/3
C.
I z dz z 2 C
2
2
x2
1.120
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
R
Evaluate the following integrals :
5
x dx
(1 x3 )1/2
(1 x 2 )dx
1/2
x (1 x 2 )3/2
1.
3.
x (1 x ) dx 4. x (1 x ) dx
5.
1
2.
1/3 3
dx
x 1 x
3
53
6.
5
7.
3
c if c > 0,
3.
ax 2 bx c = (x – ) t
2
if ax + bx + c = a (x – ) (x – ) i.e. if is real root of the
trinomial ax2 + bx + c.
The integrand in R(x, ax 2 bx c ) dx can be made
rational in several ways, which we consider in order :
ax 2 bx c t – x a
Then bx + c = t2 – 2xt
...(1)
a
bdx = 2tdt – 2 a (xdt + tdx),
or dx(b + 2t a ) = 2 dt (t – x a ) = 2dt ax 2 bx c
dx
2dt
...(2)
2
b
2t a
ax bx c
t2 – c
...(3)
b 2t a
This substitution obviously renders the proposed
expression rational.
Also
x=
When b = 0, we get
2
dt and x = t c .
,
2t a
ax 2 c t a
dx
3
7/8 1/2
For example, to find
Here x =
Assume
4
3
3 1/3
2
Integrals of the form R(x, ax bx c ) dx are
calculated with the aid of one of the three Euler
substitutions :
1.
1 x dx
2 3
3/4
function of x and a 2bx cx .
ax 2 bx c = tx ±
1/4
1/3
1 x3
dx
x2
2
2.
8.
1 x4
x (2 + 3x ) dx
10. x (1 – 2 x ) dx
11. x (1 + x ) dx
12. x (1 + 8x ) dx
Let R(x, a 2bx cx 2 ), denote a rational algebraic
ax 2 bx c = t ± a is a > 0,
11
9.
3 2
1.20 EULER’S SUBSTITUTION
1.
dx
x
dx
(p qx) 1 x
2
dx
2dt
t2 1
2
, and
2
qt
2pt q
2t
(p qx) 1 x
dx
(p qx) 1 x 2
qz p – p 2 q 2
+C
ln
p 2 q 2 qz p p 2 q 2
1
When the coefficient a is negative the preceding
method introduces imaginary numbers : we proceed
to other substitution in which they are avoided.
...(4)
2. Assume ax 2 bx c c tx
Squaring both sides, we get immediately.
ax + b = 2t c xt 2
dx(a – t2) = 2dt ( c xt ) = 2dt ax 2 bx c .
dx
2dt
2
...(5)
Hence
2
ax bx c a t
2t c b
...(6)
a t2
This substitution also evidently renders the proposed
expression rational, provided c be positive.
For example, to find
dx
And
x=
x 1 x
2
Assume 1 x 2 = 1 – xt, and we get
1 1 x2
dx
dt
C
ln
|
t
|
C
ln
x 1 x2 t
x
3. Again, when the roots of ax2 + bx + c are real,
there is another substitution.
For, let and be the roots, and the radical becomes
of the form
1.121
INDEFINITE INTEGRATION
a(x )(x ), or a(x )( x),
according as the coefficient of x2 is positive or
negative.
In the former case, assume x t x , and we
t
,
, hence x – =
get x =
2
1 t2
1 t
dx
2tdt
.
x 1 t2
Accordingly,
dx
dx
2 dt
2 ...(7)
a(x )(x ) t(x ) a
a 1 t
2
In the latter case, let
x t x, and we get
2
t
1 t2
dx
2 dt
and
2
a(x )( x)
a 1 t
transforms into
Solution
2
1 x 2x 2
Here a = 1 > 0, therefore we make the
substitution x 2 2x 2 = t – x.
Squaring both sides of this equality and reducing the
similar terms, we get
2x + 2tx = t2 – 2,
x
t2 2
t 2 2t 2
dx
dt
2(1 t)
2(1 t)2
t2 2
t 2 4t 4
.
2(1 t)
2(1 t)2
Substituting into the integral, we obtain
1+
x 2 2x 2 = 1 + t –
2(1 t)(t 2 2t 2)
(t 2 2t 2)dt
dt
=
.
(t 2 4t 4)2(1 t)2
(1 t)(1 2)2
Now let us expand the obtained proper rational fraction
into partial fractions :
I=
x=
For example, the integral
dx
Example 1. Evaluate I =
...(8)
dx
(p qx) 1 x
2
2dt
(p q)t p q ,
2
t2 1
.
t2 1
It may be observed that in the application of the
foregoing methods it is advisable that the student
should in each case select whichever method avoids
the introduction of imaginary numbers.
Thus, as already observed, the first should be employed
only when a is positive: in like manner, the second
requires c to be positive; and the third, that the roots be
real.
It is easily seen that when a and c are both negative,
the roots must be real; for the expression
on putting x =
–ax 2 bx c
is imaginary for all real values of x unless b2 – 4ac is
positive i.e., unless the roots are real.
Accordingly, the third method is always applicable
when the other two fail.
From the preceding investigation it follows that the
expression R(x, ax 2 bx c ) dx can be always
rationalized.
t 2 2t 2
A
B
D
.
(t 1)(t 2)2 t 1 t 2 (t 2)2
We find : A = 1, B = 0, D = –2.
Hence,
t 2 2t 2
dt
dt
(1 t)(1 2) dt t 1 2 (t 2)
2
2
+ C.
t2
Returning to x, we get I = ln (x +1 +
2
= ln |t + 1| +
+
2
x 2 x 2 2x 2
Example 2.
Solution
+ C.
Evaluate I =
dx
x x2 x 1
Since here c = 1 > 0, we can apply the
second Euler substitution
x 2 x 1 = tx – 1,
(2t – 1) x = (t2 – 1) x2 x =
dx = – 2
x 2 2x 2 )
t2 t 1
dt
(t 2 1)2
2
x+ x x 1
t
.
t 1
2t 1
t2 1
1.122
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Substituting into I, we obtain an integral of a rational
fraction
2t 2 2t 2
dt,
t(t 1)(t 1)2
x x2 x 1
dx
2t 2 2t 2 A
B
D
E
.
2
2
t t 1 (t 1)
t 1
t(t 1)(t 1)
1
3
; D = –3; E = – .
2
2
dt 1 dt
dt
3 dt
3
t 2 t 1
(t 1)3 2 t 1
We find A = 2; B = –
Hence, I = 2
= 2 ln |t| –
1
3
3
ln |t + 1| + C,
ln|t – 1| +
2
t 1 2
x2 x 1 1
.
x
where t =
2
2(z z 2)dz
z2 2
dx = (1 2z)2 ,
1 2z
z2 z 2
2
=
x x2
1 2z
x=
2
and
=2
=
2(z z 2)dz
(1 2z)2
2
z 2 z 2 z 2 dz
x x2 x 2
1 2z 1 2z
dx
dz
6t dt
5 2t 2
3t
2t
(x – 2) t =
.
2
1
t
1
t2
Also
Hence, I = –
2
ln
Then x =
2 5
2 5
2 dt 2t + C,
9 t 2
9
t
where t =
2
1
6 5 2t 2
dt
27
t2
z 2
1
x2 x 2 x 2
2
x x2 x 2
+C
z2 5
12z dz
,
dx =
1 z2
(1 z 2 )2
6z
5 4x x 2 = (1 – x)z = 1 z 2
x dx
=
(5 4x x 2 )3/2
z 2 5 12z
1 z 2 (1 z 2 )2
dz
216z 2
(1 z 2 )3
5 2t 2
1 t3
dx = – (1 t 2 )2
z 2 2 ln z 2 + C
In this case a < 0 and c < 0 therefore
neither the first, nor the second Euler substitution is
applicable. But the quadratic trinomial 7x – 10 – x2
has real roots = 2, = 5, therefore we use the third
Euler substitution :
Solution
xdx
Example 4. Find (5 4x x 2 )3/2 .
Solution Let 5 – 4x – x2 = (5 + x) (1 – x) = (1 – x)2 z2.
and
5 – x = (x – 2) t2 x =
dx
x x2 x 2
Let x2 + x + 2 = (z – x)2. Then
Solution
7x 10 x 2 (x 2)(5 x) = (x – 2) t.
Find
Example 3.
1
5
1 2 dz
18 z
5 2x
1
5
+ C.
=
z +C=
18
z
9 5 4x x 2
x dx
.
Example 5. Evaluate I =
( 7x 10 x 2 )
=
Example 6.
Solution
7x 10 x 2
.
x2
Evaluate
I=
dx
x x2 4
5/3
dx
x x 4
2
5/3
x
Put x x 2 4 t 1
dx dt
2
x 4
Also, x x 2 4 t
x
t2 4
2t
x2 4 t x
.
INDEFINITE INTEGRATION
2
t2 4
4
2t
1
= x – 1 x2
t
Subtracting we get,
x2 4 =
–
2
t 4 16 8t 2 16t 2 t 4
=
=
4t 2
2t
2
2 1 x2 = t +
t2 4 1
dt
So, I =
2 5/3
2t t
1 5/3
t dt 2 t 11/3dt
=
2
2/3
8/3
1
t
...(3)
1
2t
2
t 1
1 x
or,
2
3
1 t
t
2
C = t 8/3 [1 t 2 ] C ,
=
4
8 / 3
2 2 / 3
From (1), (2) and (3) we get dx =
where t x x 2 4 .
n
I= t .
Evaluate
Example 7.
Solution
2
1 x 1
dx
x x 1 x
x 3 dx
2
2
=
+
1
1
= 1 2 ln 1 1 2 + C
x
x
Solution
x2 1
C.
x
dx =
Put x + 1 x = t
1
.2x dx = dt
1
2
2 1 x
...(2)
=
1
x 1 x2
(1 1 x x 2 )2
x 1 x x
2
2
dx
Let 1 x x2 xt 1, then
2t – 1
,
1 t2
2t 2 2t 2
dt
(1 t 2 )2
1 x x 2 xt 1 =
We have t = x + 1 x 2
x 1 x2
Evaluate
...(1)
1 x2 x
dx = dt
1 x 2
= x + 1 x2 ×
(1 x 2 ) ]n + 1
1 + x + x2 = x2t2 + 2xt + 1, x =
2 n
Let I = (x 1 x ) dx
2
(tn + tn – 2) dt
1
(x (1 x 2 ))n 1 + C.
2(n 1)
Solution
x 1 x2
1
[x +
2(n 1)
Example 9.
Evaluate (x 1 x 2 )n dx.
Example 8.
t2 1
1
dt =
(2t 2 )
2
t2 1
dt
2t 2
1 t n 1
t n 1
+C
2 n 1 n 1
I=
Put 1 + x–2 = t2 – 2 x–3 dx = 2t dt
t dt
1
= 1
1 t
1
t
= t – ln(1 + t) + C
x 1 x2
= ln
x
1.123
t2 t 1
1 t2
–2t 2 t
1 t2
Putting the expressions obtained into the original
2
1 – 1 x x
integral, we find
(1 1 x x 2 )2
x 1 x x
2
2
dx
1.124
=
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(–2t 2 t)2 (1 t 2 )2 (1 t 2 )(2t 2 2t 2)
dt
(1 t 2 )2 (2t 1)2 (t 2 t 1)(1 t 2 )2
t2
1 t
dt –2t ln
C
=2
1 t
1 t2
=–
2( 1 x x 2 – 1)
x 1 x x2 – 1
ln
C
x
x – 1 x x2 1
=–
2( 1 x x 2 – 1) ln 2x 2 1 x x 2 1 C
x
S
Evaluate the following integrals :
1.
3.
dx
x x 2x 4
2
dx
(2x x )
2 3
dx
1 x 1
2.
4.
x (x x 2)
7.
x x 1
2
x dx
6.
x x 1
8.
x x 4 dx
2
2
(x 1 x 2 )15
1 x
2
dx
1.21 METHOD OF
UNDETERMINED
COEFFICIENTS
kx
To find Pn (x)e dx where Pn(x) is a polynomial of
degree n, we have to perform integration by parts n
times. We then get Q n(x)e kx, where Q n(x) is a
polynomial of degree n.
Knowing this, we need not perform integration by
parts n times. We calculate the integral using the
method of undetermined coefficients, the essence of
which is explained by the following example.
2 x
Let us find x e dx .
Let x e dx = (a2x2 + a1x + a0)ex + C
On differentiating both sides, we get
x2ex = [x2a2 + x(2a2 + a1) + a1 + a0] ex
Equating the coefficients of identical powers of x in
the polynomials on the right and left, we get
a2 = 1
2a2 + a1 = 0 a1 = –2
a1 + a0 = 0 a0 = 2
Finally, we get
2 x
2 x
x e dx = (x2 – 2x + 2)ex + C
Example 1. Applying the method of undetermined
3
2x
coefficients, evaluate I = (3x 17)e dx .
Solution
dx
5.
3
2x
Let (3x 17)e dx
dx
2
2
= (Ax3 + Bx2 + Dx + E) e2x + C.
Differentiating the right and the left sides, we obtain
(3x3 – 17)e2x = 2(Ax3 + Bx2 + Dx + E)e2x
+ (3Ax2 + 2Bx+ D)e2x
Cancelling e2x we have
3x3 – 17 2Ax3 + (2B + 3A)x2 + (2D + 2B)x + (2E + D).
Equating the coefficients of equal powers of x in the
left and right sides of this identity, we get
3 = 2A, 0 = 2B + 3A,
0 = 2D + 2B, –17 = 2E + D
Solving the system, we obtain
A=
Hence,
3
9
9
77
,B= – ,D= ,E=
2
4
4
8
3
9
9
77 2x
(3x 17)e dx 2 x 4 x 4 x 8 e C.
3
2x
3
2
Note: The method of undetermined
coefficients may also be applied to integrals of the form
P (x)sin axdx , P (x) cos axdx ,
n
n
where Pn (x) is a polynomial. In both cases the answer
is of the form Qn(x)coskx + Rn(x)sinkx, where Qn(x) and
Rn(x) are polynomials of degree n (or less than n).
Example 2. Evaluate I = (x 3x 5) cos 2xdx .
2
Solution
Let us put (x 3x 5) cos 2xdx
2
= (A 0 x 2 A1 x A 2 ) cos 2 x
1.125
INDEFINITE INTEGRATION
(B0 x 2 B1 x B2 ) sin 2 x C
Differentiating both sides :
2 + 3x + 5) cos2x
= –2(A0x2 + A1x + A2) sin2x + (2A0x + A1)cos2x
+ 2(B0x2 + B1x + B2)cos2x + (2B0x + B1)sin2x
= [2B0x2 + (2B1 + 2A0)x + (A1 + 2B2)]cos2x +
[–2A0x2 + (2B0–2A1)x + (B1 – 2A2)] sin2x
Equating the coefficients at equal powers of x in the
multipliers cos2x and sin2x, we get a system of equations:
2B0 = 1 ,
2(B1 + A0) = 3, A1 + 2B2 = 5,
–2A0 = 0 ,
2(B0 – A1) = 0,
B1 – 2A2 = 0
Solving the system, we find
( x
A0 = 0, B0 =
1
1
3
3
9
, A1 = , B1 = , A2 = , B2 =
2
2
2
4
4
2
Thus, (x 3x 5)cos2xdx
3
9
x 3
1
cos 2x x 2 x sin 2x C
2
4
2 4
2
Integrals of the form
Pn (x)
2
ax bx c
Pn (x)
ax bx c dx
1
5
1
1
,B= ,D= ,K=
3
6
6
2
Thus,
1 2
1
1 2 5
I = x x x 2x 2
6
6
2
3
where
=
dx
ax bx c , ...(1)
2
where Qn–1(x) is a polynomial of degree (n – 1) with
undetermined coefficients and K is a number.
The coefficients of the polynomial Qn–1(x) and the
number K are found by differentiating identity (1).
x3 x 1
dx .
Example 3. Evaluate I =
x 2 2x 2
Solution Here Pm(x) = x3 – x – 1.
Pm – 1(x) = Ax2 + Bx + D
We seek the integral in the form.
dx
I = (Ax2 + Bx + D) x 2 2x 2 K
2
x 2x 2
Differentiating this equality, we obtain
x3 x 1
I' =
x 2 2x 2
dx
x 2x 2
2
dx
x 2x 2
2
dx
(x 1)2 1
Example 4.
2
Q n 1 (x) ax 2 bx c K
A=
dx
where Pn(x) is a polynomial of degree n.
Put
= (2Ax + B) x 2 2x 2
x 1
K
+ (Ax2 + Bx + D)
2
2
x 2x 2
x 2x 2
Reduce to a common denominator and equate the
numerators
x3 – x – 1 = (2Ax + B)(x2 + 2x + 2)
+ (Ax2 + Bx + D)(x + 1) + K.
Equating the coefficients at equal powers of x, we get
the following system of equations :
2A + A = 1,
B + 4A + B + A = 0,
2B + 4A + D + B = –1,
2B + D + K = –1
Solving the system, we obtain
Solution
x
n(x 1 x 2 2x 2 ) C .
2
x 2 4 dx
2
2
Let x x 4 dx
x 4 4x 2
x 4 dx
2
dx
2
= (Ax2 + Bx2 + Cx + D) x 4
x2 4
.
Differentiating both sides,
x 4 4x 2
x2 4
= (3Ax2 + 2Bx + C) x 2 4
(Ax3 Bx 2 Cx D)x
2
x 4
2
x 4
.
Multiplying by x 2 4 and equating the coefficients
of identical degree of x, we obtain
A
1
1
, B = 0, C , D = 0, = –2.
4
2
1.126
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Hence,
Integration under the sign of Integration
x
=
2
2
x 4 dx
x 2 2x
x 2 4 2ln(x x 2 4 ) C
4
Differentiation under the sign of Integration
The integral of any expression of the form f(x, a),
where a is independent of x, is obviously a
function of a as well as of x. Suppose the integral to be
denoted by F(x, a), i.e. let F(x, a) = f(x, a) dx, then
d
{F(x, a)} = f(x, a). Again, differentiating both sides
dx
with respect to a, we have, since x and a are independent,
d 2 F(x,a) d f(x,a) d dF(x,a) df(x,a)
,
dadx
da
dx da
da
Consequently, integrating with respect to x, we get
d F(x,a)
d f(x,a)
dx,
=
da
da
d
d f(x,a)
dx.
i.e.
f(x, a) dx =
...(1)
dx
da
In other words, if
du
df
dx,
u = f(x, a)dx, then
da
da
provided a be independent of x; in which case,
accordingly, it is permitted to differentiate under the
sign of integration. By continuing the same process
of reasoning we obviously get
dnu
d n f(x,a)
dx,
da n
da n
...(2)
where u = f(x, a)dx, a being independent of x.
dx
1
x
tan 1 .
Consider the formula
2
2
a
a
x a
On differentiating w.r.t. the parameter a,
2adx
1
x
x
2 tan 1
2
2 2
2
a a(a x 2 )
(x a )
a
and so, on dividing by 2a, we get
dx
1
x
x
tan 1 2 2
a 2a (a x 2 )
(x 2 a 2 )2 2a 3
We now proceed to consider the inverse process,
namely, the method of integration under the sign of
integration.
If in the last section we suppose f(x, a) to be derived
function with respect to a of another function v, i.e. if
dv
f(x, a) = , then v = f(x, a) da. Also, we have
da
d
dv
vdx dx f(x, a) dx = F(x, a).
da
da
Hence vdx = F(x, a)da. In other words, if
F(x, a) = f(x, a) dx, then
F(x, a) da = f(x, a) da dx.
...(3)
Integration of Implicit Functions
Consider the integral R(x, y) dx, where R(x, y) is a
rational function of x and y.
Assume that we can find a variable t such that x and y
are both rational functions of t, say
x = (t) , y = (t).
Then, R(x, y) dx = R{ (t), (t)} '(t) dt,
and the latter integral being that of a rational function
of t, can be easily evaluated.
dx
.
Example 5. If y(x – y)2 = x then prove that
x 3y
Solution
dx
x 3y
1
ln ((x – y)2 – 1) + C
2
Put x – y = t x = t + y
=
t
yt2 = x yt2 = t + y y = 2
t 1
t
t3 t t
t3
x=t+ 2
= 2
= 2
t 1
t 1
t 1
dx =
dx =
(t 2 1)(3t 2 ) t 3 (2t) 3t 4 3t 2 2t 4
=
dt
(t 2 1)2
(t 2 1)2
t 4 3t 2
dt
(t 2 1)2
t 4 3t 2
x – 3y =
(t 2 1)
t 3 3t
(t 1) dt t 3t = t 1
2
3
2
t3
3t
2
2
t 1 t 1
1.127
INDEFINITE INTEGRATION
(t 3 3t)
t(t 2 3)dt
=
2
2
2
(t 1)(t 3)
(t 1)(t 2 3)
=
1 2t dt
2 (t 2 1)
= t2 – 1 = u 2t dt = du
1 du 1
= ln (t2 – 1) + C
2 u
2
1
= ln ((x – y)2 – 1) + C
2
T
1.
Find polynomials P and Q such that
{(3x – 1) cos x + (1 – 2x) sin x} dx = P cos x + Q sin x + C.
2.
Evaluate
3.
Evaluate
9x3 3x 2 2
dx
3x 2 2x 1
x 3 6x 2 11x 6
x 2 4x 3
Prove that, when x > a > b,
4.
dx
2
7.
8.
1.22 NON-ELEMENTARY
INTEGRALS
Given a function f(x), has it necessarily an indefinite
integral F(x)? If some functions have indefinite integrals
and others not, what property of f(x) will ensure that
the function has an indefinite integral?
In our elementary work we were not concerned with
such generalities; we concentrated on finding an explicit
formula for the integral, if we could. The one useful
general answer that is readily available to us comes
from the theory of the definite integral and is : every
continuous function f(x) has an indefinite integral F(x).
Theorem Anyfunction f(x) continuous on an interval
(a, b) has an antiderivative on that interval. In other words,
there exists a function F(x) such that F(x) = f(x).
Knowing that a given function f(x) has an indefinite integral
F(x), can we find an explicit formula for that integral?
x 2
Consider f(x) = e
. Since f is continuous, its integral
exists, and if we define the function F by F(x)
0
2
1
2
2 ax
ax
2 1
2
3
e t dt then we know (to be dealt in the next
2 3/2
2
2
2 1/2
2
2 1/2
y
3y 2x = ln y x + C.
Find a substitution to reduce the integral
R(x, y) dx when (x2 + y2)2 = a2(x2 – y2).
ax
x
ax
If y2(x – y) = x2, then prove that
dx
Use the formula e dx = a–1eax to prove that
5.
ax
x
dx
(x a)2 (x b)
1
xb
1
n
=
2
x a (a b)(x a) + C
(a b)
=
xe dx e (xa a ) + C
(ii) x e dx e (x a 2xa 2a ) + C
(iii) xe dx = ex(x – 1) + C
6. Use the integral (x a ) dx to prove that
dx
x
(x a ) a (x a ) + C
(i)
2
2
chapter) that F(x) = e x . Thus, f(x) = e x has an
antiderivative F, but it has been proved that F is not
an elementary function. This means that no matter
how hard we try, we will never succeed in evaluating
e
x2
dx in terms of the functions we know..
The functions that we have been dealing with are
called elementary functions. These are the
polynomials, rational functions, power functions,
exponential function, logarithmic functions,
trigonometric and inverse trigonometric functions,
hyperbolic and inverse hyperbolic functions, and all
functions that can be obtained from these by the five
operations of addition, subtraction, multiplication,
division, and composition. For instance, the function
x2 1
+ ln (cos x) – xesin 2x is an elementary
x3 2x 1
function.
Note that if f is an elementary function, then f is an
f(x) =
elementary function but f(x) dx need not be an
elementary function.
1.128
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Y
For instance, ( x )' = 1 / (2 x ) , (x3 – 3x2)' = 3x2 – 6x,
and (tan x)' = sec2x. But if you start with an elementary
function f and search for an elementary function F
whose derivative is to be f, you may be frustrated –
not because it may be hard to find F – but because no
such F exists.
It is not easy to tell by glancing at f whether the desired
F is elementary. After all, x tan x looks no more
complicated that x cos x, yet it is not the derivative of an
elementary function, while x cos x is.
As another example, cos x looks more complicated
than cos x2. Yet it turns out that cos x is the derivative
of an elementary function, while cos x2 is not. [It is not
hard to check that (2 x sin x 2 cos x )' is cos x ]
Some integrals can be expressed in the form of an
e mx dx
cannot be
x
obtained in terms of a finite number of elementary
functions; it however may be exhibited in the shape of
an infinite series. By expanding emx and integrating
each term separately, we have
infinite series. The integral
e mx dx
mx m 2 x 2
m 2 x2
ln x
.......
x
1
1 . 2 2 1 . 2 . 32
Even though an integral is non-elementary, we can
2
x
study its properties. Consider F(x) = e dx .
dF(x)
e x > 0 for arbitrary x, it follows that
dx
F(x) is an increasing function. The derivative has a
maximum at x = 0; hence, at x = 0, F(x) has a maximum
angle of the tangent line with the x-axis. For large
absolute values of x (positive or negative), the
dF
is very small. This means that the
derivative
dx
function is almost constant. The graph of the function
2
Since
x
F(x) = e x dx is shown in the figure (for the sake of
2
0
definiteness, the lower limit has been chosen equal to
zero).
Extensive tables have been compiled for this function
and so computations involving this integral are not
much complicated.
1
0.885
0
–1
–1
1
2
X
– 0.885
Non-elementary Integrals
The integrals which cannot be expressed using
elementary functions in finite form are called nonelementary integrals.
In all such cases, the antiderivative is obviously some
new function which does not reduce to a combination
of a finite number of elementary functions.
Such functions are said to be non-integrable in
elementary functions (non-integrable in finite form).
It is known that the following integrals are nonelementary :
ex
dx
x
sin(x )dx
cos(e )dx
x 1 dx
x
2
3
ln x dx
1
x dx
cos x
1 k sin xdx , (0<k2<1)
x dx
sin x
2
2
We have seen that for some functions f(x) it is easy to
discover an indefinite integral F(x), for others a difficult
and lengthy affair; and for certain functions f(x) we
shall find that an explicit formula for F(x) can be
obtained only if we invent new types of function in
addition to the types of function used in the definition
of f(x).
Let us discuss this issue in more detail. Suppose we
limit ourselves to using all the basic elementary
functions except logarithmic. Then many integrals
whose expressions we know should be regarded as
"inexpressible in elementary functions".
dx
dx
, 2
, cannot now
x
x 1
be expressed "in finite form". For among the remaining
basic elementary functions there is no function whose
For instance, the integrals
INDEFINITE INTEGRATION
1
1
or 2
(since we cannot
x
x 1
use the logarithmic function). But if the logarithmic
function is included it becomes possible to find these
integrals in finite form.
But there is nothing surprising in the fact that some
other integrals remain non-integrable in elementary
functions. To make them integrable it is necessary to
extend the class of basic functions which we agree to
use. This is exactly what is done in mathematical
analysis: the non-elementary functions determined by
the most important integrals inexpressible in terms of
elementary functions are thoroughly investigated and
tabulated. These new functions extend the variety of
our techniques and make it possible to express the
integrals of a number of formerly non-integrable
functions.
derivative is equal to
2 – x2
e
One of the antiderivatives of this function,
Consider the function f(x) =
2
2
e – x dx C
which vanishes for x = 0 is called the Laplace function
and is denoted by (x). Thus,
(x) = 2 e – x dx C1 if (0) = 0.
This function has been studied in detail. Table of its
values for various values of x have been complied.
2
1.129
2
2
We note that f(x) = (1 k sin x) , (0 < k2 < 1)
involves in its definition only trigonometric functions
and a square root sign; being continuous it has an
integral F(x); but there is no formula for F(x) in terms of
root signs and trigonometric functions. In order to
write down a formula for F(x), we must use a new sort
of function, namely, an elliptic function.
One of the antiderivatives of f(x) i.e.
1 k sin x dx C
2
2
which vanishes for x = 0 is called an elliptic integral
and is denoted by E(x).
E(x) =
1 k sin x dx C' if E(0) = 0
2
2
Table of the values of this function have also been
complied for various values of x.
Chebyshev's theorem
(i) Consider the integral xp(1 – x)q dx, where p and q
are rational numbers. Chebyshev proved that
there are only three cases for which the
antiderivative in question is elementary, which
are:
(A) if p is an integer,
(B) if q is an integer,
(C) if p + q is an integer.
3
2
In particular, x 3 1 x dx and x x dx are not
elemen tar y. Chebyshev's th eor em also holds
for xp(1 + x)q dx.
E
1.
2.
Two of these antiderivatives are elementary
functions; find them.
ln x dx
(A) ln x dx
(B)
x
dx
(C)
ln x
Two of these three antiderivatives are elementary.
Find them.
and
1 x 3 1 x has an elementary antiderivative. Find
the function.
1 4sin d
(C) 1 cos d
(B)
Two of these three integrals are elementary;
evaluate them
Three of these six antiderivatives are elementary.
Find them.
cos x
dx
(B)
(A) x cos x dx
x
x dx
ln x 2
dx
(C)
(D)
ln x
x
(E) x 1 x x 1 dx
(A) sin2x dx
(B) sin
(F)
(A)
3.
4.
(C) sinx2 dx
Only one of the functions x 3 1 x
2
4 4sin d
2
x dx
5.
x 1 x 1 x dx
1.130
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
U
6.
7.
Assuming that (ex/x) dx is not elementary (a
theorem of Liouville), prove that 1/ln x dx is not
elementary.
From the fact that x tan x dx is not elementary,,
deduce that the following are not elementary.
(A) x2 sec2 x dx
(B) x2 tan2 x dx
2
(C)
8.
x dx
1 cos x
From the fact that (sin x)/x dx is not elementary,,
deduce that the following are not elementary :
(A) (cos2x)/x2 dx
9.
(B) (sin2x)/x2 dx
(D) cos x ln x dx
(C) sin ex dx
Chebyshev's theorem : the integral xp(1 – x)q dx,
where p and q are rational numbers is elementary,
(A) if p is an integer,
(B) if q is an integer, or
(C) if p + q is an integer.
...(1)
(i) Deduce from (1) that
elementary.
1 x dx is not
3
(ii) Deduce from (1) that (1 – xn)1/m dx, where m and
n are positive integers, is elementary if and only if
m = 1, n = 1 or m = 2 = n.
(iii) Deduce from (1) that sin x dx is not elementary..
(iv) Deduce from (1) that sinax dx, where a is rational,
is elementary if and only if a is an integer.
are rational, is elementary if and only if p or q is an
odd integer or p + q is an even integer.
(vi) Deduce from (v) that secp x tanq x dx, where p
and q are rational, is elementary if only p + q or q
is odd, or if p is even.
(vii) (A) Deduce from (1) that (x / 1 x n ) dx, where
n is a positive integer, is elementary only
when n = 1, 2 or 4.
(B) Evaluate the integral for n = 1, 2 and 4.
2
n
(viii)(A) Deduce from (1) that (x / 1 x )dx,
where n is a positive integer, is elementary
only when n = 1, 2, 3 or 6.
(B) Evaluate the integral for n = 1, 2, 3 and 6.
(ix) (A) Using (1) determine for which positive integers
n the integral (x n / 1 x 4 )dx is elementary
(B) Evaluate the integral for n = 3 and n = 5.
c dr
(x) Let
where c is a constant. The
r r 6 c2
integral is easily evaluated by the substitution..
(A) What substitution should we recommend?
(B) Using (1), determine for which positive inte-
10. (i)
Evaluate
1
sin(x a)cos(x b) dx.
1
dx
sin(x a)cos(x b)
cos(a b)
dx
.
=
cos(a b) sin(x a) cos(x b)
1
cos{(x b) (x a)}
.
=
dx
cos(a b)
sin(x a)cos(x b)
Solution
I=
There are two values of a for which
1 a sin 2 d is elementary. What are
they?
(ii) From (1) deduce that there are two values of
(v) Deduce from (1) sinp x cosq x dx, where p and q
Problem 1.
gers,n, (c / (r n r 6 – c 2 )) dr is elementary..
a for which
1 ax 2
1 x2
dx is elementary..
cos(x b).cos(x a)
1
cos(a b) sin(x a) cos(x b)
sin(x b).sin(x a)
+ sin(x a) cos(x b) dx
1
=
{cot (x – a) + tan (x – b)} dx
cos(a b)
1
=
{ln |sin(x – a)| – ln |cos(x – b)|}+ C
cos(a b)
=
1.131
INDEFINITE INTEGRATION
sin(x a)
1
ln
+ C.
cos(a b) cos(x b)
=
cos 5x cos 4x
dx.
1 2 cos3x
Problem 2.
Evaluate
Solution
1 2 cos3x dx
=
ln(z z 2 a 2 )
cos 5x cos 4x
sin 3x cos 5x cos 4x
sin 3x sin 6x
I2
dx
3x
3x
9x
x
2sin 2 cos 2 2 cos 2 cos 2
dx
=
9x
3x
2 cos cos
2
2
= – 2 cos
3x
x
cos dx = –
2
2
cos 2x cos x dx
Evaluate
I
Solution
x2 x 1
(x 2x 3) dx .
2
(x 2 2x 3) (x 2)
(x 2 2x 3)
2
x 2x 3
dx
x2
=
(x 2x 3) dx (x 2x 3) dx
(x 2x 3) dx (x 2x 3) dx
=
(x 1) 2 dx
1
2
2
2
1 / 2(2x 2) 1
2
2
2
(2x 2) dx
dx
(x 2x 3) {(x 1) 2} dx
2
2
Denoting the integrals by I1, I2, I3 respectively,
I1 I 3
dz
x a dz (z a ) ,
2
2
I
dz
z 2 z 2 x 2x 3
2
2
(where z = x + 1, a2 = 2)
1
1
z z 2 a 2 a 2 ln(z z 2 a 2 )
2
2
2
1
(x 1) x 2 2x 3 x 2 2x 3 + C
2
1
(x 1) x 2 2x 3 + C.
2
Evaluate
Problem 4.
I=
3
dy
3
cos y sin(y )
dy
cos ysin(y ) ()
3
dy
=
dx
sin(x )cos (x ) .
Substitute x – dx = dy. Given integral
Solution
I=
sin 2x
= –
sin x + C.
2
Problem 3.
1
(x 1) x 2 2x 3 ,
2
on restoring the value of z and a2.
For I2 , we put x2 + 2x + 3 = z, so that (2x + 2) dx = dz,
3
cos y(sin y cos cos y sin )
dy
cos y(cos tan y sin )
4
sec 2 ydy
(cos tan y sin )
Now write sin + cos tan y = z2
cos sec2 y dy = 2z dz
2z sec dz
=
= 2z sec + C
z
= sec sec cos tan y + C
sin(y )
= 2 sec
+C
cos y
=
= 2 sec () .
Let a matrix A be denoted as
Problem 5.
sin(x )
+ C.
cos(x )
x
5x
A = diag. 5x ,55 ,55
integral ( det A)dx .
then compute the value of the
1.132
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Solution
5x
0
A= 0
55
0
0
x
0
5
55
5x
x
= 5x ·55 ·55
0
5x
5x
5x
· ln 5 · 5x· ln 5) dx = dt
x 5
5
I = 5 ·5 ·5 dx . Put 55 = t
x
(5
55
x
· ln 5 · 5
I = (x3m + x2m + xm) (2x2m + 3xm + 6)1/m dx, x > 0.
Solution
I = (x3m + x2m + xm) (2x2m + 3xm + 6)1/m dx
(2x3m 3x 2m 6x m )1/m
dx
x
Put 2x3m + 3x2m + 6xm = t
6m (x3m – 1 + x2m – 1 + xm – 1) dx = dt
=
=
cos x ( 1 cos2 x )
1 (cos
3/2
x)
2
cos x cos x
dx.
(1 cos3 x)
2
I=–
3
2
I = – sin–1 (cos3/2 x) + C.
3
1
2 3
ln
3 s – tan–1 t + C
3 s
Evaluate I =
3/2
2
b
cos d
a
b
cos d
a
I=
b2 2
2
a a sin b bsin
cos d
= a
(a 2 b2 sin 2 ).cos
= a
2
Substituting ax2 = b sin2
Solution
cos x.sin xdx
1 (cos x)
dx
(a dx ) b ax .
dx =
3
cos1/2 x.(–sin x) dx = dt]
2
dt
2
= – sin–1t + C
2
3
1 t
[Put cos3/2 x = t
Problem 9.
3
dx =
1
3 sin x cos x
log
2 3
3 sin x cos x
–1
– tan (sin x + cos x) + C.
cos x cos3 x
dx
1 cos3 x
Let I =
Solution
=
m1
1
{2x3m + 3x2m + 6xm } m + C.
6(m 1)
Evaluate
1 t (1/m)1
dt
=
+C
6m 6m (1 / m) 1
Problem 7.
sin xdx
cos x . sin x 1
1
sin x
2 sin xdx
2 sin xdx
=
2 2 sin x cos x
2 sin 2x
[(sin x cos x) (sin x cos x)]dx
=
2 sin 2x
sin x cos x
sin x cos x
=
dx
dx
2 sin 2x
2 sin 2x
sin x cos x
sin x cos x
=
dx
3 (1 sin 2x)
1 (1 sin 2x)
sin x cos x
sin x cos x
=
. dx
. dx
2
3 (sin x cos x)
1 (sin x cos x)2
Put sin x – cos x = s and sin x + cos x = t
(cos x + sin x) dx = ds and (cos x – sin x) dx = dt
ds
dt
I=
3 s2
1 t2
ds
dt
=
2
2
1 t2
( 3 ) (s)
= (x3m– 1 + x2m – 1 + xm – 1) . (2x3m + 3x2m + 6xm)1/m dx
I=
=
cos x
Problem 6. For any natural number m, evaluate
t1/m
dx
5x
= (x3m + x2m + xm)
1
t
55
I = 3 dt = 3 + c = 3 + C.
ln 5
ln 5
ln 5
I=
Solution
x
dx
cos x cosec x .
Evaluate
Problem 8.
d
a b sin ,
2
2
2
dividing Nr and Dr by cos2 , we get
2
1.133
INDEFINITE INTEGRATION
= a
= a
=
sec 2 d
a sec 2 b 2 tan 2
Put tan = t
2
dt
dt
a (1 t ) b t = a (a b )t a
2
a
a 2 b2
2
2 2
2
2
2
2
dt
t2
a2
a 2 b2
a 2 b 2 1 t a 2 b 2
a
= 2
.
tan
+C
a
a
a b2
x a 2 b2
. tan 1
+C
a b ax 2
a(a 2 b2 )
1
=
{since, t = tan =
Problem 10.
I=
x
b ax 2
Evaluate
}.
x2
dx
(x cos x sin x)(xsin x cos x)
Solution
Put x = tan dx = sec2 d
tan 2 sec 2 d
I = [(tan cos(tan ) sin(tan )][tan sin(tan ) cos(tan )]
=
xsin x cos x
= ln x cos x sin x + C.
Alternative :
f(x) = x cos x – sin x f ' (x) = – x sin x
f '(x) = – x sin x + cos x – sin x
g (x) = x sin x + cos x g ' (x) = x cos x
g ' (x) = x cos x + sin x – sin x
f (x) · g ' (x) + g (x) · f ' (x) = – x2
d
[f(x)·g(x)]
dx
I=–
dx =
f(x)g(x)
[ f (x) g(x) = F (x)]
= ln F(x) + C = ln f(x)·g(x) + C.
Problem 11.
I=
Solution
Evaluate
sin x sin 2x
cos x cos2x dx.
sin x(1 2 cos x)
cos x 2 cos x 1 dx.
2
Put cos x = t then – sin x dx = dt
1 2t
1 (4t 1) 1
I=–
dt –
dt
2
2
2t t – 1
2t 2 t 1
=–
1
2
d(2t 2 t 1) 1
–
(2t 2 t 1)
2
tan 2 sec2 cos2 d
[sin cos(tan ) cos sin(tan )][sin sin(tan ) cos cos(tan )]
(sec 2 1)d
=
[(sec( tan )]cos( tan )
=–
=–
(sec 2 1)d
sec(tan )·cos(tan )
Put tan – = y
dy
=–2
= – 2 cosec 2y dy
2sin y cos y
Put 2y = t
2
= – cosec t dt = – ln(cosec t – cot t) + C
2
t
1 cos t
= – ln sin t + C = – ln tan 2 + C
= – ln (tan y) + C = – ln [tan(tan – )] + C
= – ln[tan(x – tan–1x)] + C
1 x tan x
tan x x
= – ln 1 tan x x + C = ln tan x x + C
F '(x)
F(x) dx
–
dt
t 1
2 t2
2
2
1 (2t 2 t 1) –1/2 1
·
1
2
1
2
1
dt
2
22 1
1 1
t 4 16 2
= 2t 2 t 1 –
1
2 2
dt
2
1 3
t 4 4
2
= 2t 2 t 1
–
2
2
1
1
1 3
ln t t – + C
2 2 4
4 4
= – cos x cos2x
1.134
–
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
1
cos x 1
2
– C.
ln cos x cos x
2 2
4
2
2
Problem 12. Evaluate I =
Solution
cot x – tan x
dx
1 3sin 2x
(1 tan x)dx
sec 2 xdx
=
2
tan x(sec x 6 tan x)
Putting tan x = t2 and sec2x dx = 2t dt, we have
1
x 4 x2 1
1 + C =
+ C.
2
x
x
Evaluate
I=
dt
dt
t(1 t ) = t(1 t)(1 t t )
3
2
[putting t + 1/t = y]
t 1
y
1
tan–1 2 + C = – tan–1 2t C
2
2
Solution
=
I=
Evaluate I =
2
1
x x –1
1
1
ln |1 – t| – ln |1 + t + t2| + C,
3
3
where t = xecos x.
2
Solution
x4 1
x
2
4
2
x x 1
(x 1)(x 1)
dx
1
3
2
x x 2 1
x
1
1
1 x 1 2
x dx
1
2
3 t 3
dt
dt
+
1 t
1 t t2
Evaluate sin 4x.e
Problem 15.
tan x cos x
= – tan–1
C.
2
2
dt 1
I=
t 3
= ln| t | –
tan x 1
= –tan–1
C
2 tan x
Problem 13.
(1 (xe
B
Ct D
A
= t 1 t
dt
t t2
1
Comparing coefficients we get :
1
2
1
A = 1, B = , C = – , d = –
3
3
3
1
dt
t –2 dy
=–2
2
y2 4
1
t
4
t
=–2·
(1 xsin x).dx
3 3cos x .
)
x(1 x e
(1 xsin x)dx
cos x 3 .
) )
Put xecos x = t so that (xecos x . (–sin x) + ecos x) dx = dt
Let I =
Solution
1
2 1 dt
(1 t 2 )2t dt
t
2
1
t(1 t 4 6t 2 )
2
t 2 6
t
dt
2 t = t +C
Problem 14.
I=
I=
2
= x
(1 tan x)dx
tan x (1 6sin x cos x)
I=
2
1
1
1
Putting x – 1 = t 2 x x 2 dx = dt
x
x
x
dx
tan 2 x
dx
Let I = 4 sin x . cos x . cos 2x . e tan 2 x dx
= 4 tan x . cos2 x (cos2 x – sin2 x) . e tan 2 x dx
= 4 tan x . cos4 x (1 – tan2 x) e tan x dx
2
=4
tan x
(sec x) (1 – tan2 x) e
2
2
tan 2 x
dx
Put tan2 x = t 2 tan x.sec2 x dx = dt
I=4
(1 t)e t dt
(1 t)e t
.
dt
=
2
(1 t)3 2
(1 t)3
2e t (1 t)e t
dt
=2
(1 t)3
1.135
INDEFINITE INTEGRATION
1
2
1
t
dt = – 2
=2 e
et + C
3
2
(1 t)2
(1 t) (1 t)
{using ex{f(x) + f(x)}dx = ex f(x) + C}
2
=–
2e tan x
+C
(1 tan 2 x)2
such that f(0) = 1 and f (–1) = 4. If
2
= – 2 cos4 x . e tan x + C.
Problem 16.
Solution
Evaluate
Let I =
1 cos x
cos x(1 cos x) dx.
1 cos x
dx.
cos x(1 cos x)
1
2
1 y
1 cos x
=
= –
cos x(1 cos x) y(1 y) y 1 y
=
If the primitive of the function
m
1 x2
x 2009
C
2 1006 w.r.t. x is equal to
n 1 x2
(1 x )
then find (m + n) (where m, n N).
f(x) =
x 2009
dx
Solution f (x) =
(1 x 2 )1006
Put 1 + x2 = t 2x dx = dt
1004
1 (t 1)1004 dt
1 1
=
1
2
t1006
2 t
Put 1
1
1
= y 2 dt = dy
t
t
I=
1 1004
1 y1005
y dy = ·
C
2
2 1005
g(x) =
A
B
C
D
x x x 1 (x 1) dx
2
2
B
f(x) dx
D
x (x 1) = x (x 1) dx
2
2
2
2
Hence f (x) must of the form f (x) = B(x + 1)2 + Dx2
f (0) = 1 B = 1
f (–1) = 4, D = 4
f (x) = (x + 1)2 + 4x2
f (x) = 5x2 + 2x + 1
f (10) = 500 + 20 + 1 = 521
Problem 19. Evaluate
1
2
dx –
dx
cos x
1 cos x
2
dx
I = sec x dx –
2 cos2 x / 2
= sec x – sec2 x/2 dx
I = ln | sec x + tan x| – 2 tan x/2 + C.
g(x) =
I=
2
B
D
+ C ln (x + 1) –
+E
x
x 1
Since g (x) is a rational function, A = C = 0.
1 cos x
I=
dx
cos x(1 cos x)
Problem 17.
2
g (x) = A ln | x | –
1
2
–
cos x 1 cos x
=
f(x)dx
x (x 1) is a
rational function, find the value of f(10).
f(x)dx
Solution g(x) = 2
x (x 1)2
Now
Let cos x = y.
Then
1005
1005
1 x2
C
·
= 1 · t 1 C =
2010 1 x 2
2010 t
m = 1005, n = 2010 m + n = 3015.
Problem 18. Suppose f(x) is a quadratic function
1
· 2 dt
t
I=
tan 2 tan 1
1 x 1
dx
1 x 1
2
Solution Let x tan
x (0,1), (0, / 4)
x = tan4 tan = x1/4
dx = 4 tan3 sec2 d
Also, 1 x = sec
1 x 1
1
tan 2 tan
1 x 1
sec 1
= tan 2 tan 1
sec 1
= tan 2 tan 1 tan = tan
2
1.136
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
I = tan · 4 tan3 sec2 d
=
4
4
tan5 + C = (x5/4) + C.
5
5
I=
=
=
(x 2 1) x 4 2x3 x 2 2x 1
dx
x 2 (x 1)2
1 2 2
2 1
1 2 x x 2x 1 x 2 dx
x
x
2
2
x (x 2x 1)
x2
1 2 1
1
1 2 x 2 2 x x 1 dx
x
x
1
x x 2
=
=
dt
2
2
(t 2) t 2t 3 – 3 (t 2) t 2t 3
I = I1 – 3I2
...(1)
t dt
t dt
t 2t 3 and I2= (t 2) t 2t 3
2
t dt
(z 1)dz
zdz
2
2
2
Problem 22.
Solution
dz
z 2 = z 2 – z 2
2
= z 2 2 2 –ln z + z 2 4
2
2
x x (x 2x 1)(ln x 1)
dx
x 4x 1
2
(t 1)2 4
put t + 1 = z
=
I=
t(t 2)dt
t 2 2t 3
dt
(t 2)
t 2 2t 3
I1 =
Evaluate
x x (x 2x 1)(ln x 1)
dx
x 4x 1
Put xx = y xx(ln x + 1) dx = dy
y2 1
dy
I= 4
y 1
1
1
1 2
1 2
y
y
dy
dy
2
1
2
1
y 2
y y 2
y
1
x 1
y y
x xx
1
1
1
1
C
tan
tan
C.
2
2
2
2
Solution
(t 2) t 2t 3 dt
where, I1=
Problem 21.
2
=
...(2)
1
, put t + 2 =
y
I = t 2 2 t 3 – ln(t + 1 + t 2 2 t 3 )
1
t 5
+ C , where t = x +
– 3 sin–1
t2
x
1
1
put x + = t x 2 dx = dt
x
x
(t 2 2) 2t 1
dt =
(t 2)
2
t 2 2t 3
1
1 1
2 2 2 3
y y
y
1
dy
dy
=
=
2
2
2
3
1 2y 3y
1
2
y
3
3
1
1
y 3 1
5 t
–1
=
sin
sin–1
...(3)
3
2
3
2t
3
(x 1) x 4 2x3 x 2 2x 1
dx.
x 2 (x 1)
Solution
y2 .
Evaluate
Problem 20.
I=
I2 =
= t 2 2t 3 – ln (t + 1) +
dy
I=
Evaluate
e x dx
(sin e x e x cos e x )2
dx
e x (sin e x e x cos e x )2
e 2x ·e x dx
dx
= 3 x x
=
e (e sin e x cos e x )2
(e x sin e x cos e x )2
Put ex = t ex dx = dt
INDEFINITE INTEGRATION
=
100
2
=
=
+ cos(100x)(sin x)
t 2 dt
(t sin t cos t) Put t = tan
2
sin(100x)(sin x)100
+ C.
100
Problem 25. A function f (x) continuous on R
and periodic with period 2 satisfies f (x) + sin x ·
tan sec d
tan sin(tan ) cos(tan ) 2
tan 2 cos2 sec 2 d
cos(tan )cos sin(tan )sin
f (x + ) = sin2x. Find f(x) and evaluate f(x) dx .
2
...(1)
x x+
f (x + ) + sin( + x) · f (x + 2) = sin2( + x)
f (x + ) – sin x f (x) = sin2x
...(2)
cos y = sec y dy
2
2
sin 2 x f(x)
sin x
sin3x + sin2x · f (x) = sin2x – f (x)
f (x) [1 + sin2x] = sin2x (1 – sin x)
sin2x + sin x · f (x) =
2
2
e tan x
2
1
1 1 x
sec
1
x
cos
dx .
2
(1 x 2 )
1 x
Note that sec–1 1 x 2 = tan–1x
Solution
1 x2
–1
2 = 2 tan x, for x > 0
1 x
e tan 1 x
(tan 1 x)2 2 tan 1 x dx
I=
2
1 x
Put tan–1x = t
Problem 24.
Solution
tan x + C.
Evaluate sin(101x)·sin x dx
I=
99
sin(100x x)·(sin x dx
99
99
dx
99
sin(100x)
x· (sin x) dx
cos
=
u
v
+ cos(100x)·(sin x)
100
=x–
2
1
= (sin(100x) cos x cos100x·sin x)(sin x)
sin 2 x(1 sin x)
(1 sin 2 x) (1 sin x)
dx
dx
=
1 sin 2 x
1 sin 2 x
2
1
= et . t2 = e tan x
sin 2 x(1 sin x)
.
1 sin 2 x
=x–
= e (t 2t) dt
t
f (x) =
Now,
cos–1
=
sin 2 x f(x)
sin x
From (2),f (x + ) = sin2x + sin x · f (x)
From (1),– f (x + ) =
= tan y + C = tan(tan – ) + C
= tan(t – tan–1t) + C = tan (ex – tan–1 ex) + C .
Problem 23. For x > 0, evaluate
1
f(x) + sin x · f(x + ) = sin2x
Solution
dy
dx
=
2
(sec 2 1)d
=
cos 2 (tan )
Put tan – = y (sec2 – 1)d = dy
=
1.137
(1 sin x) sin x sin 3 x
dx
1 sin 2 x
1
sin x
1 sin x dx 2 cos x dx sin x dx
2
2
sec 2 x
1
sec 2 x
dx
dx
=
2 tan 2 x 1
1 tan 2 x tan 2 x
2
Put tan x = t
I1 =
=
1
1
dt
1
= · 2 ·tan ( 2t)
2
2 t2 1
2
1
tan 1 ( 2 ·tan x) = I2 =
2
Put cos x = t
I1 =
dx
sin(100x)(sin x)100 100
cos(100x)(sin x)100 dx
–
100
100
=–
dt
1
2 t
2 t = – 2 2 ln 2 t
2
sin x
2 cos x dx
2
1.138
=–
1
2 2
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
ln
1
tan 1
2
I = x + cos x –
+
Put tan x = bsin
2 cos x
2 cos x
sec2x dx = bcos d dx =
2 tan x
d
I = (a 2 b2 sin 2 )(1 b2 sin 2 )
1
2 cos x .
ln
2 2
2 cos x
Problem 26.
Put x = t4 dx = 4t3 dt
Solution
I=4
= 4
x1 4 5
x 16 dx .
Evaluate
(t 5) 3
t dt = 4
t 4 16
dt
(t 2 4)(t 2 4)
2
2
= 4t + 5 ln (t4 – 16) + 8 ·
1
t2
ln
2·2
t2
1
t
tan–1 + C
2
2
x1 4 2
= 4x1/4 + 5 ln (x – 16) + 2 ln 1 4
x 2
x1 4
– 4 tan–1
+ C.
2
Problem 27.
Evaluate
dx
a tan x b tan x , (a > b).
2
2
2
sec2 d
sec2 d
Put tan = t
dt
dt
dt
dt
= 4t + 5 ln (t4 – 16) + 8 2
2
(t 4)
(t 4)
–8·
1
d
d
(1 a 2 ) 1 b2 sin 2 a 2 b2 sin 2
= 4t + 5 ln (t4 – 16) + 8 (t 4) (t 4) dt
(t 2 4)(t 2 4)
=
= (1 a 2 ) 1 (b 2 1) tan 2 a 2 (a 2 b 2 ) tan 2
4t 3
dt
= 4 dt 5 4
dt 64 4
t 16
t 16
1
(a 2 b2 sin 2 d) (1 b2 sin 2 )
d
=
(1 a 2 ) (a 2 b2 sin 2 ) (1 b2 sin 2 )
1
t 4 5t 3
dt
t 4 16
t 4 16 5t 3 16
dt
t 4 16
= 4t + 5 ln(t4 – 16) + 64
b cos
d
1 b 2 sin 2
2
Solution
b cos d
Let I = (1 b2 sin 2 )(a 2 b2 sin 2 ) b cos
=
1
dt
(1 a 2 ) 1 (b2 1)t 2
=
1 1
tan 1 t 1 b2
(1 a 2 ) 1 b2
2
2
1
a 2 b2
1 t a b
tan
a
a
(a 2 b 2 )
2
1 1
tan 1
= (1 a 2 )
2
1 b
dt
a (a b )t
2
2
2
1 b tan
2
a 2 b2
tan 1
tan C ,
a
a a 2 b2
where tanx = bsin
1
xsinxcosx
Problem 28. Evaluate (a2 cos2 x b2 sin2 x)2 dx
Solution
xsin x cos x
(a cos x b sin x) dx .
2
2
2
2
2
Integrating by parts taking ‘x’ as the first function
sin x cos x
and 2
as second function. For
(a cos2 x b2 sin 2 x)2
the second integral put a2cos2x + b2sin2x = t
1
(b2 – a2)sinx cosx dx = dt
2
1
1
So, I = x · 2
2
2
2
2
a
cos
x
b
sin
x
2(b
a2 )
1.139
INDEFINITE INTEGRATION
1
1
1. 2
dx
2(b a 2 )
a cos2 x b 2 sin 2 x
2
= ln
x
= a 2 cos2 x b2 sin 2 x 2(b2 a 2 )
1
+
·
2
2(b a 2 )
=
sec 2 x dx
a 2 b2 tan 2 x
x
a 2 cos2 x b2 sin 2 x 2(b 2 a 2 )
1
+ 2 2 2
2b (b a )
sec 2 x dx
a
2
2
b tan x
1
x
2
2
2
2 (b a )(a cos2 x b2 sin 2 x)
=–
= ln
=
x
C.
2
2
(a cos x b sin x)
Problem 29.
Solution
I=
=
I=
Evaluate
I=
Put
1 x2
dx .
2 x2
1 x2
dx . Put x = tan
2 x2
sec3 d
=
2 tan 2
d
sin 2
cos3 2
cos2
d
cos (2 cos sin )
2
2
cos d
I = cos2 (2 cos2 sin 2 )
cos d
Put sin = y
(1 sin 2 )(2 sin 2 )
I=
(1 y)(2 y) = y 2 – y 1
dx
{f '(x)g(x) g'(x)f(x)}dx
{f(x) g(x)} g(x)(f(x) g (x))
2
f '(x)g(x) g'(x)f(x)
(g(x))2
dx
f(x) f(x)
g(x) 1 g(x) 1
f(x)
– 1 = t2
g(x)
f '(x)g(x) g'(x)f(x)
g(x) 2
dx 2t dt
2t dt
2
t
tan 1
C
I = (t 2 2)· t =
2
2
f(x) 1
1
= 2 tan 2g(x) 2 C
f(x)
g(x)
1
= 2 tan
C
2g(x)
Hence m = 2 , n = 2.
m2 + n2 = 8.
1
ln ln(ln x) dx
ln x·ln(ln x)
= g (x) + C then find the value of g (ee).
Problem 31. If
=
dy
f '(x)g(x) g'(x)f(x)
f(x) g(x) f(x)g(x) g 2 (x)
dy
+ C.
f(x) g(x)
m tan 1
C , where m, n N and
ng(x)
I=
2
x 1 x2
C is constant of integration g ( x ) 0 . Find the value
of (m2 + n2).
Solution Let
1
1
1 b tan x
= 2(b 2 a 2 ) ab tan a
2
x 2 1 x2
Problem 30. Let
b
1
a tan x
tan 1
2 2 2
2b (b a ) a
a
sin 2
y2
+ C = ln sin 1 + C
y 1
Solution
dy
f ' (x) =
f (x) = ln ln(ln x)
1
1 1
·
·
ln(ln x) ln x x
1.140
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
ln x·ln(ln x)
x f ' (x) =
= – t 2 2 – cot x – x + C
I = – cot x n(cos x cos 2x ) cosec 2 x 2
I = x ln ln(ln x) C
– cot x – x + C.
g(x)
g(ee) = 0.
Problem 32.
Problem 33.
Evaluate
that (m + n) Im, n = cosm x . sin nx + m I(m –1, n – 1).
cosec x n (cos x cos 2x )dx
2
2
Solution Let I = cosec x n (cos x cos 2x )dx
= – cot x n(cos x cos2x )
sin 2x
cot x sin x
cos 2x
dx
(cos x cos 2x )
= – cot x n(cos x cos2x )
cos x(2 cos x cos2x )
cos 2x(cos x cos2x ) dx
= – cot x n(cos x cos 2x ) + I1
where I1 =
I1 =
=
cos x(1 cos x cos 2x )
dx
cos 2x sin 2 x
cos x
(1 2sin 2 x) sin 2 x
Put sin x =
=
cos x(2 cos x cos 2x )
cos2x(cos x cos 2x ) dx
cos x(2 cos x cos 2x ) (cos x cos2x)
dx
cos2x(cos x cos 2x ) (cos x cos2x)
=–
dx cot 2 x dx
1
1
cos x dx = – 2 dt
t
t
dt
2
t
– cot x – x + C
(t 2 2) 1
2
t2
t
t dt
t 2 – cot x – x + C
2
If Im, n = cosm x . cos nx . dx, show
Solution
We have
sin nx
m
Im, n = cos
nx dx = (cosm x)
x.cos
n
u
v
m cosm – 1 x (– sin x) .
sin nx
dx
n
1
m
cosm x . sin nx +
cosm 1 x
n
n
{sin x . sin nx} dx
We have
cos (n – 1) x = cos nx cos x + sin nx . sin x
=
1
cosm x . sin x + m cosm 1 x
n
n
{cos (n – 1) x – cos nx . cos x} dx
Im, n =
m
1
cosm x . sin x +
cosm – 1 x . cos (n – 1) x . dx
n
n
m
cosm x . cos n x dx
–
n
=
=
m
m
1
cosm x . sin nx +
Im – 1, n – 1 –
I
n
n m, n
n
Im, n +
=
m
I
n m, n
1
[cosm x . sin nx + m Im – 1, n – 1]
n
mn
I
n m, n
1
[cosm x . sin nx + m Im – 1, n – 1]
n
(m + n) Im, n = cosm x . sin nx + m Im – 1, n – 1.
=
Problem 34. If In denotes zn e1/z dz, then show that
(n + 1)! In = I0 + e1/z (1 !z2 + 2!z3 + .... + n ! zn + 1).
1.141
INDEFINITE INTEGRATION
In = zn e1/z dz, applying integration by
parts taking e1/z as first function and zn as second
function. We get,
Solution
1/z
In =
n 1
n 1
e .z
1
z
e1/z 2 .
dz
(n 1)
z n 1
e1/z .z n 1
1
=
e1/z . zn – 1 dz
(n 1) (n 1)
=
I
e1/z .z n 1
n 1
(n 1) (n 1)
dx
–6t
(t 2 1)4t – (2t 2 5)2t
dt = 2
(t 1)2
(t 2 1)2
3
2t 2 5
–2= 2
2
t 1
t 1
and x – 2 =
and 1 + 7x – 10 – x 2 = 1 + (x – 2) t
3t
t 2 3t 1
t 1
t2 1
Hence, we have
=1+
2
t2 1 t2 1
–6t
· 2
· 2
dt
3 t 3t 1 (t 1)2
I n 1 e1/z .z n 1
e1/z .z n 1
= (n 1) (n 1) n n I n 2
I=
e1/z (z) n 1 e1/z .(z)n
1
I .
(n 1)
(n 1)n (n 1)n n – 2
=–
t 3t 1 = – t 3t 1 dt
e1/z .(z) n 1 e1/z .(z)n
e1/z .(z)n 1
=
(n 1)
(n 1)n (n 1)n.(n 1)
=–
t 3t 1 dt + 3
=
1
I
(n 1)n(n 1) n – 3
... ... ... ... ... ... ...
... ... ... ... ... ... ...
=
I=
Evaluate
dx
(x – 2)(1 7x – 10 – x )
2
Solution
Here, we have 7x – 10 – x2 = (x – 2)(5 – x),
therefore put 7x – 10 – x 2 = (x – 2) t
(5 – x)(x – 2) = (x – 2)2 t2
(5 – x) = (x – 2)t2
x=
2t 2 5
t2 1
2
2t 3
2
dt
2
3 5
t 2 – 4
Problem 35.
2t 3 3
= ln | t2 + 3t + 1|
3
dt
dt
dt
+
–
5 3 5 3 5
t
t –
2 2
2 3
e1/z (z) n 1 e1/z .(z) n
e1/z .(z)n 1
...
(n 1)
(n 1)n
(n 1)n....3.2
1
I
(n 1)n(n 1)...3.2 0
Multiplying both sides by (n + 1)! We get,
(n + 1) !In = (e1/z . zn + 1 . n! + e1/z . z2 (n – 1)!+ ...
+ e1/z . z3 . (2) ! + e1/z . z2 . 1!) + I0
In (n + 1) ! = I0 + e1/z (1 z2 + 2!z3 + .... + n! zn + 1).
2t dt
2
= ln | t2 + 3t + 1| +
where t =
3
2t 3 – 5
C,
ln
2t 3 5
5
5x
.
x2
Problem 36. Let f(x) and g(x) are differentiable
functions satisfying the conditions :
(i) f (0) = 2, g (0) = 1
(ii) f (x) = g (x) and
(iii) g (x) = f (x). Find the functions f (x) and g (x).
Solution f ' (x) = g(x)
g ' (x) = f(x)
adding, f ' (x) + g '(x) = f(x) + g(x)
f '(x) g'(x)
1 ln( f(x) + g(x) ) = x+ c
f(x) g(x)
Since, f(0) = 2, g(0) = 1, put x = 0 to get c = ln 3.
...(1)
Hence f(x) + g(x) = 3ex
Similarly subtraction gives
1.142
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
f '(x) g'(x)
x
f(x) g(x)
Integrating ln( f(x) – g(x) ) = –x + c
Since, c = ln(1) c = 0
...(2)
f(x) – g(x) = e–x
On adding (1) and (2) we get f(x) and on subtracting
we get g(x) :
f(1) = 1, 2
f(1) 1
( f(0) = 1 and function is injective)
Thus f(1) = 2
3e x e x
3e x e x
,and g(x) =
.
2
2
1
1
( f (1) = 2)
f(x) f = f(x) + f
x
x
Hence f(x) is of the type f(x) = 1 ± xn
f(1) = 1 ± 1 = 2
(given)
n
n
–
1
f(x) = 1 + x and f(x) = nx f(1) = n = 2
f(x) = 1 + x2
f(x) =
Problem 37. Let f be an injective function such
that f(x) f(y) + 2 = f(x) + f(y) + f(xy) for all non negative
real x and y with f(0) = 1 and f(1) = 2. Find f(x) and
show that 3 f(x) dx – x (f(x) + 2) is a constant.
Replacing y by
1
1
f(x) f + 2 = f(x) + f + f(1)
x
x
We have f(x) f(y) + 2 = f(x) + f(y) + f(xy)
...(1)
Putting x = 1 and y = 1then f(1) f(1) + 2 = 3f (1)
we get
1. Rules of integration
(A) cf(x) dx = c f(x) dx for any constant c.
(B) (f(x) + g(x)) dx = f(x) dx + g(x) dx,
(C) (f(x) – g(x)) dx = f(x) dx + g(x) dx.
Integrand f(x) Integral f (x)dx
dx = kx + C, where k is a constant
x
(ii) x dx =
+ C, where n 1
n 1
x3
x
= 3
– x (3 + x2) + C
3
= C = constant.
(xi)
cosec x cot x dx = – cosec x + C
(xii)
1 x dx = sin x + C
(xiii)
1 x dx = tan x + C
(xiv)
| x | x 1 dx = sec x + C
(i)
n 1
n
(iii)
(iv)
1
dx = ln |x| + C
x
ax
C , where a > 0
a x dx =
n a
e dx = e + C
(vi) sin x dx = – cos x + C
(v)
x
3 f(x) dx – x (f(x) + 2) = 3 (1 + x2 ) dx – x (1 + x2 + 2)
Solution
2.
1
in (1)
x
x
2
(ix)
cosec x dx = – cot x + C
(x)
sec x tan x dx = sec x + C
2
–1
2
1
–1
2
1
–1
2
We have some additional results:
1
tan x dx = n sec |x| + C
(xv)
a
1
(xvi) cot x dx = n sin|x|+ C
a
(xvii) sec x dx = n |sec x + tan x| + C
x
+C
4 2
= – n |sec x – tan x| + C
= n tan
(vii) cos x dx = sin x + C
(viii) sec x dx = tan x + C
1
3.
If f (x)dx F(x) , then
1
f(ax b)dx a F(ax b)
INDEFINITE INTEGRATION
4.
f(g(x))g(x) dx = F(g(x)) + C.
5.
[f(x)]n 1
[f(x)] f '(x)dx
+ C , n –1
n 1
6.
f '(x)
dx ln | f(x) | C
f(x)
7.
g(x)
f(x)g'(x) g(x)f '(x)
dx = ln
+C
f(x)g(x)
f(x)
8.
Taking xn common :
n
For
dx
n
1 + xn = t.
9.
Positive integral powers of sine and cosine
(i) Any odd positive power of sines and cosines
can be integrated immediately by substituting
cos x = z and sin x = z respectively.
(ii) In order to integrate any even positive power of
sine and cosine, we should first express it in terms
of multiple angles by means of trigonometry and
then integrate it.
10. Positive integral powers of secant and cosecant
(i) Even positive powers of secant or cosecant admit
of immediate integration in terms of tanx or cotx.
(ii) Odd positive powers of secant and cosecant are
to be integrated by the application of the rule of
integration by parts
m
n
11. Integrals of the form sin x cos xdx
(i) If the power of the sine is odd and positive,
save one sine factor and convert the remaining
factors to cosine. Then expand and integrate.
Odd
sin
2k 1
x cos n xdx
=
Convert to cos ine
Odd
sin m x cos 2k 1 xdx
Convert to sin e
x(x 1) , n N, take x common and put
n
1.143
Save for du
2
k
n
(sin x) cos x sin xdx
(ii) If the power of the cosine is odd and positive,
sa ve on e cosin e factor a nd convert t h e
remaining factors to sine. Then, expand and
integrate.
for du
Save
sin x (cos2 x) k cos xdx
m
=
(iii) If the powers of both the sine and cosine are
even and nonnegative, make repeated use of
the identities
1 cos 2x
1 cos 2x
sin 2 x =
and cos2 x =
.
2
2
m
n
(iv) If in the expression sin x cos x, m + n is a negative
even integer, then one should make the
substitution tan x = t (or cot x = t).
(v) If in the expression sinmx cosnx, m + n is a negative
odd integer, then one should multiply the
integrand by suitable power of (sin2x +cos2x) and
expand it into simpler integrals.
m
n
12. Integrals of the form sec x tan xdx
(i) If the power of the secant is even and positive,
save a secant-squared factor and convert the
remaining factors to tangents. Then, expand
and integrate.
Even
2k
sec
x tan n xdx
Convert to tan gents
=
Save for du
2
k 1
n
(sec x)
tan x sec 2 xdx
Here, we put tan x = t.
(ii) If the power of the tangent is odd and positive,
save a secant-tangent factor and convert the
remaining factors to secants. Then, expand and
integrate.
Odd
m
2 k 1
sec
x
tan
xdx
Convert to sec ants
= sec
m 1
Save for du
2
k
x (tan x) sec x tan x dx
Here, we put sec x = t.
(iii) If there are no secant factors and the power of
the tangent is even and positive, convert a
tangent-squared factor to a secant-squared
factor; then expand and repeat if necessary.
Convert to secants
n
tan xdx =
= tan
n 2
tan
n 2
x (tan 2 x) dx
x(sec2 x – 1)dx
1.144
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
m
(iv) If the integral is of the form sec xdx , where
m is odd and positive, use integration by parts.
(v) If none of the first four cases applies, try
converting to sines and cosines.
A sim il ar str at egy is a dopted for
cosec x cot xdx .
m
n
13. Suppose R denotes a rational function of the
entities involved. The integral of the form
(i)
R(x, b 2 a 2 x 2 )dx
(iv)
(v)
(vi)
(vii)
b
is simplified by the substitution x sin
a
(ii)
R(x, a x b )dx
2 2
2
is simplified by the substitution x
(iii)
(viii)
b
tan
a
R(x, a x b )dx
2 2
=
2
b
is simplified by the substitution x sec
a
14. When integrand involves expressions of the
form:
(ix)
dx
x a
2
a x
2
2
= ln x x a + C
2
dx
2
= sin-1
x
+C
a
a 2 x 2 dx
x
2
+
2
2
= ln x x a + C
x2 a 2
2
a 2 x2 +
x
a2
sin-1 + C
a
2
x 2 a 2 dx =
x
2
x2 a 2
a2
2
2
ln x x a + C
2
x 2 a 2 dx =
x
2
x2 a 2
a2
2
2
ln x x a + C
2
–
px q
dx ,
ax bx c
(i)
ax
ax
put x = a cos 2.
16. For integrals of the form
(ii)
x
ax
put x = a sin2.
ax bx c dx , and
(iii)
x
ax
put x = a tan2.
(px q) ax bx c dx
xa
or (x a)(b x)
(iv)
b x
put x = a cos2 + b sin2
xa
or (x a)(x b)
(v)
x b
put x = a sec2 – b tan2
15. Some Standard Integrals
dx
1
x
= tan-1 + C
(i)
2
x a2
a
a
dx
1
xa
(ii)
=
ln
+C
x2 a 2
2a
xa
dx
1
ax
(iii)
=
ln
+C
2
2
a x
2a
ax
dx
2
px q
2
2
we write (px + q) (2ax + b) + m
17. For integrals of the form
px 2 qx r
px 2 qx r
ax bx c dx , ax bx c dx , and
2
2
(px qx r) (ax bx c) dx
2
2
we express px2 + qx + r
= l (ax2 + bx + c) + m ( 2ax + b) + n
ax 2 b
dx
18. For integrals of the form
4
x px 2 1
we express the numerator in terms of
(x2 + 1) and (x2 – 1), then divide Nr and Dr by x² ,
and then put x ±
1
= t.
x
1.145
INDEFINITE INTEGRATION
19. For integrals of the form
dx
(i)
2
a cos x 2bsin x cos x bsin 2 x
dx
dx
(ii)
(iii)
2
a cos x b
a bsin 2 x
r
r
2
we divide the N and D by cos x or sin2x and then
put tanx = t or cot x = t.
20. For integrals of the form
dx
dx
(i)
(ii)
a bsin x
a b cos x
dx
(iii)
a bsin x c cos x
x
use the substitution tan = t
2
acosx + bsinx + c
dx
21. For integrals of the form
dcosx + esinx + f
we write a cos x + b sin x + c
p (d cos x + e sin x + f)
+ q(–d sin x + e cos x) + r
22. For integrals of the form
ae x be x c
dx ,
pe x qe x r
we write a e + b e + c
l (a ex + b e–x + c) + m(a ex – b e–x ) + n
cos x sin x
dx
23. For integrals of the form
f(sin 2x)
put sinx cos x = t
x
–x
cos ec x a dx
cot x a dx
2
,
2
2
,
2
sec 2 a
For the first integral write sec 2 x a
sec 2 x
2
a
sec 2 x a
cos x
sec x a
1 a cos2 x
In the first part, put u = tanx and in the second
part, put v = sinx.
For others proceed as above.
25. Formula for integration by parts
du
u.v dx = u v dx – d x . v d x dx
eax
27. (i) e . sin bx dx = 2 2 (a sin bx – b cos bx) + C
a b
ax
eax
(ii) e .cos bx dx = 2 2 (a cos bx + b sin bx)+ C
a b
ax
e [f(x) + f ¢(x)] dx = e . f(x) + C
(ii) [f(x) + xf ¢(x)] dx = x f(x) + C.
28. (i)
x
x
29. Multiple integration by parts
uv dx = uv – u'v + u"v – u"'v + .....
+ (–1) u v + (–1) u v dx.
1
n–1
2
3
(n–1)
4
n
n
n
30. Reduction formula for In =
In
n
dx
2
(x k)n 1
x
(2n 3)
I n 1 .
2k(n 1)(x 2 k)n 1 2k(n 1)
31. Reduction formula for In = sinm cosn d
cos n 1 sin m 1 n 1
+
m 1
m 1
sinm cosn–2 d
P(x)
.
Q(x)
Factor Q(x) completely into factors of the form
(ax + b)m or (cx2 + dx + e)n, where cx2 + dx + e is
irreducible and m and n are integers.
(i) For each distinct linear factor (ax + b), the
32. Consider a rational function f(x) =
sec x a dx ,
tan x a dx
ln x dx = x ln x – x + C.
In =
24. For integrals of the form
26.
The order of u and v is taken as per the order of
the letters in ILATE.
A
.
ax b
(ii) For each repeated linear factor (ax + b)m, the
decomposition must include the terms
A1
A2
Am
.
....
ax b (ax b)2
(ax b)m
(iii) For each distinct quadratic factor (cx2 + dx + e),
the decomposition must include the term
Bx C
.
2
cx dx e
(iv) For each repeated quadratic factor (cx2 + dx +
e)n, the decomposition must include the terms
decomposition must include the term
B1x C1
B2 x C 2
cx 2 dx e (cx 2 dx e)2
1.146
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Bn x C n
.
(cx 2 dx e)n
Use one of the following techniques to solve for the
constants in the numerators of the decomposition.
(A) Heaviside Cover-up Method for linear factors
(B) Method of Comparision of Coefficients
(C) Method of Particular Values
(D) Application of Limit
(e) Method of Differentiation
.....
33. Special methods for integrating a rational function
(i)
For integrals of the form
px 2 q
(x a)(x b) dx
2
2
assume x as t for finding partial fractions
P(x)
dx
(ii) For integrals of the form
Q(x)
where Q(x) has a linear factor with high index,
substitute the linear factor as 1/t.
(iii) Substitutions
2
(A)
xm
dx
(ax b)n
dx
(B)
x (ax b)
(C)
(x a) (x b)
(D)
Put ax + b = t.
m
n
dx
m
dx
x(a bx )
n
n
Put
ax b
=t
x
Put
xa
= t if m < n
xb
Put xn =
1
t
x 2m 1
dx
Put ax2 + b = t
(ax 2 b) n
(iv) Integration by parts
34. Integration by irrational functions
Substitution :
dx
(i)
Put axn + b = t2
x ax n b
dx
1
(ii)
Put x =
(a cx 2 )3/2
t
xdx
1
(iii)
Put x =
(a 2bx cx 2 )3/2
t
(E)
p1
pk
(iv) If the integrand is of the form R x, x q1 , . . . ., x q k ,
m
then put x = t , where m is the L.C.M. of the
denominators q1, q2,....., qk of the several fractional
powers.
(v) If
the
integrand
is
p1
q1
of
the
form
pk
qk
R x, (ax b) , . . . .,(ax b) , then put ax + b
m
= t , where m is the L.C.M. of the denominators
q1, q2,....., qk of the several fractional powers.
(vi) If
the
integrand
p1
ax b q1
is
of
the
form
pk
ax b q k
ax b
=
R x, cx d , . . . ., cx d , put
cx d
tm, where m is the L.C.M. of the denominators q1,
q2,..., qk of the several fractional powers.
dx
,
(vii) Integrals of the form n
(x a)p (x b)q
where p + q = 2n are solved using the substitution
xa
t.
xb
dx
35. Integrals of the form
P Q
dx
(i)
(ax b) (cx d)
(ii)
(px qx r) ax b Put ax + b = t
(iii)
(px q) (ax bx c) Put px + q = t
(iv)
(ax b) (cx d) Put x = 1/ t and then
Put cx + d = t2
dx
2
2
1
dx
2
dx
2
2
the expression under the radical should be put
equal to z2.
dx
(v)
.
2
(px qx r) (ax 2 bx c)
Case I: If px2 + qx + r breaks up into two linear
factors then we resolve 1/(px2 + qx +r) into partial
fractions and the integral then transforms into
the sum (or difference) of two integrals.
Case II: If px2 + qx + r is a perfect square, say,
(lx + m)2, then put lx + m = 1/t.
dx
,
(vi)
r
(x k) ax 2 bx c
where r N
Put x – k =
1
t
1.147
INDEFINITE INTEGRATION
(ax b)dx
(cx d) px qx r
(vii)
(ii)
2
ax 2 bx c
(dx e) px qx r dx
2
37. Integrals of the form R(x, ax 2 bx c ) dx are
calculated with the aid of one of the three Euler
substitutions :
1.
2.
ax 2 bx c = t ± a is a > 0,
ax 2 bx c = tx ± c if c > 0,
2
ax 2 bx c = (x – a) t if ax + bx + c = a
(x – a) (x – b) i.e. if a is real root of the trinomial
ax2 + bx + c.
38. The method of undetermined coefficients can be
applied to integrals of the form :
kx
ax bx c
sin x
x dx
1
dx
ln x
cos x
dx
x
1 k sin xdx ,
2
(0 < k2 < 1)
n
SINGLE CORRECT ANSWER TYPE
If f be a continuous function satisfying
1 for 0 x 1
and f(0) = 0, then f(x)
f'(lnx) =
x for x 1
can be defined as
if x 0
1
(A) f(x) =
x
1 e if x 0
if x 0
1
(B) f(x) x
e 1 if x 0
,
then F(x, a) da = f(x, a) da dx.
41. Integration of Implicit Functions If R(x, y) is a
rational function of x and y, then,
R(x, y) dx = R{f (t), y (t)} f'(t) dt, assuming that
x = f (t) , y = y (t).
42. It is known that the following integrals are nonelementary :
ex
2
dx sin(x )dx
x
x
x3 1 dx
cos(e )dx
Qn(x) are polynomials of degree n.
1.
2
where Pn (x) is a polynomial of degree n and
Qn–1(x) is a polynomial of degree (n – 1) with
undetermined coefficients and K is a number.
39. Differentiation under the sign of integration
d
d f(x,a)
dx.
f(x, a) dx =
dx
da
40. Integration under the sign of integration If
F(x, a) = f(x, a) dx,
kx
n
dx
Q n 1 (x) ax 2 bx c K
P (x)e dx = Q (x)e + C, where P (x) and
n
n
3.
(i)
n
2
Write ax + bx + c = A(dx + e) (2px + q) +
B(dx + e) + C
36. Integrals of the form xm(a + bxn)p dx
(i) p is a positive integer. Then, the integrand is
expanded by the formula of the Newton binomial.
(ii) p is a negative integer. Then we put x = tk,
where k is the L.C.M of the denominators of
the fractions m and n.
m 1
(iii)
is an integer. We put a + bxn = ta, where
n
a is the denominator of the fraction p.
m 1
(iv)
+ p is an integer. We put a + bxn = taxn,
n
where a is the denominator of the fraction p.
n
coskx + Rn(x)sinkx + C where Pn (x) is a
polynomial of degree n and Qn(x) and Rn(x)
are polynomials of degree n (or less than n).
Pn (x)
dx
(iii)
2
ax bx c
Write (ax + b) = A(cx + d) + B
(viii)
P (x)sin axdx or P (x) cos axdx = Q (x)
x if x 0
(C) f(x) x
e if x 0
if x 0
x
(D) f(x) x
e
1
if x 0
2.
The value of
6
dx
x x 1 is
13
6
x 1
6
12
(A) ln 6 x x C .
x
2
1.148
(B)
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1 x6 1 1 6 1 12
x x C
ln
6 x6 6
4
1 x6 1
1
ln 6 x 6 x 12 C
6
2
x
(D) None of these
(C)
3.
1 cos x .sin 2x.cos 2x dx is
2
4
(1 cos 2 x)3 (3 2 cos 2 x) C
10
3
2
(B)
1 cos 2 x (3 2 cos 2 x) C
5
2
(C)
1 cos2 x (3 2 cos2 x) C
5
(D) None of these
(A)
4.
5.
sin 3 2x
dx is
sin 5 x
4 2
cot 7 x
5
(B) C
(C) C
4 2
cot 5 x
5
(D) None of these
7.
1 x4 x2 1
ln
C
6 (1 x2 )2
1 1 (x 2 1)2
C
tan
6
2
x 4 x2 1
C
(C) ln
(1 x 2 )
(D) None of these
dx
1 t 1 1
2t
= ln
+
ln
+ C,
If
3
3sin x sin x 6 t 1 12
2t
then
(A) t = sin x
(B) t = tan x/2
(C) t = 2 cos x
(D) t = cos x
dx
1
If
is – f(x)
2
sin x.(2 cos x 1)
2
1
1 2 cos x
+
ln
+ C Then f(x) is
2
1 2 cos x
1 sin x
1 sin x
(A) ln
(B) ln
1 sin x
1 sin x
(B)
6.
4 2
cot 7 x
5
x3 x
dx is
1 x6
(A)
8.
(D)
ex
1 cos x
x( x x ) dx is
3
x
6
(A) tan–1 6
+ C (B) ln |x + ( x + 1)| + C
( x 1)
x
+C
(D) none of these
(C) ln 6
( x 1)6
9. Let f(x) be a function satisfying f’(x) = f(x) and f(0)
f(x)
dx is
= 2. Then
3 4f(x)
(A) 1/4 ln (3 + 8ex) + C (B) 1/8 ln (3 + 8ex) + C
(C) 1/2 ln (3 + 8ex) + C (D) none of these
ln x
dx is equal to
10.
x 1 ln x
(A) 1/3 (1 + ln x)3/2 – 1 ln x + C
(A) C
ex
1 sin x
3
x
(C)
(B) 2/3 (1 + ln x)3/2 – 1 ln x + C
(C) 2/3 (1 + ln x)3/2 – 2 1 ln x + C
(D) none of these
cos x xsin x
dx is equal to
11.
x(x cos x)
x
x cos x
(A) ln
+C
(B) ln
+C
x cos x
x
xsin x
+C
(D) none of these
(C) ln
x cos x
12.
e 2x dx
e 1 is equal to
4
x
(A) 4/21 (ex + 1)3/4 + C
(B) 4/21 (ex + 1)3/2 (3ex – 4) + C
(C) 2/21 (ex + 1)3/4 + C
(D) none of these
xn xn
, x > 1, then
n x n x n
13. If f(x) = lim
xf(x)ln x 1 x 2
1 x2
dx is equal to
(A) ln (x 1 x 2 ) x C
1 2
2
2
(B) x ln x 1 x x C
2
(C) x ln x 1 x 2 ln x 1 x 2 C
(D) None of these
INDEFINITE INTEGRATION
14.
sin (ln x) dx is equal to
2
(D) ln (x 1 x 2 ) + x + C
(A) (x/10) (5 + 2 sin (2 ln x) + cos (2 ln x)) + C
(B) (x/10) (5 + 2 sin (2 ln x) – cos (2 ln x)) + C
(C) (x/10) (5 – 2 sin (2 ln x) – cos (2 ln x)) + C
(D) (x/10) (5 – 2 sin (2 ln x) + cos (2 ln x)) + C
15.
x (1 n.x
n 1
2n
x )
e (1 x ) 1 x
n
dx is equal to
2n
1 xn
+C
1 xn
(A) ex
(C) – ex
1 xn
+C
1 xn
(B) ex
1 xn
+C
1 xn
(D) – ex
16. The value of the integral
1 xn
+C
1 xn
dx
x (1 x )
n
(A)
11/n
1
1
1 n
+C
(1 n) x
(B)
11/n
1
1
1 n
+C
(1 n) x
n 1/n
, n N is
11/n
(C) –
1 1 1
n
(1 n) x
+C
11/n
1
1
1
(D)
+C
(1 n) x n
17. Integral of 1 2 cot x (cot x cosec x) w.r.t. x
is
x
(A) 2 ln cos + C
2
x
(B) 2 ln sin + C
2
1
x
ln cos + C
(C)
2
2
(D) ln sin x - ln(cosec x - cot x) + C
18.
x.
n x 1 x
1 x2
2
1.149
dx equals
(A) ln (x 1 x 2 ) - x + C
x
x
(B)
. ln2 (x 1 x 2 ) +C
2
1 x2
x
x
(C)
. ln2 (x 1 x 2 ) +
+C
2
1 x2
(1 ln x)
19. The value of
xx 1
2
dx is
(A) sec–1(xx) + C
(B) tan–1(xx) + C
2
(C) ln x x x 1 C
(D) None of these
xk
( x )5
dx
=
a
ln
k + c, then the
1 x
( x )7 x 6
values of a and k are
(A) 2/5, 5/2
(B) 1/5, 2/5
(C) 5/2, 1/2
(D) 2/5, 1/2
20. If
cos 7x cos8x
dx is equal to
1 2 cos 5x
sin 2x cos3x
+C
(A)
2
3
(B) sin x – cos x + C
21. The value of
sin 2x cos3x
+C
2
3
(D) none of these
(C)
22. The value of
x 1
x
n 5
x2 1
. 2 dx, is equal
x
to
1
(A) x
x
x 1
x
(C)
n6
n 6
+C
x2 1
x2
(B)
n6
+C
n 6
23. The value of
+C
(D) none of these
(1 cot n 2 x)dx
is equal to
tan x cot x.cot n 2 x
1
ln | sinn x + cosn x | + C
n
1
ln | sinn – 1 x + cosn – 1x | + C
(B)
2n
(A)
(C)
n 6
1
ln | sinn – 1 x + cosn – 1 x | + C
(n 2)
1.150
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
2
ln | sinn – 1 x + cosn – 1 x | + C
n2
dx
24. The value of
is equal to
(1 x ) x x 2
(D)
(A)
2( x 1)
+C
1 x
2( x 1)
(C)
1 x
25. The value of
(A)
(B)
+C
2( x 1)
+C
1 x
(D) none of these
sec x(2 sec x)
(1 2sec x)
sin x
+c
2 cos x
2
(B)
sin x
+c
(C)
2 sin x
dx, is equal to
cos x
+c
2 cos x
cos x
(D) –
+c
2 sin x
x4 2
x
e
26. The value of
2 5/2 dx is equal to
(1 x )
(A)
e x (x 1)
+C
(1 x 2 )3/2
(B)
e x (1 x x 2 )
+C
(1 x 2 )3/2
(C)
e x (1 x)
+C
(1 x 2 )3/2
(D) none of these
27. If (sin 3 sin ) esinQ cos Q d Q = (Asin3 Q
+Bcos2 Q + Csin Q + Dcos Q + E) esin Q + F, then
(A) A = – 4, B = 12
(B) A = –4, B = – 12
(C) A = 4, B = 12
(D) A = 4, B = – 12
28. The value of
e
(x sin x cos x) x
.
4
3
cos x xsin x cos x
dx
x 2 cos2 x
1
tan x / 2 1 2
ln
+C
2
tan
x / 2 1 2
1
tan x/ 21 2
(B) 2 (sinx cosx) ln
+C
2
tan x/21 2
1
1
tan x/ 21 2
(C) 2 (sinx cosx) 2 ln tan x1 2 + C
(A) (sin x cos x)
(D) none of these
(1 x)
30. The value of
dx is equal to
x(1 xe x )2
x
1
+C
(A) ln
1 xe x (1 xe x )
xe x
1
(B) ln
+C
x
1 xe
1 xe x
xe x
1
(C) 1n
+C
1 e x 1 xe x
(D) none of these
dx
31. The value of
is equal to
x a 2 x2
1 –1 x 1
ln |x + a 2 x 2 | + C
sin
2
a 2
1 –1 x 1
ln |x + a 2 x 2 | + C
sin
(B)
2
a 2
1
x
sin–1 –ln |x + a 2 x 2 | + C
(C)
2
a
1
x 1
cos–1 ln |x + a 2 x 2 | + C
(D)
2
a 2
(A)
32. The value of
dx
(x a) (x b)
8/7
3 xa
2(a b) x b
is equal to
(A)
1
+C
(A) ex sin x + cos x . x
x cos x
1
(B) ex sin x + cos x . x cos x + C
x
1
+C
(C) ex sin x + cos x . x
x cos x
(D) none of these
3 xb
(B)
(a b) x a
29. The value of
dx
is equal to
sec x cos ecx
6/7
is equal to
2/3
+c
1/3
+c
1/7
7 xb
(a b) x a
(D) none of these
(C)
33. The value of
+c
sec x dx
sin(2x A) sin A is equal to
(A) sec A ( tan x sin A cos A ) + C
1.151
INDEFINITE INTEGRATION
2 sec A tan x.sin A cos A ) + C
1
(C)
sec A (tan x. sin A + cos A)3/2 + C
2
(D) none of these
(B)
39. The value of the integral
11
7
7
(A)
tan + C
11
2
p 1
1
2p p 1 dp = f(p) + c, then f(p) is equal
34. If I =
1
ln (p – p 2 1 ) + C
2
1 1
1
1
(B) cos p sec p + C
2
2
2
(C) ln p p 1
40.
(D) none of these
35.
3
x 2 2x
+C
(x 5 x 3 1)
(C) ln (x5 + x3 + 1 + 2x 7 5x 4 ) + C
(D) None of these
36.
e
tan x
(x sec2 x + sin 2x) dx is equal to
(A) etan x (x + sec2 x) + C
(B) etan x(x – sec2 x) + C
(C) etan x(x sec x + sin x) + C
(D) etan x(x – cos2 x) + C
37. Let x = f(t) cos t + f(t) sin t and y = – f(t) sin t +
1/ 2
f(t) cos t.
dx 2 dy 2
dt dt
Then
Then
(A) f(t) + f(t) + C
(C) f(t) + f(t) + C
38.
dt equals
(B) f(t) + f(t) + C
(D) f(t) – f(t) + C
px p 2q 1 – qx q 1
dx
2p 2q
2x p q 1
x
(A) –
(C) –
xp
x pq 1
xq
x
pq
1
C
(B)
C
(D)
xq
x pq 1
xp
x
(B) e tan –1 x + C
–1
(C) – x e tan x + C
(D) x e tan x + C
dx
= – [f(x)]1/n + c, then f(x) is
41. If 2 n
x (x 1)(n 1)/n
(B) 1 + x–n
(A) (1 + xn)
n
–n
(C) x + x
(D) None of these
2 tan x
42. e x
cot 2 x
dx is equal to
4
1 tan x
(A) ex tan x +C (B) ex tan x + C
4
4
3
(C) ex tan x +C (D) None of these
4
43. The value of the integral (x2 + x) (x–8 + 2x–9)1/10 dx
is
5 2
(A)
(x + 2x)11/10 + c
11
5
(x + 1)11/10 + c
(B)
6
6
(x + 1)11/10 + c
(C)
7
(D) None of these
x10
+C
(A)
2(x 5 x3 1)2
(B)
(1 + x + x2) d (cot–1 x) is equal to
(2x12 5x 9 )
3
tan –1 x
–1
(x x 1) dx is equal to
5
e
(D) None of these
(A) – e tan –1 x + C
1
sec–1p + C
2
11
7
7
(B)
cos + C
11
2
11
7 7
(C)
sin + C
11 2
to
(A)
(1 cos )2/7
dq is
(1 cos )9/7
pq
1
C
C
44.
e (x + x + 2x ) e x dx is equal to
x4
3
5
2
1 x2 x4
1
4
xe e + c
(B) x2 e x + c
2
2
2
4
1 x2 x4
1
e e +c
(D) x2 e x e x + c
(C)
2
2
x/2
bx
ln a a
ln b
45. x 5x/2 3x 2x 4x dx (where a, b R+)
3a
b
2a
b
equal to
1
a 2x b3x
2x 3x
(A)
a
b
ln
+k
6 ln a 2 b3
e
(A)
1.152
(B)
1
1
1
ln 2x 3x + k
2 3
2x 3x
ea b
6 ln a b a b
(C)
1
1
ln (a2xb3x) + k
6 ln a 2 b3 a 2x b3x
(D) –
46.
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
x
x
C
(B) e
2
2 3
(1 x )
1 x
1
x
x
(C) e
C
2
(1 x 2 )5
1 x
(D) None of these
1
1
ln (a2xb3x) + k
6 ln a 2 b3 a 2x b3x
cos ec2 x 2005
dx is equal to
cos2005 x
(A)
cot
C
(cos x)2005
(B)
(C)
–(tan x)
C
(cos x)2005
(D) None of these
50. If I =
tan x
C
(cos x)2005
(A)
3
3cos x 8
+C
(3 4 cos x)2
3 8cos x
(B) 16 (3 4 cos x)2 C
47. If xf(x) = 3f2(x) + 2, then
sin 2x
(3 4 cos x) dx, then I equals
(C)
2x 2 12xf(x) f(x)
dx equals
(6f(x) x)(x 2 f(x))2
3 cos x
C
(3 4 cos x)2
3 8cos x
(D) 16 (3 4 cos x)2 C
(A)
1
C
x 2 f(x)
(B)
1
C
x 2 f(x)
(C)
1
C
x f(x)
(D)
1
C
x f(x)
(A)
1
[sec2x – tan x] + C
2
1
ln f(x) + C, then
2(b a 2 )
(B)
1
[x sec2x – tan x] + C
2
(C)
1
[x sec2x + tan x] + C
2
(D)
1
[sec2x + tan x] + C
2
48. If f(x) sin x cos x dx =
51.
2
f(x) is equal to
1
(A) 2 2
a sin x b 2 cos 2 x
(B)
1
a sin x b2 cos2
2
2
1
(C) 2
2
a cos x b2 sin 2 x
(D)
52.
1
a 2 cos2 x b2 sin 2 x
2 sin x
+C
(B) log
2 sin x
1 sin x
+C
(C) log
2 sin x
(D) None of these
1
1 2x 2
ex
dx is equal to
2
(1 x 2 )5
1 x
1
x
x
C
(A) e
2
(1 x 2 )3
1 x
cos x dx
(1 sin x)(2 sin x) is equal to
1 sin x
+C
(A) 2 log
2 sin x
49. The value of integral
x sin x sec3x dx is equal to
53.
cos2 x.dx
is equal to
sin x.cos3x
INDEFINITE INTEGRATION
| Csin x |
(A) ln
1 4sin x
| Csin x |
(C) ln
54.
(B) ln
2
(D) None of these
1 4sin 2 x
x nx
(x 1)
2
3/2
dx equals
nx
(A) sec–1 x –
x2 1
nx
(B) sec–1 x +
2
x 1
nx
(C) cos–1 x –
x2 1
(D) none of these
55.
56.
57.
| C cos x |
cos 2x
+C
(A)
sin x
+C
(b a cos x)
(B)
cos x
+C
(b a sin x)
(C)
sin x
+C
(b a sin x)
(D)
cos x
+C
(b a cos x)
x 1
x x 1 dx is equal to
(A) ln |x –
x 2 1 | – tan–1 x + C
(B) ln |x +
x 2 1 | – tan–1 x + C
(C) ln |x –
x 2 1 | – sec–1 x + C
(D) ln |x +
x 2 1 | – sec–1 x + C
sin (ln x) dx is equal to
2
x
1 e dx is equal to
x
x
1 e 1
1 ex 1
– (2x + 1) 1 e x + C
1 ex 1
(D) none of these
sin 2x
= dx = a cot–1(b tan2x) + c, then
59. If
4
sin x cos 4 x
(A) a = 1, b = – 1
(B) a = – 1, b = 1
(C) a = – 1, b = – 1
(D) none of these
(C) ln
(A) g–1 (x)
(B) x f–1 (x) – g(f–1(x))
(C) xf–1 (x) – g–1(x)
(D) f–1(x)
f(x) '(x) (x)f '(x)
dx is
61. The value of
(f(x)(x) 1) f(x)(x) 1
2
1 1 ex
+C
1
a b cos x
(A) ln
1 ex 1
60. If f(x)dx = g(x), then f (x) dx is equal to
+C
(b a cos x) dx is equal to
x.e
1 ex 1
(B) (2x + 1) 1 e x – ln
+C
(A) x/10 (5 + 2 sin (2 ln x) + cos (2 ln x)) + C
(B) x/10 (5 + 2 sin (2 ln x) – cos (2 ln x)) + C
(C) x/10 (5 – 2 sin (2 ln x) – cos (2 ln x)) + C
(D) x/10 (5 – 2 sin (2 ln x) + cos (2 ln x)) + C
58.
1.153
+ (2x + 1) 1 e x + C
(A) sin–1
f(x)
(x)
(B) cos–1 f 2 (x) 2 (x)
(C) tan–1 [f(x) . f(x)]
(D) none of these
(1 x cos x)
dx is equal to
62.
x(1 x 2 e 2 sin x )
(A) log |x esin x| + 1/2 log |1 – x2 e2sinx| + C
(B) log |x esin x| – 1/2 log |1 – x2 e2sinx| + C
(C) log |x esin x| – 1/2 log |1 + x2 e2sinx| + C
(D) none of these
dx
63.
is equal to
tan x.cos 2x
| Csin x |
| Ccos x |
(A) ln
(B) ln
cos 2x
cos 2x
| Csin x |
(C) ln
(D) none of these
1 4sin 2 x
64. If the anti-derivative of
through (1, 2) is
the value of m is
(A) 1
(C) 5
x3
1 2x 2
which passes
1
(1 + 2x2)1/2 (x2 – 1) + 2. Then
m
(B) 3
(D) 6
n
65. If In = cot x dx and I0 + I1 + 2 (I2 + I3 + .......... + I8)
1.154
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(where u = cot x) Then the value of l is
(A) 1
(B) –1
(C) 2
(D) –2
u 2 u3
u9
.....
+ I9 + I10 is l u
2
3
9
MULTIPLE CORRECT ANSWER TYPE FOR JEE ADVANCED
66. If
x 7/6 x5/6
dx
1/2
1/3
x1/3 x 2 x 1 x1/2 x 2 x 1
z3 3z B 3z C
log z c where
=–A
2
D
3
1/6
1
z = x 1 1 , then
x
(A) A = 6
(B) B + C = 4
(C) B + C = 3
(D) B + C + D = 5
1 sin x
67.
dx equals
cos x
(A) ln (1 + sin x) + C
x
(B) 2 ln cos + C
4 2
1 x2
equals
1 x
3
- 1 x2 + C
1
3
2
(C) x2 1 x 2 3
(D) none of these
(B) x2 1 x 2 -
69.
1
x2 1
+
+C
x
x
1
x2 1
+ cos -1 + C
x
x
(A) sin -1
(B)
(C) sec -1 x -
72. If I =
2 3/2
(A)
x 1 1
.
dx equals
x 1 x2
x2 1
+C
x
x2 1
+C
x
71. A curve g(x) = x27 (1 + x + x2)6 (6x2 + 5x + 4) dx is
passing through origin, then
37
27
(A) g(1) =
(B) g(1) =
7
7
2
1
(C) g(–1) =
(D) g(–1) =
7
7
x
x
(D) 2 ln cos sin + C
2
2
x3 dx
70.
(D) tan -1 x 2 1 -
x
(C) 2 ln sin + C
4 2
68.
1 2
ln (sin x sec x) + C
2
1 2
ln (cos x cosec x) + C
(D)
2
(C)
then
sin x sin 3 x
dx = P cos x + Q ln |f(x)| + R
cos2x
1 x2 + C
(A) P = 1/2, Q = –
3
4 2
1 x2 + C
(B) P = 1/4, Q = –
1
, f(x) =
2
2 cos x 1
2 cos x 1
3
2 cos x 1
2 cos x 1
3
n (tan x)
sin x cos x dx equal
1 2
ln (cot x) + C
(A)
2
1 2
ln (sec x) + C
(B)
2
3
(C) P = 1/2, Q = –
4 2
(D) P = – 1/2, Q = –
73. If
, f(x) =
3
4 2
, f(x) =
2 cos x 1
2 cos x 1
e x 1
2x dx = AF (x – 1) + BF (x – 4)
(x 5x 4)
2
INDEFINITE INTEGRATION
+ C and F(x) =
(A) A = – 2/3
(C) A = 2/3
74. If
ex
dx, then
x
(B) B = 4/3 e3
(D) B = 8/3 e3
x2 x 1 x
e dx = exf(x) + c, then
(x 2 1)2
80.
sin–1 x cos–1x dx = f–1(x) [Ax – xf–1(x) – 1 x 2 ]
+ 2x + C, then
(A) f(x) = sin x
(C) A =
4
1 sin x dx is equal to
1
cos–1(sin2x) + C
2
(C) tan–1 (sin2 x)+ C
(B) –
nx
(D)
(mn).2 x.3x
+C
mn2 nn3
n(tan x)
sin x cos x dx equal
2
sec 2x 4 dx equals
82.
x 2 cos2 x
cosec2 x dx is equal to
1 x2
(A) cot x – cot–1 x + C
(B) C – cot x + cot–1 x
(C) –tan–1 x –
cosec x
+C
sec x
1
1 x7
dx = a ln |x| + b ln |x7 + 1| + c, then
x(1 x 7 )
(A) a = 1
(B) a = – 1
(C) b = – 2/7
(D) b = 1/7
2
sin 4x2 dx = K e 5x (A sin 4x2
(B) K = 1/82
(D) None of these
2 . 3 dx when m, n N is equal to
mx
2 mx.3mx
C
n(2 m.3n )
84. If
cos x
+C
(D) cos–1
2
79.
(C)
1
(A) sin–1 (sin2 x) + C
2
+ B cos 4x2) + C. Then
(A) K = – 1/82
(C) A = 5
e(mn2 nn3)x
+C
mn2 nn3
dx
1 x3
b
c
= a ln
+ 3 +
+ d,
83. If
x3
x 4 (1 x3 )2
x
(1 x3 )
then
(A) a = 1/3, c = 1/3
(B) b = –1/3, c = – 1/3
(C) a = 2/3, b = –1/3
(D) a = 2/3, b = 1/3
4
5x 2
(B)
(D) e n tan x – cot x + C
sin x cos x
78. If xe
2 mx 3mx
+C
mn2 nn3
(A) C – 1/2 cot (2x + /4)
(B) 1/2 tan (2x – /4) + C
(C) 1/2 (tan 4 x – sec 4 x) + C
(D) None of these
2
dx
76.
= f(g(x)) + c, then
x 2 ax 1
(A) f(x) is inverse trigonometric function for |a| > 2
(B) f(x) is logarithmic function for |a| < 2
(C) g(x) is quadratic function for |a| > 2
(D) g(x) is rational function for |a| < 2
77.
81.
(B) f(x) = cos x
(D) A =
(A)
(A) 1/2 ln2 (cot x) + C
(B) 1/2 ln2 (sec x) + C
(C) 1/2 ln2 (sin x sec x) + C
(D) 1/2 ln2 (cos x cosec x) + C
(A) f(x) is an even function
(B) f(x) is a bounded function
(C) The range of f(x) is (0, 1]
(D) f(x) has two points of extrema
75.
1.155
85. If e
x
b ln (x 2 1)
cx
dx
x2 1
c x
e ln (x2 + 1) + K , then the values of b and c
2
can be
(A) b = 1, c = 2
(B) b = 1/3, c = 1/2
(C) b = 1/2, c = 1
(D) b = 2, c = 3
=
1.156
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Comprehension - 1
In a certain problem the differentiation of product
(f(x).g(x)) appears. One student commits mistake and
differentiates as
df dg
.
but he gets correct result if
dx dx
1
f(x) = x3 and g(4) = 9, g(2) = –9 and g(0) = – .
3
3
(x 3)3
(B)
4
(x 3)3
9
27
(D)
3
(x 3)
(x 3)3
87. The derivative of f(x – 3) . g(x) with respect to x at
x = 100 is
(A) 0
(B) 1
(C) –1
(D) 2
(C)
f(x).g(x)
88. lim
will be
x 0 x(1 g(x))
(A) 0
(C) 1
(B) –1
(D) 2
Comprehension - 2
Consider a differentiable function f : R R which
1
x = f(x) for all x R and f(1) = 2,
satisfies f2
2
f(0) 0.
89. If R satisfying = 1, then for all x,
f(x) . f(x) =
(A) 2f(x)
(B) f(x)
(C) abf2(x)
(D) None of these
90. The value of xlim
(A) 1
(C) 0
f(x) . ln 2 is equal to
4x
(A) 2 x2 1 C
(B) 2 2x C
(C) 2 2x2 1 C
(D) 2 2x2 C
Comprehension - 3
If A is square matrix and e A is defined as
eA = I + A +
86. The function g(x) is
(A)
91.
f(x) x 2
is
f(x) 2 x
1
2
(D) None of these
(B)
A 2 A3 ... 1 f(x) g(x) ,
where
2 g(x) f(x)
2! 3!
x x
A=
and 0 < x < 1, I is an identity matrix.
x x
92.
g(x)
f(x) dx is equal to
(A) ln (ex + e–x) + C
(C) ln (e2x – 1) + C
(B) ln (ex – e–x) + C
(D) None of these
93. (g(x) + 1) sin x dx is equal to
A)
ex
(sin x – cos x)
2
(B)
e 2x
(2 sin x – cos x)
5
ex
(sin 2x – cos 2x)
5
(D) None of these
94. dx is equal to
(A) – cosec–1 (ex) + C
(B) – sec–1 (ex) + C
(C) + sec–1 (ex) + C
(D) None of these
(C)
Comprehension - 4
Let n be a positive integer or zero and let
n = dx (a > 0) We can find the reduction formula as
n = –+ a2B n–2 , where A and B are constants. Also
2
2 3/2
I1 = –(a – x ) .
1.157
INDEFINITE INTEGRATION
95. A must be equal to
(A) n + 1
(C) n + 2
96. B must be equal to
n 1
(A)
n2
n2
(C)
n 1
97. The value of integral
equal to
2
(B) n – 1
(D) n
101. Assertion (A):
x
0
2
Reason (R): The function
(B)
(C)
a 4
64
(D) none
x
cos 8x . cos 16 x dx =
dx = fog(x) + C, where f and
2
98. The nature of the function y = f(x) + (g(x)) is
(A) even
(B) odd
(C) neither even nor odd
(D) one-one
99. If f(x) g(x) dx = ax g(x) + b(1 + x )
2 1/2
+ c(1 + x ) + d, then a + c is equal to
1
1
(A)
(B)
2
3
(C) 0
(D) 1
3
g(x)
1
(B)
3
(D) 3
Assertion (A) and Reason (R)
(A) Both A and R are true and R is the correct
explanation of A.
(B) Both A and R are true but R is not the correct
explanation of A.
(C) A is true, R is false.
(D) A is false, R is true.
cos 32 x
+ C.
1024
Reason (R) : sin 2 n x dx
104. Assertion (A):
= sin 1
cos2 n x
C .
2n
1 cos e c x dx
1
cos 1 (1 2 sin x) + C.
2
sin x +
Reason (R): 1 cose c x =
2 3/2
dx = ax x 1 x 2 ag(x) x then a
is equal to
1
(A)
2
(C) 2
x
a
103. Assertion (A): sin x . cos x . cos 2x . cos 4x .
1 x2
g are some functions and C is an arbitrary
constant.
100. If e
2
= 5x tan x + C
a f(x) f '(x) log e dx a f(x) C .
a 4
16
Comprehension - 5
ln(x 1 x 2 )
x
Reason (R): If a > 0, then
a 4
32
5 (sec x tan x ln 5) dx
102. Assertion (A):
2
a x dx must be
(A)
Let
80
2
9x 4
has a
x 5 3x 2 x
non-elementary antiderivative.
n
n2
n 1
(D)
n2
4
5
1
(B)
a
9x 4
x 3x x dx n 23
1 sin x 1 sin x
=
sin x 1 sin x
=
cos x
1 1 sin x 2
4
2
105. Assertion (A):
cos x
sin x 1 sin x
for all x in the domain.
x
tan (ln x) tan ln tan (ln 2)
2
dx
x
sec(ln x)
ln
C
is equal to x tan (ln 2)
sec ln x
2
Reason (R) : We have tan (A + B) tan A tan B
= tan (A + B) – tan A – tan B and
= ln sec (ln x) + C.
tan (ln x)
dx
x
1.158
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
dx
x dx a ln xk
106. Assertion (A): If
+c,
7
xk 1
x x6
5
the value of (a + k) is 9/2.
Reason (R) : The given integral reduces to the
f '(x)
dx where f(x) (x 5/2 1) .
form
f(x)
1
2
107. Assertion (A): If
dx = ln(f(x)) + c then
f(x)
1
f(x) = x.
2
1
x
,
dx
Reason (R): When f(x) =
f(x)
2
2
= dx = 2 ln | x | + c
x
108. Assertion (A): For – 1 < a < 4,
x 2(a 1)x a 5 = ln |g(x)| + c, where
2
and c are constants.
Reason (R): For – 1 < a < 4,
1
is a continuous function.
x 2 2(a 1)x a 5
109. Assertion (A): If the primitive of f(x) = sin x
+ 2x – 4, has the value 3 for x = 1, then there are
exactly two values of x for which the primitive of
f(x) vanishes.
Reason (R): cos x has period 2.
{f(x) (x) – f (x) (x)}
110. Assertion (A):
f(x) (x)
2
1 (x)
ln
C.
2 f(x)
(h(x))n 1
Reason (R): (h(x)n h(x) dx =
+ C.
n 1
{ln (x) – ln f(x)} dx =
MATCH THE COLUMNS FOR JEE ADVANCED
111. Column-I
Column-II
2x
(A) If
1 4 dx = k sin–1(f(x)) + C, then k is greater than
(B) If
( x) x dx = a ln x 1 c, then ak is less than
(C)
x
( x)5
xk
7
6
k
x4 1
(P) 0
(Q) 1
m
x(x 1) dx = k ln |x| + 1 x n, where n is the constant (R) 3
2
2
2
of integration, then mk is greater than
x
dx
k tan–1 m tan C, then k/m is greater than (S) 4
2
5 4 cos x
112. Column-I
Column-II
(D)
(A) Let f(x) = x
sin x
(1 + x cos · n x + sin x) dx and
(P) rational then the value
2
f
of f() is
2 4
(B) Let g(x) =
1 2 cos x
(cos x 2) dx and g(0)
2
= 0 then the value of g(/2) is
(C) If real numbers x and y satisfy (x + 5)2 + (y – 12)2
(Q) irrational
(R) integral the minimum
1.159
INDEFINITE INTEGRATION
= (14)2 then value of (x 2 y 2 ) is
(D) Let k (x) =
(x 2 1)dx
x 3x 6 and k(–1) = 12 then the
3
3
3
value of k (–2) is
113. Column-I
(A)
dx
sin x sin 2x
(S) prime
Column-II
(P) 2cos 2 x cos x 1
1
2 2
1
2
cos x + 1 + 4 cos2 x cos x
(B)
sin x sin 2x
cos x cos 2x dx
(Q)
2 tan x 3
sin x 2 cos x
(D)
sin x cos x dx
2
2
tan x
4
4
(R)
1
|+C
2
1
1
log (1 – cos x_ + log
6
2
2
log | 1 + 2 cos x | + C
3
(1 + cos x) –
(C)
log | 4
1
log tan 2 x 1 tan 4 x + C
2
2
(S) log (2 + tan x
3
–1
tan
2
tan x
+C
2
114. Column-I
Column-II
(A) If x 2 d(tan x) = x f(x) + c, then f(1) is equal to
(P) 0
x
x
(B) If 1 3 tan x(tan x sec x) dx = a log cos sin + C
2
2
(Q) –2 then a is equal to (0 < x <
–1
(R)
2x
2x
(C) If x 2 e dx = e f(x) + c, then the minimum value of
(D) If
x4 1
b
dx = a log | x | + 2
+c
x(x 2 1)2
x 1
)
2
4
(S) 1 f(x) is equal to
(T)
1
8
then a – b is equal to
115. If x (0, 1) then match the entries of column-A with column-B considering 'c' as an arbitrary constant of
integration.
1.160
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Column-I
Column-II
tan 2 tan
(A)
1
cot 2 tan
(B)
1
(C)
(D)
1 x 1
dx
1 x 1
(P)
4 3/4
x +C
3
1 x 4 x
dx
1 x 4 x
(Q)
4 5/4
x +C
5
(R)
2 3/4
x +C
3
(S)
2 5/4
x +C
5
1 1 1
1 tan sin
2
1
1 1 1
1 tan sin
2
1
x tan 2 tan 1
x
x
dx
x
x
1 x 1
1 x 1
d
tan cot sec cos ec
1.
Evaluate
2.
Evaluate x x 1 ln x (1 + ln x) dx.
3.
Evaluate
4.
Evaluate
5.
Evaluate
6.
(x 2)2 tan 1 (2x) 12x3 3x
dx
(4x 2 1)(x 2)2
1 x2
dx
.
2
1 x 1 x2 x4
7.
Evaluate
8.
Prove that
(ii)
x dx
(1 x 4 )2
a 2 x 2 dx .
x 2 b2 x
e x dx
x4
Evaluate
sin x dx
(a cos x b sin x) .
2
2
2
2
dx
x (x 2x 1) , by the
2
substitution z = x + (x 2 2x 1) a nd
sh ow tha t the va lue is 2 tan–1{x +
(x 2 2x 1) } + C.
11. Evaluate
x7
(1 x2 )5 d x
9.
10. Integrate
dx
.
4
(1 x ){(1 x 4 )1/2 x 2}1/2
Evaluate the following integrals :
(i)
1 x 1
dx
1 x 1
sin 3 xdx
(cos x 3cos x 1) tan (sec x cos x)
4
2
1
12. Let f(x) be a polynomial of degree three such that
f(0) = 1, f(1) = 2, x = 0 is a critical point but f(x) does
not have local extremum at x = 0. Then evaluate
f(x)
x 2 7 dx .
3x3 5x 2 7x 9
dx
13. Evaluate
2x 2 5x 7
2e 5x e 4x 4e3x 4e 2x 2e x
14. Evaluate
(e 2x 4) (e 2x 1)2
2
1
e x dx .
= e 1 1 1
3
2
3 x
2x
3. 2 .1
x
2x
x
1
1 e
C
tan
=
2
2x
2(e 1)
INDEFINITE INTEGRATION
15. Determine the coefficients A, B so that
dx
A sin x
dx
B
2
a
b
cos
x
a
b
cos x .
(a b cos x)
16. Evaluate the following integrals :
(x 1)4 dx
dx
(1 x )
(i)
(x 2x 2)
(iii)
5 3x 6x 2 5x3 x 4
dx
x 5 x 4 2x3 2x 2 x 1
2
4
(ii)
3
3
2
x 8x x 2x 1
dx .
(x 2 x)(x3 1)
17. Evaluate the following integrals :
e x (x 2 5 x 7)
(i)
dx
(x 3) 2
(iv)
2
1 1 x x
dx
e tan x
2
1 x
(ii)
(iii)
(1 cos 2x )e x dx
1 sin 2 x
n x
dx
(1 n x)2
18. Evaluate the following integrals :
1
dx
(i)
sin x sin x(1 sin x)
(iv)
(ii)
a sin x
cos x cos x a sin x dx.
2
2
2
19. Evaluate the following integrals :
(i)
(ii)
x2 4
dx
6
x 2x 4 x 2
(x 2 1)2 dx
.
(1 x) (1 x 2 )3
20. Evaluate the following integrals :
7cos x 3sin x 5
dx
(i)
3cos x 4sin x 5
3 2 cos x 4sin x
dx .
(ii)
2sin x cos x 3
21. Evaluate the following integrals :
sin 2x dx
(i)
(a b cos x)2
(ii)
a bsin x
b a sin x dx.
2
2 4
1.161
22. Evaluate the following integrals :
m.x m 2n 1 n.x n 1
dx
x 2m 2n 2x m n 1
m.x m n n
dx
(ii)
x m n 1 x
23 Evaluate the following integrals:
(i)
(i)
x3 dx
(a cx 2 )4
(ii)
dx
x(a bx )
n 2
x 2 dx
(1 x 2 )dx
(iii)
(iv)
.
(1 x 2 )3
x(1 x 2 x 4 )
24. Evaluate the following integrals:
x2 3
x 2 2x 2
dx (ii)
dx
4
2
x 2x 9
x 4 2x 2 4
1
x2 5
dx (iv)
dx .
(iii)
4
2 2
4
4
x a x a
x 2x 2 4
25. Evaluate the following integrals:
dx
(i)
7/8
(x 2) (x 3)9/8
dx
(ii)
(x 1) x 2 x 1
(i)
(iii)
2x 2 7x 11
(x 2) x 4x 8
2
1
dx .
(x 1) 4 (x 1)
26. Evaluate the following integrals :
(iv)
1
xea tan x dx
x(cos x)e dx (ii)
(i)
(1 x 2 )3 / 2
27. Evaluate the following integrals :
x
2
cos 4 x dx
dx
sin 3 x(sin 5 x cos5 x)3/5
tan x
4
(ii)
dx
2
3
2
cos x tan x tan x tan x
28. Evaluate the following integrals :
(i)
x3dx
(1 x 4 ) dx
(1 x 4 )1/2
29. Evaluate the following integrals :
sin x
dx
(i)
(1 sin x)
(i)
(a bx )
2 3/2
(ii)
(ii)
sin x
sin x cos x dx
1.162
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
30. Evaluate the following integrals :
2x 1
dx
(i)
2
(4x 2x 1)3
(ii)
(iii)
2 – x – x2
dx
x2
x (1 2x)
dx
x4
(i)
2.
3.
(x 4r 1 x 2r 1 )
dx
x6r 1
x3 (x 2 1)
dx
Evaluate
x10 1
Evaluate
5.
6.
7.
Evaluate
(i)
tan x . ln(1 + x )dx
(ii)
x
–1
2
tan 1 x
4
dx
35. Evaluate the following integrals :
cos2x.sin 4x
(i)
cos 4 x(1 cos2 2x)
(ii)
3cot 3x cot x
tan x 3tan 3x dx
dx
Evaluate (1 x 2n ){(1 x 2n )1/n x 2 )1/2 .
9.
Evaluate
1
(x a) (x b) dx
(x 2x 3)
2
sin d
2
dx
11. Evaluate
12. Evaluate
1
x a 2 x2 C
4
x 2dx
.
(1 x 4 )(1 x 4 )1/2
26 cos 4sin 4sin cos 12 sin 26 .
2
x 2x x 7
10. Evaluate
n
1 x 4 x2
8.
m.x m 3n 1 n.x 2n 1
dx
x3m 3n 3x 2m 2n 3x m n 1
2
1 2 cos
tan
Evaluate
2 sin . sec2 .d
x
Prove that
x x 2 a 2 dx
n
= 1 (2x 2 a 2 ) n x x 2 a 2
4
Evaluate
(1 x 4 )
3
4.
1 x4
dx
Evaluate
2
cos (x x 1) dx
(ii)
32. Evaluate the following integrals :
dx
(i)
[x x(1 x)]2
x3 x 1
(ii)
dx
x 2 2x 2
1.
(iv)
dx
11
33. Evaluate the following integrals :
(x 1)dx
(i)
2
(x x 1) x 2 x 1
x 2 dx
(ii)
(4 2x x 2 ) 2 2x x 2
34. Evaluate the following integrals :
sec x
cosec x cot x
cosec x cot x (1 2sec x) dx
1
x
1 x2
dx
1 x2
31. Evaluate the following integrals :
(iv)
(iii)
(sin 3/2 cos3/2 )d
sin .cos .sin( )
3
3
1/3
9ax 2 (x 2 24a 2 ) x 2 3a 2 dx
13. Use an appropriate substitution to obtain an
integrand that is a rational function of a single
variable or of trigonometric functions. Do not
evaluate
1.163
INDEFINITE INTEGRATION
(i)
(ii)
[(4 x ) 5] 4 x dx
2 3
2
(x 2 5)7
x 3 x 5 dx
2
14. Evaluate
2
dx
sin(x )cos (x )
3
15. Evaluate
cos x dx
(1 cos sin x) (1 cos2 sin x 2 cos sin x)
2
16. Evaluate
m.x m 3n 1 n.x 2n 1
dx
x3m 3n 3x 2m 2n 3x m n 1
1
2
2
17. Evaluate sin (x x a )dx
18. Evaluate
(1 cos sin x) (1 cos2 sin x 2 cos sin x)
2
(sin x cos x)dx
(sin x cos x) sin x cos x sin x cos x
2
20. Evaluate
21. Evaluate
2
1 cos
cos (1 cos )(2 cos ) d
(1 x 4 )1/2 dx
.
1 x4
1 ax 2
dx
.
22. Evaluate
1 ax 2 1 2cx 2 a 2 x 4
23. Evaluate x·
24. Evaluate
2 sin(x 2 1) sin 2(x 2 1)
dx
2sin(x 2 1) sin 2(x 2 1)
cos x cos3x
4 3cos x cos3x dx
25. Prove that
x 6 7x 5 15x 4 32x3 23x 2 25x 3
dx
(x 2 x 2)2 (x 2 1)2
=
1
3
x2 1
2
ln 2
+C
x x 2 x 1
x x2
2
cos x dx
19. Evaluate
x 5 x3 1
dx
(x 6 1)
x 4 4x3 11x 2 12x 8
dx
27. Evaluate
(x 2 2x 3)2 (x 1)
f (x)
28. Prove that ef(x) · xf (x)
dx
{f (x)}2
1
= ef(x) · x
c. Hence or otherwise
f (x)
evaluate
x 4 cos3 x xsin x cos x
exsinx + cos x ·
dx.
x 2 ·cos2 x
29. Explain the following apparent difficulties :
dx
x
(a)
2 3/2
(a cx )
a(a cx 2 )1/2
yet, when a becomes nearly zero, the denominator
on the right-hand side becomes nearly zero, while
that on the left hand remains finite.
dx
(b)
2 ln( x a x b )
(x a)(x b)
yet if a and b are positive, and x is less than either
of them, the square root on the left hand side is
real, but those on the right hand side are
imaginary.
26 Evaluate
( 4 x 2 )3 1
x 2 ax R
,
x 2 ax R
find the relation between the integrals
dx
xdx
,
.
R
R
31. Show that if the roots of Q(x) = 0 are all real and
distinct, and P(x) is of lower degree than Q(x), then
P()
R(x) dx
ln | x |, the summation
Q'()
applying to all the roots a of Q(x) = 0.
Hint: The form of the partial fr action
corresponding to a may be deduced from the facts
that
P()
Q(x)
Q(), (x – ) R(x)
Q()
x
30. If R = (x2 + ax)2 + bx, and u = ln
32. If (x2 + y2)2 = 2c2(x 2 – y2), then prove that
dx
1
x2 y2
–
ln
.
xy
y(x 2 y 2 c2 )
c2
33. Show that
dx
(x x )y , where y = ax + 2bx + c,
2
0
2
1.164
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
may be expressed in one or other of the forms
axx 0 b(x x 0 ) c yy 0
1
ln
,
–
y0
x x0
axx 0 b(x x 0 ) c
1
tan 1
,
z0
yz 0
2
according as ax 0 2bx 0 c is positive and
equal to y20 or negative and equal to – z 20 .
34. Evaluate
(x )dx
{p(x )2 q(x )2}{r(x )2 s(x )2}1/2
by means of the substitution t = (x – )2 /(x – )2,
showing that the integral becomes
1
dt
2( ) (pt q) (rt s) Show how to obtain
the value of the integral
(x 2)dx
.
2
(7x 36x 48) (x2 2x 1)
x a
35. (i) By means of the substitution u
a x
find the integral of (a4 – x4) / (a4 + a2x2 + x4)3/2
(ii) By means of the substitution 1 – x = xy2
–1
1.
4e x 6e x
2x
dx = Ax + B log (9e – 4) + C, then
9e x 4e x
A = ......, B = .........and C =........
[IIT - 1990]
If
4.
B. Multiple Choice Questions with ONE
correct answer :
2.
(3x 1)
(x 1) (x 1) dx equal to
3
(A)
(B)
(C)
5.
1
1
–
+c
2(x 1)
(x 1)
1
1
log |x + 1| +
log |x – 1|
4
4
1
1
–
+c
2(x 1)
(x 1)2
(D) None of these
cos3 x cos5 x
dx is
sin 2 x sin 4 x
[IIT - 1995]
–1
(A) sin x – 6 tan (sin x) + c
–1
(B) sin x – 2 (sin x) + c
The value of the integral
2
p–q
xq
+c
xp
1
+c
(x – p)(x – q)
(D) None of these
–
3.
x–p
+c
x–q
(C)
1
1
log |x – 1| –
log |x + 1|
4
4
–
2
p–q
(B) –
[IIT - 1992]
1
1
–
+c
2(x 1) (x 1)2
dx
(x – p) (x – p)(x – q) is equal to [IIT - 1996]
(A)
1
1
log |x + 1| –
log |x – 1|
4
4
–
–1
(C) sin x – 2 (sin x) – 6 tan (sin x) + c
–1
–1
(D) sin x – 2 (sin x) 5 tan (sin x) + c
A. Fill in the blanks :
6.
If
dx
A
(sin x 4)(sin x – 1) = tan x – 1
2
–1
+ B tan f(x) + C then
[IIT - 1997]
4 tan x 1
2
1
, f(x) =
(A) A = , B = –
5
5
5 15
x
4 tan 1
2
1
2
(B) A = – , B = –
, f (x) =
5 15
5
15
2
2
4 tan x 1
(C) A = , B = –
, f(x) =
5
5
5 15
x
4 tan 1
2
2
2
(D) A = , B = –
, f(x) =
5
5 15
15
cos x sin x
cos x sin x (2 + 2 sin 2x) dx is equal to
[IIT - 1997]
1.165
INDEFINITE INTEGRATION
7.
(A) sin 2x + c
(C) tan 2x + c
dx
(B) cos 2x + c
(D) None of these
(2x 7) x 7x 12 is equal to[IIT - 1997]
2
(A) 2 sec (2x – 7) + c (B) sec–1 (2x – 7) + c
(C) 1/2 sec–1 (2x – 7) + 2 (D) None of these
x
cosec x log tan 2 dx is equal to [IIT - 1998]
x
(A) sin x log tan + c
2
x
(B) sin x log tan – x + c
2
x
(C) sin x log tan + x + c
2
(D) None of these
dx. Then, for an arbitrary constant C, the value of
J – I equals
[IIT - 2008]
(A)
e 4x e 2x 1
1
log 4x 2x
+C
2
e e 1
(B)
e 2x e x 1
1
log 2x x
+C
2
e e 1
(C)
e 2x e x 1
1
log 2x
+C
x
2
e e 1
(D)
e 4x e 2x 1
1
log 4x 2x
+C
2
e e 1
–1
8.
9.
x2 1
x 2x 2x 1 dx =
3
4
[IIT - 2006]
2
12. The integral
sec 2 x
Ú (sec x + tan x)
9/ 2
dx equals (for
some arbitrary constant K)
[IIT - 2012]
Ï1
1
2¸
Ï1
1
2¸
Ï1
1
¸
Ï1
1
2¸
1
(A) - (sec x + tan x)11/2 ÌÓ11 - 7 (sec x + tan x) ˝˛ + K
1
(A)
2x 4 2x 2 1
+c
x2
(B) (sec x + tan x)11/2 ÌÓ11 - 7 (sec x + tan x) ˝˛ + K
(B)
2x 4 2x 2 1
+c
x3
2
(C) - (sec x + tan x)11/2 ÌÓ11 + 7 (sec x + tan x) ˝˛ + K
(C)
2x 4 2x 2 1
+c
x
(D) (sec x + tan x)11/2 ÓÌ11 + 7 (sec x + tan x) ˛˝ + K
(D)
2x 4 2x 2 1
+c
2x 2
1
1
C. Subjective Problems :
13. Evaluate
x
for n 2 and
10. Let f(x) =
(1 x n )1 / n
f
f ...
f ) (x). Then x
g(x) = (
n 2
g(x) dx equals
f occur n times
[IIT - 2007]
1
1
1
(1 nx n ) n + K
(A)
n( n 1)
1
1
1
(1 nx n ) n + K
n 1
1
1
1
(1 nx n ) n + K
(C)
n( n 1)
(B)
1
(D)
1
1
(1 nx n ) n + K
n 1
11. Let I =
4x
2x
4x
ex
e 2x 1
[IIT - 1978]
x 2 dx
[IIT - 1979]
(a bx)2
15. Evaluate the following integrals : [IIT - 1980]
14. Evaluate
2
x dx
1
(A) 1 sin x dx (B)
2
1 x
16. Evaluate
(e
17. Evaluate
(x 1)e x
dx
(x 1)3
log x
+ sin x) cos x dx. [IIT - 1981]
18. Evaluate the following
ex
e e 1 dx, J = e
sin x
dx
sin x cos x
[IIT - 1983]
dx
x (x 1)
19. Evaluate the following
2
4
3/4 [IIT - 1984]
1 x
dx [IIT - 1985]
1 x
1.166
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
sin 1 x cos 1 x
dx [IIT - 1986]
sin 1 x cos 1 x
cos 2x)1/2
[IIT - 1987]
21. Evaluate
dx
sin x
20. Evaluate
22. Evaluate ( tan x cot x ) dx
23. Find the indefinite integral
1
ln(1 6 x
3 x 4 x 3 x x dx
24. Find the indefinite integral
cos sin
cos 2. In
d
cos – sin
(x 1)
25. Evaluate
dx
x(1 xe x )2
26. (A)
(B)
5x 4 4x 5
dx
(x 5 x 1)2
3x 1
dx.
(x –1)3 . (x 1)
x4
dx
(x – 1)3 (x 2 1)
(C)
[IIT - 1989]
[IIT - 1992]
[IIT - 1994]
30. For any natural number m, evaluate
(x + x + x ) (2x + 3x + 6) dx, x > 0.
m
1/m
32. Consider the statements:
[IIT - 2012]
P : There exists some x such that f(x) + 2x
2
= 2 (1 + x )
Q : There exists some x such that
2 f(x) + 1 = 2x (1+x)
Then:
(a) both P and Q are true
(b) P is true and Q is false
(c) P is true and Q is true
(d) both P and Q are false
33. Which of the following is true?
[IIT - 2012]
(a) g is increasing on (1, )
(b)
(c)
(d)
[IIT - 2002]
CONCEPT PROBLEMS—A
1.
(i) x – 2x2 + 3x3
2 5/2 2 3/2
x x 5x
(ii)
5
3
2
[IIT - 1996]
2m
2
x
x3 3x 2
dx
[IIT - 1999]
(x 2 1)2 (x 1)
2x 2
1
29. Evaluate sin
dx
4x 2 8x 13
[IIT - 2001]
m
2
2(t 1)
g(x) =
ln t f (t) dt for all x (1, )
t 1
1
2m
E. Comprehension
Let f(x) = (1 – x) sin x + x for all x , and let
1/2
3m
2
31. Let F(x) be an idenfinite integral of sin x.
Assertion (A): The function F(x) satisfiesf
F(x + ) = F(x) for all real x.
2
2
Reasons (R): sin (x + ) = sin x for all real x.
[IIT - 2007]
(A) Both A and R are true and R is the correct
explanation of A.
(B) Both A and R are true but R is not the correct
explanation of A.
(C) A is true, R is false.
(D) A is false, R is true.
[IIT - 1996]
1 x dx
27. Integrate the following :
x
1 x
[IIT - 1997]
28. Integrate
D. Assertion & Reasoning :
g is decreasing on (1, )
g is increasing on (1, 2) and decreasing on
(2, )
g is decreasing on (1, 2) and increasing on
(2, )
4
(x 1)5/4
5
1
4
(iv) (x / 2 7)
2
(iii)
INDEFINITE INTEGRATION
2.
(B) F(0) – G(0) =
8
3
(iii)
1 x
(e + 1)
2
3
(x + 2)4 – 12
(ii) 4t2 – 9t – 16
6. (i)
4
1
(iii) x3 + 5x – 1.
3
7. 12
2x . ex
+C
(ii) ex + e-x + c
8. (i)
1 n 2
a px q
e ax b
C
(iv)
a>0
(iii)
p.ln a
a
2 3/2
1
x C
(ii) – C
9. (i)
3
x
x2
x m 1
C
C, m 1
(iii) –
(iv)
2
m 1
a mx . b nx
C
10. (i)
m n a nnb
2 2x
32x
6x
2.
C
(ii)
2 n 2 2 n 3
n 6
e 4x
C
(iii)
(iv) x2 + C
4
(ax b)n 1
C
(ii)
11. (i)
x C
a(n 1)
n(3 2x)
C
(iii)
2
4.
(B)
(iv)
12. (i)
13. (i)
14. (i)
1
2
sec 1
5x
2
1
2 3x
tan 1
+C
3 3
3
PRACTICE PROBLEMS—A
16. (i)
1
25x
32 x
cos
sin
2 log | 1 2 x |
5
3
5
2
3x 1 + C
3
1
(ii) (7 – 4x)4 – log|3 – 7x| + cot(4x+3)
16
1
1 3x
+ tan
+C
6
4
19. No, we cannot. for instance, f(x) = 1 + cosx is a
periodic function and F(x) = (1 cos x)dx = x +
sinx + C is a non-periodic function.
21. y = 3 ln x + 1
22. (a) f(t) = 2 t – 1 if t > 0
(b) f(t) = t –
1 2 1
t + if 0 t 1
2
2
1 3 1
t + if |t| 1
3
3
(d) f(t) = t if t 0; f(t) = et – 1 if t > 0
(c) f(t) = t –
1
23.
x3
– x + tan–1x + C
3
(iii) – cotx – x + C
x
+C
2 3
Y
1
a
x n(a bx) C
b
b
180
sin xº + C
x
x3 3 2
x 3x n x C
3
x
(ii) 2x + 3 ln (x - 2) + C
(iii) x – tan–1x + C
(iv)
1 1 2x 1
sec
C
4
2
15. (i) 2cosec 1 2 + C (ii) 2sin–1
(iii)
1.167
–2 –1 0
1
2
3
4
5
X
–3
(ii)
1
tan(ax b) C
a
(iv) 2 tan
x
–x+C
2
C (ii) sec–1|x + 1| + C
24. Yes, one
25. f(x) = cx + d – 2/9 sin 3x, for constants c and d.
67
26.
5
27. 6x5 – 15x4 + 10x3 + 1
1.168
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
CONCEPT PROBLEMS—B
1.
2.
(i) ln x + 2 tan-1 x + C
1 x3
tan 1 x + C
(ii)
2 3
1 2
(iii) x – 2x + C
2
4 2x 1 3x
2 .2 + C
(iv)
ln 2
3
(i)
1 sin 3x
3sin x C
4 3
3
sin 2x sin 4x
x
C
8
4
32
(iii) tan x - cot x - 3x + c
(iv) a secx – bcosecx + C
(ii)
3.
(i)
1 sin 5x
sin x C
2 5
cos2 x cos 4x cos x.cos3x cos 2x
+C
6
12
24
1 cos2x cos 4x
C
(iii) –
2 2
8
(iv)
(ii)
1
1
1
1
cos9x cos10x cos11x cos12x+C
10
11
12
9
4.
5.
6.
x
x
x
2 cos sin C (ii)
+C
2
2
2
1
1
1
(iii) – cos 2x –
cos 4x +
cos 6x + C
8
16
24
1
cos 8x + C
(iv) 64
(i) tanx – cotx + C
3
1
(ii) – cotx – x – sin2x + C
2
4
3
1
(iii) tanx – x + sin 2x + C
2
4
3
1
(iv) – cotx – x – sin 2x + C
2
4
x
(i) tan + C
2
(ii) tan x - sec x + C
1
(iii) (tan 3x + sec 3x) + C
3
(i)
7.
8.
9.
(iv) (2sinx + x) + C
1
1
(i)
(tan x + x) + C
(ii)
sin 2x + C
2
2
(iv) - (cot x + tan x) + C
(iii) tan x - x + C
1
(i) [(– sin x– sin2x)] + C
2
cos 2x
C
(ii) x
4
(iii) tanx + C
(iv) – cotx + secx – cosx + C
cos3x
1
(i) –
+C
(ii)
(x - sin x) + C
3
2
1
x
(iv) – x cos 4x + C
(iii) 2 cos + C
2
PRACTICE PROBLEMS—B
1
2x
tan-1
+C
10
5
2
2
(ii)
(x + 1)3/2 + x3/2 + C
3
3
(iii) tan x - tan-1 x + C
9
3
(iv) secx – 3tanx + x – sin 2x + C
2
4
1
11. (i) ln x – 4 + C
4x
1
(ii)
[(2x + 3)]3/2 – (2x – 3)3/2] + C
18
x2
–x+C
(iii)
2
4
2
x x +C
(iv)
x 3
10. (i)
x3 x 2 3x 7
ln (2x + 1) + C
3
2
2
4
(ii) x2 + x + C
2
(iii) 2x1/2 - x3/2 + C
3
12. (i)
(iv)
2(1 x)3/2
C
3
x2
–xC
(ii) – x + C
2
(iii) x + 2 ln |x – 2| + C (iv) 6 2x C
14. (i) – 5 cosec x + 3sec x + C
3sin 4x
1
5x
+C
(ii)
4
8
13. (i)
INDEFINITE INTEGRATION
1
x 3
x
1
1
cos 3 – cos +
cos
4
2 4
2 28
28
1
5x
+
cos
+C
20
2
x– 6
1
tan
(iv) log
2 + C
4
2
15. (i) sec x - cosec x + C (ii) – 2 cos x + C
cos8x
(iii) sin 2x + C
(iv) –
+C
8
x
x
x
16. (i) 2 2 sin – 2 log sec tan + C
2
2
2
(ii) cosec a [log sinx – log sin (x + a)] + C
(iii) –cot 2a log [|cosec (x – a)| + log |sin (x + a)|] + C
(iii) –
CONCEPT PROBLEMS—C
2.
4.
They are both right.
(i) ln (ln x) + C
5.
(i)
(tan x )2 C
(iii) sin(sin x) + C
6.
2
(i)
1 x (3x2 + 4x + 8) + C
15
2 –1 3/2
sin (x ) + C
(ii)
3
(iii) x + log |x + 1| +
7.
1
+C
x 1
(27 e 3x )4 3
C
(iv)
4
(i) –ln(1 + e–x) + C
(iii) ex+1/x + C
8.
(ii) ln ln lnx + C
1
5
(ii) tan x
5
1
(iv) sin(x2+3x+2) + C
3
(ii) 2 ln(ex/2 + e–x/2) + C
1 log x2
(iv) x e C
4
(i) (3/2) 3 1 sin 2x + C(ii) sec ex + C
(iii) –1/3 cos 3x + C
(iv) log [log (tan x)] + C
3
1
(2x + 1)5/2 + (2x + 1)3/2 + C
10
6
1 3
(x + 3x + 6)2/3 + C
(ii)
2
(i)
2
+C
3(4 3 x )
5x
55
(iv)
+C
(log 5)3
1
10. (i) – log |1 – sin 2x| + C
2
(ii) – cot (1 + log x) + C
(iii) 2 (tan 1 x 3) + C
1
1
(iv) ln tan x + C
x
1
ln (a cos2x + b cos2 x) + C
11. (i)
2(b a)
(ii) ln|ln sin x| + C
(iii) ln|ln(secx + tanx)| + C
(iv) ln|a sin x + b cos x + c| + C
(tan x x)2
+C
2
1
1
+C
(ii) 2 2
2b a b2 tan 2 x
(iii) ln |ln sec x| + c
12. (i)
(iv) 2 sin(tan 1 x) C
13. (i) x + 2e–x – 1/2 e–2x + C
(ii) 2 tan1/2 x + 2/5 tan5/2 x + C
(iii) 1/2 (sin–1 x2)2 + C
(iv) 1 + cos5/2 x + C
14. (i) 1/6 (x + log x)3 + C
(ii) – (4x + 1) / [8(2x + 1)2] + C
(iii) –3 ln |2 + sin x| –
PRACTICE PROBLEMS—C
9.
(iii)
1.169
x
(2 sin x)2
+ 4(2 + sin x) + C
2
x
x e
(iv) – + C
e x
15. (i)
2 (x x 2 )3/2
+C
3
x3
1/4
1
(ii) 1 4
x
C
1
(iii) ln 1 1 x 2 C
(iv)
(x 4 x 2 1
+C
x
1.170
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1/5
1
16. (i) – 1 5
x
5.
+C
(ii)
PRACTICE PROBLEMS—D
(i)
1
1
cos5x – cos3x + C
2
3
(ii)
2 cos3/2
– 2 cos1/2 + C
5
(iii)
tan 2
ln | tan | C
2
(i)
(iv) ln
+ 2 tan–1 cos(x / 2) + C
1
1
cosec5 x + cosec3 x + C
5
3
3
cot 2 x
tan 4 x
+ 3log tan x + tan2 x +
+C
2
2
4
(ii) –
1
1
+C
3
3tan x tan x
tan 2
2 tan Q 1
+C
5
5
5
3/5
13/5
(iii) cos x cos x + C
3
13
6.
1/2
(ii) 2 tan x + C
2
4
(iv) – cosx + cos3x –
4.
2
2
4
7.
3
1
cos5x + cos7x + C
5
7
2
tan x (5 tan 2 x) C
5
1
2
3
5
(ii) C cot x cot x cot x
5
3
4 2
cot 5 x
(iii) C
5
2
(iv) cos x cos x C
3
2
1
(i) sec x – sec3x + sec5x + C
3
5
2
2
tan5/2 x + tan9/2 x + C
5
9
1
1
(iv) tan7 x + tan5x + C
7
5
(iii)
(i)
1 1
(ii) tan3x tan 2 x + C
3 5
(iii) tanx (1 +
2 2
1
tan x – tan4x) + C
3
5
1
3
(iv) – cot x cosec3 x– cot x cosec x
4
8
+
1 3
cot x + C.
3
5
1
3
1
x + sin x +
sin 2x –
sin3 x + C
(i)
16
2
32
24
1
1
sec33x – 3 sec 3x + C
(ii)
9
(iv) tan x – 2 cot x –
(iii) sin x dx 3 7 sin x 11 sin x sin 3 x + C
3.
4
1 cos(x / 2)
+
cos(x / 2)
1 cos(x / 2)
(i) –
(iv) –
2.
sin 3 x
– 2 sin x – cosec x + C
3
2
(cot x)3/2 + C
3
(iii) ln |tan x| + C
(ii) log |(1 + x2 + 1 x 4 )/x| + C
1.
(i)
3
x
ln|tan |+C
8
2
PRACTICE PROBLEMS—E
1.
2.
x3
+C
12(4 x 2 )3/2
x2
4 4x x 2
+C
4
3.
4.
5.
6.
1 –1 x
sin + C
4
a
2
2x a
+C
tan –1
a
a
1
x
x
tan 1
+C
54
3 18(9 x 2 )
y 2 49
y
sec 1 C
7
7
7
INDEFINITE INTEGRATION
x
7.
1 x2
C
2 –1
sin (x/a)3/2 + C
3
1 –1
sin (x/a)3 + C
9.
3
1
10. sec–1x2 + C
2
x3
+C
12(4 x 2 )3/2
12. 2 tan Q – sec Q + Q + c where sin Q = x
11.
13.
4 4x x 2
+C
1
x
x
tan 1
+C
14.
54
3 18(9 x 2 )
2
15.
( )
t 2 7 7
7
3
t n t t 2 C ,
2
3
6
3
where t = x – 1
1
a2
2
11.
(x + a) 2ax x 2 –
ln x a 2ax x + C
2
2
1
x cos
tan 1
+C
12.
sin
sin
1 2 sin x
C
13. ln
4 2 sin x
1
3 sin x
tan 1
14.
+ C.
3
3
10.
8.
x2
x
+C
x
15. ln cos x 2 cos2 x 4 cos x 1 + C.
16.
2.
3.
1 3x 2
ln
+C
12 3x 2
2x 1
1
ln
+C
2x 2
3
x4
ln
+C
x5
4.
2
2x 1
+C
tan–1
3
3
5.
1
2
ln x x x 1 + C
2
1 ln x 1
ln
+C
4 ln x 3
17. (a) 2sin–1 ( x / 2 ) + C; – 2 sin–1 ( x / 2 / 2 )
+ C ; sin–1 (x – 1) + C
PRACTICE PROBLEMS—G
PRACTICE PROBLEMS—F
1.
1.171
1.
2.
3
1
+C
x 1
2
ln|x + 3x – 10| + C
–2 ln|x – 2| + 3 ln|x – 3| + C
ln (x + 1) +
4.
1
1
2x 1
+C
ln |x2 + x + 3| +
tan–1
11
2
11
5.
3
1 x 1
ln |x2 + 2x – 3| + ln
+C
2
2 x3
x 1
+C
6. – 2 3 2x x 2 + 3 sin–1
2
7.
2 3x 2 5x 1 + C
8.
3 x 2 4x 3 n (x 2) x 2 4x 3 C
1 2
3
(x + x + 1) 3/2 – (2x + 1) 1 x x 2
3
8
6.
xa
+C
sin–1
a
7.
3
3
1
2
ln x 4 x 2 x 1 + C
2
9.
8.
1
41 2
4x 3
(4x 3) 4 3x 2x 2
sin 1
+C
8
32
41
9
ln (2x +1 + 2 x 2 x 1 ) + C
16
10. (x2 + 2x – 3)3/2 + (x + 1) (x2 + 3x – 2) – 4
9.
1
(5x 4) 5x 2 8x 4
10
2
5 5
ln{(5x 4) 5(5x 2 8x 4)} + C
–
2
ln x 1 x 2x 3 + C
11.
1
3
(2x2 + 2x + 1)3/2 + (2x + 1) 2x 2 2x 1
6
8
1.172
3
8 2
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
ln{(2x 1) 2(2x 2 2x 1)} + C
3.
1
x2 1
1
x 2 3x 1
n 2
tan 1
+C
2
x
4 3
x 3x 1
2 ln x
+ C.
5
4.
1
x2 1
1
x 2 3x 1
tan 1
ln 2
+C
2
x
4 3
x 3x 1
5.
1
t
sec –1
+ C,
2
2
6.
x
sin 1 2
+C
x 1
7.
1 1
3x
tan x 2
+ C.
54
x 9
8.
1
1
2
ln x x x x 2 3 + C
9.
1
x2 a 2
tan 1
+
2 2a
2ax
12. 1 4 ln x ln 2 x 2sin 1
1
x x 2 16 + 8ln |x +
2
13.
x 2 16 | + C
1
x
x 1
15
2
2
ln | 2x 3x 1| ln
+C
14.
2 8
8
x 1
1
x 1
(3 x) 1 2x x 2 2sin 1
C.
2
2
15.
2
2
16. x x 2x 5 5ln x 1 x 2x 5 C
x2
–2x + ln|x2 – 2x + 2| – 2tan–1(x – 1) + C
2
17.
1
15
1
2
(2x+5) x x 1 + ln{(x+ )
4
8
2
18.
+
1
x 2 2ax a 2
n 2
C
4 2a
x 2ax a 2
x2 x 1 } + C
1
(x 6) x 2 4 4 ln(x x 2 4) C
2
19.
10.
20. (x2 + 5x + 36) x 2 4x 7
11.
2
+ 112 ln x 2 x 4x 7 C
21.
–
1
(3x2 + 3x + 1) x 2 x
3
3 2x 1 2
1
x x ln(2x 1 2 x 2 x ) + C
2
4
8
x
– ln x x 2 2 + C.
2
Hint. Divide the numerator by the denominator
term by term.
22. sin–1
where t = x + 1/x
PRACTICE PROBLEMS—H
1.
1
x 3
1
x
ln
+C
1
6
x 3
x
2.
x 3
1 1 x x2
1
ln
tan 1
2
2 +C
4 1 x x
2 3
1 x
1
x 2 ax a 2
n 2
+C
2a
x ax a 2
y 2
1
y
1
+
tan–1
ln
+C
y 2
2 2 2
2
1
where y = tan x –
tan x
12. 2 ln
13.
x 2 ax 1 x 2 bx 1
C
x
1
1 x2 x4 x 3
ln
C
1 x2
3
PRACTICE PROBLEMS—I
1.
(i)
1
2 3
ln
1 3 tan x
+C
1 3 tan x
1
cot x
1
2
C
(ii) n
1
2
cot x
2
INDEFINITE INTEGRATION
(iii)
2.
(iv)
1 tan x 2
C
n
5 2 tan x 1
(i)
3 tan x
tan
C
2
2 3
(ii)
(iii)
x
2 tan 1
1
2
(iii) ln
+C
3 2 tan x 4
2
(iv)
1
1
1
2 3
1
2 3
(iv) –
3.
1
+C
3(3tan x 4)
ln
3 tan x
+C
3 tan x
ln
1 3 tan x
C
1 3 tan x
1
+C
1 tan x
6.
(i)
2 –1 1
1
tan
tan x C
5
5
2
(ii)
2 1 tan x / 2
tan
C
3
3
4.
(i)
1
x
tan tan 1 + C
2
1
tan 1
(ii)
3
4 tan
x
1
2
+C
3
1
x
(iii) tan 1 2 tan + C
2
1 x 1
1 3
(iv) 5 tan 2 2 cos 5 C
5.
(i)
(ii)
1
2 5
ln
5 tan 1
+C
5 tan 1
2 tan 1 1 tan ln(cos 2) + C
2
3
3
1
2 2
ln
1 2 sin x
+C
1 2 sin x
(i)
1 tan 1 tan –1 (tan 2 ) + C
2
2 2
(ii)
tan 2 x 1
1
–1
tan
+C
2
2 tan x
(iii)
1
1
tan x
+
tan–1
+C
2 tan x 2 2
2
(iv)
(a b) (a b)sin 2
–
+C
2(ab)3/2
4(ab)3/2
where tan
7.
1
tan(x / 2) – 2 – 3
(iii) 3 ln tan(x / 2) – 2 3 C.
5tan x
2
1
2
(iv) tan
+C
3
3
1.173
8.
b
tan
a
5
12
x+
ln | 3 cos x + 2 sin x | + C
13
13
3
2
x+
ln|2sin x + 3cos x| + C
(ii)
13
13
40
9
x+
ln |5sinx + 4cosx| + C
(iii)
41
41
(iv) 2x + ln|3 + 4 sin x + 5 cos x| + C
1
sin x cos x 4 / 3
(i)
ln
+C
24 sin x cos x 4 / 3
(i)
1
tan–1 (sin x + cos x) + C
2
(iii) sin 2x + c
x 1
x
1
(iv) ln tan – tan2 + C.
2
4
2
2
1
1
x + ln |cos x + sin x| + C
9. (i)
2
2
a
b
x– 2
(ii) 2
ln|b cos x + a sin x| + C
2
a b
a b2
(iii) (1/2) {sec x + ln |sec + tan x|} + C
tan x
10. (i) tan–1
2 tan 2 x
(ii)
+
tan 2 x tan x 2
1
ln
+C
tan 2 x tan x 2
2
(ii) 2 tan–1 sec x 1 + C
1.174
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(iii) –2 tan–1 cos e c x 1 + C
(iv) –ln |cot x +
(ii)
2 ln | 2 cos x
11. (i)
2
2 cos x 1 | + C
1
+C
2(1 2 tan x)
+
11. (i)
cot 2 x 1 | +
(iv)
CONCEPT PROBLEMS—D
4.
1 2
1
(x + 1) tan–1 x – + C
2
2
(i)
x ln x
– ln (x + 1) + C
x 1
(1 x)2
n(1 x) 2 C
4
x
x
(iv) x tan 2n sec C
4 2
4 2
(iii)
5.
(i) x sec–1 x – ln(x +
(ii)
x2 1 ) + C
1 2 –1
1
1
x sin x – sin–1x + x 1 x 2 + C
2
4
4
1 1
1
x
x(1 x) + C
(iii) x sin
2
2
(iv) 2x tan–1x – ln(1 + x2) + C
PRACTICE PROBLEMS—J
9.
(iii)
1 2 1 x
x ln
ln(1 x 2 ) x 2 + C
3
1 x
(iv)
12. (i)
1
x
1
ln x ln
C
x 1 x 1
2(x 1)2
2 x sin x cot 1 (ln 2)
1 (ln 2)2
C
3x sin 3sin 3x (ln 3)cos3x
(ii)
+C
9 (ln 3)2
x4
ln x 1
(ln x)2
C
4
2
8
(ii) (
ln(x 1 x 2 ) 2x + C
2
tan
1
5
sin
8
2 + C
.
tan
(iii)
9 5 4 cos 27
3
(iv) tan–1 (tan x – cot x) + C
2.
x n x x 2 a 2 x 2 a 2 C
2
2
2
(ii) x ln (x 1 x ) 2 1 x
tan x 2
tan x
tan–1
+C
–
2
4(tan
x 2)
8 2
2 2 tan x
3
1
ln sec x tan x 2 C
2
(i) tanx . ln cos x + tanx – x + C
x
(ii) – cosx . ln tanx + ln tan + C
2
(iii) 2{x ln(1 + x2) – 2x + 2 tan–1x} + C
(iv) xex ln(xex–1) + C
10. (i) – cot x ln(sec x) + x + C
(ii) sin x ln(cosec x + cot x) + x + C
(iii) –cosx ln (secx + tanx) + x + C
(iii)
1 x
1
e {(cos x + sin x}– (cos 3x + 3 sin 3x)}+ C
4
5
e 2x
(2 – sin 2x – cos 2x) + C
8
x
13. (i)
[sin(ln x) – cos(ln x)] + C
2
e x 2 sin 2 x cos 2 x 4 sin 4 x cos 4 x
+C
(ii)
2
5
17
x
x
(iii) x tan–1 – ax + a tan–1 + C
a
a
2
1 4
x
1
(iv) 4 (x 1) tan x 3 x + C
(iv)
14. (i) x sec–1x – ln(x +
x2 1 ) + C
x
1
ln(1 – x2) + C
2
(ii)
1 x
2
sin–1x +
1
1 x2
+C
(iii) 2 cos–1x +
2x
2x
1
(iv) x cos 1 x 1 x 2 + C
2
15. (i)
(ii)
x cos 3x
1
1
sin 3x sin x C
3
36
4
1 2
1
(x – x sin 2x – cos 2x) + C
4
2
INDEFINITE INTEGRATION
1
1
(3x sin 3x + cos 3x) + (x sinx+cosx) + C
18
2
16. (i) (x3 – 3x2 + 6x – 6)ex + C
(ii) x3 sin x + 3x2 cos x – 3.2. x sin x – 3.2.1.cosx + C
1
1
1
4
(iii) (n x) (n x) nx C
4
2
8
(iii)
7.
8.
PRACTICE PROBLEMS—K
1.
(i) – ex cosx + C
(iii) ex sec x + C
2.
(i) – ex cot
3.
(ii) ex tanx + C
x
+C
2
(iii)
1 2x
e cot 2 x + C
2
(i)
1
e .
C
1 x2
(iii)
ex
+C
(1 x ) 2
9.
(ii) ex tanx + C
(iv) e x
x
2
C
1 x
x 1
(ii) e 2
+C
x 1
x x 3
C
(iv) e
x 1
x
x
x
4.
5.
6.
e ( x 1)
x 1
(i) e x
(ii)
)+C
+C
x 1
x2
ex
x x2
(iv) e
+C
(iii)
2 +C
x2
1 x
(i) ex ln (secx + tanx) + C
11. ex
x 2 3x 2 3x
2 3x
e xe
e C
3
9
27
2.
e 3x 3
11
2
2
x x x C
3
3
9
3.
1
3
x(2x2 – 3)sin 2x + (2x2 – 1) cos 2x + C
4
8
4.
–
6.
2 3ln 1 1 C
x 2
1 xn
+C
1 xn
2.
3.
(i) In = xnex – nIn–1
(ii) In = x (ln x)n – nIn–1
tan 3
– tan Q + Q + C
3
1
1
(ii)
+ ln|sin Q| + C
4 tan 4 2 tan 2
cos
1
ln tan C
(iii)
2
2
2sin 2
(i)
(iv)
sin cos3 2
5
cos
6
4
5
(sin cos ) + C
16
1 + ln tan + C
(i)
cos
2
1.
ex
(cos2x – sin 2x + 2) + C
5
1 x3 x 2 2 x 13 3x
e + C.
3
9
3
3
3/2
PRACTICE PROBLEMS—M
1
(ii) e x log x C
x
(i) x tan x + C
(ii) x tan(ln x) + C
1
(iii) x sin x C
3
18x 2 6x 13
sin(6x 2)
72
6x 1
1
1
cos(6x 2) x3 x 2 x C
–
72
2
4
3 (x 2 7x 1)(2 x 1)2 / 3
4
9
27
–
(2x–7)(2x+1)5/3 +
(2x+1)8/3+ C
40
320
1
x 1
sin 2x ln (1 + tan x) – + ln (cos x + sin x) + C
2
2 2
1
10. 1/9 1 2
x
PRACTICE PROBLEMS—L
5.
1.175
2
x
x
x
2
3 x 3x nx 9 2 3x C
4.
sin cos sin 4 sin 2 1
(ii)
3 – 12 – 8 16 + C
2
5.
(iii)
1
cos
3
–
ln tan C
2
cos 2sin 2
2
(i)
1 1 1 x2
1 x2
ln
–
C
2
x
2x 2
(ii)
–x3
3
x
x a tan –1 + C
a
2(a 2 x 2 ) 2
1.176
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
x3
cx 2
–
(iii) 2
2
2 +C
a (a cx ) 3 5(a cx )
–(2a 2 3x 2 )
+C
3(a 2 x 2 )3/2
x
4n 5
14. In =
+
I ;
4(n 1)(1 x 4 )n 1 4(n 1) n–1
x
I2 =
4 (1 x 4 )
(iv) ln(1+x) +
5.
x 1
x 1 2
1
1
1
x
x
3
tan
n
+ C
+ 2 2
1
2
4
2
x 2
4
x
1.
2.
3.
(iii) x - tan–1 x + ln
2 + 1
(iii) ln
ln
2
1 sin
2
4.
(iv)
(iv)
1 2
x + ln
2
(i)
1
1
ln|x3 + 1| – ln|x2 – x + 1|
3
2
x2 2x 3
+C
6.
7.
1
ln | ln x 2 | C
14
1 cos x
1
1
cos x
+C
ln
–
tan–1
1 cos x 4 3
3
8
(i)
(ii)
1 2 sin x 1
1 sin x
1
ln
– ln
+C
1 sin x
1 2 sin x 2
2
(iii)
sec x
2
x
1
ln
–
– +C
tan x 2 tan x 2 5
3
(i)
x2 x
1
ln x 1
2
4(x 1)(x 1) 2
1
1
ln(x2 + 1) + tan–1 x + C.
4
4
3 1
x
3
x 1
ln
+ C.
(ii) 8 tan x
4(x 4 1) 16 x 1
1
2x 1
tan–1
+C
3
3
1
(ii) ln | x | – ln(x2 + 1) + C
2
+ 2 tan–1 2 (x+1) + C
1
1
ln | ln x | ln | ln x 5 |
10
35
ex 1
(iv) ln x
+C
e 2
+
1
1
1 x 1
(iii) –
4 (2x 2 4x 3) 4 x 1)2 1
2
2 sin 1
2
C
2 sin – 1
2
1 x2
+C
x
x
ln|1 – cosx| + ln|1 + cosx| + C
10
2
2
(ii) – ln|3 + 2 cosx| + C
5
(i) –
1 sin
PRACTICE PROBLEMS—N
1
1
1
1
x
C
(i)
4
x 2x 1 2x 1
(ii) 4 ln|x – 1| – 7 ln|x + 3| + 5 ln|x – 4| + C
4
1
1
(i) – x2 – 3x – ln(1 – x)6 –
+C
1 x 2(1 x)2
2
1
1
+C
(ii) –2 ln|x| + 2 ln|x – 2| –
x2 x
1
1
2
x
ln|x – 1| –
ln |x2 + 4| + tan–1 + C (ii)
(i)
5
10
5
2
x
2 tan 1
tan 1 x C
2
1
2 x1/3 1
tan–1
+ C.
3
3
+
(iv)
1
1
ln (1 + x1/3) – ln (x2/3 – x1/3 + 1)
3
6
(iii)
15x 5 40x 2 33x 15
tan 1 x + C.
48
48(1 x 2 )3
x 1
(iv) x x 2 2x 2 + 2 ln (x2 – 2x + 2)
8.
+ 3 tan–1 (x – 1) + C
13x 159
53 1 x 3
C
(i) 8( x 2 6x 13) 16 tan
2
(ii)
INDEFINITE INTEGRATION
x2 1
1
x
n ( x 2 2 )
tan 1
C
2
2
2
2
1
ln x 2 1 C
(iii) 2 2
x ( x 1)
(iv)
9.
2 x 3 1 1
1
ln 3 – 3 –
+C
3 x 3x 3 x 3 1
(i) ln x 2 3 + tan–1 x + C
3
2 x 1 1
1
ln 3 3
C
(ii)
3
3
x
3x 3(x 1)
3.
+
3
x
2x x
2x C
4
3
2
4
1 –1 x
tan + 2
+C
2 x 4
2
5
x 1
tan–1
(ii) ln x 2 2x 3 +
2
2
10. (i) ln (x2 + 4) +
4.
5.
(iii)
1
tan–1(x 2 ) + C
2
a4
x2
a 2 ln a 2 x 2 C
2(a 2 x 2 ) 2
1
(iv) –
– tan–1(x – 2) + C
x2
2.
(i)
1
{ln|x2 – 2| –ln(x2 + 1)} + C
6
2 x2 1 x4 x2 1
3
tan –1 2
– ln
+C
2
x4
x 3 4
(i)
2b
x
(a 2bx)
+C
ln
2
2
a
bx
a
a x(a bx)
x
– tan–1x + C
2
(ii)
2
(x 2)2
3
1
tan –1 x –
C
ln 2
25
25
5(x 2)
x 1
(i) –
1
1
3
4
4x (x 1) 4x
3
1 x(1 x 2 ) 1 1 x
C
– ln
8 (1 x 2 )2 16 1 x
2 3x 2
3
tan 1 x + C
2
2x(1 x ) 2
1
1
n(x 4 1) n(x8 x 4 1)
12
14
(i)
+
1
4 3
tan 1
2x 4 1
+C
3
1 2x 6 3x 2 3 x 2 1
ln 2
C
(ii)
4 x 4 1
2 x 1
1 x2
ln
tan –1 x + C
2 x2
(ii) tan–1x –
(iv)
(iv)
PRACTICE PROBLEMS—O
(i)
2 tan–1
(iii)
1 1 x 1 3(x 1)
18(x1)
2
C
(iii) 648 tan 3 2
x 2x10 (x 2x10)2
1.
(iii)
2
3 1
1
x 2 2x 1
1 x 1
tan
n
+C
+ 4 2 2
2x 4 2
x 2 2x 1
x
x
1
1 x
(ii) 216(x 2 9) 36(x 2 9)2 648 tan 3 C
x
+C
– 5 tan–1
5
1
1
x
2
(iv) ln(x2 + 2) –
tan–1
+C
2
(x
2)2
2
2
2
x2
I
1
2 ln
2
2 +C
2a(a bx ) 2a
a bx
(iii)
2
57x 4 103x 2 32 57 1
(iv) C
tan x
8
8x(x 2 1)2
(ii)
1
xn
1
C
ln
2
n
na
a bx
na(a bx n )
x
2
2x 1
tan 1
(iii) x 2 x 1
– 2 ln(x2 + x + 1)
3
3
4
1.177
PRACTICE PROBLEMS—P
1.
2x 1
ln
x
2x 1 1
2x 1 1
(i)
C
(ii)
2(a bx)3/2 2a(a bx)1/2
–
+C
3b2
b2
(iii) (x a)(x b) + (a – b)
ln( x a x b ) + C
1.178
2.
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(i) 4x + 6x1/6 + 24x1/12 + 2 ln |x1/12 – 1| + C
5
1
2t 1
(ii)
ln (t2 + t + 1) –
tan–1
3
2
3
1/4
+ ln
1/3
(t 2)4/3
x 1
1/3 + C where t =
(t 1)
x
9
2
(iii) – log (x2 + 4) + log (x4 + 9) + C
10
5
3.
(i)
(ii)
2
q
(px q)
+C
2
p
(px q)
(i)
(1 x 6 ) 1
1
2x 2
(iii)
4.
+C
1
2
(1 x 2 )5/2 (1 x 2 )3/2 (1 x 2 ) C
5
3
(iv) –
4.
1 x6 1
1
ln
6
(iii)
(ii) 2 tan–1
1
1 x2 1
+C
x 2 1 + 2 ln
|x|
(iii)
5.
2.
(i)
(ii)
1
ln
2
x 1 2
+C
x 1 2
6.
3.
(i)
(x 2 1)
+C
1
x 1
1
x (2x) 1
tan 1
|+ C
ln |
2
(2x) 2 2
x (2x) 1
1
ln
2 2
2 2x 2 x
2 2x 2 x
+ ln (x + x 2 1 ) + C
4 x2 2
1
2(x 1)
+C
sin 1
x
x4
8 3
1
ln
8
1 (4x 2 4x 5)
+C
8
2x 1
(i)
2 x 2 8x 2 12x 7
C
15 x 1
(x 1)2
ln
(iii) ln
x 2
1
sin –1
+C
2
1 x
(x 2 1)
1
1 x2
tan–1
+C
2
3
3x
x 2 4x 3
1
2 sin 1
C
x 1
(x 1)
x
3
(ii) 2 (x 1) 1 tan 1
2
2(x 1) 2 2
1 3x 1
(ii) sin
+C
(1 x) 5
(iv)
5x
+C
tan –1
12 9x 2
5 3
(iv)
(i) ln |x| – ln|1 + 2x| + (9x 2 4x 1) + C
(iii)
x 1 3 1
– tan–1 x 1 + C
x 1 3 2
x 2 2x 4 6 (x 1)
+C
ln
(iii) –
2 6 x 2 2x 4 6 (x 1)
PRACTICE PROBLEMS—Q
1.
ln
1
( x 1 1)2
2
2 x 1 1
+C
tan 1
x 2 x 1
3
3
4x 3 1
C
4x 3 1
(i)
–1
1
(ii)
(ii) 2( 4 5 x 1)2 4 ln(1 4 5 x ) + C
(iii) 2(x + 2)1/2 – 4(x + 2)1/4 + 4 ln{1 + (x + 2)1/4}+ C
1
ln
2
4 3
(ii) –
2 tan 1 x 1 + C
(iv) ln
(i)
1
x 1
+C
2
x 1 2 tan
–
(x 2) 2(x 1)
+C
(x 2) 2(x 1)
x 2 2x 4 1
x 2 2x 4 1
1
2(x 2 2x 4)
tan–1
+C
2
x 1
PRACTICE PROBLEMS—R
1.
2(1 x3 )1/2 (x3 2)
C
9
1.179
INDEFINITE INTEGRATION
2.
3.
4.
5.
2x1/2
+C
(1 x 2 )1/2
3.
3
x
2 3 x 3
C
3 ln
3
3
2
1 x 2(1 x )
13
1
(1 x 3 )8 3 (1 x3 )5 C
8
5
u 1
1
3
1 2u
C,
ln
tan 1
2
5
5
3
u u 1
6.
C
3
3
1 x
1
2 1 x x
tan 1
x
3
x 3
1
ln
3
3
3
3
1 x3 x
3
7.
1 1 x4 1 1 x4 1 1 x4
C
10 x 4 3 x 4 2
x4
8.
3 u13 3 3 u10 3 3 u 7 3 u 4
12
C,
10
7
4
13
where u = 1 + 4 x
32 5/4 144 13/4
72 21/4 108 29/4
9.
x +
x +7
x +
x +C
5
13
7
29
1 2
12
16
10. 3x4/3 – x1/2 x – x3/2 + C.
7
17
4 11
16
(1 + x7/8)3/2(3x7/8 – 2) + C.
11.
105
|t–2|
1
1 t
1
– ln
12.
2
3
3 t 8 36 (t 2t 4) 12 3
tan–1
t 1
+ C, where t = x–1 (1 + 8x3)1/3.
3
PRACTICE PROBLEMS—S
1.
2 n x 2 2x 4 x
3
2
x 2x 4 x 1
2
3
n x 2 2x 4 x 1 C
2
2.
1 1 x2
1 x
2 tan 1
C
x
1 x
C
(x 1 x 2 )15
C
15
5.
1
ln
2
6.
1 3
x (x 2 1)3 C
3
7.
x2 x 2
1
x 1 ln x x 2 1 C
2 2
3
8.
2
[(x + 2)3/2 – (x – 2)3/2] + C
3
(x 2 x 2) x 2
(x 2 x 2) x 2
+ C.
PRACTICE PROBLEMS—T
(1 x3 )2 x 3 1 x3 x 2
5
2x x 2
4.
where u 3 1 x 5
3
x 1
1.
2.
3.
P(x) = 2x + 2; Q(x) = 3x – 3.
3x 2 x 1
3x 2 2x 1 C
3
1 2
(x – 14x + 111) x 2 4x 3
3
– 66 ln 2x 1 2 x 2 4x 3 C
7.
8.
Hint: Putting y = tx, we obtain x = 1/{t2(1 – t)},
y = 1/{t(1 – t)}
Hint : Put x2 + y2 = t(x – y), from where we obtain
x = a2t(t2 + a2)/(t4 + a4), y = a2t(t2 – a2)/(t4 + a4).
CONCEPT PROBLEMS—E
1.
2.
(a) and (b)
(a) Not elementary (b) 2 sin q + C (c)
2 1 cos C
3.
4.
5.
(a) and (b)
1 x 3 1 x
(a), (d) and (f)
PRACTICE PROBLEMS—U
9.
(iii) Hint : let u = sin2 x
2
(vii) (b) n = 1 : (1 + x)3/2 – 2 (1 + x)1/2 + C;
3
1
2
n = 2 : 1 x C; n = 4 : ln (x2 + 1 x 4 ) + C
2
1.180
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(ix) (a) n is odd. (b) n = 3 :
1
2
1 x 4 + C;
1
x2
1 x 4 – ln (x2 + 1 x 4 ) + C
4
4
10. (ii) a = 0, – 1
n=5:
OBJECTIVE EXERCISES
1. D
2. C
3. B
4. C
5. A
6. D
7. B
8. C
9. A
10. C
11. A
12. B
13. D
14. C
15. B
16. B
17. B
18. A
19. A
20. A
21. C
22. C
23. A
24. B
25. A
26. B
27. B
28. C
29. C
30. B
31. A
32. C
33. B
34. C
35. A
36. D
37. C
38. C
39. A
40. C
41. B
42. B
43. A
44. D
45. B
46. D
47. A
48. A
49. A
50. B
51. B
52. B
53. A
54. A
55. A
56. D
57. C
58. D
59. B
60. B
61. D
62. B
63. A
64. D
65. B
66. AC
67. ABCD
68. AC
69. ACD
70. C
71. AC
72. AC
73. AD
74. ABC
75. AD
76. ABD
77. ABD
78. AC
79. BC
80. ACD
81. ABC
82. BCD
83. BC
84. AC
85. AC
86. C
87. A
88. A
89. B
90. A
91. D
92. A
93. B
94. C
95. C
96. D
97. A
98. A
99. B
100. A
101. C
102. C
103. B
104. C
105. A
106. D
107. A
108. D
109. B
110. A
111. (A) – (P,Q), (B) – (P), (C) – (RS), (D) – (PQ)
112. (A) – (Q), (B) – (P), (C) – (P,R), (D) – (PRS)
113. (A) – (Q), (B) – (P), (C) – (S), (D) – (R)
114. (A) – (R), (B) – (Q), (C) – (T), (D) – (P)
115. (A) – (Q), (B) – (Q), (C) – (Q), (D) – (Q)
REVIEW EXERCISES for JEE ADVANCED
1.
1/2 (sinq – lnq - q) + C
3.
6
(tan 1 2 x)2
– 3 ln|x – 2| +
+C
x2
4
4.
1
1 x2 x 4 x 3
ln
C
1 x2
3
5.
x
+C
sin –1
4 1/ 2
(1 x )
6.
(i)
1
1
8(1 x 2 )4 2(1 x 2 )
9.
1
3
C
2 3
2(1 x )
4(1 x 2 )2
1
1
4
ln |1 x |
+C
4
1 x4
(ii)
7.
2. xx(x lnx – 1) + C
tan –1
a 2 x2 a
a 2 x2
–1 b
C
tan
a x 2 b2
x 2 b2 b
1
(a b 2 )
2
ln
a 2 b 2 cos x a 2 cos 2 x b 2 sin 2 x
+ C, if a > b,
b2 a 2
sin 1
cos x + C, if a < b.
2
2
b
b a
11. ln |tan–1 (sec x + cos x)| + C
1
x3 7 2
(x – 14) + ln |x + x 2 7 | + C
3
1
(32x2 – 20x – 373) 2 x 2 5x 7 +
13.
64
3297
ln 4x 5 2 4x 2 10x 14 C
128 2
15. A = –b/(a2 – b2), B = a/(a2 – b2)
12.
16. (i)
3
5x3 15x 2 18x 8
C (ii)
tan–1(x + 1) –
8
8(x 2 2 x 2)2
15x 5 40x3 33x 15
tan 1 x C
2 3
48
48(1 x )
INDEFINITE INTEGRATION
(iii)
17. (i)
(iii)
2
x4 x2 1
3
1
1 2 x
tan
C
(iv) 2 tan 2
x4
x 3 4
x 1
3 7x 2 x 2
ln
C
3
2
( x 1)2
2(x x x 1)
(iv) ln
1.181
x3 x 2 x
3
2
2x 1
+C
–
tan–1
( x 1)2
3
x 1
3
24. (i)
x 2 2x 3
1
ln 4
+C
x 2x 3
4
e x ( x 2)
)+C
x3
(ii) xe tan x C
(ii)
x2 1
1
tan–1
+C
3
3
e x sec x
+C
2
(iv)
x
+C
1 lnx
(iii)
1
2a 3
1
1 sin x
1 sin x
2 tan–1
+C
sin x
2 sin x
(put u = 1/sinx)
1 x 2 a 2
1
x 2 3ax a 2
C
ln 2
tan
2
ax 2 3 x 3ax a
cos2 x a 2 sin 2 x
–1
(ii) tan
+C
a
(iv)
18. (i)
19. (i)
2
4
3x
11
x 1
log
+C
2
x 2(x 1) 4
x 1
20. (i) x + ln|3 cos x + 4 sin x + 5| – (1/5) tan(x – a)/2
+ C where a = tan–1 (4/3)
21. (i)
C–
–
2
a
C
ln | a b cos x |
(ii)
2
a
b
cos
x
b
cos x
b a sin x
xn
22. (i) –
xmn 1
C
1
(ii) x + n ;
x
ln|| + c = ln |xm+n + 1| – n ln|x| + C.
(ii)
1/8
25. (i)
8 x3
5 x2
+C
2
1
t 1 3
t
(ii) – ln 2
2 4 + C,
where t = 1/(x + 1)
2
2
(iii) 2 x 4x 8 n (x 2) x 4x 8
5
1
1
1
n
c
2
2
(x 2)
4
(x 2)
m
23. (i)
x 2 6x 2
ln 2
+C
x 6x 2
(ii)
1 1 x
1 x 2 1
.
.
tan 1 x C
2 (1 x 2 )2 4 x 2 1 4
x2 2
3
tan–1
–
2x
4 2
8 2
7
1
a
2
C
4c (a cx 2 )2 6c 2 (a cx 2 )3
1
xn
1
l
n
C
2
n
na
a bx
na(a bx n )
1 (1 x 2 )
1
1 x
ln
(iii) x
+C
8 (1 x 2 )2 16 1 x
(iv) 2y2 + 4y + 4ln (y – 1), where x + 1 = y4
26. (i)
ex
((3 – 5x) cos 2x + (4 + 10x) sin 2x
50
– 25(x + 1)) + C
1
ea tan x (ax 1)
C
(ii)
(1 a 2 )(1 x 2 )1/ 2
27. (i) –1/2 (1 + cot5x)2/5 + C
(ii) –2 tan–1 1 tan x cot x + C
28. (i)
x
2a bx 2
+C
2
2 1/2 + C (ii)
(1 x 4 )1/2
b (a bx )
29. (i)
x
2 1 sin x 2 ln tan + C
4 8
1.182
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(ii) ln (1 + t) – 1/4 ln (1 + t2) +
t 2 2t 1
ln 2
2 2
t 2t 1
– 1/2 tan–1 t2 + C where t =
cot x
2x 1
30. (i)
4x 2 2x 1
1
35. (i) ln
+C
(ii)
2
(ii) –
ln
2
2–x–x
+
4
x
2x 1
4 – x 2 2 2 – x – x2
+C
– sin–1
3
x
(iii) –2/5
(1 2x)5/2 4 (1 2x)3/2
–
+C
3
x 5/2
x3/2
1 x 2 2x
+ ln |sec t + tan t| + C
32. (i)
1
1
ln |x2n – 1| –
ln |x4r + x2r + 1| + C)
3r
6r
2.
1
1
1
1
1
ln x –
ln x x 5 x x 5
x 20
5
5 1 2x
+C
5 1 2x
5
3
1 x4
4 C
x
+ ln(1 + x2) – n 1 x
2
8.
+C
1
n 1
k 1 k(b a) n k (x a) k , n ³ 2.
1 2
1
(x + 2x + 3)3/2 – (x + 5) x 2 2 x 3
3
2
x2 2x 3 ) + C
2
6.
6 2 2x x 2
1
–
ln
+C
6 2 2x x 2
6
34. (i) x tan–1x . ln(1 + x2) + (tan–1 x)2 – 2x tan–1x
– 6 ln(x + 1 +
5.
x 1 2
2 2x x 2
tan–1
3 3
(1 x) 2
1
xb
log
xa
(b a)n
+
4.
x
1
(iv) sin 4
C
1 x4
2(x 1)
33. (i)
+C
3 x2 x 1
(ii) sin–1
3.
5
x 2 2x 2 + 2
1
1 x4
4 +3
x
3 tan x
+C
3 tan x
2
x 1 5 5
x 2
1
C
2
+
ln
1 5 5
4 5
x
x 2
ln (x + 1 + x 2 2x 2 ) + C
1
(iii) –
10
1
ln
3
1.
t
cos t
2(3 4x)
2
ln
2 +
5(1 x x ) 5 5
1 2 5x 1
(ii) x
6 6
3
–
TARGET EXERCISES for JEE ADVANCED
– ln |x + 1 x 2 | + C
31. (i) sin–1 (1/2 sec2 1/2 x) + C
(ii) 1/2 t cos t – 1/2 sin t –
x
4 cos 4 x
+ sec2x + C
1 cos2 2x
4
1 x 2 – 2x
1
(iv) –
ln
2
2
1
tan 1 x
1 x 1
–
–
ln
2 +C
2
6x
6 x
3x3
(ii)
9.
n
1 x
C – m n
1
2 x
x cot–1 (x2 – x + 1) + 1/2 ln (x2 – 2x + 2)
– 1/2 ln (x2 + 1) + C
x
+C
sin –1
2n 1 / 2n
(1 x )
1
4 2
ln
(1 x 4 )1/2 x 2
1 x2
4 1/ 2
+
2
10.
1 tan–1 (1 x )
x 2
4 2
+C
1
ln [25 tan2 (Q/2) + 4 tan (Q/2) + 13]
50
2
INDEFINITE INTEGRATION
–
11.
2
25tan( / 2) 2 C
tan–1
321
25 321
2
cos .tan sin
cos
2
cos cot .sin C
sin
12. 3a x +
2
x x 2 3a 2 3a
–
ln |x +
2
2
8cos3 1
7
(ii) 5 5
sec tan15 d
5sec 3 5 tan
2
14.
2
+C
(cos cos sin sin sin cos )
15.
sin sin x
1
sin 1 1 cos sin x C
sin
16. C –
17.
n
1 x
mn
1
2 x
2
x2 a 2
1
sin sin x C
1
sin sin 1 cos sin x
19. cosec–1(1 + sin2x) + C
18.
20. cos ec 1 2 cos2 + C
2
21.
22.
1
2 2
ln
x 2
1 x4 x 2
1
tan–1
+C
1 x2
2 2
1 x4
1
x 2(c a) 1 2cx 2 a 2 x 4
ln
C
1 ax 2
2(c a)
1
when c > a.
sin–1
2(a c)
x 2(a c)
,when a c .
2
1 ax
26
1
x 1 1
n
n(x 2 x 1)
6
x 1 6
+
1
1
2x 1
+C
n(x 2 x 1)
tan 1
3
3
3
x 1
x2
2
ln | x 1 | C
–
tan–1
2
2(x 2x 3) 4
2
28. exsinx + cosx · x
1
C
x cos x
29. If R = (x2 + ax)2 + bx, and u = ln
x 2 ax R
, find
x 2 ax R
the relation between the integrals
dx
xdx
R, R.
31. The form of the partial fraction corresponding to a
may be deduced from the facts that
2
1 a –1 1
a2
2 –
t t sin t +
2
2 1 t
2
1
1
ln t t 2 1 + C where t = x +
2
sin–1 (cos3/2 x) + C
3
27.
6
x2 1
C
2
24.
x 2 3a 2 | + C
64 cos 5 d
13. (i)
23. ln sec
1.183
Q(x)
P()
Q(a), (x – a) R(x)
x
Q()
34. Last part, a = 2 and 7x2 – 36x + 48 – p(x2 – 4x + 4)
must be a square : p = 3 and from this q = 4, b = 3.
Also r = 2, s = –1 from x2 – 2x – 1 r(x2 – 4x + 4) +
s(x2 – 6x + 9).
35. (i) a –1(u 2 – 1) –1/2 ; the integral reduces to
1
udu
2
a (u 1)1/2
(ii) The integral reduces 2
2y2 3
dy
1 3y 2 y 4
1
1
2
dy
1
1
2
2
y (3 5) y (3 5)
2
2
2
1
(1 5) = 1 (3 5) ; integrals
2
2
in inverse tangent form.
Moreover,
1.184
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
PREVIOUS YEAR'S QUESTIONS
(JEE ADVANCED)
3 35
, , any real value
2 36
2. A
3. C
4. B
5. D
6. A
7. B
8. D
9. D
10. A
11. C
12. C
1
x
13. log |sin x – cos x| + + C
2
2
1.
14.
(1 x1/6 )3 3
(1 x1/6 )2 3(1 x1/6 )
+ 6
3
2
25. – log
x
x
15. (a) 4 sin cos C
4
4
(1 x)5/2 2(1 x)3/2
1 x C
(b) 2
5
3
1
16. x sin x + cos x – cos 2x + C
4
1
1 4
18. 1 4 C
x
19. 2 1 x cos 1 x x 1 x C
2
2
1
x] x C
20. [ x x (1 2x)sin
21.
2 cot x cos2 x 1
2 log
cot x
– log (cot x+
cos x 1 ) + C
2
22.
tan x cot x
2 tan 1
+C
2
23.
2 2/3 12 7/12
x x 2x1/2
3
7
1 xe x
1
C
x
xe
1 xe x
x
3
3
x –1
+C
tan–1 x –
–
In
4
4x(x – 1) 16
x 1
8
1
1
1
x 1
log
–
–
(b)
2 +C
4
x – 1 2(x – 1) 2(x – 1)
1
x2
+ x + 2 ln (x – 1) –
(c)
4
2
1
In (x2 + 1) – tan–1 x + C
2
1
xe x
(d)
+C
x + log
1 xe
1 xe x
1 1 x
+C
27. 2cos–1 x – 2 log
x
26. (a)
1
a2
C
a bx 2a log | a bx |
3
a bx
b
ex
+C
17.
(x 1)2
12 5/12
x 3x1/3 6x1/6 12x1/12
5
sin 2 cos sin 1
ln
ln sec 2 C
24.
2
cos sin 2
+ 12 log |x1/2 + 1|
1
1
2
28. – log | x 1 | log(x 1)
2
4
3
x
tan 1 x
C
2
1 x2
–1 2x 2 3
2
29. (x + 1) tan
log(4x + 8x + 13) + C
3 4
m 1
1 (2x3m 3x 2m 6x m ) m
30.
+C
6
m 1
31. D
32. C
33. B
C H A P T E R
2
DEFINITE
INTEGRATION
2.1
INTRODUCTION
The definite integral is one of the basic concepts of
mathematical analysis and is a powerful research tool
in mathematics, physics, mechanics, and other
disciplines. Calculation of areas bounded by curves,
of arc length, volumes, work, velocity, path length,
moment of inertia, and so forth reduce to the
evaluation of a definite integral.
There are various problems leading to the notion of
the definite integral : determining the area of a plane
figure, computing the work of a variable force, finding
the distance travelled by a body with a given velocity,
and many other problems.
In the previous chapter we dealt with integration as
the inverse process of differentiation. The concept of
integration first arose in connection with determination
of areas of plane regions bounded by curves and an
integral was recognized as the limit of a certain sum. It
was only later that Newton and Leibnitz established
an intimate relationship between the processes of
integration and differentiation, known now as the
fundamental theorem of integral calculus which we
shall discuss in this chapter. A definite integral will be
defined as the limit of a sum and it will be shown how
a definite integral can be used to define the area of
some special region.
A definite integral may be described as an analytical
substitute for an area. In the usual elementary treatment
of the definite integral, defined as the limit of a sum, it
is assumed that the function of x considered may be
represented graphically, and the limit in question is
the area between the curve, the axis of x, and the two
bounding ordinates, say at x = a and x = b.
The Area Problem
Let us understand the problem of finding the area of
a curvilinear trapezoid. Consider a nonnegative
continuous function y = f(x), x [a, b].
Y
B
A
0
a
b X
The figure AabB bounded by a segment of the axis of
abscissas, segments of vertical lines x = a and
x = b, and the graph of the given function is called a
curvilinear trapezoid.
In other words, a curvilinear trapezoid is the set of
points in the plane whose coordinates x, y satisfy the
following conditions: a x b, 0 y f(x).
Let us find the area of this curvilinear trapezoid. To
this end we partition the closed interval [a, b] into n
subintervals of equal length
[a, x1], [x1, x2], ....., [xn–1, b],
using for this purpose the points
ba i
xi = a +
, i = 0, 1, ...., n.
n
2.2
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
We then denote by mi and Mi the least and greatest
values (respectively) of the function f(x) on the interval
[xi–1, xi], where i = 1,....., n.
Y
Example 1. Let us show that, according to the
given definition, the area of the right angled triangle
with the vertices at points (0, 0), (a, 0) and (a, b) is
equal to
formula.
B
Solution The given triangle is a curvilinear
trapezoid for the function
b
f(x) = x , x [0, a]
a
A
0
a
x1
x2
x3
b
1
ab, that is, it is computed by the known
2
X
Y
b
The curvilinear trapezoid AabB is thus separated into
n parts. Obviously, the area of the ith part is not less
than mi(xi – xi–1) and is not greater than Mi(xi – xi–1).
Consequently, the area of the entire curvilinear
trapezoid AabB is not less than the sum
n
mix1 + ....... + mnxn = m i x i
i 1
where xi = xi – xi –1 and is not greater than the sum
n
M1x1 + ....... + Mnxn = M i x i
i 1
Denoting respectively these sums by sn and Sn, we
see that SAabB satisfy the inequalities
sn SAabB Sn,
Here sn represents the area of the stepped figure
contained in the given curvilinear trapezoid, and Sn
the area of the steplike figure containing the given
curvilinear trapezoid.
If the interval [a, b] is divided into sufficiently small
subintervals, i.e. if n is sufficiently large, then the areas
of these figures differ but slightly from each other as
also from the area of the curvilinear trapezoid.
Consequently, we may assume that the sequences
(sn) and (Sn) have one and the same limit and this limit
is equal to the area of the figure AabB.
This assertion is obtained in the assumption that the
curvilinear trapezoid under consideration has an area,
but the latter notion is not yet defined. And so, the
above reasoning leads us to the following definition.
Definition Let there be given a continuous
nonnegative function f(x), x [a, b]. Then, if the limits
of the sequences (sn) and (Sn) exist and are equal to
each other, their common value is called the area of
the cuvilinear trapezoid.
{(x, y) | a x b, 0 y f(x)}.
0
a
X
ai
, i = 0, 1,.....,n, we divide the
n
a
interval [0, a] into n equal parts of length . Then
n
b
mi = f(xi –1) = (i 1)
n
b
Mi = f(xi) = i
n
and therefore
n
n (i 1) a ab . (n 1)n ab 1 1
sn =
n n2
2
2 n
i 1 b
Using the points xi =
n
n a ab (n 1)n ab 1 1
Sn = (i) 2 .
n n
2
2 n
i 1 b
Hence, it is obvious that
lim s n lim Sn ab
n
n
2
Thus, it has been proved that the area of the given
1
triangle is equal to ab .
2
Example 2. Find the area of the figure bounded
by a portion of the parabola y = x2 and segments of
the straight lines x = 0 and x = a, where a > 0.
Solution Proceeding as in previous example, we
partition the interval [0, a] into n subintervals each
a
of length , using points xi = a i , i = 0, 1,...., n.
n
n
DEFINITE INTEGRATION
2.3
n
Y
Hence, it follows that the limit lim f ( i )x i
n i 1
b
exists, does not depend on the choice of points xi and
is always equal to the area of the figure AabB.
n
Thus, lim f ( i )x i = SAabB
Then
a
0
n
a x 2 a 3 (i 1)2
i1 3
,
n i 1
n i 1
Sn =
a x 2 a 3 i2
i 3
n i 1
n i 1
n
We have i 2
i 1
2.
X
n
sn =
n
n
Sn =
n(n 1)(2 n 1)
6
a 3 . n(n 1)(2 n 1) a 3 1 1 1 1
6
3 n 2 n
n3
a 3 . (n 1)n(2 n 1) a 3 1 1 1 1
6
3 n 2 n
n3
and therefore
sn =
3
lim s n lim Sn a
n
n
3
Thus, the area of the given figure is equal to
Here we would like to note several points.
Note:
1 a3
.
3
Let us return to the curvilinear trapezoid AabB,
and proceeding in the usual manner, partition the
interval [a, b] into n parts of equal length, making
use of points xi, i = 0, 1, ...., n. On each subinterval
[xi–1, xi] we choose arbitrarily some point and
denote it by i .
If mi and Mi are respectively, the least and the
greatest values of the function f on th e
subinterval [xi -1, xi], then, obviously, mi f(i) Mi,
where i = 1, ...., n. Let us now multiply either of
these inequalities by xi = xi – xi–1 and add
termwise the inequlities thus obtained.
Then we get the following inequality :
n
n
n
m i x i f ( i )x i M i x i
i 1
i 1
In the previous examples we divided the interval
[a, b] into n equal subintervals. It can be proved
that formula (1) remains true also for the case
when [a, b] is separated into n parts of arbitrary
lengths but such that the greatest of these lengths
tends to zero as n .
The Definite Integral
Hence,
1.
...(1)
n i 1
i 1
Consider the function f(x) defined on the interval [a, b].
As before, we divide the interval [a, b] into n equal
subintervals by means of points
ba i
xi = a +
, i = 0, 1, ...., n.
n
On each of these subintervals [xi – 1, xi], i = 1, ...., n we
choose one point denoting it by i where i [xi–1 , xi]
Then the sum
n
f(1)x1 + .....+f(n)xn = f ( i )x i
i 1
where xi = xi – xi –1 , is called an integral sum of the
function f.
Obviously, this sum depends both on the manner the
interval [a, b] is subdivided and on the choice of points i.
n
Definition. If the limit lim f ( i )x i exists and is
n i 1
independent of the coice of points i , then the function
f is said to be integrable on the interval [a, b] and the
limit is called the definite integral or simply the integral
of the function f(x) with respect to x over the interval [a,
b
b] and is denoted as
f(x)dx
a
(read as " the integral of f(x)dx from a to b").
The symbol is called the integral sign, the function
f(x) the integrand, x the variable of integration, the
expression f(x)dx the element of integration. The
numbers a and b are called the lower and upper limits
of integration.
Thus, according to the definition,
b
a
n
f(x) dx lim
n
f( )x
i
i 1
i
2.4
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
The interval [a, b] is called the interval of integration.
The word 'limit' here has nothing to do with the word
limit as used in differential calculus. It only signifies
the 'end points' of the interval of integration.
We have a theorem that every continuous function is
integrable, but integrability extends to a class of
functions wider than the class of continuous functions.
1.
Note:
The above definition of a definite integral is a
special case of the more generalized definition as
given below.
Let f(x) be a bounded function defined in the
interval (a, b), and let the interval (a, b) be divided
in any manner into n sub-intervals (equal or
unequal) of lengths 1, 2, ......, n. In each subinterval choose a perfectly arbitrary point (which
may be within or at either end points of the
interval), and let these points be x = 1, 2, ..... n.
Let Sn
n
f ( ) .
r
r
r 1
2.
Now, let n increase indefinitely in such a way that
the greatest of the lengths 1, 2, ...... , n tends to
zero. If, in this case, Sn tends to a definite limit
which is independent of the way in which the
interval (a, b) is sub-divided and the intermediate
points 1, 2, ......, n are chosen, then this limit,
when it exists, is called the definite integral of f(x)
from a to b.
The process of forming the definite integral shows
b
that the symbol
f(x) dx is a certain number..
a
Its value only depends on the properties of the
integrand and on the numbers a and b
determining the interval of integration.
Geometrical interpretation of definite
integral
b
In general,
f(x) dx represents an algebraic sum of
a
areas of the region bounded by the curve y = f(x), the
x-axis and the ordinates x = a and x = b. Here algebraic
sum means that area which is above the x–axis will be
added in this sum with + sign and area which is below
the x–axis will be added in this sum with – sign.
So, value of the definite integral may be positive, zero
or negative.
This is because the value of f(x) in the integral sum is
considered without modulus sign. The area above the
x-axis enter into this sum with a positive sign, while
those below the x-axis enter it with a negative sign.
Y
0
A3
A1
a
A2
A4
b X
b
f(x) dx = A1 – A2 + A3 – A4
a
where A1, A2, A3, A4 are the areas of the shaded region.
b
Hence, the integral
f(x)dx represents the "net
a
signed area" of the region bounded by the curve
y = f(x), x–axis and the lines x = a, x = b.
4
Example 3.
Solution
Evaluate
(2x 3)dx .
1
y = 2x – 3 is a straight line, which lie
3
3
below the x–axis in 1, and above in , 4
2
2
Y
E (4, 5)
D
x=4 X
C
(3/2, 0)
A
x = –1
B
(–1, –5)
1 5
25
5 =
2 2
4
1 5
25
Area of CDE = 5 =
2 2
4
(using formula from geometry)
4
25 25
(2x 3)dx =
0.
So,
1
4
4
Example 4. Evaluate the following integrals by
interpreting each in terms of areas :
Now area of ABC =
1
1 x dx
(c) (x 1)dx
(a)
0
3
2
2
(b) (x 2)dx
1
0
Solution
(a) We sketch the region whose area is represented
by the definite integral, and evaluate the integral
using an appropriate formula from geometry.
2.5
DEFINITE INTEGRATION
Since f(x) = 1 x 2 0 , we can interpret this integral
as the area under the cure y = 1 x 2 from x = 0 to
x = 1. We have y2 = 1 – x2, that is x2 + y2 = 1, which
shows that the graph of f is the quarter-circle with
radius 1.
Y
2.2
DEFINITE INTEGRAL AS
A LIMIT OF SUM
Let us use rectangles to estimate the area under the
parabola y = x2 from 0 to 1.
Y
(1, 1)
y = 1- x 2
1
y = x2
S
1
0
Therefore,
1
0
X
1 x 2 dx = area of quarter-circle
1
(1)2 .
4
4
(b) The graphs of the integrand is the line y = x + 2,
so the region is a trapezoid whose base extends
from x = –1 to x = 2. Thus,
2
(x 2)dx = (area of trapezoid)
1
=
1
15
(1 + 4)(3) =
2
2
Y
X
We first notice that the area of S must be somewhere
between 0 and 1 because S is contained in a square
with side length 1, but we can certainly do better
than that. Suppose we divide S into four strips S1, S2,
S3, and S4 by drawing the vertical lines x =
and x =
y=x
S1
0
–1 0
1 2
X
(c) The graph of y = x – 1 is a line with slope 1 as
shown in the figure. We compute the integral as
the difference of the areas of the two triangles :
3
1
1
(x 1) dx = A1 – A2 = (2 . 2) – (1 . 1) = 1.5.
0
2
2
Y
(3, 2)
y=x–1
A1
–1
1
Y
2
1
S2
1
4
1
2
(1, 1)
(1, 1)
S4
S3
3
4
1 X
0
1
4
1
2
3
4
1 X
(a)
(b)
We can approximate each strip by a rectangle whose
base is the same as the strip and whose height is the
same as the right edge of the strip [see Figure (b)]. In
other words, the heights of these rectangle are the
values of the function f(x) = x2 at the right end points
of the subintervals
0, 1 , 1 , 1 , 1 , 3
3
4 4 2 2 4 and 4 ,1 .
1
Each rectangle has width and the heights are
4
2
3 X
1
1
,x=
4
2
3
as in Figure (a)
4
Y
4 y=x+2
3
2
0 A2
1
0
2
2
1 , 1 , 3
, and 12.
4 2 4
If we let R4 be the sum of the areas of these
2.6
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
approximating rectangles, we get R4
2
=
2
2
1 1 1 1 1 3 1 2 15
, . . .1
4 4 4 2 4 4 4
32
= 0.46875
From the Figure (b) we see that the area A of S is less
than R4, so A < 0.46875
Instead of using the rectangles in Figure (b) we could
use the smaller rectangles in Figure (c) whose heights
are the values of f at the left endpoints of the
subintervals. (The leftmost rectangle has collapsed
because its height is 0). The sum of the areas of these
approximating rectangles is
0.2734375 < A < 0.3984375
So, we can say that the true area of S lies somewhere
between 0.2734375 and 0.3984375.We could obtain
better estimates by increasing the number of strips.
For the region S in previous example, we now show
that the sum of the areas of the upper approximating
1
.
3
Rn is the sum of the areas of the n rectangles in Figure.
Each rectangle has width 1/n and the heights are the
values of the function f(x) = x2 at the points 1/n, 2/n, 3/n,
..., n/n; that is, the heights are (1/n)2, (2/n)2, (3/n)2, ...,
(n/n)2.
rectangles approaches
Y
Y
(1, 1)
(1, 1)
2
y=x
y=x
2
Z
0
1
4
1
2
(c)
3
4
2
1 X
0 1
n
2
2
1 1 1 1 1 3
7
1
L4 = . 02 + . . .
4 4 4 2 4 4
32
4
= 0.21875
We see that the area of S is larger than L4, so we have
lower and upper estimates for A :
0.21875 < A < 0.46875
We can repeat this procedure with a larger number of
strips. Figure (d), (e) shows what happens when we
divide the region S into eight strips of equal width.
Y
Y
(1,1)
(1,1)
0 1
1 X
8
(d) Using left endpoint
1
1 X
8
(e) Using right endpoint
y=x
2
By computing the sum of the areas of the smaller
rectangles (L8) and the sum of the areas of the larger
rectangles (R8), we obtain better lower and upper
estimates for A :
1
X
Thus, Rn
2
2
2
11 12 13
1n
= ...
nn nn nn
nn
=
1. 1 2 2 2
(1 + 2 + 3 + .... + n2)
n n2
=
1 2 2 2
(1 + 2 + 3 + ... + n2)
n3
2
(n +1)(2n +1)
6n 2
Thus, we have
1)
lim R = lim (n 1)(2n
n
n
n
6n 2
1 n 1 2n 1
= nlim
6 n
n
1 1
1 1
1
1 2 = . 1 . 2
= nlim
6
n
n 6
3
It can be shown that the lower approximating sums
also approach 1/3, that is,
=
1n 2n + .....
1
1 1
lim L n lim (0)2
n
n n
n n
2
2
2.7
DEFINITE INTEGRATION
2
1 n 1
+
n n
(n 1(2n 1) 1
.
= nlim
3
6n 2
From Figures it apears that, as n increases, both Ln
and Rn become better and better approximations to
the area of S. Therefore, we define the area A to be
the limit of the sums of the areas of the approximating
rectangles, that is,
1
A = nlim
Rn = nlim
Ln = .
3
If for a function f(x) the limit exists, then we say the
function is integrable on the interval [a, b].
It can be shown that, when f(x) is a continuous
function, the above limit always exists. Hence, if a
function f(x) is continuous on an interval [a, b], then it
is integrable on that interval.
Right end estimation
Considering the sum of areas of rectangles using the
heights at the right end points of the subintervals, we
have
Rn = hf(a + h) + hf (a + 2h) + .........+ hf(a + nh) and
Definite integral as a limit of sum
Let f(x) be a continuous real valued function defined
on the closed interval [a, b] which is divided into n
parts as shown in the figure.
b
a
n
f(x) dx = lim
n
ba
n f a b n a r
r 1
It follows from theorems that for a continuous function f,
b
f(x)dx .
lim R = lim L =
n
n
n
n
a
That is, we may compute the integral using either the
left end estimation or the right end estimation.
Note:
1.
If a = 0, b = 1, then
1
0
2.
The points of division on the x-axis are
a, a + h, a + 2h ..........a + (n – 1)h, a + nh,
ba
where
= h.
n
Left end estimation
Let Ln denotes the area of these n rectangles.
Then, Ln = hf(a) + hf(a + h) + hf(a + 2h) + ........
+ h f(a + (n – 1)h)
Clearly, Ln represents an area very close to the area of
the region bounded by curve y = f(x), x–axis and the
ordinates x = a, x = b.
b
f(x) dx = lim L
Hence,
n
a
n
b
f(x) dx = lim h [f (a) + f (a + h) + f (a + 2h)
a
n
+ ..... + f (a + (n – 1)h)]
n 1
= nlim
= lim
n
h f(a rh) , where nh = b – a.
r0
n 1
ba
n f a (b na) r .
r0
n 1
f(x) dx = lim
1
r
n f n
n r 0
From the definition of definite integral, we have
(x)
1
r
f
n n
n
r (x)
lim
b
f(x)dx , where
a
(i) is replaced by sign,
r
is replaced by x,
n
1
is replaced by dx,
(iii)
n
(iv) To obtain the limits of integration, we use
( x)
(x)
a = lim
and b = nlim
.
n
n n
(ii)
pn
For example, nlim
1
r
n f n = f(x) dx
p
0
r 1
1
since nlim
= 0,
n
pn
lim = p.
n
n
4
Example 1.
limit of sums.
Calculate I =
(1 x)dx as the
1
2.8
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Solution We divide the interval [–1, 4] into n equal
parts. On each subinterval
5( i 1)
, 1 5 i
[xi–1, xi] = 1
n
n
the continuous function 1+ x attains the least value at
the left endpoint of the interval and the greatest value
at the right end point.
Therefore
Y
y = x2
n
5(i 1) 5
.
Ln f 1
n n
i 1
n
–2 –1 0
5
5 25
= (i 1) 2 ( i 1)
n
n
n i 1
i 1
Since f(xi) = x i2 and xi =
n
n
n
5i 5
5 5 25
Rn f 1 . = .i. 2 i
n n i 1 n n n i 1
i 1
Hence
n
n
25 i (i 1)
Rn – Ln = 2
n i 1 i 1
X
n
4
(1 x)dx exists.
1
25
n i lim
2
i 1
n
4i 2 2
8i 2
8
3
3
n
n n n n i
2
i 1
3
i 1
2
i 1
2
8 2n 3n n 4
3 1
2 2 ,
3
6
3
n n
n
2
To calculate it as the limit of sums, we can consider
any of the sequence of sums.
Here, we use right end estimation.
I= 1 (1 x)dx nlim
2i
, it follows that
n
4
3 1 4
8
2 2 .2 .
and so nlim
Rn= nlim
3
n n 3
3
4
This means that the integral I =
n
Rn =
=
25
25
0 as n .
= 2n
n
n
2
1
n
n
25n(n 1) 25
2
2n 2
8
.
3
We conclude that 0 x dx
2
Evaluate
Example 3.
4
(5 – x) dx.
1
The function f, defined by f(x) = 5 – x, is
linear and decreasing on the interval [1, 4]. Its graph
is shown in the figure.
Solution
Y
25
4
(1 x)dx 2 .
Thus,
1
Example 2.
Evaluate
2
0
x2 dx.
y=5̶x
Solution The function f to be integrated is defined
by f(x) = x2, and the interval of integration is [0, 2].
2
0
x2 dx = nlim
Rn
The partition {x0,... , xn} which subdivides [0, 2] into
n subintervals of equal length is given by
ba
2
2i
xi = a +
i=0+ i ,
n
n
n
for each i = 0, ...., n.
ba
2
= , i = 1, ...., n.
Moreover, xi – xi – 1 =
n
n
P
1
y=1
4
X
The partition {x0, . . ., xn} subdivides the interval
4 1 3
, and the
[1, 4] into subintervals of length
n
n
points are given by
3
xi = 1 + i,
i = 0, . . . , n.
n
DEFINITE INTEGRATION
3
,
n
We shall compute the integral as
In addition, xi – xi – 1 =
i = 1, ......, n.
3
Evaluate (x 3 – 6x) dx using
Example 4.
0
limit of sum and interpret the result.
Solution
4
(5 x) dx = lim Ln.
n
1
2.9
Y
3i
We have xi = 1 + and f(xi) = 5 – xi, and so
n
3i
3i
f(xi) = 5 – 1 = 4 – .
n
n
5
3
y=x– 6x
0
A2
A1
3X
3
Since xi – xi – 1 = , we get
n
n
Ln =
3i 3
3
4 n n .
(x – 6x)
i 1
n
n
n
4 3i 3 12 9i
=
n n i1 n n 2
i 1
n
= nlim
12 n 9i
=
2
i 1 n
i 1 n
3
0
12 n
9 n
1 2
i
=
n i 1
n i 1
n
3i 3
f(x )x lim f n n
i
n
i 1
3
= nlim
n
n
3i
3
i 1
3i
n 6 n
i 1
n
3 27 3 18
i i
n i 1 n3
n
9 1
12
9 n(n 1)
n 2
=
= 12 – 1 .
2 n
n
2
n
But it is easy to see that
= nlim
9
1
9
1
lim 12 1 12 7 ,
2 n
2
2
and we finally conclude that
81 n(n 1) 2 54 n(n 1)
lim
= n 4 2 2
2
n
n
n
1
(5 x) dx = nlim
L =7 .
n
1
2
This answer can be checked by looking at the figure.
The value of the integral is equal to the area of the
shaded region P, which is divided by the horizontal
line y = 1 into two pieces : a right triangle sitting on
top of a rectangle.
4
1
9
(3 . 3) = , and that of
2
2
the rectangle is 3 . 1 = 3. Hence
The area of the triangle is
9
1
+ 3=7 .
1
2
2
The excessive lengths of the computations in the
above examples make it obvious that some powerful
technique is needed to streamline the process of
evaluating definite integrals.
4
(5 x) dx = area(P) =
81 n 3 54 n
i 2 i
= nlim
n 4
n i 1
i 1
81 1 2
1
lim
= n 4 1 n 27 1 n
81
27
=
– 27 = –
= – 6.75
4
4
This integral can be interpreted as net signed area
because f takes on both positive and negative values.
It is the difference of areas A1 – A2, where A1 and A2
are shown in the figure.
3
(x 6x)dx = A1 – A2 = – 6.75
3
0
Example 5.
Express the following limit as a
1
1
1
... .
definite integral : nlim
n 1 n 2
6n
Solution
1
1
1
...
lim
n2
6n
n n 1
2.10
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
1
1
...
= nlim
n 1 n 2
n 5n
n
= lim
n
1
1 5n 1
=
lim
lim
n n
n
r
r 1 n r
r 1 1
5n
n
Since the lower limit of r is 1, the lower limit of
1
0
n
integration = nlim
Since the upper limit of r is 5n, the upper limit of
5n
5.
n n
integration is lim
Hence, the given limit is equivalent to
Example 6.
1
1 x dx .
0
Use the definition of the integral as
the limit of a sum to evaluate
Solution
5
sin x dx
0
Here a = 0, b = , = nh, f(x) = sinx.
n
sin x dx lim ( / n) sin(r / n)
0
n
r 1
lim sin sin 2 ... sin n
n n
n
n
n
n 1 n
sin
. sin
n
2
n
2n
lim
n n
sin
2n
1
2( / 2n)
1
. sin 1 sin
lim
n sin( / 2n)
n 2
2
= 2. 1 sin sin = 2 sin2 = 1 – cos.
2
2
2
b
Example 7. Show that a dx ln (0 < a < b)
x
a
using limit of sum where the interval (a, b) is divided into
n parts by the points of division a, ar, ar2, ....... , arn–1, arn.
Solution We have arn = b, i.e. r = (b/a)1/n.
Evidently as n , r = (b/a)1/n 1, so that each of
the intervals a(r – 1), ar(r – 1) ..... 0. Now, by
definition,
n
b1
b1
x dx lim f (x )x
a
n
i
i 1
1
ar
k 1
k 1
(ark – ark–1)
lim (r 1) lim n(r 1)
n
n
lim n[(b / a)
1/ n
n
1]
b
(b / a)1/n 1
ln .
n
a
1/ n
lim
Evaluate the integral
Example 8.
2
dx
x.
1
Solution We subdivide the interval [1, 2] into n
parts so that the points of division xi (i = 0, 1, 2, ..., n)
form the geometric progression :
x0 = 1; x1 = q; x2 = q2; x3 = q3; ....; xn = qn = 2.
Here q = n 2 .
The length of the ith subinterval is equal to
xi = qi + 1 – qi = qi(q – 1),
and qn – 1(q – 1) 0 as n , i.e. as q 1.
Now let us choose the right hand endpoints of the
subintervals as the points xi + 1 = qi + 1.
Forming an integral sum :
n 1
n
1
Rn =
i 0 q
i
i 1 q (q – 1) = q (q – 1) =
1
1 n(2 n 1)
1
2n
1
lim R n lim
n
n
n(2 n 1)
1
2n
n 2
2
dx
n 2.
x
1
and so,
b
Example 9.
Evaluate the integral
dx
x (a < b).
a
2
Solution
1
1
1
1
....
Sn=h 2
2
2
2
(a h) (a 2h)
(a n 1h)
a
1
1
1
< h (a h)a a(a h) (a h)(a 2h)
....
(a n 2 h) (a n 1h)
1
2.11
DEFINITE INTEGRATION
1
1 1
1 1
1
= (a h) a a (a h) (a h) (a 2h)
1
1
....
(a n 2 h)
(a n 1h)
1
1
Thus Sn < a h a nh h
...(1)
We know that 2 r <
small h > 0.
1
> 2[ r h
rh r
h
r]
h
>2 a h a
a
h
2 a 2h a h
ah
h
2 a 3h a 2h
a 2h
h
2 a nh a (n 1) h
a (n 1) h
1
1
.......
h
a (a h ) (a h )(a 2h )
1
1 1
1
= a (a h) (a h) (a 2h)
1
1
....
(a nh)
(a n 1h)
1
1
Thus Sn > a a nh
...(2)
We have b = a + nh
From (1) and (2), using Sandwich Theorem, we get
b dx
1
1
1 1
lim
lim
2 = n Sn = h 0 a h
bh a
a x
b
Evaluate the integral
b> 0
Solution
r
1
=
rh r
>
r for sufficiently
Substituting r = a , a + h , a + 2h , ....... we have
Similarly, Sn >
Example 10.
2 r
h
rh +
b
dx
a
x
, a> 0,
By definition
1
1
1
1
Sn = h a a h a 2h ...... a (n 1)h
On addition,
1
1
h
....
a
ah
2 a nh a
a n 1h
1
lim S 2 b a
= 2 a ba
n
a =2 b
a
n
Similarly, by considering 2 r
can prove that lim S 2 b
n
Hence,
b
dx
a
x
n
r r h , we
a .
= nlim
S = 2
n
b a .
A
1.
2.
3.
4.
Compute the area of the figure bounded by a
portion of the straight line y = x and segments of
the straight lines y = 0 and x = 3.
Evaluate
0
2
5.
4
rule to show that finally that sin xdx 2 .
0
(iii) (2x 5)dx
(v) cos x dx
(i)
4 x 2 dx ,
Find the area of the curvilinear trapezoid defined
by the graph of the function y = ex on the interval
0 x 1.
n
k
lim
sin
sin xdx = n
First show that
n k 1
n
0
n
k
cot
sin
Use the fact
and L'Hospital's
n
2
n
k 1
Evaluate the following integrals as limit of sums:
1
2
(x 2 x)dx
0
/2
6.
0
3
3
(x 1)dx
(iv) (x e )dx
(vi) sin x dx
(ii)
0
1
2
2x
0
b
a
Let f(x) denote a linear function that is
nonnegative on the interval [a, b]. For each value
of x in [a, b], define A(x) to be the area between
the graph of f and the interval [a, x].
1
(a) Prove that A(x) = [f(a) + f(x)] (x – a).
2
(b) Use part (a) to verify that A(x) = f(x).
2.12
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
A
7.
Use appropriate formulas from geometry to
evaluate the integrals.
1
3
(x 2 1 x )dx (b) (4 5x)dx
(c) (1 3 | x |)dx
8. E va lua te th e i nt egr al 10x x dx by
(a)
2
1
0
2
2
10
2
0
9.
compl et i ng t h e squa re a n d a ppl yi n g
appropriate formulas from geometry.
Prove that the area of the curvilinear trapezoid
defined by a portion of the parabola y = 1 – x2
2
(0 x 1) is equal to .
3
10. Determine a region whose area is equal to the
n
i
limit lim tan . Do not evaluate the limit.
n i 1 4 n
4n
6
11. Express the integral
x
1 x dx as a limit of
5
2
integral sums. Do not evaluate the limit.
n
2
2i
5 n
n
n
i 1
2
12. Evaluate lim
13. Explain why
1 100 i
f
is an estimate of
100 i 1 100
1
f(x) dx .
0
2.3
RULES OF DEFINITE
INTEGRATION
We have a number of rules regarding the definite
integral
1.
b
a
14. Prove using the concept of definite integral as
limit of sum that
a n
ka
a n 1
ka
cos sin a
cos
,0<a<
n k 1
n
n k 0
n
2
15. Form the integral sum sn by dividing the interval
[a, b] into n parts by the points xi = aqi (i = 0, 1, 2,
b
and pass to the limit to
a
compute the following definite integrals:
.., n), where q = n
b
(i)
x dx
2
a
b
dx
(ii)
x , where 0 < a < b
(iii)
a
b
x dx.
a
16. Let A denote the area between the graph of
f(x) = x and the interval [0, 1], and let B
denote the area between the graph of f(x) = x2
and the interval [0, 1]. Explain geometrically
why A + B = 1.
17. Let A denote the area between the graph of
f(x) = 1/x and the interval [1, 2], and let B denote
the area between the graph of f and the interval
1 , 1
2 . Explain geometrically why A = B.
2. Order of integration
a
a
b
b
f(x) dx = – f(x) dx
b
When we defined the definite integral f(x) dx, we
f(x) dx .
If a = b, then x = 0 and so
a
a
f(x) dx = 0. This is
a
natural also from the geometric standpoint.
Indeed, the base of a curvilinear trapezoid has
length equal to zero. Consequently, its area is
zero too.
implicitly assumed that a < b. But the definition as a
limit of sum makes sense even if a > b.
Notice that if we reverse a and b, then x changes
from (b – a)/n to (a – b)/n.
a
b
b
a
Therefore, f(x) dx = – f(x) dx.
2.13
DEFINITE INTEGRATION
Example 1.
(a)
Evaluate the definite integrals :
sin x dx
(b)
Y
1
(5 x) dx
y=f(t)
4
Solution
(a) Because the sine function is defined at x = , and
the upper and lower limits of integration are
A
sin x dx = 0.
equal, we can write
a
b
t
4
(b) This integral is the same as (5 x) dx except
1
that the upper and lower limits are interchanged.
Because the integral has a value of 7.5, we can
write
1
4
4
1
(5 x) dx = – (5 x) dx = –7.5.
3. Dummy variable
b
The definite integral f(x) dx is a number which
a
depends only on the form of the function f(x) and the
limits of integration, and not on the variable of integration, which may be denoted by any letter. In fact,
we could use any letter in place of x without changing
the value of the integral
b
b
b
a
a
a
4. Homogeneous Property
b
b
a
a
cf(x) dx = c f(x) dx, where c is any
constant.
This property says that the integral of a constant
times a function is the constant times the integral of
the function. In other words, a constant (independent
of x) can be taken in front of an integral sign.
b
For example, 2f(x) dx = 2
a
Y
b
f (x) dx
a
y = f(x)
f(x) dx = f(t) f(t) dt = f(u) du.
Because the variable of integration in a definite
integral plays no role in the end result, it is often
referred to as a dummy variable.
Whenever you find it convenient to change the
letter used for the variable of integration in a
definite integral, you can do so without changing
the value of the interval.
This result should not be surprising, since the area
under the graph of the curve y = f(x) over an interval
[a, b] on the x-axis is the same as the area under the
graph of the curve y = f(t) over the interval [a, b]
on the t-axis (See figure).
a
b
X
Y
y = 2f(x)
a
b
X
5. Additivity Property
b
b
b
[f(x) + g(x)] dx = f(x)dx + g(x) dx
Y
a
y=f(x)
a
and
b
a
A
a
b
X
a
[f(x) – g(x)] dx =
b
a
f(x) dx –
b
a
g(x) dx
This property says that the integral of a sum is the
sum of the integrals. It says that the net signed area
under f + g is the area under f plus the area under g.
The figure helps us understand why this is true in
view of how graphical addition works.
2.14
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
= 5.
3
The next property tells us how to combine integrals of
the same function over adjacent intervals :
Y
=4+3.
f+g
7. Additivity with respect to the interval
of integration
f
g
a
0
bX
b
b
b
a
a
a
[f (x) + g(x)dx = f (x)dx + g(x)dx
In general, this property follows from the fact that
the limit of a sum is the sum of the limits :
b
[f(x) g(x)]dx lim
n
n
a
[f(x ) g(x )]x
i
i
i 1
n
n
f(x i )x g(x i )x
= nlim
i 1
i 1
n
= lim
n
n
i 1
b
=
f(x i )x lim
f(x) dx +
a
b
c
a
n
g(x )x
i
f(x) dx +
b
f(x) dx =
c
f(x) dx
This theorem reflects the additive property of area.
If an interval is decomposed into two intervals, the
sum of the areas of the two parts is equal to the
area of the whole.
Note that c may or maynot lie between a and b,
Property 2 allows us to conclude that this property
is valid not only when c is between a and b but for
any arrangement of the points a, b, c, but the
function must be integrable in the desired intervals.
It is easy to prove for the case where f(x) 0 and a < c <
b. This can be seen from the geometric interpretation
of the following figure
Y
i 1
y = f(x)
g(x) dx.
a
b
a
The above two properties can be combined into
one formula known as the linearity property.
6. Linearity Property
For every real c1 and c2, we have
b
b
b
a
a
a
[c1f(x) + c2g(x)] dx = c1 f(x) dx + c2 g(x) dx.
Example 2.
evaluate
1
0
Solution
Use the properties of integrals to
(4 + 3x2) dx.
Using linearity property of integrals, we
have
1
(4 + 3x2) dx =
0
1
0
=
1
4 dx + 3
0
1
0
We have
1
0
4 dx +
1
3x2 dx
0
x2 dx
Example 3.
0
x2 dx =
1
1
1
0
0
0
1
. So,
3
(4 + 3x2) dx = 4 dx + 3 x2 dx
b X
8
10
8
Solution
10
f(x)dx.
10
f(x) dx =
8
10
8
f(x)dx = 17 and
By the rule of additivity, we have
f(x) dx +
8
10
0
0
f(x) dx =
10
0
10
f(x) dx –
0
f(x) dx
8
f(x) dx
0
= 17 – 12 = 5
4 dx = 4(1 – 0) = 4 and we found in one
1
Given that
f(x)dx =12,find
So,
of the previous examples that
c
0 a
The area under y = f(x) from a to c plus the area from
c to b is equal to the total area from a to b.
Example 4.
If for every integer n,
4
dx = n2, then find the value of
Solution
We have
n 1
n
f(x)dx.
2
f(x) dx = n 2
n 1
n
f(x)
DEFINITE INTEGRATION
Putting n = 2, 1, 0, 1, 2, 3 we get
1
0
1
2
1
0
2
3
4
1
2
3
f(x) = 4 , f(x) dx = 1 , f(x) = 0,
f(x) dx = 1, f(x) dx = 4 , f(x) dx = 9
4
Hence,
f(x) dx = 4 + 1 + 0 + 1 + 4 + 9 = 39.
1
Comparision properties of integral
Next, we have a comparison theorem which tells us
that if one function has larger values than another
throughout [a, b], its integral over this interval is
also larger.
1.
2.
If f(x) 0 for a x b, then
Domination Law
b
a
f(x) dx 0.
If f(x) g(x) for for every x in [a, b], then
3.
4.
b
a
f(x) dx
b
a
g(x) dx.
If f(x) < g(x) for every x in [a, b], then
b
b
a
a
f(x)dx g(x)dx .
Max-Min Inequality
If m f(x) M for a x b, then
m (b – a)
b
a
If f(x) 0, then
b
a
f(x) dx M(b – a)
f(x) dx represents the area under
2.15
maximum values of f on the interval [a, b]. In this case
the inequality says that the area under the graph of f
is greater than the area of the rectangle with height m
and less than the area of the rectangle with height M.
Since m f(x) M, Property 2 gives
b
a
b
m dx
f(x) dx
a
b
M dx
a
Evaluating the integrals on the left and right sides,
we obtain
m(b – a)
b
a
f(x) dx M(b – a)
This inequality is useful when we want to find a rough
estimate of the value of an integral.
Example 5. Use Max-Min Inequality to estimate
1
e
x2
0
dx.
2
Because f(x) = e x is a decreasing
function on [0, 1], its absolute maximum value is
M = f(0) = 1 and its absolute minimum value is m = f(1)
= e–1. Thus by Max-Min Inequality,
Solution
1
e–1(1 – 0) e
0
or e–1
1
e
x2
0
x2
dx 1(1 – 0)
dx 1
Since e–1 0.3679, we can write
1
0.367 e
0
x2
dx 1
The result is illustrated in the figure.
Y
1
the graph of f, so the geometric interpretation of
Property 1 is simply that areas above the x-axis are
positive.
Property 2 says that a bigger function has a bigger
integral. It follows from Property 1 because g – f 0.
y=1
y = e –x
2
y = 1/e
Y
M
y = f(x)
m
b X
0 a
Max-Min Inequality is illustrated in the figure for
the case where f(x) 0. If f is continuous we could
take m and M to be the absolute minimum and
0
1 X
The integral is greater than the area of the lower
rectangle and less than the area of the square.
Example 6.
Prove that 4
3
1
3 x 3 dx 2 30
Solution Since the function f(x) = 3 x3 increases
on the interval [1, 3].
M = maximum value of 3 x3
= 3 33 = 30
2.16
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
From definite integral as limit of sum, we can evaluate
m = minimum value of 3 x3
= 3 13 = 2
1
m = 2, M = 30 , b – a = 2
Hence, 2.2
3
4
3
1
1
2
(1 + x2) dx to be equal to 6.
Placing this value in (1), we get f(c) = 2.
Therefore, 1 + c2 = 2 so c2 = 1.
Thus, in this case there happen to be two numbers
c = ± 1 in the interval [–1, 2] that work in the Mean
Value Theorem for Integrals.
3 x 3 dx 2 30
3 x 3 dx 2 30
Y y=1 + x2 (2, 5)
Mean Value Theorem for Integrals
If f is continuous on the interval [a, b], there is atleast
one number c between a and b such that
b
a
f(x) dx = f(c) (b – a)
Proof : Suppose M and m are the largest and smallest
values of f, respectively, on [a, b]. This means that
m f(x) M when a x b
b
a
m dx
b
a
m(b – a)
b
a
f(x) dx
b
M dx
a
[ Domination Law ]
f(x) dx M(b – a)
b
1
f(x) dx M
a
ba
Because f is continuous on the closed interval [a, b]
and because the number
m
b
1
f(x) dx
ba a
lies between m and M, the Intermediate Value Theorem
says there exists a number c between a and b for
which f(c) = I, that is,
I=
b
1
f(x) dx = f(c)
a
ba
b
a
1
favg= 2
–1 0
1 2
X
For a nonnegative function the Mean Value Theorem
for Integrals has a simple geometrical interpretation.
It asserts that the area of the curvilinear trapezoid
corresponding to the function f is equal to the area of
the rectangle whose base is equal to the base of the
trapezoid, and the altitude to one of the values of the
integrable function.
Note: The formula in the theorem holds true
not only for integrals in which the lower limit of
integration is less than the upper one, but also for
those in which the lower limit exceeds the upper one.
Example 7. Let f be the function defined by
1 1
1 x, when 0 x 2 , 2 x 1
f (x)
0, when x 1
2
1
3
Show that f (x)dx but there is no point c in [0, 1]
0
2
f(x) dx = f(c) (b – a).
1
The Mean Value Theorem for Integrals does not
specify how to determine c. It simply guarantees the
existence of atleast one number c in the interval.
Since f(x) = 1 + x2 is continuous on the interval
[– 1, 2], the Mean Value Theorem for Integrals says
there is a number c in [–1, 2] such that
2
(–1, 2)
such that f (x)dx f (c) .
0
Solution Here f is bounded and has only one point of
discontinuity, i.e. x =
1
.
2
1
To find f (x)dx , we consider the partition
0
(1 + x2) dx = f(c)[2 – (–1)]
In this particular case we can find c explicitly.
...(1)
1 2
n
0, , , ... , obtained by dissecting [0, 1]
n n
n
into n equal parts.
2.17
DEFINITE INTEGRATION
1
f (x)dx lim
n
n
0
n 1 n
1
r
r 1
lim 1 1 1 .... 1 n
n n
n
n
1 n(n 1)
lim n
n
n
2
3n 1 3
.
2n
2
But there is no point in [0, 1] at which f takes this
value. The only likely candidate for the point c is the
point x = 1/2 but f(1/2) = 0. This does not contradict
the Mean Value Theorem for Integrals as the given
function is discontinuous.
lim
n
Average value of a function
b
1
f(x) dx.
a
ba
1
= 6 = 2.
3
Another frequently seen averaging process is the
root mean square (r.m.s.) of f over [a, b], defined as
follows :
b
If f is integrable on the interval [a, b], the average
value of f on this interval is given by the integral
favg = =
From the Mean Value Theorem for Integrals we
conclude that this average value is definitely attained
by continuous functions at some c [a, b].
Hence, f(c) =
Example 8. Find the average value of the
function f(x) = 1 + x2 on the interval [–1, 2].
Solution With a = –1 and b = 2 we have
b
2
1
1
f(x)dx
(1 x 2 )dx
favg =
ba a
2 (1) 1
2
[f (x)] dx
fr.m.s =
a
ba
One can show that favg fr.m.s., the equality holding
only for constant functions.
B
1.
Given that
1
1
x dx 3 . Use this fact and the
2
properties of integrals to evaluate
2.
Y
0
1
4
(5 6x )dx
2
0
y = g(x)
Write as a single integral in the form
b
2
5
1
1
2
2
2
0
f(x) = f(x)dx f(x)dx f(x)dx
a
3.
7
X
1
2
2
1
1
If the function f is integrable in a closed interval
containing a, b, c and d, prove that
b
c
d
d
a
b
c
a
f(x)dx f(x)dx f(x)dx f(x)dx .
5.
6
Find [f(x) 2g(x)]dx if
f(x)dx 5 and g(x)dx 3
4.
4
2
2
The graph of g consists of two straight lines and
a semicircle. Use it to evaluate each integral.
(a)
(c)
2
g(x)dx
0
7
0
g(x)dx
(b)
6
g(x)dx
6.
Replace the symbol * by either or so that
the resulting expressions are correct. Give your
reasons.
1
1
(a)
x dx * x dx
(b)
x dx * x dx
(c)
x dx * x dx .
2
2
0
1
2
1
3
1
3
0
1
3
1
2
3
1
3
2.18
7.
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Find out which integral is greater :
1
1
2 dx or 2 dx ?
(ii) 2 dx or 2 dx?
(iii) ln x dx or (ln x) dx?
(iv) ln x dx or (ln x) dx?
8. (a) If 7 f(x) dx = 7, does f(x) dx = 1 ?
(b) If f(x) dx = 4 and f(x) 0, does
f(x)dx = 4 = 2 ?
(i)
0
2
x2
x3
0
2
x2
1
2
1
1
4
1
4
3
3
x3
2
2
2
1
1
0
1
0
0
1
9.
0
If f is continuous on [a, b], f(x) 0 on [a, b],
and f(x0) > 0 for some x0 in [a, b], prove that
b
f(x)dx 0 .
a
10. Assume that f is integrable and nonnegative
on [a, b]. If
b
f(x)dx = 0, prove that f(c) = 0 at
a
each point of continuity of f.
11. If favg really is a typical value of the integrable
function f(x) on [a, b], then the number favg
should have the same integral over [a, b] that f
does. Does it ? That is, does
b
b
f dx = f(x)dx ?
avg
a
a
12. It would be nice if average values of integrable
functions obeyed the following rules on an
interval [a, b] :
(a) (f + g)avg = favg + gavg
(b) (kf)avg = k favg , (any number k)
(c) favg gavg if f(x) g(x) on [a, b].
Do these rules ever hold ? Give reasons for
your answers.
B
Y
9
9
0
0
f(x)dx = 37 and g(x)dx = 16, find
13. If
9
[2f(x) 3g(x)]dx
0
14. Suppose
5
2
2
2
f(x)dx = 4,
5
2
1
f(x)dx = 3,
(b)
(c)
2 3
f(x)dx = – 3
5
16. Evaluate the integral
5
(f(x) g(x))dx = 9
2
2
f(x)dx
0
7
f(x)dx
5
4 5
6 7 8 9 X
‒2
‒
‒3
2
(c) f(x) g(x) on the interval –2 x 5
15. The graph of f is shown. Evaluate each integral
by interpreting it in terms of areas.
(a)
0
g(x)dx = 2. Which, if any, of the following
statements are true ?
(a)
y = f(x)
3
2
(b)
(d)
5
f(x)dx
0
9
f(x)dx
0
0
(1 9 x )dx by
2
3
interpreting it in terms of areas.
if 3 x 0
x 1
17. Let f(x) =
Evaluate
2
1 x if 0 x 1
1
f(x)dx by interpreting the integral as a
3
difference of areas.
18. Determine whether the value of the integral is
positive or negative.
1 x 4
4
x3
(i)
dx
(ii) 2
dx
3 3 x
| x | 1
2.19
DEFINITE INTEGRATION
19. Draw the graph of the function
f(x) = x(x – 2) (x – 4) = x3 – 6x2 + 8x, and indicate the
region P+ defined by the inequalities 0 x 3 and
0 y f(x), and the region P– defined by 0 x 3
and f(x) y 0. Let P = P+ P–, and suppose that
2
3
1
f(x) dx 4 and f(x) dx 2 . Find area
0
0
4
(P+), area (P–), and area (P).
20. Use the properties of integrals to verify the
inequality without evaluating the integrals.
2
2
24. Find the average value of
x 4, 4 x 1
f(x) =
on [–4, 2], using
x 2, 1 x 2
graph of f (without integrating).
25. Suppose that f and g are continuous on [a, b],
b
a b, and that (f(x) – g(x)) dx = 0. Show that
a
f(x) = g(x) atleast once in [a, b].
26. The inequality sec x 1 + (x2/2) holds on
(–/2, /2). Use it to find a lower bound for the
1
x3 2dx
value of sec x dx .
0
27. Let f be a function that is differentiable on
[a, b]. In the chapter of derivatives, we defined
the average rate of change of f over [a, b] to be
f ( b) f (a )
and the instantaneous rate of
ba
change of f at x to be f(x). Here we defined the
average value of a function. For the new
definition of average to be consistent with the
f ( b ) f (a )
old one, we should have
= average
ba
value of f on [a, b] Is this the case ?
28. Is it true that the average value of an integrable
function over an interval of length 2 is half the
function's integral over the interval ?
FIRST FUNDAMENTAL
THEOREM OF CALCULUS
where f is a function defined on [a, b] and x varies
between a and b. Observe that g depends only on x,
which appears as the variable upper limit in the integral.
The Fundamental Theorem of Calculus is appropriately
named because it establishes a connection between
the two branches of calculus : differential calculus and
integral calculus. Differential calculus arose from the
tangent problem, whereas integral calculus arose from
a seemingly unrelated problem, the area problem. The
Fundamental Theorem of Calculus gives the precise
inverse relationship between the derivative and the
integral. It was Newton
and Leibnitz who exploited this relationship and used it
to develop calculus into a systematic mathematical
method. In particular, they saw that the Fundamental
Theorem enabled them to compute areas and integrals
very easily without having to compute them as limits of
sums as we did before.
If x is a fixed number, then the integral f(t) dt is a
a
definite number.
1
5 x dx
1
x 1dx
21. Show that the value of
1
sin(x ) dx cannot
2
0
possibly be 2.
22. Given that, when x > 0, the function f(x) is positive
and is strictly decreasing, prove that
f(n + 1)
n 1
f(x)dx f(n)
n
23. Find the maximum and minimum values of
x 3 2 for 0 x 3, and use these values to
find bounds on the value of the integrals
3
0
2.4
x
If we then let x vary, the number
x
f(t) dt also varies
a
and defines a function of x denoted by g(x). If
happens to be a positive function then g(x) can be
interpreted as the area under the graph of f from a to
x, where x can vary from a to b. Think of g as the “area
so far” function (See figure).
Y
y = f(t)
Area function
The First Fundamental Theorem deals with functions
defined by an equation of the form
g(x) =
x
a
f(t) dt
...(1)
area = g(x)
0
a
x
b t
2.20
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
The number g(x) for the given x is represented in the
figure by the area of the figure ABxa. g(x) changes as
x varies on [a,b].
Theorem. If a function f is integrable on a closed
x
a
f ( t)dt = algebraic sum of areas
f(x)
a
x
Let us discuss the question of notation. the
independent variable in the upper limit is usually
denoted by the same letter (say x) as the variable of
integration. thus, we write
g(x)
x
f(x)dx
a
however the letter x in the element of integration only
serves to designate the auxiliary variable (the variable
of integration) which runs over the values ranging
from the lower limit a to the upper limit x as the integral
is formed. If it is necessary to evaluate a particular
value of the function g(x), for instance, for x = b, i.e.
g(b), we substitute b for x in the upper limit of the
integral but do not replace by b the variable of
integration. Therefore, it is more convenient to write
x
g(x) =
f(t)dt
a
denoting the variable of integration by some other
letter (in this case by t).
However, for simplicity, we shall often denote by the
same letter both the variable of integration and the
independent variable in the upper limit bearing in mind
that in the upper limit and under the integral sign they
have different meanings.
Now consider the graph of a bounded piecewise
continuous function f with a point of discontinuity c.
Let us take an arbitrary value x [a, b]. We shall be
again interested in the definite integral of f on [a, x].
Let us denote it by g(x).
Y
B
A
Hence, g(x) =
x
f(t)dt
a
x x+h b
X
a
continuous at any point x [a, b].
Proof : Let us take an arbitrary point x and assign an
increment h to it (shown in figure is a positive h).
We have
|g(x + h) – g(x)| =
=
xh
x
xh
x
f(t) dt
f(t) dt
a
a
f(t) dt M | h |
(M | f(t) |, t [a, b]).
We have obtained the inequality
| g(x + h) – g(x) | M | h |,
it follows that
lim [g (x + h) – g(x)] = 0,
h 0
i.e. g is continuous at the point x.
It should be underlined that x may turn out to be either
a point of continuity, or a point of discontinuity of f,
but all the same, the function g(x) is continuous at this
point.
Example 1. If f is the function whose graph is
shown in the figure and g(x) =
x
a
f(t) dt, find the
values of g(0), g(1), g(2), g(3), g(4) and g(5). Then
sketch a rough graph of g.
f(t)
2
1
0
–1
1
2
3
4
t
5
–2
0
First we notice that g(0) = f(t) dt = 0.
Solution
0 a
x
f(t)dt is
interval [a, b], then the function g(x) =
0
From Figure we see that g(1) is the area of a triangle :
1
1
g(1) =
f(t) dt = (1 . 2) = 1
0
2
To find g(2) we add to g(1) the area of a rectangle :
DEFINITE INTEGRATION
g(2) =
2
0
1
2
0
1
f(t)dt +
f(t) dt =
f(t) dt
=
= 1 + (1 . 2) = 3
1
g(3) = g(2) +
f(t) dt = 3 + (1 . 2) = 4
2
2
For t > 3, f(t) is negative and so we start subtracting
areas :
4
1
g(4) = g(3) +
f(t) = 4 – (2 × 1) = 3
3
2
5
1
g(5) = g(4) +
f(t) dt = 3 – × 2 × 1 = 2
4
2
We use these values to sketch the graph of g.
3
x h
a
f(t)dt
x h
a
f(t)dt = x
x
h
g
1
0
1
2
3
g (x ) =
4
X
5
mh
x
0
f ( t )d x
x2
g(x) =
t dt =
0
2
Notice that g(x) = x, that is , g = f. In other words, if g
is defined as the integral of f, then g turns out to be
an antiderivative of f, atleast in this case.
And if we sketch the derivative of the function g
shown in the figure by estimating slopes of tangents,
we get a graph like that of f.
x
First Fundamental Theorem of Calculus
If f is continuous on [a, b], then the function g defined
by g(x) =
a
f(t) dt a x b
...(1)
is continuous on [a, b] and differentiable on (a, b),
and g(x) = f(x).
Proof : If x and x + h are in (a, b), then
g(x + h) – g(x) =
xh
x
xh
f(t) dt Mh
x
Notice that, because f(t) is positive for t < 3, we keep
adding area for t < 3 and so g is increasing up to x = 3,
where it attains a maximum value.
For x > 3, g decreases because f(t) is negative.
If we take f(t) = t and a = 0, then, we have
x
x
f(t) dt – f(t) dt
a
f(x)
b t
x x+h
For now let us assume that h > 0. Since f is continuous
in [x, x + h], the Extreme Value Theorem says that
there are numbers u and v in [x, x + h] such that
f(u) = m and f(v) = M, where m and M are the absolute
minimum and maximum values of f on [x, x+ h]. By
Max-Min Inequality, we have
2
a
0
3
...(2)
Y
4
f(t) dt
and so, for h 0,
g(x h) g(x) 1 x h
f(t) dt
h
h x
Y
2.21
that is, f(u)h
xh
x
f(t) dt f(v) h
Since h >0, we can divide this inequality by h:
1 x h
f(t) dt f(v)
h x
Now we use (2) to replace the middle part of this
inequality :
f(u)
g(x h) g(x)
f(v)
...(3)
h
Inequality 3 can be proved in similar manner for the
case h < 0.
Now we let h 0. Then u x and v x, since u and
v lie between x and x + h.
Therefore
f(u)
lim f(u) = lim f(u) = f(x) and
h0
ux
lim f(v) = lim f(u) = f(x)
h0
v x
because f is continuous at x. We conclude, from (3)
and the Sandwich Theorem, that
g(x h) g(x)
= f(x)
g(x) = lim
h 0
h
...(4)
2.22
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
If x = a or b, then equation 4 can be interpreted as a
one-sided limit.
Using Leibnitz notation for derivatives, we can write
the First Fundamental Theorem (FTC1) as
d x
f(t) dt = f(x)
...(5)
dx a
when f is continuous. Roughly speaking, equation 5
says that if we first integrate f and then differentiate
the result, we get back to the original function f.
This theorem is called the theorem on differentiating
the definite integral with respect to its upper limit.
For a continuous function the derivative of the
integral with respect to its upper limit is equal to the
function itself.
The antiderivative of a continuous function
It follows from the First Fundamental Theorem that
any continuous function has an antiderivative
(primitive) which is definite integral with variable upper
limit of the given function.
Theorem The function f(x) continuous on the
closed interval [a, b] has an antiderivative on this
interval. One of the antiderivatives is a function
F(x) =
x
f(t)dt
...(1)
a
For example,
x
F(x) =
e dt is an antiderivative of f(x) = e ,
t2
x2
0
since f(x) is continuous and FTC1 ensures that F(x) = f(x).
Note: An integral with a variable upper limit
is defined for any function f(x) integrable on [a, b].
However, for the function F(x) of form (1) to be an
antiderivative for f(x), it is essential that the function
f(x) be continuous. Thus, out of the definite integral we
have constructed a function, which we call an indefinite
integral of the integrand. Loosely speaking, an indefinite
integral is the definite integral with a varying upper end
point. The theorem is stated generally for indefinite
integrals based at an arbitrary point x0 in [a, b], namely
x
f(t) dt . Of course, for a given x the two indefinite
integrals f(t) dt and f(t) dt just differ by a
constant, namely ± f(t) dt .
0
x0
x
x
x0
a
x0
a
Example 2.
g(x) =
x
Find the derivative of the function
1 t 2 dt .
0
Since f(t) = 1 t 2 is continuous, the
First Fundamental Theorem of Calculus gives
g(x) = 1 x 2
Solution
Example 3.
t
If F(t) = 0
1
dx, find F (1),
x 1
2
F (2), and F(x).
Solution The integrand in this example is a
1
.
x2 1
continuous function f defined by f(x) =
By the First Fundamental Theorem,
F(x) = f(x) =
1
.
x2 1
1
1
,
12 1 2
1
1
.
F(2) = 2
2 1 5
In particular, F(1) =
d x4
sec t dt.
dx 1
Solution Here we neet to use the Chain Rule in
conjunction with FTC1.
Let u = x4. Then
d u
d x4
sec t dt =
sec t dt
1
dx 1
dx
d u
du
sec t dt
=
(by the Chain Rule)
du 1
dx
du
(by FTC1)
= sec u
dx
= sec (x4) . 4x3.
dy
of the
Example 5. Find the derivative
dx
Example 4.
Find
implicit function
x
/2
3 2 sin 2 zdz
y
cos tdt 0 .
0
Solution Differentiate the left side of the equation
with respect to x,
d y
dy
d x
0
cos tdt
3 2 sin 2 zdz +
0
dy
dx
/
2
dx
2.23
DEFINITE INTEGRATION
3 2 sin 2 x cos y
dy
3 2 sin 2 x
.
dx
cos y
Example 6.
f(x) =
x cos(t 2 )dt
= cos(x2).
0
Therefore, applying L'Hospital's rule, we obtain
dy
0
dx
x
cos(t) dt lim cos(x ) 1
lim
Let
{(a 1)(t t 1) (a 1)(t t 1)}dt
x0
x
1
Note that an antiderivative of cos(x2) is not an
2
2
4
2
0
f(x) = (a – 1) (x2 + x + 1)2
– (a + 1) (x2 + x + 1) (x2 – x + 1).
Now, f(x) = 0
(a – 1) (x2 + x + 1)
– (a + 1) (x2 – x + 1) = 0
x2 – ax + 1 = 0.
For distinct real roots D > 0 i.e. a2 – 4 > 0
a2 > 4 a (–, – 2) (2, )
Example 7.
If
1
t2 f(t) dt = 1 – sin x, where x
sin x
2
0, , then find the value of f 1 .
3
We have
Solution
sin x
t2 f(t) dt = 1 – sin x
1
Differentiating both sides, we get
– sin2x f(sin x) cos x = – cos x
f(sin x) = cosec2x =
1 , x 0,
2
sin 2 x
z
1
f = 3.
3
x
Example 8.
cos(t )dt
Find lim
2
0
x
Solution The limit is in indeterminate form 0/0.
The integral with a variable upper limit
x0
cos(t )dt has derivative
2
x
elementary function, i.e.
cos(t) dt cannot be
2
0
expressed in terms of elementary functions. This
however, has not prevented us from calculating the
required limit.
x x sin t dt
.
Example 9. Evaluate lim
x3 x 3 3
t
x x sin t
lim
dt
x3 x 3 3
t
Solution
= 3 lim
x sin t
dt
F(x) F(3)
t
= 3 lim
x
3
x3
x 3
x3
[ applying L'Hospital's Rule and FTC1]
3
sin 3
= sin 3.
3
Example 10. Find
= 3F(3) = 3
x
(ln(t 1 t ) ln(1 t)dt
lim
2
0
x 1
Solution The limit is in (/) form. Therefore,
applying L'Hospital's Rule and FTC1, we get
x
f(z) = 12 , z (0, 1).
0
2
x0
x
Find the value of ‘a’ for which f (x) = 0 has two distinct
real roots.
Solution Differentiating the given equation, we get
x
2
0
lim ln x 1 x 2 ln(1 x)
x
1
x 1 x2
1 2 1
lim ln
x
ln
ln 2 .
1 x xlim
x
1 1
x
Example 11. Find the critical points of the
function f(x) if
(i) f(x) = 1 + x +
x
1
(ln2 z + 2 ln z) dz
x 1
(ii) f(x) = x – ln x + 1 2 2 cos 4t dt
t
2.24
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(x 1)( x 1)( x 2)( x 2).2 x
2 ex
– + – + – +
–2 –1 0
1 2
From the sign scheme of f(x), it is clear that f(x) has
points of maxima at x = –1, 1(as sign changes from
+ve to –ve) and points of minima at –2, 0, 2 (as sign
changes from –ve to + ve) at x = 1.
Solution
(i) f(x) = 1 + x +
=
x
(ln2 z + 2 ln z) dz
2
f(x) = 0 + 1 + ln2 x + 2 ln x
For critical points f(x) = 0
1 + ln2 x + 2 ln x = 0
(ln x + 1)2— = 0 ln x = –1
1
.
e
x = e–1 =
Example 14.
x 1
(ii) f(x) = x – ln x + 1 2 2 cos 4t dt
t
1 1
f(x) = 1 – – 2 – 2 cos 4x – 0
x x
= – 1 – 2 cos 4x
For critical points f(x) = 0
cos 4x = –
Hence, g'(0) =
( x > 0)
(t 1) (e – 1) (t–2) (t + 4)
t
0
dt, then find the points of local minima of f(x).
Here f (x) = (x + 1) (e –1) (x – 2)(x+ 4)
x
Sign scheme for f(x) :
+
–
–4
–1
+
–
+
0
2
Clearly x = –1 and x = 2 are the points of
local minima.
Example 13.
of
x
0
2
Find the points of maxima/minima
2
t 5t 4
dt.
2 et
Let f(x) =
Solution
f(x) =
4
x 2 t 2 5t 4
0
2 et
x 5x 4
2e
x2
dt
that
x
0
.2x – 0
when y = f (x)
x 4 3x 2 13
= 121 = 11.
If f(x) is a continuous function such
f(t) dt as x , show that every line
y = mx intersects the curve y2 +
Solution
x
f(t) dt = awhere a R+.
0
We have to show that there exists some x
such that m2x2 +
x
f(t) dt = a (a R+)
0
...(1)
Consider the function
g(x) = m2x2 +
x
0
f(t) dt
Since f is a continuous function, therefore g is also a
continuous function. Also, g(0) = 0 and g(x) as
x . Thus, by intermediate value theorem, there
must be some x (0, ), such that g(x) = a (a R+).
Hence, for every real m, there exists some a (a R+)
that satisfies equation (1).
Example 16. Consider the function
f(x) = cos x –
2
x 3x 2 13
34 27 13
Example 15.
x
Solution
4
If g (x) is
When y = 0 then x = 3.
n
x = , , n N
6 2 6
If f(x) =
dt
4
3
1
dy
=
dx
1
g'(y) = dy dx =
1
2
= cos
2
3
x
t 3t 2 13
the inverse of f(x) then find g'(0).
Solution
2
n
4x = 2n ±
or x =
3
2 6
Example 12.
Let f(x) =
x
(x – t) f(t) dt.
0
Show that f(x) + f(x) = – cos x.
2.25
DEFINITE INTEGRATION
We have f(x) = cos x –
Solution
x
= cos x – x
f(t) dt +
0
x
x
(x – t) f(t) dt
0
t f(t) dt
0
= – sin x –
x
x
f(t) dt + xf(x)
k
x
k
x
= f(x) – p f ( t ) (sin px cos pt – sin pt cos px) dt
k
k
x
2
dx
x
k
+ sin2 px f(x) + p cos px f ( t ) sin pt dt
f(t)sin p(x t)dt is a solution of the
2
x
On differentiating again,
d2y
x
2
2 = cos px . f(x) – p sin px f ( t ) cos pt dt
dx
k
x
differential equation
= cos px . f(t) cos pt dt + sin px f ( t ) sin pt dt
Example 17. Show [assuming (f(t) to be
continuous for all values of t considered in the problem]
that, when p and k are constants,
d y
Differentiating again w.r.t. x, we have
f(x) = – cos x – f(x)
i.e. f(x) + f(x) = – cos x. Hence proved.
1
y=
p
0
f(t) dt
0
1 x
f(t) (sin px cos pt – cos px sin pt) dt
p k
x
x
1
sin px f(t)cos pt dt cos px f(t)sin pt dt
k
k
p
x
1
dy
= sin px . f(x) . cos px + cos px f(t)cos pt dt
p
k
dx
x
1
– cos px . f(x) sin px + sin px f(t)sin pt dt
p
k
=
Differentiating w.r.t. x, we have
f(x) = – sin x – xf(x) –
Solution y =
= f(x) – p f ( t ) sin p(x – t) dt = f(x) – p2 y..
+ p2y = f(x).
k
C
1.
find F" (2).
d g(x)
f(t) dt = f[g(x)]g(x).
dx a
Prove that
x
2.
Suppose f(t)dt = x2 – 2x + 1. Find f(x).
3.
Let f be continuous on [a, b] and
1
x
a
f(t) dt = 0
for all x [a, b], then prove that f(x) = 0 for all x [a, b].
4.
Find the following derivatives :
d b
d
sin(x 2 )dx
(i)
(ii)
da
dx a
d
(iii)
dx
(v)
d
dx
x2
1 x 2 dx
0
x3
dx
x2
1 x
2
d
(iv)
dx
(vi)
5.
Find the derivative of y =
6.
If F(x) =
d
dx
x2
sin(x )dx
2
a
x3
dt
x2
1 t2
x3
dt
t2
x t4
2
u3
du .
2
1 3X 1 u
1
x
t2
1
1
f(t)dt where f(t) =
1 u4
du
u
7.
8.
13 t 3 sin 2t
d 2y
dt .
2 if y =
x
dx
1 3t
Let f ' be continuous in [a, b]. State under what
Find
d x f ( t )dt x d f (t )dt
a
.
dt
dx a
2
9. Find an antiderivative F of f(x) = x sin(x2) such
that F(1) = 0.
10. Let f(t) be a function that is continuous and
conditions
satisfies f(t) 0 on the interval 0, 2 . Suppose
it is known that for any number x between 0 and
, the region under the graph of f on [0, x] has
2
area A(x) = tan x.
2
(b) Differentiate both sides of the equation in
part (a) and deduce the formula of f.
x
(a) Explain why 0 f (t) dt tan x for 0 x
2.26
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
11. (a) Over what open interval does the formula
x dt
F(x) =
represent an antiderivative of
1 t
f(x) = 1/x ?
(b) Find a point where the graph of F crosses
the x-axis.
12. (a) Over what open interval does the formula
x
1
dt represents an
F(x) = 1 2
t 9
1
antiderivative of f(x) = 2
?
x 9
(b) Find a point where the graph of F crosses
the x-axis.
13. Suppose that f has a positive derivative for all
values of x and that f(1) = 0. Which of the
following statements must be true of the function
g(x) =
x
f(t)dt ?
(a) g is a differentiable function f x.
(b) g is a continuous function of x.
(c) The graph of g has a horizontal tangent at
x = 1.
(d) g has a local maximum at x = 1.
(e) g has a local minimum at x = 1.
(f) The graph of g has an inflection point
at x = 1.
(g) The graph of dg/dx crosses the x-axis at
x= 1
x
14. Let F(x) = 0
t 3
dt for –x
t2 7
(a) Find the value of x where F attains its minimum
value.
(b) Find intervals over which F is over increasing
or only decreasing.
0
C
15. Find
d2
dx 2
sin t
g(x)
1
0
16. If f(x) =
g(x) =
x
0
cos x
0
1
1 u 4 du dt
1 t3
[1 sin(t 2 )]dt , find f '(/2).
x
t
1 t dt.
2
9 t dt has inverse for
2
4
x 0. Find the value of (f –1)'(0).
19. Suppose x and y are related by the equation
y
1
dt.
x= 0
1 4t 2
d2y
is proportional to y and find
Show that
dx 2
the constant of proportionality.
20. Let f be a function such that f(x) > 0.
Assume that f has derivatives of all orders and
that lnf(x) = f(x)
(i) f(0),
(iii) f(2)(0).
x
f(t)dt. Find
0
(ii) f (0),
(1)
f(t) sin {K(x – t)} dt, then prove that
x2
(a)
f(t)dt = x cos x,
(b)
4
x2
18. The function f (x) =
x
0
d2y
+ K2y = K f(x).
dx 2
22. Find f(4) if
dt , where
17. Find f(2) if f(x) = eg(x) and g(x) =
21. If y =
0
f (x)
0
t 2 dt = x cos x.
23. If f is a continuous function such that
x
x
f(t)dt xe e f(t)dt
2x
0
t
0
for all x, find an explicit formula for f(x).
24. If x sinx =
x2
f(t) , where f is a continuous
0
function, find f(4).
25. Find the critical points of the function
1
x 1
2 3 x
1
2
x
cos
2t
t
(i) f(x)=
dt
3
2 1 2 2
x
(ii) f(x) =
(sin 2t 2 cos 2t a)dt
0
2
2
DEFINITE INTEGRATION
26. Let g(x) =
x
0
Y
f(t)dt , where f is the function whose
2.27
b
c
graph is shown.
Y
4
X
a
2
f
1
0
1 23
29. Find f(/2) from the following information.
(i) f is positive and continuous.
(ii) The area under the curve y = f(x) from
5 6 7t
‒2
a2 a
sin a + cos a.
2
2 2
30. Find the interval [a, b] for which the value of the
x = 0 to x = a is
(a) Evaluate g(0), g(1), g(2), g(3), and g(6).
(b) On what interval is g increasing ?
(c) Where does g have a maximum
value ?
(d) Sketch a rough graph of g.
b
integral
x
27. Let g(x) =
f(t)dt where f is the function whose
0
graph is shown.
(a) At what values of x do the local maximum
and minimum values of g occur ?
(b) Where does g attain its absolute maximum
value?
(c) On what intervals is g concave down?
(d) Sketch the graph of g.
Y
f
3
2
1
0
2
4
6
t
8
–1
–2
(2 x x )dx is a maximum.
2
a
31. If f is a differentiable function such that
x
f(t)dt = [f(x)] for all x, find f.
2
0
xe
32. Evaluate xlim
x2
x
e dt .
t2
0
1/ t
1
t
lim
33. If 0 < a < b, find t 0 [ bx a(1 x)] dx .
0
34. Let f be a function such that f (x) is continuous,
f(x) 0, f(0) = 0, f(0) = 0, and f(0) > 0. The graph
of f is shown below. Find the limit as
x 0+ of the quotient
Area under curve and above [0, x]
Area of triangle OAP
Y
P=(x, f(x))
x
28. The figure shows the graphs of f, f', and
f(t)dt .
0
Identify each graph and explain your choices.
2.5
SECOND FUNDAMENTAL
THEOREM OF
CALCULUS
The examples in the section of definite integral as limit
of sum show that the direct evaluation of definite
integral as limit of sum involves great difficulties. Even
when the integrands are very simple (kx, x2, ex), this
O
A(x, 0) X
method involves cumbersome computations. The
finding of definite integrals of more complicated
functions leads to still greater difficulties. The natural
problem that arises is to find some practically
convenient way of evaluating definite integrals. This
method, which was discovered by Newton and
Leibnitz, utilizes the relationship that exists between
integration and differentiation. The Newton-Leibnitz
Formula yields a convenient method for computing
2.28
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
definite integrals in cases where the antiderivative of
the integrand is known.
index must be subtracted from the one
corresponding to the upper index. We also have
The Newton-Leibnitz Formula
(a) [cF(x) ]ab = c[F(x) ]ab
If f is continuous on [a, b] and F is any antiderivative
of f on the interval [a, b], that is a function F exists
(b) [F(x) + G(x) ]ab = F(x) ]ab + G(x) ]ab
b
such that F(x) = f(x), then
f(x) dx = F(b) – F(a).
a
This formula is known as the Second Fundamental
Theorem of Calculus (FTC2).
Proof : Let F(x) =
x
f(t) dt.
a
2.
x
F '(x)dx = F(x) – F(a)
a
which can also be written as
...(1)
x
dF(x) = F(x) – F(a)
We know from the First Fundamental Theorem that
F(x) = f(x), that is F is an antiderivative of f.
If G is any other antiderivative of f on [a, b], then we
know that F and G differ by a constant:
G(x) = F(x) + C, for a x b.
...(2)
If we put x = a in the formula for F(x), we get
a
F(a) =
f(t) dt = 0
a
So, using (2) with x = b and x = a, we have
G(b) – G(a) = [F(b) + C] – [F(a) + C]
= F(b) – F(a)
...(3)
0
3.
b
=
f(t) dt using (1) and (3)
a
Thus, the difference F(b) – F(a) is independent of the
choice of antiderivative F, since all antiderivatives differ
by a constant quantity, which disappears upon
subtraction anyway.
In short, the method to evaluate
4.
b
f (x) dx is as follows:
a
Thus,
b
a
f(x)dx = F(b) – F(a).
Note:
1.
We also have two notations :
b
F(b) – F(a) = [F(x)] ab or F(b) – F(a) = F(x) a
We have thus come to a modification of the
Newton-Leibnitz formula which makes it possible
to state the basic theorem in the following way:
The increment of a function on an interval is equal
to the definite integral of the differential of the
function over that interval.
With the help of the Second Fundamental
Theorem, the value of a definite integral can be
obtained much more easily than by the tedious
process of summation. This also establishes the
existence of the limit of the sum.
The Fundamental Theorems establish a
connection between the integration as a particular
kind of summation, and the integration as an
operation inverse to differentiation. The
connection between antiderivatives and definite
integrals is as follows : FTC1 says that if f is
continuous, then
First we evaluate the indefinite integral f ( x) dx by
the usual methods, and suppose the result is F(x).
Next we substitute for x in F(x) first the upper limit and
then the lower limit, and subtract the second result
from the first.
(c) [F(x) – G(x) ]ab = F(x) ]ab – G(x) ]ab .
Since F(x) is an antiderivative of F'(x), we have
x
f(t) dt is an antiderivative
a
of f. FTC2 says that
b
f(x) dx can be found by
a
5.
evaluating F(b) – F(a), where F is an antiderivative
of f.
While evaluating a definite integral, arbitrary
constant need not be added in the expression of
the corresponding indefinite integral.
6.
The indefinite integral f(x) dx is a function of
b
x, where as definite integral
f(x) dx is a
a
b
The expression a is called the sign of double
substitution.
We can use any notation. It indicates that the
value of the function corresponding to the lower
number. Given f(x) dx we can find
b
but given
b
f(x) dx ,
a
f(x) dx we cannot find f(x) dx .
a
2.29
DEFINITE INTEGRATION
7.
The effectiveness of the Fundamental Theorem
depends on having a supply of antiderivatives
of functions.
From the above theorem it is clear that the definite
integral is a function of its upper and lower limits
and not of the independent variable x.
It should be noted that if the upper limit is the
independent variable, the integral is not a definite
integral but simply another form of indefinite
integral.
Thus, suppose
x
f(x) dx = F(x) then f(x) dx =F(x) – F(a)
a
=F(x) + constant = f(x) dx.
Assume f is continuous on an open interval I
and let F be any primitive of f on I. Then, for
each a and each x in I, we have
F(x) = F(a) +
8.
x
f(x) dx.
a
Proper t i es of a fun ct ion deduced from
properties of its derivative:
If a function f has a continuous derivative f
on an open interval I, the Second Fundamental
Theorem states that
f(x) = f(a) +
x
f (t)dt
a
For every choice of points x and a in I. This
formula, which express f in terms of its derivative
f, enables us to deduce properties of a function
from properties of its derivative.
Suppose f is continuous and nonnegative on I. If
x > a, then
9.
x
f (t)dt 0, and hence f(x) f(a).
a
In other words, if the derivative is continuous
and nonnegative on I, the function is increasing
on I.
Also, the indefinite integral of an increasing
function is concave up. That is, if f is continuous
and increasing on I, f is concave up on I. Similarly,
f is concave down on those intervals where f is
continuous and decreasing.
The Newton-Leibnitz Formula may be applied to
all indefinite integrals, though care is necessary
to ensure that (i) f(x) is continuous in [a, b], and
(ii) F(x) is continuous in [a, b].
For example,
2
dx
x [ln x ] i.e. ln 2
2
1
1
because the integrand, x–1, is not continuous at
x = 0; neither is lnx
Although we stated the Newton-Leibnitz
Formula specifically for continuous functions,
many discontinuous functions are integrable
as well. We treat the integration of bounded
piecewise continuous functions later and we
shall explore the integration of unbounded
functions in the section of improper integrals.
10. Sign of definite integral : Let us consider
dx
1
x 2 x . This equation follows from the earlier
found value of the derivative
d 1
x 1 Is the sign of the integral correct
dx
x2
1
here ? Can the integralof a positive function, 2 ,
x
be negative ? Any doubt is due to the fact that
formula is not written in a proper fashion. If we
dx
1
write it as 2 C then we cannot say
x
x
that the sign of the integral is alway negative
since this also depends on the sign and value of
the quantity C. Actually, all statements
concerning the sign of the integral are referred to
the definite integral. Let us take
b
1
1
1 1 1 1 b a
dx
.
2
a x
x a b a a b
ab
When 0 < a < b or a < b < 0 , the integral is positive,
as it should be.
b
Example 1.
3
Evaluate the integral ex dx.
1
The function f(x) = ex is continuous
everywhere and we know that an antiderivative is
F(x) = ex, so the Second Fundamental Theorem gives
Solution
3
e dx = F(3) – F(1) = e3 – e.
x
1
Notice that FTC2 says we can use any antiderivative
F of f. So we may as well use the simplest one, namely
F(x) = ex, instead of ex + 7 or ex + C.
2.30
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Evaluate
Example 2.
3
1
(3 – 2x + x2) dx.
The function f defined by f(x) = 3 – 2x + x2
is continuous and has antiderivative F given by
Solution
1 3
x.
3
Therefore, by the Fundamental Theorem
F(x) = 3x – x2 +
3
1
cos(m n)x cos(m n)x
if m 2 n 2
2(m n) 2(m n)
cos2nx
if m n
4n
Also,
1
cos mx cos nxdx 2 cos(m n)xdx cos(m n)xdx
1
20
)=
.
3
3
sin(m n)x sin(m n)x
if m 2 n 2
2(m n) 2(m n)
=
.
x sin 2nx
if m 2 n 2
2
4n
An extremely fundamental consequence of the above
results are the following so called "orthogonality
relations" :
Evaluate
Example 3.
/2
0
cos 2 x dx .
1
2 cos 2 x dx
2
1
1
1
(1 cos 2 x)dx = x + sin 2x.
2
4
2
/2
/2
1
1
cos2 x dx x sin 2x
0
4
2
0
1
sin .
4 4
4
Solution
Now cos x dx
2
Note: One should remember the values of the
following definite integrals:
/2
(a)
/2
sin x dx =
0
/2
cos x dx = 1
0
/2
2
0
/2
2
0
/2
3
0
/2
3
0
/2
4
0
4
sin mx cos nxdx 0 for all m and n
cos mx cos nxdx if m n
0 if m n
where we assume that both m and n are positive.
Also, when m and n are unequal integers, we have
cos mx cos nxdx 0,
0
When m = n, we have
2
2
0
Similarly, with the same condition, we have
2
0 cos nxdx 2 .
3
Fourier Integrals
Example 4.
We have sin mx sin nxdx
1 cos(m n)xdx cos(m n)xdx
2
if m 2 n 2
if
sin nxdx 2 , when n is an integer..
0
sin( m n )x sin( m n )x
2(m n ) 2(m n )
= x sin 2 nx
4n
2
sin x dx cos x dx 16 .
(d)
0 if m n
0
sin x dx cos x dx 3
(c)
sin mx sin nxdx if m n
sin mxsin nxdx = 0, and
sin x dx cos x dx 4
(b)
=
1
(3 – 2x + x2) dx = F(3) – F(1)
= (9 – 9 + 9) – (3 – 1 +
and sin mx cos nxdx 2 sin(m n)xdx sin(m n)xdx
=
m n
Evaluate
2
| 2x 1| dx.
0
Solution Using the definition of absolute value, we
rewrite the integrand as follows.
(2x 1), x 1
2
|2x – 1| =
1
2x 1,
x
2
2.31
DEFINITE INTEGRATION
Using this, we can rewrite the integral in two parts.
2
0
1/2
| 2x 1| dx = 0
–(2x – 1) dx +
2
1/2
(2x – 1) dx
2
= [–x2 + x] 10 /2 + [x2 – x] 1 / 2
Calculate
5
(| x 3 | | 1 x |) dx.
1
We can represent the integrand as
Solution
3
5
3
(|x – 3| + |1 – x| dx + (|x – 3| + |1 – x|) dx
3
= 2 dx +
1
5
(2x – 4) dx
3
a
Example 6. Compute the integral
0
1 cos 2x
dx
2
1 cos 2x 2 cos2 x cos x
2
2
cos x, 0 x
2
=
cos x, x
2
Therefore,
0
/2
( sin x) / 2 = (1 – 0) + (0–(–1)) = 2.
Note: If we ignore the fact that cosx is
negative in , we get a wrong result:
2
cos xdx sin x 0 .
0
0
The definite integral of f(x) may however exist, if f(x)
possesses no antiderivative at all in the interval (a, b).
The definite integral depends only upon the difference
between two particular values of F(x), which may often
be found by some special device when the form of
F(x) is unknown.
CAUTION
Let us evaluate
/2
1 cos 2x
dx
cos xdx
( cos x)dx
0
/2
2
= sin x 0
x
interval (a, x), then the function F(x) = f(x)dx,
a
and the function F(x) which satisfies the differential
dy
f (x ), are identical, except for an
equation
dx
arbitrary additive constant.
Thus, whenever f(x) possesses an antiderivative,
which it certainly does when it is continuous, the
Second Fundamental Theorem enables us to evaluate
b
= 4 + 8 = 12.
a
the value of the integral is F(b) – F(a).
The question as to whether an antiderivative exists,
and the question of the existence of an integral of the
function f(x) in (a, b) are entirely independent
questions. The First Fundamental Theorem however
shows that when f(x) is a continuous function in the
f(x) dx if it exists.
= 2x| 13 + (x2 – 4x)| 35
Solution
f(x) dx, for if F(x) is the antiderivative in question,
x 1,
4 2x,
1 x 3, We get
f(x) = 2,
2x 4,
x 3.
1
From the Fundamental Theorems we obtain the
important result that whenever f(x) is a continuous
function it possesses an antiderivative, and knowledge
of the antiderivative is equivalent to ability to evaluate
b
1 1
1 1 5
= –l (0 + 0) + (4 – 2) – = .
4 2
4 2 2
Example 5.
Existence of an integral
/4
0
x tan x dx
To apply the Fundamental Theorem, it is necessary to
find a function F such that F '(x) = x tan x. The
mathematicians have proved that there is such a
function F but it is not an elementary function. That
is, F is not expressible in terms of polynomials,
logarithms, exponential, trigonometric functions, or
any composition of these functions. We are therefore
blocked, since the fundamental theorem of calculus is
b
f (x) dx only if f is "nice"
of use in computing
a
enough to be the derivative of an elementary function.
Note: The fact that a given fun ction
(theoretically) possesses an antiderivative does not
mean that rules for obtaining its actual value are
necessarily known.
2.32
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
under a velocity curve.
Isolation of roots
v
If f is a continuous function in [a , b] and
b
f (x)dx = 0 then the equation f(x) = 0 has
v(t)
a
A1
atleast one root lying in (a , b).
Example 7. Let a, b, c be non-zero real numbers
such that
0
A3
t1
t2
A2
t
1
(e–x + ex) (ax2 + bx + c) dx
0
=
2
Displacement =
(e–x + ex) (ax2 + bx + c) dx
0
Then show that the quadratic equation ax2 + bx + c = 0
has at least one root in (1, 2).
Solution Let f(x) = (e–x + ex) (ax2 + bx + c)
We have
2
1
2
0
f(x) dx =
1
2
0
1
f(x) dx +
f(x) dx
Distance =
2
1
f(x) dx > 0 (< 0).
Since, the integral is zero, f(x) = (e–x + ex) (ax2 + bx + c)
must be positive for some values of x in [1, 2] and
must be negative for some values of x in [1, 2]. As
e–x + ex 2, it follows that if g(x) = ax2 + bx + c, then
there exists some [1, 2] such that g() > 0 and
g() < 0. Since g is continuous on R, there exists some
between and such that g() = 0. Thus ax2 + bx + c
= 0 has at least one root in (1, 2).
1
t1
2
3
t2
v(t) dt = A + A + A
1
t1
2
3
The acceleration of the object is a(t) = v'(t), so
t2
a(t)dt = v(t ) – v(t ) is the change in velocity
2
t1
1
Example 8. A particle moves along a line so that
its velocity at time t is v(t) = t2 – t – 6 m/s.
(a) Find the displacement of the particle
during the time period 1 t 4.
(b) Find the distance travelled during this time
period.
Solution
(a) Displacement = s(4) – s(1)
=
Finding displacement using velocity
If an object moves along a straight line with position
function s(t), then its velocity is v(t) = s'(t), so
t1
v(t)dt s(t ) s(t )
2
1
is the net change of position, or displacement of the
particle during the time period from t1 to t2.
If we want to calculate the distance travelled during
the time interval, we have to consider the interval when
v(t) 0 (the particle moves to the left). In both cases
the distance is computed by integrating v(t), the
speed, Therefore
t1
v(t) dt = total distance travelled
t2
v(t)dt A A A
from time t1 to time t2.
f(x) dx = 0
If f(x) > 0 (< 0) x [1, 2], then
t2
t2
The figure below shows how both displacement and
distance travelled can be interpreted in terms of areas
4
1
v(t)dt
4
(t t 6)dt
2
1
4
t3 t2
9
= 6t = .
2
3 2
1
This means that the particle moved 4.5 m towards the
left.
(b) Note that v(t) = t2 – t – 6 = (t – 3)(t + 2) and so
v(t) 0 on the interval [1, 3] and v(t) 0 on
[3, 4]. Thus, the distance travelled is
4
3
1
1
3
v(t) dt [ v(t)]dt v(t) dt
1
3
4
t3 t2
t3 t2
= 6 t 6 t
3 2
1 3 2
3
61
=
10.17 m
6
Now, consider some more examples.
2.33
DEFINITE INTEGRATION
x
Example 9. Evaluate the function F(x) = 0 cos t
dt at x = 0, /6, /4, /3 and //2.
We could evaluate five different definite
integrals, one for each of the given upper limits.
However, it is much simpler to fix x(as a constant)
temporarily and apply the Fundamental Theorem once,
to obtain
Solution
x
cos t dt = [sin t] 0x = sin x – sin 0 = sin x.
0
Now using F(x) = sin x, we obtain the results shown in
the figure.
Y
1
F =
6 2
Y
F(0) = 0
x=0
2
Y F 4 = 2
x=
4
t
3
Y F =
3
2
x=
3
t
sin xdx
0
2 – 0
– cos x 02
2
0
0
2
1
Find the mean value of f(x) =
x 1
over the interval [0, 2].
1
Solution f(x) = x 1
Mean value of f(x) over [0, 2]
2 1
1
1
1
dx = [ln (x + 1)] 20 = ln 3.
=
(2 0) 0 x 1
2
2
Example 11.
Example 12. Find the average value of the function
f(x) = x over the interval [1, 4], and find all numbers
in the interval at which the value of f is the same as the
average.
b
1
Solution favg= b a a f (x)dx
x=
6
t
2
t
4
4
1 2x3 / 2
1
xdx
=
3 3 1
4 1 1
1 16 2 14
1.6
=
3 3 3 9
The x-values at which f(x) = x is the same as the
average satisfy x = 14/9, from which we obtain
x = 196/81 2.4 (Figure).
Y
Y F 2 = 1
y x
2
14
favg
9
1
1
x=
2
Example 10. Find the average of sin x for x in
(a) [0, /2]
(b) [0, 2].
Solution
(a) The average for the interval [0, /2] is
b
sin xdx
a
( / 2) 0
cos 0 / 2
/2
1
2
0.64
/2
(b) The average for the interval [0, 2] is
2
3
4
X
196
81
t
Example 13. Let the voltage e(t) in an electrical
circuit be given by e(t) = Esint volts, where E and
are constants.
(a) What is the average voltage over the interval of
time [0, p/] ?
(b) What is the root mean square of the voltage over
[0, p/] ?
Solution
/
/
E sin tdt E cos t
(a) eavg=
0
0
2E
2.34
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(b) er.m.s =
/
2
2
E sin tdt
0
. E2
2
2E
/
0
/
t sin 2t
E 2 sin 2 tdt E 2
2
4 0
Example 14.
E2
2
t
t 2
=
2 t
t
1 1
1
f(x) = a
dx and a
dx = 2 then find f(x).
f(x)
f(x)
x
x
1
2f(x) f(x) = 2
f(x)
Integrating both sides,
{f(x)}2 = 2x + c
f(x) =
f(x) = (2x c)
f(x) dx 2
1
c = 0 then f(x) =
2x .
If f(x) = min {|x – 1|, | x |, |x + 1|}, then
1
evaluate f(x)dx .
1
Here, f(x) = min {|x – 1|, |x|, |x + 1|},
Graphically it can be shown as :
Solution
y=x+1
y= –x
1, 1
2 2
y=x
y= –x+1
(1,0)
(–1,0)
1
f(x)dx = area of the triangles
1
=2
1
1
1
1
f(x)dx = 2 2 1 2 2 .
0
0 t 1
,
t 1
f(x) =
1/2 3
3/2 2
dx +
0
1/2
(2 x) dx +
3 1 3
1
0 1 + 2 (2 – 0)
2 2 2
8
=
51
3 7
4 =
.
2 8
8
2
2 dx
0
39
, then find a, b and c.
2
Solution We have f(x) = ae2x + bex + cx .
f(0) = 1 a + b = 1
...(1)
f (x) = 2 ae2x + bex + c
f (log 2) = 2 ae2 log 2 + belog 2 + c = 8a + 2b + c
...(2)
f (log 2) = 31 8a + 2b + c = 31
log 4
39
[f(x) cx] dx =
0
2
log 4
39
(ae2x + bcx) dx =
0
2
log 4
a
39
2x
x
e be
=
2
2
0
a e 2 log 4 be log 4 a b 39
=
2
2
2
39
a
a
(16) + b (4) b =
2
2
2
15
39
a+3b=
15 a + 6 b = 39
2
2
log 4
0
Y
(0,1)
,
=
2c 2
Example 15.
, 1 t 0
Example 17. If f(x) = ae2x + bex + cx satisfies the
conditions f(0) = 1, f (log 2) = 31, and
1
dx 2
and f(1) = a
f(x)
3/2
...(2)
a
2
...(1)
1
1
But
t 1
,
3 / 2 , 3 / 2 x 1 / 2
1 / 2 x 0
Hence, f(x) = 2 x ,
2
,
0x2
1
dx
a f(x)
Differentiating both sides w.r.t.x.
Since f(x) =
Solution
3/2 t x
Let g(t) = |t – 1| – |t| + |t + 1|
Solution
If a positive function f satisfies
2
max (|t – 1| – |t| + t + 1|).
given by f(x) =
since
Evaluate : 3 /2 f(x) dx, where f(x) is
Example 16.
X
[f(x) cx] dx =
2.35
DEFINITE INTEGRATION
5 a + 2 b = 13
Solving (1), (2) and (3) we get,
a = 5, b = 6, c = 3 .
(tan x + tan x) d (x - [x]) , wh er e [x]
0
denotes the greatest integer function.
Solution Let I =
I=
(tann x + tann 2 x) d (x [x])
0
[x]
4
Now, 0 < x
Then
/4
/4
/4
tan n 1 x
=
tan x sec x dx =
0
n 1 0
1
1
=
0=
.
n 1
n 1
/4
n2
the equation
2
Find the value of which satisfies
Example 19.
a
sin x dx = sin 2, [0, 2].
/2
We have
Solution
a
/2
sin x dx = cos 2
cos = sin 2
cos (2 sin + 1) = 0
cos = 0 or sin =
1
2
3x 4 x 5 dx a 2 .
Solution (3 x + 4 x – 5) d x a – 2
2
3
0
a
2
0
a3 + 2 a2 – 5 a a3 – 2
2 a2 – 5 a + 2 0
2 a2 – 4 a – a + 2 0
(2 a – 1) (a – 2) 0
1
a 2.
2
x
f(x) =
Solution
| t 1 | dt
0
(1 t)dt, 0 x 1
0
= 1
x
(1
t)dt
(t 1)dt, 1 x 2
0
1
x2
x 2 , 0 x 1
=
2
x x 1, 1 x 2
2
The range of the function f(x) is [0,1].
x
Example 22.
x
Let f(x) = 2 | t 1 | dt. Define f(x) in
the interval 2,1 .
If – 2 x – 1 then
Solution
x
t2
f(x) = – (t + 1) dt = – t
2
2
2
x
4
x2
x2
– x + – 2 = –
– x.
2
2
2
If – 1 < x 1 then
3
7 11
,
or =
.
,
2 2
6
6
These are the value of d belonging to [0, 2].
Example 20. Find the range of values of 'a'
a
0
=–
=
for which
| t 1 | dt , where 0 x 2.
(tan n 2 x (sec2 x 1) dx
0
Find the range of the function
x
n 2
n
Example 21.
f(x) =
Find the value of
Example 18.
/4
...(3)
3
x
1
2
2
f(x) = |t + 1| dt =
=
1
x
|t + 1| dt + |t + 1| dt
1
x
– (t + 1) dt + 1(t + 1) dt
2
1
x
2
2
= t t + t t
1
2
2 2
2
1
x
1
=
x – – 1
2 2
2
2
= x + x + 1.
2
Thus, the definition of f(x) is as follows :
x2
x,
2 x 1
f(x) = 22
x x 1, 1 x 1
2
2.36
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 23.
x
1 1 t dt , x 2
Let f(x)= 0
.
5x 1
, x2
Then show that
(i) f(x) is not continuous at x = 2,
(ii) the right derivative of f(x) does not exist at x = 2.
Solution
x
(1 + 1 t) dt, x > 2
= (2 t) dt + t dt
= (2 t) dt + t dt
(i) f(x) =
0
1
x
0
1
1
x
0
1
x
1
t2
t2
= 2 t
2 0 2 1
2
1
1 1
x
= x2 + 1.
= 2 0
2
2
2 2
1 2
x 1 , x2
We have f(x) = 2
5 x 1 , x 2
lim f(x) = lim f(2 + h)
x2
h0
1
(2 + h)2 + 1 = 3,
2
lim f(x) = lim f(2 h)
= hlim
0
x2
h0
= hlim
0 5(2 h) + 1 = 11
Also f(2) = 11
lim
xlim
2 f(x) x 2 f(x)
Hence f(x) is not continuous at x = 2.
f (2 h) f (2)
(ii) f (2+ ) = hlim
0
h
1 2 h2 1 11
2
= hlim
0
h
2
2
h
20 lim h 2 4 h 16
= hlim
=h0
=
0
2h
2h
Hence the right hand derivative of f(x) at x = 2 does
not exist.
Alternative : Since, f(x) is discontinuous from the
right, it cannot have right hand derivative at x = 2.
Example 24. Find the minimum value of the function
2
f(x) =
x t dt .
0
2
We have f(x) = | x – t | dt
0
For x 2, we have
Solution
2
t2
f(x) = (x t) dt = xt – = 2x – 2.
2 0
0
For x < 0, we have
2
2
2
(t x) dt = t xt = 2 – 2x.
0
0
2
For 0 x < 2, we have
f(x) =
2
x
2
0
x
f(x) = (x t) dt + (t x) dt
0
2
(x t)2 (t x)2
=
2 x 2 x
2
x 2 ( 2 x ) x 2 – 2 x 2.
2
2
Thus, we have
x0
2 2x,
2
f(x) = x 2x 2, 0 x 2 and
2x 2,
x2
x0
2,
f(x) = 2 x 2, 0 x 2
2,
x2
Now, we have f(x) = 0 at x = 1, where f(x) changes
sign.
f(x) strictly decreases in (0, 1) and strictly increases
in (1, 2)
Hence, f(x) attains minima at x = 1, and its minimum
value, is f(1) = 1 – 2 + 2 = 1.
Example 25. Find the range of the function
=
1
f(x) =
sin xdt
(1 2t cos x t ) .
1
2
sin xdt
1
Solution
We have f(x) =
sin x (t cos x)
2
1
1
sin x 1 t cos x
tan
sin x
sin x 1
1 cos x
1 cos x
– tan–1 sin x
= tan–1
sin x
= tan–1 (tan x/2) + tan–1 (cotx/2).
=
2
2.37
DEFINITE INTEGRATION
Case I : When 0 < x <
1
x
x
0 < and 0 <
2 2
2 2 2
x
f(x) = tan–1 (tan x/2) + tan–1 tan
2 2
= x /2 + /2 – x/2
= /2
Case II : When < x < 2
x
x
<
and – < 0
2 2
2 2 2
f(x) = x/2 – /2 – x /2 = –/2
Hence, the range of f(x) is , .
2 2
Example 26. A cubic function f(x) vanishes at
x = –2 and has relative minimum / maximum at x = – 1 and
1
x = . If
3
1 2 2
1
f(x) = a (x + 1) x – a x x –
3
3
3
Now integrating w.r.t x we get,
x3 x 2 x
– +c
f(x) = a
3 3
3
Since f(–2) = 0, we have a – 8 4 2 c 0
3 3 3
2a
– 2a c 0
c=
3
3
3
2
x
x – x 2a
Thus, f(x) = a
+
3 3
3
3
x3 x2 x 2
f(x) = a 3 3 – 3 3
We also have
a
3
1
1
f(x)dx =
1
2sin x·sin x dx , independent of and.
0
Solution
=
(x3 + x2 – x + 2) dx =
1 1
2 sin x·sin x dx
2 0
1 1
2 sin x·sin x dx
2 0
1
1 sin( ) sin( )
2
Now
2 tan and 2 tan
2( ) tan tan
2( ) tan tan
sin( )
2( – ) =
cos cos
sin( )
and 2( + ) =
cos cos
=
...(1)
sin( )
sin( )
and
2( )
2( )
we get I = (cos · cos ) – (cos · cos ) = 0.
Example 28. Prove that
Substituting the value of
n 1
n 1
b a
x n 1 ((n 2)x 2 (n 1)(a b)x nab
=
,
(x a)2 (x b)2
2(a b)
a > 0, b > 0.
Solution
x n 1 (( n 2 ) x 2 ( n 1)( a b ) x n a b
(x a )2 (x b)2
x n 1 [n(x a)(x b) x(2x a b)]
dx
=
a
(x a)2 (x b)2
b
nx n 1
x n (2x a b)
dx
= a
2
2
(x a)(x b) (x a) (x b)
14
3
I=
1 sin( )x sin( )x
= 2
0
14
3
1
1
a 2 4 14
3 a=3
3 3
f(x) = x3 + x2 – x + 2.
Example 27. Let , be the distinct positive
roots of the equation tan x = 2x then evaluate
1
14
f(x)dx =
, find f(x).
3
1
Solution Given f(x) is a cubic polynomial.
Therefore, f(x) is a quadratic polynomial.
Also f(x) has relative minimum / maximum at
1
x = – 1 and x = .
3
1
Hence, – 1 and are roots of f(x) = 0.
3
4
3
2
a x x x 2x 14
3
2
3 4
1 3
b
2.38
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
b
xn
d
xn
dx =
= a
(x a)(x b) a
dx (x a)(x b)
b
bn
an
b n 1 a n 1
=
–
=
.
(b a)2b 2a(a b)
2(a b)
cos x
2
Example 29.
Let I =
a cos x b sin x dx and
0
sin x
dx , where a > 0 and b > 0.
J=
0
a cos x b sin x
Compute the values of I and J.
b
bI – aJ = ln
a
From (1) and (2)
a
b2I – abJ = b ln(b/a)
a2I + abJ =
2
Adding, we get
I=
2
aI + bJ =
Solution
and bI – aJ =
2
0
2
...(1)
...(2)
1 a
b
b ln
a 2 b2 2
a
b
2
and abI – a2J = a ln (b/a)
Subtracting, we get
Again abI + b2I =
1 b
b
a ln
.
2
a b 2
a
Alternative: We can convert a cos x + b sin x into a
single cosine say cos(x + ) and put x + = t.
b cos x a sin x
dx
a cos x b sin x
J=
2
bI – aJ = ln a cos x bsin x 0
2
D
1.
Prove the results
1
2
1
2 n 1 1
x n dx
and x n dx
,
0
1
n 1
n 1
n W, and use them to evaluate the following:
(a)
2.
3.
2
2
6.
3x 2x 1 dx (b) (t t t)dt .
1
2
3
2
0
For the function f(x) = 1 + 3x n 3 find the
antiderivative F(x), which assumes the value 7
for x = 2. At what values of x does the curve F(x)
cut the x-axis ?
If f(1) = 12, f' is continuous, and
4
f (x)dx 17 , what is the value of f(4) ?
1
4.
5.
For f(x) find such an antiderivative which attains
the given magnitude y = y0 at x = x0.
Suppose that the function f is defined for all x
such that |x|>1 and has the property that
1
for all such x.
f(x) =
x x2 1
(a) Explain why there exists two constants A
and B such that
f(x) = sec–1 x + A if x > 1 ;
f(x) = – sec–1 x + B if x < –1.
(b) Determine the values of A and B so that
f(2) = 1 = f(–2). Then sketch the graph of
y = f(x).
7.
(a) Compute the area under the parabola y = 2x2
on the interval [1, 2] as the limit of a sum.
2
(b) Let f(x) = 2x2 and note that g(x) = x3 defines a
3
function that satisfies g'(x) = f(x) on the interval
[1, 2]. Verify that the area computed in part (a)
satisfies A = g(2) – g(1).
2
(c) The function defined by h(x) = x3 + C for any
3
constant C also satisfies h'(x) = f(x). Is it true
that the area in part (a) satisfies A = h(2) –
h(1) ?
(a) Prove that if f is continuous on [a, b],
then
b
[f(x) f ]dx 0
a
avg
(b) Does there exist a constant c favg such
that
b
[f(x) c]dx 0 ?
a
8.
Find the mean value of the function on each of
the indicated closed intervals:
(a) f(x) = x on [0, 1], [0, 10], [0, 100],
(b) f(x) = 10 + 2sin x + 3cos x on [–, ]
(c) f(x) = sin(x + ) on [0, 2]
9.
(a) Find favg of f(x) = x2 over [0, 2].
DEFINITE INTEGRATION
(b) Find a number c in [0, 2] such that f(c) = favg.
(c) Sketch the graph of f(x) = x2 over [0, 2] and
construct a rectangle over the interval
whose area is the same as the area under
the graph of f over the interval.
10. Construct the graph of the function
x
f (t)dt for a = 0, a = 4, a = 8. The function
F(x) =
a
f(x) is shown below:
2.39
12. Find a cubic polynomial P for which
0
P(0) = P(–2) = 0, P(1) = 15, and 3 P(x) dx = 4.
2
13. Find the number K, L and M such that the function
Kx 2 L Mx
satisfies the
x 1
conditions f(2) = 23, f'(0) = 4 and
of the form f(x) =
37
0
(x 1)f(x)dx 6
1
Y
x 2 for 0 x 1
14. Given the function f(x) = x for 1 x 2
1
2
4
–1
6
8
10 X
f(x)dx .
Compute
0
15. Find the abscissas of the points of intersection
of the graph of the functions
x
F1(x) =
11. Find all values of c for which
c
x(1 x)dx 0
(b) | x(1 x) | dx 0
(a)
(2t – 5) dt, F (x) = x – 5x + 6.
2
2
3
16. If w'(t) is the rate of growth of a child in kg per
0
c
10
year, what does
0
w '(t) dt represent ?
5
D
17. Evaluate the following integrals :
(c) Where does F have its maximum value?
Its minimum value ?
(d) Sketch the graph of F.
Y
(i)
(2 sec x x 2) dx
2
4
3
0
x
1 sec x dx
(iii) | x 2x 3 | dx
3
(ii)
f
0
2
2
0
0
(iv)
2
sec 2 (sec 1 x)dx
2
x x2 1
10 X
–5
2
18. If f(x) = |2x – 1| + |x – 1| then evaluate f(x) dx .
2
x
x
20. Show that | t | dt
0
1
x | x | for all real x and
2
x
f(t)dt , where f is the function
express F(x) = | t | dt in a piecewise form that
whose graph is shown in the accompanying
figure.
(a) Find F(0), F(3), F(5), F(7), and F(10).
(b) On what subintervals of the interval
[0, 10] is F increasing ? decreasing?
does not involve an integral.
21. Write the equation of the tangent lines to the
19. Let F(x) =
0
1
x
graph of the function F(x) =
(2t 5) dt at the
2
points where the graph cuts the x-axis.
2.40
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
x
f(x) = x – nx +
1
33. Evaluate the integrals
z 2 2 cos 4z dz
2
23. Find the greatest and the least value of the function
x
| t | dt on the interval 12 , 21 .
F(x) =
1
24. A function f is defined by
cos t cos(x t)dt
f(x) =
0 x 2
0
x
5 / 3
(6cos u 2sin u)du
2 cos t dt , if 0 x .
34. Find all real values of x such that
x
1 x
2
3
0 (t t)dt 3 2 (t t )dt
Draw a suitable figure and interpret the
equation geometrically.
35. In each part, the velocity versus time curve is
given for a particle moving along a line. Use
the curve to find the displacement and the
distance traveled by the particle over the time
interval 0 t 3.
Find the minimum value of f .
25. On the interval [5/3, 7/4] find the greatest value
of the function F(x) =
1
x
22. Find the critical points of the function
v
1
(a)
–1
1
2 3
t
1
2
t
26. Find a function f and a number a such that
x
6+
f(t)
v
2
1
t dt 2 x for all x > 0.
a
27. If f(x) is a non-negative continuous function such
that f(x) + f (x + 1/2) = 1, then find the value of
(b)
–1
2
f(x)dx.
0
28. Solve the inequality
x 2 x 12 –
x
/2
0
0
dz x cos 2xdx.
29. Find all the numbers a (a > 0) for each of which
a
(2 – 4x + 3x ) dx a.
2
0
30. Find all solutions of the equation
0
cos (x + 2) dx = sin belonging to the interval
[2, 3].
31. Find the values of A, B and C for which the
function of the form f(x) = Ax2 + Bx + C satisfies
the conditions
1
f(1) = 8, f(2) + f(2) = 33,
7
f(x) dx = 3 .
0
32. Find all the values of from the interval
[–, 0] which satisfy the equation
sin +
2
cos 2x dx = 0
3
36. Find the mean value of the velocity of a body
falling freely from the altitude h with the initial
velocity v0.
37. Suppose that the velocity function of a particle
moving along a line is v(t) = 3t 3 + 2. Find the
average velocity of the particle over the time
interval 1 t 4 by integrating.
38. The cross section of a trough has the form of a
parabolic segment. Its base a = 1m, depth h = 1.5 m.
Find the mean depth of the trough.
1
dx
39. If a is positive and I = 1
1 2ax a 2
then show that I = 2 if a < 1 and I =
2
if a > 1.
a
40. If p, q are positive integers, show that
cos px sin qx dx
0
2q / (q 2 p2 ), if (q p)is odd,
=
if (q p) is even.
0,
41. Solve the following equations :
x
dx
(i)
2
2
12
x x 1
DEFINITE INTEGRATION
(ii)
x
dx
ln 2
ex 1
6
43. A honeybee population starts with 100 bees and
increases at a rate of n'(t) bees per week.What
x
1.5x 1
2 28
(iii) 8t 3 t 4 dt
log
x 1 x 1
1
42. If oil leaks from a tank at a rate of r(t) litres per
120
minute at time t, what does r(t )dt represent?
0
2.6
2.41
INTEGRABILITY
Theorem. If a function f is continuous at every point
of an interval [a, b], then we know that f is integrable
over [a, b].
Note: If a function is integrable in a closed interval I,
then it is integrable in every closed subinterval of I.
However, it is important to realize that a function does
not have to be continuous to be integrable and that
there are many simple discontinuous functions which
can be integrated.
A piecewise-continuous function does not have an
antiderivative on any interval containing a point of
discontinuity. Hence, we need to modify the Newton Leibnitz Formula and the definition of antiderivative
to allow integration of piecewise-continuous
functions.
1 for x 0
Let f(x) = sgnx = 0 for x 0, x [ 1, 1]
1 for x 0
This function is a piecewise-continuous on the closed
interval [–1,1], but has no antiderivative.
Indeed, any function of the form
x C1 for x 0
F(x) = x C
for x 0
2
where C1 and C2 are arbitrary numbers, has an
antiderivative equal to sgn x for all x 0. But even "the
best" of these functions, i.e. the continuous function
F(x) = |x| + C (if C1 = C2 = C), does not have an
antiderivative for x = 0.
Therefore, the function sgn x ( and, in general, every
piecewise-continuous function) does not have an
antiderivative on any interval containing a point of
discontinuity.
Here is an extended definition of an antiderivative
which is suitable for integration of piecewisecontinuous functions.
15
does 100 +
n '(t)dt represent ?
0
44. The linear density of a rod of length 4m is given
by (x) = 9 + 2 x measured in kilograms per
metre, where x is measured in metres from one
end of the rod. Find the total mass of the rod.
Extended definition of antiderivative
Definition The function F(x) is an antiderivative of
the function f(x) on the closed interval [a, b] if :
(i) F(x) is continuous on [a, b], (ii) F '(x) = f(x) at the
points of continuity of f(x).
Remark The function f(x) continuous on [a, b] is a
special case of a piecewise-continuous function.
Therefore for a continuous function the extended
definition of an antiderivative coincides with the old
definition since F'(x) = f(x) x [a, b] and the
continuity of F(x) follows from its differentiability.
Here is an example of a function which has an
antiderivative in the "new" sense and has no
antiderivative in the "old" sense.
The function f(x) = sgnx had no antiderivative in the
"old" sense on [–1, 1] whereas in the "new" sense the
function F(x) = |x| is its antiderivative since it is
continuous on [–1,1] and F'(x) = f(x) for x 0, i.e.
everywhere except for the point of discontinuity x = 0.
Hence,
x
sgn t dt = |x|.
0
Under the integral sign, there stands a bounded
function discontinuous at the point x = 0. The integral
as a function of the upper limit F(x) = |x| is a continuous
function, at the point x = 0. But the derivative F(0) is
not existent, and this does not contradict FTC1 which
guarantees the existence of the derivative F(x) only if f
is continuous at the point x.
It is now clear that a function f(x), piecewisecontinuous on the closed interval [a, b], has an
antiderivative on this interval in the sense of the
x
extended definition. The function F(x) =
one of the antiderivatives.
f(t)dt is
a
Example 1. Find an antiderivative of the piecewisecontinuous function
1 for | x | 1
f(x) = 0 for | x | 1 , x R.
2.42
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
One of the antiderivatives is an integral
with a variable upper limit, and we can take any number,
say, x = –2, as the lower limit of integration. Thus
Solution
0
f(t)dt = x 1
F(x) =
2
2
for
for
for
x
x 1,
1 x 1,
x 1,
Y
2
–1
y = F(x)
0
1
X
The significance of the extended definition of the
antiderivative is clear if we consider the following
result which retains the Newton -Leibnitz Formula with
the "new" definition of the antiderivative for
piecewise-continuous functions.
Newton-Leibnitz Formula
For piecewise-continuous function the NewtonLeibnitz formula
b
f(x)dx F(b) F(a) ,
holds true, where F(x) is an antiderivative of the
function f(x) on [a, b] in the sense of extended
definition.
2
sgn xdx | x |
Example 2.
1
2
1
Calculate I =
= 2 – 1 = 1.
cos x
0
1 sin 2 x
dx
The integrand f(x) is not defined at the
point x = /2. We divide the closed interval [0,] in
two: [0, /2] and [/2, ]. Setting f(/2) = 1 on the first
interval, we obtain an integral of the continuous
function f(x) = 1;
Solution
/2
I1 =
1dx / 2
0
I2 =
(1)dx x
/2
/2
/2
The final result is I1 + I2 = 0.
Alternative : We use the extended definition of the
antiderivative. The function F(x) which satisfies this
definition has the form
for 0 x / 2
x
F(x) =
x
for / 2 x
Indeed, F(x) is continuous on [0,] and F'(x) = f(x)
x [0,], x /2, i.e. F'(x) = f(x) at the points of
continuity of f(x). (Recall that x = /2 is a point of
discontinuity of f(x).)
In accordance with the Newton-Leibnitz formula valid
for piecewise-continuous functions and for the
extended definition of the antiderivative, we obtain
I=
f(x)dx F(x) ( x)
0
0
x
x x0 0
Theorem Let f(x) be bounded in a x b and
continuous in a < x < b. Let there be finite limits f(a +),
f(b –), and let F(x) be a function for which
d
F(x) f(x) when a < x < b
dx
and let there be finite limits F(a +), F(b –). Then
b
a
For example,
On the second interval we set f(/2) = –1 and again
obtain an integral of the continuous function f(x) = –1:
f(x)dx = F(b –) – F(a +).
a
Theorem If f is bounded on [a, b] and is continuous
at every point of [a, b] except possibly at the endpoints,
then f is integrable over [a, b].
Bounded on an interval I means that for some finite
constant M, |f(x)| M for all x in I.
Example 3. Let f be the function defined by
sin , if x 0
f ( x) x
if x 0 .
0
Show that f is integrable over [0, 2].
Solution This function is continuous everywhere
except at 0, and its values oscillate wildly as x
approaches 0. Since |f(x)| 1 for every x, the function
is bounded on every interval. It therefore follows that
f is integrable over [0, 2].
DEFINITE INTEGRATION
CAUTION
The following two examples show that a formal
application of the Newton-Leibnitz formula (i.e. the use
of this formula without due account of the conditions
of its applicability) may lead to errors in the result.
1
(i) Consider an integral
dx
2 x.
0
Taking the function F(x) = x as an antiderivative
of the function f(x) = 1 /(2 x ) and using formally
the Newton-Leibnitz formula, we obtain
1
dx
2 x x 1.
1
0
0
However, this result is incorrect since the
function f(x) = 1 /(2 x ) is unbounded on [0, 1]
1
and, consequently, the integral 0
exist.
(ii) Consider an integral I =
1 d
dx
does not
2 x
0
At first sight the function tan
1 1
x
may seem to
be an antiderivative of the integrand function
d tan 1 1
and then, from the Newton-Leibnitz
dx
x
formula, we obtain
1
x 1 4 4 2
However, this result is incorrect since the function
d tan 1 1
tan–1(1/x) is not an antiderivative of
dx
x
on the interval [–1, 1]. It can be seen that the
function has a discontinuity of the first kind at
the point x = 0 whereas the derivative must be
continuous at all points according to the definition
itself.
To calculate the integral I, we note that the
integrand function is as follows :
I = tan
1 1
Extending the definition of this function to the
point x = 0 by continuity, we get a continuous
function
1
, x [–1, 1]
f(x) =
1 x2
The function F(x) = – tan–1x is an antiderivative
of f(x) and therefore, from the Newton-Leibnitz
formula, we have
1
4 4
2
Note that we can also construct an antiderivative
for f(x) using the function tan–1(1/x), namely,
tan 1 1
for 1 x 0
x
G(x)
for x 0
2
tan 1 1 for 0 x 1
x
From the Newton Leibnitz formula we obtain
I = tan 1 x
1
1
I = G(x) 1
1 1
dx tan x dx
1
for x 0,
d tan 1 1 1 x 2
dx
x
is not defined for x 0
2.43
3
4 4
2
Note: Comparatively few functions have
antiderivatives, a great many more have integrals.
A function may have an antiderivative but not an
integral and vice-versa.
For example the function f defined by
f(x) = 2x sin(1/x2) – (2/x) cos(1/x2), x 0
and f(0) = 0 has an antiderivative F(x) = x2sin(1/x2) but
in the interval [–1, 1] there is no integral since f is
unbounded in the neighbourhood of zero.
Further, the function defined by (x) = 1 for x 0 and
(0) = 0 is integrable being bounded and having only
one point of discontinuity at x = 0 but has no
antiderivative in the actual sence.
The next theorem shows that we can, without
changing the value of the integral, assign arbitrary
(bounded) values to the integrand at any finite number
of points.
Theorem Let g(x) be zero except at a finite number
of points in (a, b) and at these points let g(x) take finite
values. Then g(x) is integrable over (a, b) and
b
g(x)dx 0 .
a
2.44
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Corollary. Let f(x) be bounded and integrable in (a, b).
Let g(x) f(x) except at a finite number of points x1,....xk,
at which the (finite) values of g(x) are X1, ...., Xk. then
g(x) is also integrable in (a, b) and
b
b
a
a
g(x)dx f(x)dx .
Theorem Let f(x) be continuous in (a, b) except for
a discontinuity of first kind at x = c, where a < c < b.
then f(x) is integrable over each of the intervals (a, c), (c,
b), (a, b) and
b
c
b
a
a
c
f(x)dx f(x)dx f(x)dx
Note: The theorem can be extended to cover
any finite number of discontinuities of first kind in the
range of integration, including discontinuities of first
kind at the end points a and b.
Theorem
(a) If f has a countable number of discontinuities
of the first kind in [a, b], then f is integrable on
[a, b].
(b) If f is not bounded on [a, b], then f is not
integrable on [a, b].
Thus, a piecewise-continuous function (a
function which has a countable number of points
of discontinuity of the first kind on the closed
interval [a, b]) is integrable on this interval.
If the conditions of the theorem are fulfilled then
the value of the integral
b
f(x)dx does not
a
depend on the values of f(x) at the points of
discontinuity.
We therefore often raise and solve the problem of
calculation of the integral of a function which is
not defined either at a finite number of points of
the interval [a, b] or on the set of points which
can be covered by a finite number of intervals of
an arbitrarily small length. In that case we assume
that the definition of the function f(x) is completed
arbitrarily at these points but the function remains
bounded on the interval [a,b] and consequently,
integrable.
For example, strictly speaking, the integral
1 sin x
dx
...(1)
0
x
does not exist since at the point x = 0 the function
sin x
is not defined.
x
1
However, the integral
f(x)dx , where
0
sin x for x 0
, (m is an arbitrary
f(x) = x
m
for x 0
number), exists and is independent of the choice of m.
We therefore assume that integral (1) also exists and
1
is equal to f (x)dx .
0
It often happens that the integrand has a discontinuity
which is simply due to a failure in its definition at a
particular point in the range of integration, and can be
removed by attaching a particular value to it at that point.
In this case it is usual to suppose the definition of the
integrand completed in this way. Thus, the integrals
1/2 sin mx
1/2 sin mx
dx,
dx
0
0
x
sin x
can be treated as ordinary definite integrals, if the
integrands are regarded as having the value m when x = 0.
dx
Let us now evaluate
cos x(1 tan x) .
0
2
2
1
Here f(x) = cos2 x(1 tan 2 x )
for 0 x / 2, / 2 x ,
1
= is not defined for x / 2
Extending the definition of this function to the point / 2,
say, by continuity, i.e. setting f(/2) = 1, we get f(x) 1
x [0, ], and, consequently, the required integral is
equal to .
Now, consider the function
f(x) = 0 where x is an integer,
f(x) = 1
otherwise, in the interval (0, m), where m is a positive
integer. This function is integrable, since it is bounded
and its discontinuities are finite in number, being
situated at the points x = 1, 2, ...., m.
Note that the function f(x) = 1/x2 is not integrable
on any interval containing x = 0. Indeed, even if we
extend f to be defined at 0, say by setting
1 / x 2 , x 0
f(x) =
x0
0,
f would still be unbounded on any interval
containing x = 0, so the theorem tells us that f is
not integrable across any such interval.
DEFINITE INTEGRATION
Example 4. Which of the following integrals exist?
2
1
sin dx
0
x
(a)
1
(c)
/2
0
(b)
ln x
1 x dx
1
tan x dx
This is the same as asking whether or
not each function is integrable over its proposed
interval of integration.
Let us suppose therefore, that f is bounded and
continuous on the open interval (a, b). We may choose
values f(a) and f(b) completely arbitrarily, and the
resulting function will be integrable over [a, b]. Further
Solution
b
more, the integral a f dx is independent of the choice
of f(a) and f(b). Hence, if f is bounded and continuous
on (a, b), we shall certainly adopt the point of view
that f is integrable over [a, b] and, equivalently, that
b
f dx is defined.
a
(a) Following this convention, we see that the
function sin
1
x
is bounded and continuous on
1
(0, 1), and so sin dx exists.
0
x
(b) Using L'Hospital's rule , one can easily show that
ln x
lim
1.
x 1 1 x
Hence,
ln x
is bounded and continuous on
1 x
(1, 2), and so
(c) Here
1
2 ln x
1 x dx exists.
1
lim tan x , and we therefore
x ( /2)
conclude that tan x is not integrable over the
interval 0, . We now give an example of a
2
nonintegrable function.
Consider the Dirichlet function on the interval
[0,1]. It is equal to 1 at rational points and to zero
at irrational points. Therefore, if we take irrational
points as points i in integral sums, then
n
n
i 1
i 1
n
n
n
i 1
i 1
i 1
2.45
f ( i )x i 0.x i 0
and consequently, the limit of these sums is equal
to zero as n . And if rational points are chosen
as i, then
f ( i )x i 1.x i x i 1
and, hence, the limit of these sums is equal to 1.
Thus, in the case of Dirichlet function the limit of
integral sums on the interval [0, 1] depends on
the choice of points i and this means that the
Dirichlet function is not integrable on [0, 1].
Theorem Let each of the functions f(x) and g(x) be
bounded and integrable over [a, b]; and let c be a
given constant. Then each of the functions
(i) cf(x)
(ii) f(x) + g(x)
(iii) f(x)g(x)
is bounded and integrable over [a, b].
Example 5. Using examples, prove that the sum,
the product and the quotient of two non-integrable
functions can be integrable.
Solution Consider the Dirichlet function,
0 if x is an irrational number,
D(x) = 1 if x is a rational number
The function D(x) is non-integrable. The function
f(x) = 2 + D(x) is not integrable either.
2 if x is an irrational number,
f(x) =
3 if x is a rational number
Indeed, if we assume that f(x) is integrable, then the
difference of two integrable functions f(x) – 2 = D(x)
must be integrable, but this contradicts the fact that
D(x) is non-integrable.
Now, let g(x) = f(x).
Since g(x) = f(x), it follows that g(x) is non-integrable.
Let us consider the function
1/2 if x is an irrational number
1
h(x) = g(x) = 1/3 if x is a rational number
This function is not integrable either. The reason is
similar to the reason for the non-integrability of the
Dirichlet function.
We set up the sum, the product and the quotient of
non-integrable functions:
F1(x) = f(x) + (–g(x)) 0,
F2(x) = f(x) h(x) 1,
F3(x) = f(x)/g(x) 1.
2.46
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Being constant, the functions F1, F2 and F3 are
integrable on any closed interval [a, b]. However, the
integrability of a sum or a product does not imply the
integrability of the summands or factors.
Example 6. Using examples, prove that the
product of the integrable function f(x) by the nonintegrable function g(x) may be (a) an integrable, (b) a
nonintegrable function.
Solution
(a) Let us consider, for example, an integrable
function f(x) 0 and a nonintegrable Dirichlet
function D(x) on [a, b]. Since f(x) D(x) 0, it follows
that f(x)D(x) is an integrable function on [a, b].
(b) Let f(x) 2 and g(x) = D(x) on [a, b]. Then f(x) g(x)
= 2D(x) is a function nonintegrable on [a, b].
Integrals of piecewise continuous functions
If a function f(x) has a countable number of points of
discontinuity of the first kind in an interval [a, b], the
integral of this function is defined as the sum of the
ordinary integrals taken over the subintervals into
which the interval [a, b] is broken up by all the points
of discontinuity of the function. Denoting these points
as c1, c2,...ck (a < c1 < c2 < ...<ck < b) we write
b
a
f(x)dx
c1
a
f(x)dx
c2
c1
f(x)dx .....
b
f(x)dx
ck
In the first integral on the right hand side the value of
the function f(x) at the point c1 is understood as its left
hand limit f(c1–), and in the second integral as the right
hand limit f(c1+). The values of the function f(x) at the
other points of discontinuity are understood similarly.
Under the conditions assumed, in every closed
subinterval of integration the integrand is continuous.
The geometrical meaning of the integral under
consideration is clear from the figure.
Y
[c1, c2], ...., [ck, b] lying between the subsequent points
of discontinuity.
Let us find the integral of the function
1 x, 1 x 0
2
0 x 2 over [–1, 3].
f(x) = x ,
1, 2 x 3,
We have
3
0
2
1
1
0
3
f (x)dx = (1 x) dx + x dx + (1)dx
0
2
2
2
x3
x2
3
= x x2
2 1 3 0
3 8
19
1 .
2 3
6
Thus, the Newton-Leibnitz Formula applies to
bounded piecewise continuous functions with the
=
d x
restriction that a f(t)dt is expected to equal
dx
f(x) only at values of x at which f is continuous.
0 x 1
2 1 x 2,
,
Example 7. If f ( x)
2 x 3,
3
1 3 x
4
find
f(x)dx .
0
Solution
The graph of the function f is shown in
the figure.
Y
3
2
1
–1 0
1
2
4
3
5
X
–1
a
c1
c2
c2
bX
The integral is equal to the sum of the areas of the
trapezoids having as bases the subintervals [a, c1],
Since f is a bounded piecewise continuous function,
it is integrable over any closed interval. In particular,
4
2
3
4
0
0
2
3
f dx f dx f dx f dx
= 2 . (2 – 0) + 3 . (3 – 2) + (–1) (4 – 3)
= 4 + 3 – 1 = 6.
2.47
DEFINITE INTEGRATION
Let f be the function defined by
Example 8.
x
x 0,
2
0 x 2,
f ( x) 2 x
2 x 5 2 x .
3
3
2x2 3 = 0 x =
2
2x 3 = 1 x =
2
2
5
2x2 3 = 2 x =
The graph of f is drawn in the figure.
Y
2x 3 = 3 x =
2
2x2 3 = 4 x =
2
2
1
–1
2
1
X
4
3
2
3
7
2
[2x 3] dx
2
1
3/ 2
=
2
[2x2 3] dx
[2x2 3] dx +
1
–1
3/ 2
5/ 2
+
–2
3
[2x2 3] dx
[2x2 3] dx +
2
The function is continuous except at 0
and at 2, and is bounded on any bounded interval.
Thus f is integrable over the interval [–1, 3] and that
Solution
3
0
2
3
1
1
0
2
3
3
3
2
2
2
2
3
f dx (2x 5)dx (x 5x) 0 .
2
2
1 4
13
f 0
.
Hence,
1
4 3
12
3
Evaluate
dx + (3)
dx
2
7/ 2
3
dx + (4)
3
2
7/ 2
dx
3
5
2
1 + 0 +
= ( 1)
2
2
+2 3
7
5
3 + 4 2
+3
2
2
3
2
=9
2
Example 10.
5
3
2
Evaluate
2
0
7
2
7
.
2
[x2 – x + 1] dx, [ . ] is
the greatest integer function.
[2x 3] dx
2
1
At x = 1, value of 2x2 3 = 2 (1)2 3 = 1
and at x = 2, value of (2x2 3) = 2 (2)2 3 = 5.
The integers between 1 to 5 are 0, 1, 2, 3, 4
Solution
dx + (1)
3/ 2
2
Example 9.
dx + (0)
5/ 2
x3
4
f dx (2 x )dx 2x
.
0
0
3
3
0
Similarly, f(x) = 2x – 5 for all x in [2, 3] except that f(2) = –2
2
5/ 2
2
3/ 2
+ (2)
0
For all x in [0, 2], we have f(x) = 2 – x2 except that
f(0) = 0. Hence
2
7/ 2
1
x
1
.
f dx
x dx
1
1
4 1
4
0
[2x2 3] dx
[2x2 3] dx +
3
For all x in [–1, 0], we have f(x) = x2, and so
0
2
7/ 2
+
= ( 1)
f dx f dx f dx f dx .
4
5/ 2
Solution
Let I =
2
0
[x2 – x + 1] dx
Let f(x) = x2 – x + 1 f(x) = 2x – 1
For x > 1/2, f(x) > 0 and for x < 1/2, f(x) < 0.
The values of f(x) at x = 1/2 and 2 are 3/4 and 3.
Between them we have integers 1, 2.
2.48
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Hence we solve x2 – x + 1 = 1, 2
1 5
We get x = 1, x =
2
Also the values of f(x) at x = 0 and 1/2 are 1 and 3/4.
There is no integer between them.
I=
+
1/2
[x2 – x + 1] dx +
0
1
[x2 – x + 1] dx
1/2
2
1 5
2
[x2 – x + 1] dx + 1 5 [x2 – x + 1] dx
2
1
=0+0+1
1 5
2
1
1. dx + 2
2
1 5
2
1 dx
1 5
1 5 5 5
1 2 2
.
=
=
2 2
2
Alternative: Graphical Method
Y
3
21 / 3
1
[x 3 1]dx
1
31 / 3
21 / 3
[x 3 1]dx ...
2
[x 1]dx
3
71 / 3
Now if x 1, 2 3 , then x 3 [1, 2) or [x3 – 1] = 0
and so on.
21/ 3
Therefore I =
1
0.dx
31/ 3
21/ 3
2
1.dx ......... 1/ 3 6.dx
7
= [31/3 – 21/3] + 2[41/3 – 31/3] + 3[51/3 – 41/3]
+ 4[61/3 – 41/3] + 4[61/3 – 51/3] + 6[2 – 71/3]
= 12 – [71/3 + 61/3 + 51/3 + 41/3 + 31/3 + 21/3].
/2
2 x
Evaluate / 4 sin x dx,
where [.] denotes the greatest integer function.
Example 12.
/2
2x
Here, I = / 4 sin x dx.
1 2x
2x
1 = 0.
Also x
4
2
2
Solution
/2
2
y=x – x + 1
so that I =
[sin x]dx 0 .
/4
Example 13. Find the mean value of the function
on each of the indicated closed intervals:
(a) f(x) = cosx on [0, 3/2],
(b) f(x) = sgn x on [–1,2].
2
1
3/4
Solution
1/2 1 1 + 5 2
2
It is clear from the figure that
X
2
[x2 – x + 1] dx = Area of the bounded region
0
1 5
1 5
= 0 + 2 1 × 1 + 2 2 × 2
1 5 5 5
= 3 – 2 = 2 .
2
Evaluate [x 1]dx
3
Example 11.
1
where [.]denotes the greatest integer function.
Solution 1 x 2 1 x 3 8 0 x 3 1 7
2
So I =
=
[x 1]dx
1
3
2 3 / 2
2
cos xdx
3 0
3
Note that the continuous function cos x assumes
the value favg = –2/(3), namely, cos = –2/(3), at
(a) favg =
2
the point = cos–1 of the closed interval
3
[0, 3/2]
1 2
1
sgn xdx .
3 1
3
In this case the discontinuous function sgnx does not
assume the value favg = 1/3 on the closed interval [–1, 3].
(b) favg =
Example 14. Find the average value of f(x) = x[2x]
sgn (x – 2) on [1, 3], where [ . ] denotes the greatest
integer function.
Solution The average value of f(x) over [1, 3]
=
3
1
f(x) dx
(3 1) 1
2.49
DEFINITE INTEGRATION
n(n 1)
(replacing n = [x])
2
[ x]([ x] 1)
.
= [x]x –
2
1 3
x[2x] sgn (x – 2) dx
2 1
2
1 15
x[2 x]sgn(x 2)dx x[2 x]
=
.5
1
1
2
sgn(x – 2) dx
=
= nx –
+
2.5
2
x[2x] sgn(x – 2)dx+
3
2.5
x[2x]sgn(x 2)dx
2.5
1
4x dx
= 2x dx + 3x dx +
1.5
3
2 1
1.5
2
3
1
f(x) = {x} on , 1.
2
Solution By definition, the mean value of
function is
favg =
1
1
{x} dx
1 1/ 2 1/ 2
2 1
(x – [x]) dx
3 1/2
1
0
= 2 1/2 (x 1)dx 0 x dx
3
=
0
1
x2
x2
2
x
=
1/2 2 0
3 2
1 1 1 7
= 2 8 2 2 12 .
3
f(x)=
0
1
n
0
1
x
e dt , so
{t}
0
t
0
t 1
x
t 1
1
t 2
2
[loga x]
dx where a > 1, and [ . ] denotes the
e 1
then find the value of [a] .
2
x·a
[loga x]
1
dx
1
2
(a ·a
= ln a ·
(a ·a ) dt = ln a · a dt
t
0
1
{t}
[t ]
·a t ) dt = ln a ·
t
0
1
2t
0
0
(as {t} = t if t (0, 1) )
1
I = ln a ·
1
ln a a 2 t
·
=
= (a2 – 1)
2
2 ln a
x
...+ n 1[x]dx + n [x]dx
= 0 + 1(2 – 1) + 2 (3 – 2) + ....
.... + (n – 1) [n – (n – 1)] + n(x – n)
= [1 + 2 + 3 .... + (n – 1)] + nx – n2
n(n 1)
= nx +
– n2
2
1
2
1
3
= [x] dx + [x] dx + [x] dx + ...
0
0
dt =
Put logax = t at = x
x
2
x
t
a
0
1
t [t ]
t
0
1
Solution
Then, [ x]dx = 0 [x]dx + n [x]dx
n
x
e
e dt if x [0,1]
e dt e dt if x [1, 2]
e dt e dx e dt if x [2,3]
greatest integer, is
1
dt (x >0), where[x]
ex 1
if x [0, 1]
x 1
f(x) (e 1) (e 1) if x [1,2]
x 2
1) if x [2,3]
2(e 1) (e
Clearly f(x) is continuous but not differentiable
at x = 1 and 2. Thus, f is continuous but not
differentiable in [0, 3).
Also f (2) = 2(e – 1) + 0 = 2(e – 1).
Example 18. If the value of definite integral
1
[x]([x] + 1).
Solution Let n be an integer such that n x < n + 1
i.e. [x] = n.
x
Solution We have f(x) =
a
x
t [t ]
0
denotes greatest integer less than or equal to x. Show
that f is continuous but not differentiable in [0, 3)
and f (2) = 2(e – 1).
x·a
[x]dx = x[x] – 2
Prove that
Example 16.
e
x
15
+ 2.5 5xdx =
.
4
Example 15. Find the mean value of the function
x
Let f(x)=
Example 17.
1 2
e 1
(a – 1) =
a=
2
2
[a] = 1.
e
(a
0
t [t ]
·a t ) dt
2.50
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Alternative : x (1, a) logax (0, 1)
[logax] = 0
1
a
Example 20. A function f is defined on [0, 1] by
f(x) = 1/2n for 1/2n+1 < x 1/2n, n W, and f(0) = 0.
e 1
x dx = 2 (a2 – 1) = 2 a = e .
I=
1
[a] = 1.
Example 19. Prove that for any positive integer K,
sin 2Kx
= 2 [cos x + cos 3x + ... + cos (2K – 1)x]
sin x
/2
sin 2Kx . cot x dx = .
Hence, prove that
0
2
Solution
To prove: sin 2K x
= 2 sin x [cos x + cos 3x + cos 5x + .. .. + cos (2K – 1) x]
R.H.S.
= (sin 2x) + (sin 4x – sin 2x) + (sin 6x – sin 4x) +
....... + (sin 2Kx – sin (2K–2)x) = sin 2Kx
L.H.S.
Now,
/2
=
/2
=
0
0
/2
=
0
/2
0
sin 2Kx . cot x dx
sin 2Kx
. cos x dx
sin x
2 cos x [cos x + cos 3x +...+cos(2K – 1)x]dx
[(1 cos 2x) + (cos 4x + cos 2x)+....
/2
/2
0
0
(cos2nx) dx = 0 n I and n 0
sin 2K x . cot x dx =
/2
1 dx + 0 = 2 .
0
x
f (t)dt , where x lies between 1/2m and 1 2m–1.
0
It is easy to see that f is discontinuous at
x = 1/2n, since we have
Solution
1
1 1
1
f n n 1 and f n n .
2 2
2 2
Thus, the points of discontinuity
1, 1 , 12 , 13 , .......
2 2 2
Now
x
0
f (t) dt
1
x
1/2
fdx
m
1/2m
1/2
fdx
m 1
1/2m 1
1/2m 2
x
m
m 1
dx
1/2
m
fdx ...
m 1
m 1
1 / 2 2m 1
1
2
2
1
4
x
1
1
m 1 2m 1
2
2
3.2 2m 1
1
mx1
.
2
3. 2 2 m 2
x
m 1
1
2m 1
E
1.
Is the function f(x) = 1/x integrable on the closed
intervals (i) [1, 2] and (ii) [–1, 1] ?
are
1/2
1
1
dx m 2 m 1 dx ...
m
1/2 2
1/2
1/2
2
2
1
1
1 1
1
m 1 x m m m m 1
2
2 2 2
2
1 1
1
m 1 m 1 m 2 ......
2 2
2
1
1
1
1
1
m 1 x m 2m 1 2m 3 2m 5 ...
2
2 2
2
2
+ (cos 2K x + cos (2K – 2) x)] dx
But we know that,
Calculate
(a) f(x) = cosx
x / | x |, x 0
(b) f(x) =
x0
x,
2.
Is the function f(x) = tanx.cotx integrable on the
closed intervals (i)[/6, /4] and (ii) [–1, 1] ?
3.
Is the function f(x) = e–1/x integrable on the closed
intervals (i) [–3, –2] , (ii) [–1, 0] and (iii) [–1, 1]?
1 / x 2 , x 0
(c) f(x) =
x0
0,
4.
In each part, determine whether the function f
is integrable on the interval [–1, 1].
sin 1 / x, x 0
(d) f(x) =
x0
0,
DEFINITE INTEGRATION
5.
Is each of the following integrals defined ? Give a
reason for your answer.
1 sin x
1 / 2 tan 2 x
dx
dx
(a)
(b)
0 x
0
x
1/ e 1
11
dx
(c) 0 dx
(d) 0
ln x
x
(e)
6.
7.
8.
e
ln x dx
0
State whether or not each of the following
functions is integrable in the given interval. Give
reasons for each answer.
(i) f(x) = |x – 1|, [0, 3].
1 if x rational
(ii) F(x)
1 if x irrational, [0,1].
If | f | is integrable in an interval [a, b], show that f
need not be integrable in [a, b].
Use the Second Fundamental Theorem of
Calculus to evaluate the integral, or explain why
it does not exist.
(i)
5
2
5
3
x dx
3 /2
(iii)
1/ 2
9.
6
1 t2
(ii)
2
cos e c d
2
2.51
b
10. Draw the graph of f, and evaluate a f(x)dx in each
of the following examples.
1 if x 0,
(a) f ( x) 5 if 0 x 2, and [a, b]=[–3, 3].
3 if 2 x ,
x2
if x 0,
(b) f(x)
2
if 0 x ,
2 x
and [a, b] = [–2, 2].
(c) f(x) = n if n x < n + 1 where n is any integer,
and [a, b] = [0, 5].
11. Prove that
b
|x|
x dx | b | | a | .
a
12. Evaluate the following integrals, where [x]
denotes greatest integer less than or equal to x.
3
(i)
[x]dx
0
9
[ t ]dt
(iii) x d ([x] – x)
1
(iv) [x] x dx
2
(ii)
0
3
0
3
1
dt
Is this computation correct?
1
1
1
dx
1
– 2 2x 1 2 ln | 2x 1 | – 2 = 2 ln 3 – 2 ln 3 = 0
1
b
b
a
x
a
[x]dx [x]dx a b .
1
1
14. Show that x – [x] – dx {x}({x} – 1)
2
2
13. Show that
0
where [x] and {x} are integral and fractional parts
of x, respectively.
E
15. Determine whether or not each of the following
functions is integrable over the proposed interval.
Give a reason for your answer.
x2 x 2
1
,[0, 1]
(a) cos , [0, 1]
(b)
x 1
x
x2 x 2
x2 x 2
,[0, 2]
, [0, 2]
(c)
(d)
x 1
x 1
1 1
16. Prove that the function f(x) = for x 0,
x x
f(0) = 0, is integrable on the closed interval [0, 1].
17. Prove that the function f(x) = sgn sin is
x
integrable on the closed interval [0, 1].
18. A number is dyadic if it can be expressed as the
quotient of two integers m/n, where n is a power
of 2. (These are the fractions into which an inch is
usually divided.)
if x is dyadic
0
Let f(x) =
3 if x is not dyadic
Why does f not have a definite integral over the
interval [0, 1]?
19. Prove that the sum of an integrable and a non
integrable function is a nonintegrable function.
20. Find out whether the following functions are
integrable on the closed interval [0, 1]
2.52
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(a) f1(x) = x,
(c) f1(x) + g1(x),
(e) f2(x) = x ,
(b) g1(x) = 1/x
(d) f1(x) g1(x)
(f) f2(x) g1(x).
for 2 x 0,
1
21. f(x) = D(x ) for 0 x 2
where D(x) is the Dirichlet function,
0 if x is irrational
D(x) = 1 if x is rational , Is the function
f(x) integrable on the closed intervals [–2, 2],
[–2, –1], [–1, 1], [1, 2] ?
b
22. Let there exist | f (x) | dx . Does it follow that
a
the function f(x) is integrable on the closed interval
[a, b] ?
23. Show that the function
xnx , 0 x 1
1 x
x0
0 ,
f(x) = 1 ,
x 1
is integrable on the interval [0, 1].
24. Can one assert that if a function is absolutely
integrable on the interval [a, b], then it is integrable
on this interval ?
25. Find the mean value of the function on each of
the indicated closed intervals.
(a) f(x) = sinx on [0, ], [0, 2], [, + 2],
[, + ],
(b) f(x) = sgn x on [–2, –1] [–2, 1], [–1, 3], [–2, 2],
[1, 2].
(c) Is the mean value of the function on each
intervalone of the values of the function on
2.7
IMPROPER INTEGRAL
Up to now, when speaking of definite integrals we
assumed that the interval of integration was finite and
closed and that the integrand was continuous or
piecewise continuous. It is this particular case to which
the theorems for existence the definite integral stated
in the previous section applies. However, it often
becomes necessary to extend the definition of the
definite integral to an infinite interval of integration or
to an unbounded integrand function.
We define integrals with infinite intervals of
that interval ? Explain why in some cases the
answer in positive and in the other negative.
2
26. Calculate
f(x)dx , where
0
2
for 0 x 1 employing two
f(x) = x
2
x for 1 x 2
techniques :
(i) using the antiderivative of f(x), constructed
on the whole closed interval [0,2],
(ii) Dividing the interval [0, 2] into the intervals
[0, 1] and [1, 2].
27. (a) If n is a positive integer, prove that
n
[t]dt = n(n – 1)/2.
0
(b) If f(x) =
x
[t] dt for x 0, draw the graph of
0
f over the interval [0, 4].
2
28. (a) Prove that
[t ]dt = 5 – 2 3 .
3
(b) Compute
2
0
[t ]dx .
2
3
29. (a) If n is a positive integer, prove that
n
[t] dx = n(n – 1) (2n – 1)/6.
(b) If f(x) = [t] dx for x 0, draw the graph
2
0
x
2
0
of f over the interval [0, 3].
x
(c) Find all x > 0 for which
[t] dx = 2(x – 1).
2
0
1
30. Let f(x) = 12 and F(x) = – . Find F(1) – F(–1).
x
x
1
Does
f(x) dx = F(1) – F(–1)? Explain
1
integration or infinite discontinuities within the
interval of integration as improper integrals.
Integrals with infinite limits
Let y = f(x) be a continuous function in an infinite
interval [a, ), i.e. for x a . Then we can take the
integral of the function f(x) over any finite interval
b
[a, b] (b > a) : I(b) =
f(x)dx
a
Now let us make b grow indefinitely. Then, there are
two possibilities, namely, I(b) either has a limit as b
or has no limit, which justifies the following definition :
DEFINITE INTEGRATION
1.
The improper integral
a
f(x)dx of the function
f(x) over the interval [a, ) is the limit of the integral
b
a
:
the line x = 1. One might think that, since S is infinite
in extent, its area must be infinite, but let’s take a
closer look.
Y
f(x)dx lim
b
f(x)dx
...(1)
b a
a
x=1
If the limit exists the improper integral
f(x)dx
integral
f(x)dx is said to be divergent. In this
a
b
case
f(x)dx either tends to infinity or has no
a
limit at all.
If an antiderivative F(x) of the integrand f(x) is
known, it can easily be checked whether the given
improper integral is convergent or divergent:
If lim F(x) = F() exists, then, by the Newtonx
Leibnitz formula,
a
0
a
is said to be convergent. If the limit does not exist
equality (1) is meaningless, and the improper
1
2
x
1
area = 1–
b
y=
f(x)dx as b provided that this limit exists
f(x)dx lim[F(b) F(a)] F() F(a)
b
1
Y
X
The area of the part of S that lies to the left of the line
x = b (shaded in Figure) is
b
1
1
1
dx 1
A(b) =
2
1 x
x 1
b
Notice that A(b) < 1 no matter how large b is chosen.
b
1
lim 1 1.
We also observe that blim
A(b)
=
b
b
The area of the shaded region approaches 1 as b ,
so we say that the area of the infinite region S is
equal to 1 and we write
b 1
1
dx
lim
dx 1.
2
b 1 x 2
1 x
b
and the integral is convergent; if this limit does
not exist the integral is divergent.
It is easy to see the geometric meaning of an
improper integral :
2.53
Evaluate
Example 1.
(1 x)e dx
x
0
Solution Integrating by parts with u = 1 – x and v = e–x
yields
(1 x) e dx = – e (1 – x) – e dx
x
–x
–x
= –e–x + xe–x + e–x + C = xe–x + C
0 a
If the integral
b
X
b
f(x)dx expresses the area of a region
a
bounded by the curve y = f(x), the x-axis and the
ordinates x = a, x = b, it is natural to consider that the
improper integral
f(x)dx expresses the area of an
a
unbounded (infinite) region lying between the
curves y = f(x), x = a, and the x–axis.
Consider the infinite region S that lies under the
curve y = 1/ x2, above the x-axis, and to the right of
Thus,
b
(1 x)e dx lim[xe ] lim e
0
x
b
x b
0
b
b
The limit is an indeterminate form of type /, so
we will apply L'Hospital's rule by differentiating
the numerator and denominator with respect to b.
This yields
1
(1 x)e x dx lim b 0
b e
0
An explanation of why this integral is zero can be
obtained by interpreting the integral as the net
signed area between the graph of y = (1 – x) e–x and
the interval [0, ).
2.54
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Evaluate
Example 2.
[2e–x] dx, where [ . ]
0
denotes the
greatest integer function.
= n!,
[2e ]dx
Let I =
Solution
=
ln 2
x
Evaluate
x3
(x 1)(x 1) dx.
2
2
x3
(x 1)(x 1) dx
2
2
x3
b
(x 1)(x 1) dx
= blim
2
2
b
2 2x 1
dx
= blim
2 x 1
x2 1
x
ln 2
ln 2
0
ln 2
[2 ln(x – 1) – ln(x2 + 1) – tan–1x ]2b
= blim
b
( x 1)2
ln 2
tan 1 x
= blim
x 1
2
= (ln 2 – 0) + 0 = ln 2.
x2
Example 3. Evaluate 0 (1 x 2 )2 dx .
Solution Put x = tan. dx = sec2 d
2
As x increases from 0 to , increases from 0 to .
x
Solution
1. dx + 0 .dx
I
x
0
[2e ] dx + [2e ] dx
0
e dx .
0
Example 5.
dy
= – 2e–x < 0 x [0, ]
dx
2e–x is decreasing function x [0, ]
0 < 2e–x 2 x [0, )
1
For x > ln 2 ex > 2 e–x <
2
2e–x < 1 0 2e–x < 1
[2e–x] = 0
I=
e dx 1 .
since
x
0
Let y = 2e–x
= n(n – 1) (n – 2) .... 2.1
1
2
0
tan 2 sec 2 d
sec 4
1
2
1
sin d 4 .
2
0
Note: Thus, sometimes an infinite integral
can be tranformed into an ordinary definite integral by
a suitable substitution, we must see that the
transformation is legitimate.
Show that
Example 4.
e x dx n! , n being
x
n
0
a positive integer.
Solution Let In denote the given integral.
I n lim
e x dx
x
n
0
lim e x x n n e x x n 1 dx
0
0
[integrating by parts]
n lim
e x
0
x
n 1
dx , since lim e . n 0
= n In–1 = n(n – 1) In–2
= n(n – 1) (n – 2) In–3, .....
(as before)
( b 1)2
1
lim
tan 1 b – ln + tan –1 2
= b ln 2
5
b
1
+ ln 5 + tan –1 2 1.14.
=0–
2
Notice that we combined the logarithms in the
antiderivative before we calculated the limits as
b . Had we not done so, we would have
encountered the indeterminated form
lim (2 ln (b – 1) – ln (b2 + 1)) = – .
b
The way to evaluate the indeterminate form, of
course, is to combine the logarithms, so we would
have arrived at the same answer in the end. But our
original route was shorter.
2. The improper integral over the interval (–, b] is
defined similarly
b
b
a a
f(x)dx lim f(x)dx F(b) F()
where F(–) is the limit (if it exists) of the antiderivative
F(x) as x – .
0
Evaluate xe x dx.
Example 6.
Solution
0
0
xe dx = lim xe dx
x
a a
x
DEFINITE INTEGRATION
We integrate by parts with u = x, dv = ex dx so that
du = dx, v = ex :
0
0
xe dx = xex] – e dx = –tea – 1 + ea
x
0
a
a
x
a
We know that ea 0as a , and by L’Hospital’s
Rule we have
a
a
lim
=
= alim
(–ea) = 0
a
a e
a e a
lim aea = lim
a
Therefore
0
xe dx = lim (–aea – 1 + ea)
x
t
= – 0 – 1 + 0 = –1.
3. If f(x) is a function continuous in (0, ), the
improper integral can be defined for the whole interval
(–, ) by definition,
0
0
f (x)dx f (x)dx f (x)dx
If both integrals on the right hand side are convergent,
the integral
f (x)dx is said to be convergent.
If atleast one of the integrals on the right hand side is
divergent the equality has no sense, and the integral
on the left is called divergent.
Here we have a simpler notation. If an antiderivative
F(x) is known, then
1
1
2
Since 1/(1 + x2) > 0, the given improper integral can be
interpreted as the area of the infinite region that lies
under the curve y = 1/(1 + x2) and above the x-axis
(see figure).
Y
y=
1
dx
tan 1 x .
1 x 2
2 2
x
1
2
2
.
Here both F(–) and F() are equal to infinity and the
integral is divergent.
It should be noted that all the simplest properties of
definite integrals enumerated are extended without any
changes to improper integrals provided all the integrals
on the right hand sides of the equalities are convergent.
Example 7.
Evaluate
area =
1 + x2
0
where F() and F(–) are respectively, the limits (if
they exist) of F(x) as x and x – . If atleast one
of these limits does not exist the improper integral is
divergent.
1
dx.
1 x2
1
1 x dx 2 2
We must now evaluate the integrals on the right side
separately :
t
1
1
1
t
= lim tan x]0
0 1 x 2 dx = lim
t 0 1 x 2
t
= lim
(tan–1 t – tan–1 0) = lim
tan–1 t =
t
t
2
0
0 dx
1
lim tan 1 x]0t
1 x2 dx = tlim
t 1 x 2
t
= tlim
(tan–1 0 – tan–1 t) = 0 –
2 2
Since both of these integrals are convergent, the
given integral is convergent and
1 x dx 2 n (1 x )
1
0
Solution 1 x 2 dx = 1 x 2 dx + 0 1 x 2 dx
f (x)dx F() F()
2.55
X
Evaluate
Example 8.
dx
.
e ex
x
x
e dx
dx
=
e x e x e2x 1
Solution
x
c e dx
e x dx
lim
=
0 e2x 1 c 0 e2x 1
ec
= clim
0
du
(by the substitution u = ex)
u 1
2
ec
= lim tan 1 u lim (tan –1(ec) – tan –1 (1))
1
c
c
tan 1 (e c )
= clim
4 2 4 4
Similarly,
0 e x dx
e x dx
lim
=
2x
e
1 c e e 2x 1
0
2.56
= clim
du
lim tan 1 u
c
ec u 1
ec
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
1
2
1
lim tan –1 (ec)
4 c
tan (e ) =
= clim
4
c
–0=
=
4
4
0 e x dx
e x dx
dx
0 e 2x 1
e x e x
e 2x 1
Thus,
Integral of a function with infinite
discontinuity
We now proceed to the definition of the improper
integral for functions with infinite discontinuities. Let
y = f(x) be a continuous function for all x [a, b)
(i.e. a x < b) having an infinite discontinuity at the
right end point x = b of the interval [a, b]. It is clear that
the ordinary definition of the definite integral is
inapplicable here. Let us first take the ordinary integral.
I() =
b
a
f(x)dx where > 0 and then make tend
to 0.
Then I() either tends to a finite limit or has no finite
limit (in the latter case it either tends to infinity or has
no limit at all).
b
The improper integral
a
f(x)dx of a function f(x)
continuous for a x < b and unbounded as x b is
the limit of the integral
b
a
f(x)dx for 0 ( 0)
provided this limit exists:
b
f (x)dx lim
b
0 a
a
b
The integral
f(x) dx of the function can also be
a
evaluated as follows :
b
a
f(x)dx = lim
cb
b
f(x)dx
a
1
Example 9.
Evaluate
dx
1 x .
0
0
= – lim 2
Example 10.
Evaluate
b 1
b
dx
0
1 x
lim 2( 1 b 1) 2 .
b 1
/2
cos x
0
1 sin x
dx.
Solution The integrand is discontinuous at x =
/2
cos x
0
1 sin x
= lim
u /2
u
dx = lim 0
u /2
u
(1 sin x)
1/2
0
.
2
cos x
dx
1 sin x
(–cos x) dx
u
1/2
= lim 2(1 sin x)
u /2
0
= lim – 2[(1 – sin u)1/2 – 1] = 2
u / 2
/2
Example 11. Prove that
3
sin cos
0
1 / 2
d 8 .
5
1
Solution The integrand is continuous in 0 < 2 ,
1
. The integral can only be
2
but tends to as
defined as lim
1
2
0 0
1
sin 3 cos 2 d
...(1)
Put cos = , –sin d = d
when = 0, = 1;
Thus (1) is given by
1
1
,0
2
1
lim (1 2 ) 2 d
0
f (x)dx, 0 .
1 x |0b = –
= .
4 4 2
dx
1
1 x = blim
1
Solution
1
8
1 2
lim 2 2 5 / 2 = 2 2 = .
0
5
5 5
5. Similarly, if the function f(x) has an infinite
discontinuity only at the left end point x = a of the
interval [a, b] we put
b
b
f(x)dx lim f(x)dx where > 0 provided
a
0 a
this limit exists.
If the limit does not exist, the integral is divergent.
2.57
DEFINITE INTEGRATION
Y
1
dx.
Example 12.
2
x2
Solution We note first that the given integral is
Find
5
0
improper becuase f(x) = 1/ x 2 has the vertical
asymptote x = 2. Since the infinite discontinuity
occurs at the left end point of [2, 5]
Example 14. Prove that I
= lim 2( 3 t 2 ) = 2 3 .
t 2
Thus, the given improper integral is convergent and,
since the integrand is positive, we can interpret the
value of the integeral as the area of the shaded region
as shown in the Figure.
Y
1
ln x dx.
0
We know that the function f(x) = ln x has a
Solution
vertical asymptote at 0 since xlim
ln x = – . Thus,
o
the given integral is improper and we have
t
ln x dx = lim
t0
1
t
lnx dx
t
ln x dx = x ln x ] 1t –
1
t
Therefore,
1
0
x
1/ 2
(1 x)1/ 2 dx
2
cos ec sec 2 sin cos d 2( )
I1=
2
1
Note: The above is an example of how a
convergent improper integral, such as
1
x
1/ 2
(1 x)1/ 2 dx
...(1)
1
t 0
1
1
Since 2 as 0 and 1 0 as 0, I = .
2
02 2d
dx
lim t ln t = lim ln t = lim ln t2 = lim (–t) = 0.
t 0 1 / t
(1 x)1/ 2 dx .
may, when we use a substitution, (here we use
x = sin2) become a proper integral, such as
= 1 ln – t ln t – (1 – t) = – t ln t – 1 + t
To find the limit of the first term we use L’Hospital’s rule :
t 0
1/ 2
1
, and let 1, 2
2
correspond to the values, , 1 – of x. Then
dx = 2sin cosd and
0
Now we integrate by parts,
1
1
–
Evaluate
–
Example 13.
–
1
0
Put x = sin2, where 0 < <
area = 2 3
2 3 4 5 X
–
–
0
1
x
Solution The integrand both when x 0
and when x 1. We consider I as the limit, when and
0, of
I1 =
1
x–2
y=
y = ln x
2
1
area = 1
5
dx
dx
5
lim
= lim 2 x 2 ]t
t 2
x 2 t 2 t x 2
5
X
1
1/ t
t 0
ln x dx = tlim
(–t ln t – 1 + t)
0
= –0 –1 + 0 = –1
The figure shows that the area of the shaded region
above y = ln x and below the x-axis is 1.
...(2)
The solution that merely puts x = sin2 in (1) and at
once gets (2) gives the right answer; but since (1) is,
in fact, an improper integral, this solution proves
nothing.
6. If the function f(x) has an infinite discontinuity at
an intermediate point x = c of the interval [a, b]
(i.e. a < c < b) then, by definition
b
a
f(x)dx
c
a
f(x)dx
b
f(x)dx
c
If both integrals on the right hand side of the equality are
2.58
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Y
b
convergent, the integral
f(x)dx is also convergent.
a
If atleast one of these integrals on the right is divergent
the integral is divergent. For example,
2
dx
1 3
2
dx
0
1 3
2
2
dx
2
x 3 x 3 x 33 2
0 3
0
3
3
0
x
x
Note : One should not be puzzled by the use of the
term improper integral to denote something which
has a definite value such as 2 or /2. The distinction
between an improper integral and a definite integral is
similar to that between an infinite series and a finite
series, and no one supposes that an infinite series is
necessarily divergent.
x
f(t)dt we defined
Recall that the definite integral
a
as a simple limit, i.e. as the limit of a certain finite sum.
The improper integral is therefore the limit of a limit, or
what is known as a repeated limit.
Evaluate
Example 15.
1
dx
1
2
x .
Solution Since inside the interval of integration
there exists a point x = 0 where the integrand is
discontinuous, the integral must be represented as the
sum of two terms :
dx
2
1 x
1
dx 1 dx
2
2
1 x
0 x
0
1 dx
1 dx
dx
lim
lim
2
2
2
1 x
1 0 1 x
2 0 2 x
We calculate each limit separately:
1
1 dx
1 0
x
–1
1 1
= – lim =
1 0
1 1
Thus, the integral diverges on the interval [–1, 0] :
lim
2 0
1
dx
2
2
–1 0
1
x
Here both integrals on the right hand side are divergent,
and consequently, the given integral is also divergent.
In other cases, an integral with the integrand function
becoming infinite on the interval of integration can
yield a finite result.
It should be noted that if we had begun to evaluate
the given integral without paying attention to the
discontinuity of the integrand at the point x = 0, the
result would have been wrong :
1 dx
1 1
1 1
–
– –2 ,
2
1 1
1 x
x –1
which is impossible.
4
dx
Example 16. Evaluate 0 (x 1)2
Solution The integrand is discontinuous at x = 1,
which is inside (0, 4).
u
dx
1
lim
2
0
u 1
u 1
x 1 0
(x 1)
1
1
(1) = lim
1 =
= ulim
1 u 1
u 1 u 1
4
dx
Hence, 0
is divergent.
(x 1)2
4
dx
at all.
We do not have to consider ulim
u
1
(x 1)2
4
dx
to be convergent, both
For 0
(x 1)2
u
4
dx
dx
lim
lim
2 and
2 must exist.
u 1 0 (x 1)
u 1 u (x 1)
lim
4
lim 1 x 2 = – lim 1 1
1 0
2
3
1
2
y = 1/x
Note:
1.
If the function f(x), defined on the interval [a, b],
has within this interval, a finite number of points
of discontinuity a1, a2 ...., an, then the integral
of the functionf(x) on the interval [a, b] is defined
as follows:
x = – lim 1 12 =
2 0
And this means that the integral also diverges on the
interval [0, 1]. Hence, the given integral diverges on
the entire interval [–1, 1].
b
a
f(x) dx
a1
a
f(x)dx
a2
a1
f(x)dx ...
b
f(x)dx
an
2.59
DEFINITE INTEGRATION
if each of the improper integrals on the right side
of the equation converges. But if even one of
these integrals diverges, then
2.
b
f(x)dx too is
a
called divergent.
Let 0 f(x) g(x) for a x < b. Assume that
lim f(x) = and lim g(x) = . (See figure). It is
xb
xb
not hard to show that, if
then so does
hard to show that, if
g(x) dx coverges, so does
a
a
a
f(x)dx (and, equivalently, that, if f(x)dx
does not converge, then neither does g(x)dx .
a
b
g(x)dx converges,
Y
a
y = g(x)
b
f(x)dx and, equivalently, if
a
y = f(x)
b
f(x)dx does not converge, then neither
a
does
b
g(x)dx . A similar result also holds for
O
a
a < x b, with lim replacing lim .
x a
a
As an example, consider
xb
1
Y
1
For x 1,
y=g(x)
y=f(x)
O
Since
b
X
dx
. For 0 x < 1,
1 x4
4
2
1 – x = (1 – x)(1 + x)(1 + x ) < 4(1 – x) and
1 1
1
.
4 1 x 1 x4
1 1 dx
Since
does not converge, neither
4 0 1 x
1 dx
does
converge.
0 1 x4
1
dx
Now consider 0 2
.
x x
1
1
.
For 0 < x 1, 2
x x
x
1 1
1
dx
Since
dx converges, so does
.
2
0 x
0 x + x
Assume that 0 f(x) g(x) for x a. Assume also
that lim f(x) = lim g(x) = 0. (See figure). It is not
1
As an example, consider 0
3.
x
Ú
x
x
1
a
Ú
dx
4
x 2x 6
x 4 2x 6
.
1
.
x2
converges, so does
2
X
dx
dx
1
4
x 2x 6
.
Example 17. Classify each of the following
integrals as proper or improper. If improper,
determine whether convergent or divergent, and, if
convergent, evaluate it.
a
1
dx
x
1
1
1
1 1
(a)
(c)
sin x dx
(b) a
x2
dx
1 1
(d) 1 dx
x
a
Solution
(a) Since
1
x
takes on arbitrarily large values near 0,
1
dx is not a proper integral.
x
For every t in [0, 1],
we know that
1
t
1
0
1
1
dx 2 x 2(1 t ) .
x
t
Since lim 2(1
t 0
t ) exists, we get
2.60
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
dx lim 2(1 t ) 2 .
0 x
t 0
Hence, (a) is a convergent improper integral
with value 2.
1
(b) The values of 2 also increase without bound
x
as x approaches zero, and (b) is therefore not a
proper integral. For every t in (0, 1],
1
1
1
1
1.
t x
xt t
1 1
1
dx lim 1 ,
However, tlim
t x2
0
t0 t
and, since the limit does not exist, the improper
integral is divergent.
1 1
dx
2
1
1 for all nonzero x, the function f
x
1
defined by f(x) = sin is bounded on [0, 1]. It is
x
also continuous at every point of that interval.
We now assign a value, say 0, to f(0), and it follows
that f is integrable over [0,1], and hence,
1
1
sin dx is a proper integral.
0
x
1 1
dx is an improper integral.
(d)
1 x
It follows from the definition that
1 1
0 1
11
dx
dx
dx ,
0 x
1 x
1 x
and that both integrals on the right should be
convergent.
However,
11
11
dx lim
dx lim (ln 1 ln t )
0 x
t0
t0 t x
0 1
dx is divergent.
and similarly
1 x
1 1
dx is divergent.
We conclude that
1 x
(c) Since sin
CAUTION
1
Failure to note the discontinuity of the function
at
x
0 can result in the following error :
1 1
x dx ln x
1
1
1
0 0 0.
3
0
convergent.
1
3
Let
Solution
(x 1) (x 3) dx
0
1
1
1
(x 1) (x 3) dx is
Example 18. Find whether
1
3
(x 1) (x 3) dx (x 1) (x 3) dx
0
1
and both integrals on the right should be convergent.
However, it easy to show that neither is convergent. A
partial fractions decomposition yields
1
1 1
1 1
, and so
(x 1) (x 3)
2 x 1 2 x 3
1
1
1
dx ln x 1 ln x 3 c
(x 1) (x 3)
2
2
1 x3
ln
c.
2 x 1
In particular, therefore,
1
t
1
1
dx lim
dx
0 (x 1) (x 3)
t 1 0 (x 1) (x 3)
1 t3 1
lim ln
ln 3 ,
t
1
2 t 1 2
which is sufficient to establish that the integral is
divergent.
Example 19. If the value of the definite integral
(2x 3 1) dx
can be expressed in the
6
3
2
1 x 2x 9x 1
I=
form
A
C
A
C
cot 1 where
and
are rationals in
B
D
B
D
their lowest form, find the value of (A + B2 + C3 + D4).
(2x x 2 ) dx
4
2
1 x 2x 9 x
r
r
2
(dividing N and D by x )
I=
Solution
(2x x 2 )dx
2
I = 1 x2 1 9
x
Put x2 +
1
1
= t 2 x 2 dx = dt
x
x
As x 1, t 2 and as x , t
I = 2
1
dt
1 t
dx = tan
2
t 9
3
3 2
2.61
DEFINITE INTEGRATION
Hence, A = 1, B = 3, C = 2, D = 3
(A + B2 + C3 + D4) = 1 + 9 + 8 + 81 = 99.
1
1 1 2
1 2
= cot
= tan
3 2
3 3
3
F
1.
Which of following integrals are improper ? Why?
1
2.
2
1
/2
(a)
x e dx
(c)
x 5x 6 dx (d) x 5 dx
(b)
1
0
x
2
9.
sec xdx
1
0
2
0
2
Find the error in the following steps:
4.
11
use your sketch to show that
dx
1 x
2
0
xe
2 x2
(2 x) 1 x 2
...(2)
...(3)
3
...(4)
b
...(5)
...(6)
b
[ln( x 2)]3 lim [ln x]3
= blim
b
...(7)
= – .
0
0
sin d and cos d are
indeterminate.
1 x2
dx
e dx
2 0
7.
Evaluate the following integrals :
0
dx
1
Here is an argument that ln 3 equals – .
Where does the argument go wrong ? Give
reasons for your answer.
1
ln 3 = ln 1 – ln
3
b2
1
ln
ln
= blim
...(1)
b
3
10. Show that
Show that
11.
(i)
(ii)
xe
(iii)
dx
0 x ln 2 x
(iv)
x 2x 2
3 x2
0
dx
dx
2
b
f(x)dx may not equal lim f(x)dx .
b
Show that
dx
1 x 2 (x 1)
(iv) 1
dx
, and
1 x2
1 y
dy
y
1
6.
1
e
3
x 2 dx
(x 3) (5 x)
x
x
b
1 1 dx
= blim
3 x 2
x
1 1
dx
= 3
x2 x
1
1
=
dx
–
dx
3 x2
3 x
t
Sketch the region whose area is 0
0
5
b
Compute
(ii)
b
1
dx
tan x dx .
(a) lim
(b) tlim
1 0
t 0 t x
2
How does the result give insight into the fact
that neither integrand is integrable over the
interval [0, 1] ?
5.
(iii)
1
e e dx
dx
3
ln x 2
= blim
x 3
= lim [ln(x– 2) – lnx ]b
1
–1 1
– 1 – – 1 –2.
dx
=
2
–1 x
1 1
x –1
x
1
0 ex
(i)
2
ln(x 1)dx
(d)
4 x4
Evaluate the following integrals :
1
0
Explain why each of the following integrals is
improper.
3.
1
2x 1 dx
(b)
1
dx
1 2x 1
sin x
dx
2
1 x
(c)
(a)
8.
2xdx
2
b
2xdx
x 1 diverges and hence that
0
2
x 1 diverges. Then show that
lim
b
2xdx
x 1 0 .
b b
2
2.62
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
12. (a) Show that if f is even and the necessary
integrals exist, then
0
f (x)dx 0
integrals exist, then
13. For each x > 0. let G(x)
f (x)dx 2 f (x)dx
(b) Show that if f is odd and the necessary
=
e
xt
0
dt. Prove that xG(x) = 1 for each x > 0.
F
ln x
1
14. Given that
(1 x) x dx is a convergent
0
improper integral, prove that
ln x dx
(1 x) x
0
0.
0
/2
sin ln tan d
ln tan d ,
16. Prove the following :
4
1
dx
3.4 3
(i) 0
2/3
(4 x)
4
dx
63 2
(ii) 0
(x 2)2/3
dx
1
1 b
(iii) 0 a 2 e x b 2 e x ab tan a
1
dx
2 3
(iv) 1/2 4
x 1 x2
17. Evaluate the integrals
b x 2 dx
xdx
(i)
(ii)
3
4
0 (1 x)
0 (1 x)
and show that they converge to finite limits as
b .
1
2
a, b > 0.
19. Prove that
2
0
cos2 d
,
a cos2 b 2 sin 2 2a(a b)
2
sin 2y
sin 2 x
sin x
sin 2
dx
dy
dx .
0
0
x
y
x2
Deduce that
x
1
sin 2 x
sin x
dx
dx
0
0
x
x2
dx
1
(i)
(1 x) dx 2 (1 x) 2 .
(ii)
x 2 dx
dx
1
=–
4
0 1 x
0 1 x4
25
22. Find
2
0
2
0
2
f(x) dx , where
0
1
for 0 x 1
4 3
x
f ( x)
1
for 1 x 2 .
3
4
( x 1)
b
dx
,
23. Prove that a
{(x a)(b x)}
b
xdx
1
(a b),
a
2
{(x a)(b x)}
(i) by means of the substitution x = a + (b – a) t2,
(ii) by means of the substitution (b – x)/(x – a) = t, and
(iii) by means of the substitution x = a cos2t
+ b sin2t.
24. Prove that
2
(i)
(ii)
dx
1
(x 1) x 1 3 .
1
2
1
dx
1
1
(1 x)(2 x) x(1 x) 2 6 .
0
x n 1
dx = cosec n
0 1 x
for 0 < n < 1. Verify that this equation is correct for
n = 1/2.
26. If a and b are positive, then prove that
dx
,
(i)
0 (x 2 a 2 )(x 2 b 2 )
2ab(a b)
25. It can be proved that
1
0
18. Evaluate 0
cos nx tan x dx 0 .
b
1
21. Show that
2
0
are convergent improper integrals, prove that their
values are 0,
respectively..
4
20. Prove that, as n ,
15. Given that
/2
(It may be assumed that the integrals are
convergent)
2.63
DEFINITE INTEGRATION
x 2 dx
.
2
0 (x a 2 )(x 2 b 2 )
2(a b)
(ii)
a2
(iii) 0 1 2 dx = a.
x
27. Prove that
dx
n
(i)
2 , n 1
n
1
n 1
x x2 1
2.8
When evaluating a definite integral by substitution,
two methods are possible. One method is to evaluate
the indefinite integral first and then use the Newton
Leibnitz Formula.
Another method, which is usually preferable, is to
change the limits of integration when the variable is
changed.
While integrating an indefinite integral by the
substitution of a new variable, it is sometimes rather
troublesome to transform the result back into the
original variable. In all such cases, while integrating
the corrsponding integral between limits (i.e.,
corresonding definite integral), we can avoid the
tedious process of restoring the original variable, by
changing the limits of the definite integral to
correspond with the change in the variable.
Substitution Theorem
b
g(b)
a
g(a)
f(g(x)) g(x) dx =
f(u) du
Proof : Let F be an antiderivative of f. Then, F(g(x))
is an antiderivative of f(g(x)) g(x), so by the Second
Fundamental Theorem, we have
a
f(g(x)) g(x) dx = F(g(x))] ab = F(g(b)) – F(g(a))
But applying FTC2 a second time, we also have
g(b)
g(a)
f(u) du = F(u) ] gg (( ab)) = F(g(b)) – F(g(a)).
Therefore in a definite integral the substitution should
be effected in three places (i) in the integrand, (ii) in
the differential and (iii) in the limits.
Example 1.
(iii)
(1 x ) dx 0
(iv)
(1 x) dx 2 4 .
1
x
–x
x ln x
2 2
0
x
1
1
2
0
Evaluate
Solution Using the substitution, we have u = 2x + 1
and dx = du/2. To find the new limits of integration we
note that when x = 0, u = 1 and when x = 4, u = 9
91
4
u du
2x 1 dx =
Therefore,
1
0
2
1 2 3/2 9 1
26
= . u ]1 = (93/2 – 13/2) = .
2 3
3
3
e ln x
dx
Example 2. Calculate 1
x
Solution Let u = ln x because its differential
du = dx/x occurs in the integral. When x = 1, u = ln 1 = 0;
when x = e, u = ln e = 1. Thus
1
e ln x
1
u2
1
dx
udu .
1 x
0
2 0 2
Note: When computing the definite integral
from this formula we do not return to the old variable.
2
If g is continuous on [a, b] and f is continuous on the
range of u = g(x), then
(1 e )(1 e ) 1
SUBSTITUTION IN
DEFINITE INTEGRALS
b
dx
(ii)
4
0
2x 1 dx
Let f (0)= 0 and
Example 3.
f '(2t) e
f (2t )
0
dt 5 ,
then find the value of f (4).
2
Solution
We have
f '(2t) e
Put ef (2t) = y
2 f ' (2t) ef (2t) dt = dy
f (4)
1 e
dy 5
2 ef (0)
ef (4) – ef (0) = 10
Hence, f (4) = ln 11.
Now
f (2t )
0
ef (4)
ef (0)
dt 5
dy 10
ef (4) = 10 + e0 = 11
CAUTION
If we take f(x) =
1
0
x dx
x , then
/2
0
sin u cos u du is true.
2.64
But
1
0
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
x dx
9 / 2
0
Example 6.
(sin u) cos u du
is absurd since is x is not defined for negative x. On
the other hand both the statements.
1
/2
x dx sin u cos u du and
2
0
1
2
2
9 /2
0
sin 2 u cos u du are valid.
2
Example 4.
x sin x dx
2
1
1
4
1
1
4
1
1
(cos1 cos 4)
2
Note that here we are not allowed to put x =
t , since
t 0, and in the given integral x attains negative
values as well.
Note: The derivative of a differentiable
function need not be integrable. Examples to illustrate
this are not however very easy to construct. if we
assume that g (t) is continuous (the usual
assumption), the integral
b
f(g(x)) g(t)dt certainly exists.
a
Evaluate
1
sin 1 x
0
2
(1 x ) dx
Put sin–1x = d
1
dx
(1 x 2 )
0 and 1 are the limits of x ; the corresponding limits of
where = sin–1x are found as follows :
When x = 0, = sin–1 0 = 0
Solution
When
x = 1, = sin–1 1 = .
2
/2
1
I d 2
0
2 0
/2
Let u = x3 – 17.
Solution
1
2 .
8
Note: Of course this example can be worked
out by first finding the indefinite integral in terms of x
and then substituting the limits.
du
.
3
x 2 dx
1 du 1
ln |u| + c
x3 17 3 u 3
1
ln |x3 – 17| + c.
3
x 2 dx
1
ln | x3 17 ||20
0 x 3 17
3
1
1
ln | 8 17 | ln | 17 |
3
3
1
1
1
9
= ln 9 – ln 17 = ln
.
3
3
3
17
/2
dx
Example 7. Compute
0
1 cos x
Solution To find the indefinite integral of the
Finally,
Example 5.
3
=
1
x sin x dx 2 sin ydy 2 ( cos y)
2
0
So,
We put y = x2. Then
Solution
2
Compute
x 2 dx
x 17 .
Then du = 3x2dx, or, equivalently, x2 dx =
0
x dx
2
Compute the definite integral
2
1
let us take advantage of the
1 cos x
x
= t. To be more precise, let us
substitution tan
2
function f(x) =
change the variable of integration in the following
manner:
x = 2tan–1t.
From trigonometric formulae it follows that
2
1 t2
1
1
1 t
2
2
2
1 cos x 1 1 t
2
1 t 1 t
1 t2
We then compute dx :
2 dt
dx =
1 t2
Finally, since tan
0
tan 1
= 0, tan
, we have
2.2
4
2
to take 0 and 1 as the limits of integration in the new
integral. Thus,
1 1 t2
1
/2
dx
2
dt dt 1 .
2
0
0
0
1 cos x
2 1 t
Example 8. Show that
1/2
dx
1
0 (1 2x 2 ) 1 x 2 2 ln(2 3) .
DEFINITE INTEGRATION
Solution Put x = sin. Then dx = cos d
Also when x = 0, = 0, and
I
2 5
is negative on the interval ,
, we obtain
3 6
1
, = .
2
6
when x =
/6
0
cos d
cos 2 cos
0
sec 2 d
/6
1
1
= ln tan
2
4
0
1
5
1
ln tan ln tan = 1 ln( 2 3 ) .
2
12
4 2
Example 9. Compute the integral
3 /2
dx
x 1 x2
Solution Apply the substitution x = sin t (the
given function is not monotonic), dx = cos t dt. The
new limits of integration t1 and t2 are found from the
equations
x 1 x2
1/ 2
1/ 2
3
1
= sin t;
= sin t. We may put t1 =
2
6
2
, but other values may also be chosen, for
3
5
2
and t2 =
.
instance, t1 =
6
3
In both cases, the variable x = sin t runs throughout
and t2 =
1 3
the entire interval 2 , 2 .
dx
3 /2
/6
2.65
5 / 6 dt
cos tdt
5 / 6 sin t( cos t)
2 / 3 sin t
2 / 3
5 / 6
tan 5
12 n 2 3 .
= n tan t
n
2 2 / 3
3
tan
3
5
Note: If we take t1 = 6 , t2 = 3 , we need to
split the interval of integration at t =
, since
2
1 x 2 | cos t | and cos t changes sign at t = .
2
dx
b
Example 10.
Evaluate
x {(x a)(b x)}
a
Put x = acos2 + bsin2
dx = 2(b – a)sin cos
Solution
2(b a)sin cos d
(a cos2 b sin 2 )(b a)sin cos
/2
sec 2
=2 0
a b tan 2
2 /2
b.sec 2 d
I=
/2
0
=
b
0
( a )2 ( b tan )2
/2
=
2 . 1 tan 1 b tan
b a
a 0
=
2 . .
(ab) 2
(ab)
Example 11. Find the value of the definite integral
e2010 1
1 ln x
1
dx
e
x
x .
ln
x
ln
ln x
Let us show that the results of the two integrations
will coincide. Indeed,
3 /2
1/ 2
dx
x 1 x
2
/3
/ 3 dt
cot dt
t
n tan
/ 6 sin t cos t
/ 6 sin t
2 /6
/3
n tan n 2 3
.
6
12
3
On the other hand, taking into consideration that cost
= n tan
Solution
2010
1
Substituting ln x = t, we get
1 t
1 t t (ln t) dt
2010 ln 2010
= 2009
1
1
du (u = t – ln t)
u
2.66
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Evaluate
Example 12.
2 1
2
0
dx
(2x 1) x 2 x
.
Let x2 + x = t2 (2x + 1) dx = 2t dt
Solution
2t dt
(2x + 1)
dx =
=2
=
2t dt
0
12
12
0
2
1/2
12
1
· 2 tan 1 2 t 0 = tan–11 = .
4
2
Alternative:
Quadratic substitutions
1
t
Put 2x + 1 =
A substitution of the type
y = x2 – 2x + 10
...(1)
fails at the outset to give x uniquely in terms of y and
must be replaced by one or other of
1
dt
2t 2
dx = –
x=1+ y9 ,x=1– y9
at least for the purpose of substituting for x in terms of y.
For x 1 we use the second form; for x 1 we use the
first. In order to use (1) in an integral
If x = 0 then t = 1
1
1
– then
2 2
If x =
2 11 =
1
t
2
f(x)dx
1
t=
2
2
we consider the integral in the form
Hence, the integral becomes
I= –
3
1
3 1/2
u
(1 u 2 )du =
16
u 1
16 1
3
3 1
2 (1 1)
.
16 2
32
2 12
dt
dt
= 4 0 2
2
t
1 4
4(t ) 1
0
2 dt
(2x 1) t = 4x 4x 1
2
2
3 /3 1 cos
sin d
16 0
sec
Now we substitute u = cos so that du = –sin d.
1
When = 0, u = 1; when /3, u = .
2
3 3 /2
x3
3 1/2 1 u 2
dx
du
Therefore, 1
2
3/2
16 1
(4x 9)
u2
1
( 2 1) ( 2 1) 1
= t = .
4
4
2
When x = 0 then t = 0
12
=
=
Hence, I =
3 /3 tan 3
3 /3 sin 3
d
d
16 0 sec
16 0 cos2
=
2 - 1 Ê 2 - 1 + 1ˆ
Á
˜
¯
2 Ë 2
t2 =
x3
dx
(4x 2 9)3 / 2
3
/ 3 27 tan 3
8
sec 2 d
=
0
27sec 3 2
3 3 /2
= 3. 0
2 -1
then
2
When x =
and 4x 2 9 9 tan 2 9 = 3 sec
When x = 0, tan = 0, so = 0;
when x = 3 3 /2, tan = 3 , so
1
2
1
2t
1 t
2
·
t
dt = – sin t 1
1
2t 2
3 3 /2
Example 13.
Find 0
2
= .
4
x3
dx.
(4x 2 9)3/2
3
3
Solution Let x = 2 tan , which gives dx = 2 sec2 d
1
3
f(x)dx
2
f(x)dx
1
and use in each integral the substitution appropriate
to its range of values of x.
Suppose for example that I =
7
(x 6x 13)dx.
2
0
We find by direct integration that I = 48.
Now let us apply the substitution y = x2 – 6x + 13,
which gives x = 3 ± (y 4) .
2.67
DEFINITE INTEGRATION
Since y = 8 when x = 1 and y = 20 when x = 7, we appear
to be led to the result
20 dx
1 20 ydy
I = y dy
.
8
dy
2 8 y4
The indefinite integral is (1/3) (y – 4)3/2 + 4 (y – 4)1/2,
80
, which is wrong,
and so we obtain the value ±
3
whichever sign we choose. The explanation is to be
found in a closer considerationof the relation between
x and y. The function x2 – 6x + 13 has a minimum for
x = 3, when y = 4. As x increases from 1 to 3, y decreases
from 8 to 4, and dx/dy is negative, so that
dx
1
–
.
dy
2 y4
As x increases from 3 to 7, y increases from 4 to 20,
and the other sign must be chosen. Thus
20
y
y
I = ydx
dy
dy,
1
8
4 2 y4
2 y4
a formula which will be found to lead to the correct
result. Similarly, if we transform the integral
0
7
4
dx by the substitution x = sin –1 y, we must
observe that dx/dy is (1–y2 ) –1/2 or – (1–y2 ) –1/2
according as 0 x < /2 or /2 < x .
Example 14. Transform the integral
3
(x 2) dx
2
0
by the substitution (x – 2)2 = t.
Solution A formal application of the substitution
throughout the interval [0, 3] would lead to wrong
result, since the function x has two branches :
x1 = 2 – t ; x2 = 2 + t . The former branch cannot
attain values x > 2, the latter values x < 2.
To obtain a correct result we have to break up the
given integral in the following way :
3
2
3
(x 2) dx (x 2) dx (x 2) dx
2
0
2
0
and to put x = 2 – t in the first integral, and
x = 2+ t in the second. Then we get
2
dt
0
1
4
(x 2) dx t 2 t 2
I2 =
(x 2) dx t 2 t 2 tdt 3
2
0
3
2
4
2
1
0
0
dt
1
1
0
tdt
1
3
(x 2)3
1 8
3.
(x 2) dx
0
3
3 3
0
3
2
Discontinuous substitutions
The example that follows shows that a formal
application of the formula for a change of variable
(without due account of the conditions for its
applicability) may lead to an incorrect result.
Example 15. Since the integrand of the integral
I
1
1
1 x dx is positive, it follows that I > 0.
2
1
However, if we make the change of variable x = 1/u,
then I
1
1
1
1
1 x dx 1 u du I
2
1
2
1
This implies I = 0. Explain.
Solution The graph of u = 1/x has a discontinuity at
x = 0. Thus the indicated change of variable does not
satisfy the requirements of the Substitution Theorem.
To make the change of variables properly, we first
break up the interval of integration, i.e.,
I
1
0
1
1
1 x dx 1 x dx .
2
1
2
0
Now if we let x = 1/u, we obtain
I
1
du
1 1 u 2
The integrals
1
1 1 u 2
1
1
1 u du
2
du and
1 u du
1
1
2
can be evaluated to get the correct result.
However, by putting x = tan, in I
2
2
I1 =
8 1 3
which is a correct result. It can
3 3
be easily verified by directly computing the initial
integral :
Hence, I =
8
3
we get I
4 d
4
easily..
2
Note: While applying Newton-Leibnitz
Formula students are advised to check continuity of
antiderivatives before putting the limits of integration.
A discontinuous function used as an antiderivative
may lead to wrong result.
2.68
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 16. Find a mistake in the following
evaluation of the integral :
dx
dx
2
2
2
0 1 2 sin x
0 cos x 3sin x
2
sec xdx
1
tan 1 ( 3 tan x) 0 .
=
2
0
0 1 3 tan x
3
(The integral of a function positive everywhere turns
out to be zero).
Solution The Newton-Leibnitz formula is not
applicable here, since the antiderivative
1
tan 1 ( 3 tan x) has a discontinuity at
F(x) =
3
the point x = because :
2
1
lim F(x) lim
tan 1 ( 3 tan x)
3
x
x
2
2
1
tan 1 ()
and,
3
2 3
lim F(x) lim 1 tan 1 ( 3 tan x)
3
x
x
1 0 0
.
2
3 2
3
=
Now, we have an alternative solution for above
integral.
dx
I= 0
1 2 sin 2 x
sec 2 xdx
0 1 3 tan 2 x
=
using
If f(x) =
Put
1
= 1 tan ()
3
2 3
The correct result can be obtained in the following way:
dx
2
2
0 cos x 3sin x
1
dx
2
2
0 cot x 3 sin x
1
1
tan 1 ( 3 cot x)
.
0
2
2
3
3
3
It can also be found with the aid of the function
1
1
F(x) = 3 tan ( 3 tan x ) . For this purpose divide
the interval of integration [0, ] into two subintervals,
=
0,
2 and 2 , and take into consideration the
above indicated limit values of the function F(x) as
x
. Now,,
2
dx
0
=
2
1
3
0, f (2a x) f (x)
f(x)dx = a
2 0 f (x)dx, f (2a x) f (x)
sec 2 x
, we have f( – x) = f(x)
1 3 tan 2 x
/2
0
sec 2 xdx
1 3tan 2 x
2 dt
3 0 1 t2
3 tan x = t 3 sec x dx = dt
2
2
=
3
(tan–1(t)) 0 =
2
0
.
32
3
dx
1 2sin x 3 .
2
0
Example 17.
Calculate I =
2
dx
1 0.5cos x
0
1
Solution The integrand function f(x) = 1 0.5 cos x
is continuous on the closed interval [0, 2] and,
consequently, has an antiderivative. An appropriate
change of variable for finding an antiderivative of the
function f(x) is t = tan(x/2). However, when we seek
the integral I, such a change of variable does not
satisfy the conditions of the Substitution Theorem.
The change of variable t = tan(x/2) is permissible for
each of the intervals 0 x < and < x 2.
2dt
1 t2
, dx =
and we obtain an
1 t2
1 t2
antiderivative
cosx =
cos x sin x
2
..(1)
(1) reduces to I = 2
=
2
2a
0
=
2
/2
0
( 3 tan x)
dx
dx
2
2
2
/ 2 cos x sin x
cos x sin x
2
/2
0
1
3
tan 1 ( 3 tan x)
/2
dx
dt
C
(x) = 1 0.5cos x 4 3 t 2
t tan(x/2)
2.69
DEFINITE INTEGRATION
4 tan 1 1 tan x C
2
3
3
For any constant C, the function (x) is an
1
antiderivative of f(x) =
on the intervals
1 0.5 cos x
[0, ) and (, 2].
Since it has a discontinuity of the first kind at the point
4
x = , i.e. (+) – (–)= 3 , it follows that
1
(x) is not an antiderivative of f(x) =
on
1 0.5 cos x
the whole closed interval [0, 2].
However, using (x), we can now easily construct an
antiderivative for f(x) on the whole interval [0, 2] we set
4 tan 1 1 tan x
for 0 x ,
3
2
3
2
for x
F(x) =
3
4 tan 1 1 tan x 4 for x 2
2
3
3
3
We have thus taken C = 0 on [0, ], extended the definition
of (x) (for C = 0) to the point x = by continuity from the
4
left and have taken C = 3 on (, 2].
We have got a function F(x) whose derivative is equal
to the function f(x) at all points of the interval [0, 2],
the point x = inclusive, i.e. F(x) is an antiderivative of
f(x) on [0, 2].
From the Newton-Leibnitz formula we have
2
4 0 4
.
I = F(x) 0 = F(2) – F(0) =
3
3
=
Note: We could have divided the integral I
into two integrals I =
0
f(x)dx
2
for 0 x
for x
and on [, 2] the function
4
tan 1 1 tan x
2
3
3
F2(x) =
2
3
2
I = F1 (x) 0 F2 (x) = F1() – F1(0) + F2(2) – F2 ()
2 0 0 2 4
.
3
3
3
=
Prove that
Example 18.
I=
/2
for x
/2
cos2 d
.
cos 4 sin 2 6
2
0
0
/2
d
sec 2 d
2
2
1 4 tan 0 (1 tan )(1 4 tan 2 )
Now put tan = x
dx
(1 x )(1 4x )
I= 0
=
2
2
1 1
4
dx = .
2
2
0
3 1 x 1 4x
6
dx
Evaluate
Example 19.
=
0
(5 4 cos x)
0
(5 4 cos x)
0
2
x
dx
2
2
x
x
5 1 tan 2 4 1 tan 2
2
2
sec 4
Put tan
x
1 2x
= t sec dx dt
2
2
2
I=
9 t2 8
(1 t 2 )2dt
2
dt
2 2
2 2
0
0 (9 t )
(9 t )
2
dx
I=
Solution
1
dt
9 t dt 16 (9 t )
=2 0
2
2 2
0
for 0 x 2
Solution
f(x)dx and use
the fact that the antiderivative of f(x) on [0,] is the
function
4
tan 1 1 tan
2
3
3
F1(x) =
2
3
(F1(x) results from (x) for C = 0 when the definition of
(x) is extended to the point x = by continuity from
the left and F2(x) from the right). In that case, applying
the Newton-Leibnitz formula to each of the integrals,
we obtain
3
=
3 0
t
1 tan 1 t
2
3 0
18(9 t ) 54
= 2. 1 tan 1 t 16
2 16
1 4 5
.
. .
3 2 54 2
3 27 27
2.70
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
/2
1
1
2 2008
du
2 2008 ln u
=
–
2
2
(2009) u(1 u)
(2009) 1 u
Prove that
Example 20.
dx
1
( cos x sin x ) 4 3
0
/2
I = 0
Solution
dx
cos 2 x(1 tan x ) 4
a = 2, b = 2008 and c = 2009
(a + b + c) = 4019.
Example 22.
1
dz dt and
2 z
t = 1;
z = t = .
z= 0
1
2(t 1)
1
dt = 2
I= 1
4 dt
1 t3
t
t4
2 = t; then
ln(cot x)
If 0
(sin x)
(cos x)2009
2009
2
·(sin 2x)
2008
I=
=
Solution
=
ln(cot x)
2009
(cos x)
ln(cot x)
2008
4
= 0
1 (cot x)
2009 2
·
2
2009 2
a
1
ex
dx
1 a
(tan
e tan 1 e x ) (e 2x 1)
a
Put tan–1ex = t
I=
0
a
sin 1 e x cos 1 e x e x
dx
1 a
tan
e tan 1 e x e 2x 1
2
a b ln a
(where a, b, c are in their lowest terms)
c2
then find the value of (a + b + c).
4
dx
=
I = (sin x)
Show that the value of the definite
(sin 1 e x sec 1 e x ) dx
1 a
1 x
x
x , (a R) is
(tan e tan e )(e e )
independent of a.
Solution We have
integral
= 2 1 2 13 –2 1 1 = 1 – 2 1 .
2 3
3t 1
3 3
– 2t
Example 21.
4
·(sin 2x)2008 dx
2008
1
a bln a
=
c2
Let 1 +
1
u u 1 du
1
sec xdx .
(1 tan x ) 4
Put tan x = z then sec2x dx = dz and
x = 0 z = 0; x = z = .
2
dz
I= 0
(1 z )4
= 0
1
2 2008
u
2 2008 ln 2
l
n
=–
=
( 2009) 2 u 1
( 2009) 2
2
/2
22008
(2009)2
=–
2
tan 1 ea
0
ex
dx = dt
e2x 1
dt
(t tan 1 e a )
tan 1 ea
ln (t tan 1 ea )
0
2
1 a
1 a
ln 2 .
= ln 2 tan e ln tan e =
2
2
2008
(sin x) (cos x)
(sin x)2009 (sin x)2009
dx
1
Example 23.
Evaluate
x d(ln x)
2
1
Solution Here the limits are the values of ln x. Hence,
2 2008
4
ln(cot x) 2009
= – (2009) 0 (1 (cot x) 2009 ) 2 ·
Let (cot x)2009 = u,
2008
I=
2
(2009)
1
2009(cot x) 2008
dx = du
sin 2 x
ln(u)
(1 u) du
2009(cot x) 2008
dx
( sin 2 x)
2
ln x = – 1 x = 1/e
ln x = 1 x = e
Thus the given integral is equal to
e
1
x 2 · dx
x dx
1/ e
1/ e
x
e
e
2
2
4
= x
= e – 12 e 2 1 .
2 1 / e 2 2e
2e
2.71
DEFINITE INTEGRATION
G
1.
Applying substitution, calculate the following
integrals :
0.75
dx
1
xdx
(a)
,
(b) 0
1 5 4x
(x 1) x 2 1
(c)
2.
ln 2
0
e x 1dx
(d)
1
sin 1 x
0
x(1 x)
Evaluate the following integrals :
1 2x 3
(i)
2
0 5x 1
1
4
9.
cos x
(iii) 0
dx
4 3 sin x
3 2
3
2
x
cos
x 2 dx
(iv) 0
2
1
b
f(x)f '(x)dx = [f(b)] – [f(a)] .
2
2
a
1
(1 + 5x – x5)4 (x2 – 1) (x2 + 1) dx.
5.
Evaluate
6.
If f is continuous and
0
9
3
f(x)dx 4 , find
0
xf(x )dx
2
0
7.
Meet says, " 0 cos 2 d is obviously positive.
"Avni claims, "No, it's zero. Just make the
substitution u = sin ; hence du = cos d. Then I
get
cos d = cos cos d
= 1 u du 0 Simple."
2
0
0
0
0
2
= 14 u du = – 14 u du,
which is obviously negative." Who is right?
Why is it impossible to use the substitution
x 1 x dx ?
3
2
2
10. Verify the result of transforming the integrals
can take, as the new limits of integration, the
numbers
(a) and /2 ,
(b) 2 and 5/2
(c) and 5/2.
Calculate the integral in each case when this
change of variable is permissible.
If f ' is continuous on [a, b], show that
2
2x2 dx = 1–2 x · 2x dx
3
1 x 2 dx by changing a variable x = sint, we
0
1
–2
x = sin t in the integral
Find out whether, when calculating the integral
1
Avni asserts that 1– 2 2x2 dx is obviously positive.
"After all, the integrand is never negative and
–2 < 1." "You are wrong again," Meet replies,
"It's negative. Here are my computations. Let
u = x2; hence du = 2x dx. Then
5
1
4.
8.
dx
(ii) 5x x 1 dx
3.
(a) Who is right? What is the mistake?
(b) Use the identity cos2 = (1 + cos2/2 to
evaluate the integral without substitution.
(4x – x 1 / 16) dx, and cos xdx
2
0
2
0
by the substitutions 4x2 – x + 1/16 = y, and x = sin -1y
respectively.
11. Make sure that a formal change of the variable
t = x2/5 leads to the wrong result in the integral
2
2
5
x 2 dx . Find the mistake and explain it.
12. Is it possible to make the substitution x = sec t in
1
x 1 dx .
13. Given the integral 1 x dx . Make the
the integral I =
2
0
1
2
0
substitution x = sin t. Is it possible to take the
numbers and as the limits for t?
2
dx
. Make sure that
14. Given the integral
2
0 1 cos x
2 cos x
1
1
the functions F1(x) = 2 cos
and
1 cos2 x
1 tan x
1
F2(x) = 2 tan
2 are antiderivatives for the
integrand. Is it possible to use both antiderivatives
for computing the definite integral by the NewtonLeibnitz formula? If not, which of the
antiderivatives can be used?
2.72
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
G
15. Evaluate the following definite integrals by
finding antiderivatives:
8 sin x 1
dx
(i)
3
x 1
(ii)
/4
0
cos 2x 4 sin 2x dx
1 e 2
ln(1 t)
dt
1 t
1
1 sin
x
dx
(iv) 0
1 x2
16. Evaluate the following definite integrals by
finding antiderivatives :
1/ 2
dx
(i)
2
0
(1 2x ) 1 x 2
(iii)
18. Find the value of 'a' such that
19. The integral
dx
(ii)
(iii)
/2
0
(x )( x)
3
17. Evaluate the following definite integrals by finding
antiderivatives :
sin x
4
4
(i)
dx
0
sin 2x 2(1 sin x cos x)
(ii)
(iii)
(iv)
2.9
4
2
0
x2 4
dx
x4
x 2 dx
3
(x 3)(5 x)
INTEGRATION BY PARTS
FOR DEFINITE
INTEGRALS
The formula for integration by parts in case of definite
2 dz
0.
1 z2
(1 z ) 5 3
1 z2
2
1 f (x)dx
0
1 x
b
/2
f (x)dx
2
f (a cos2 bsin 2 )d .
0
(x a)(b x)
2
/2
0
f (sin ) d,
a
1 5
2
x2 1
1
ln 1 x dx
1
x
x4 x2 1
1 2e 2x xe x 3e x 1
dx
x
2
x
2
0 (e 1) (e x 1)
21. Evaluate
22. If
1
, where K is a constant. Find
(e 1)(e 2)
the value of K.
23. Suppose that the function f, g, f and g are
continuous over [0, 1], g (x) 0 for x [0, 1],
=K–
2009
and g (1) = 1. Find
2
the value of the definite integral,
2
5
x
z . We have
2
Find the mistake.
20. Establish the following :
sin x
x
cos x dx
4
x
sin x
0
0
(x 2)2
2
dx
2
(b > a)
1 (x 2) dx
(iv)
0
5 3cos x
dx
4 5sin x
3
dx
2
5 3 cos x is readily taken with
the aid of the substitution tan
x
x
0
0
dx
a
e 4e 5 = ln 3 2 .
f (0) = 0, g(0) = , f (1) =
f(x)·g'(x){g2 (x) 1} f '(x)·g(x){g2 (x) 1}
dx .
0
g2 (x)
1
integrals is as follows :
b
u(x)v(x)dx
a
u(x) u(x)dx
u(x) v(x)dx dx
b
b
a
a
2.73
DEFINITE INTEGRATION
Let I =
Solution
=
u
0
/4
0
/4
1
–
2
8
1
–
= +
2
8 8
/ 4
e
e
ln 2
2
f (ln 2) = 6, f '(ln 2) = 4 and
0
ln 2
Solution
''(x) dx
e·f
I=
1/ e
0
Dividing the integral I into the sum of
integrals over the closed intervals [1/e, 1] and [1, e] (to
get rid of the absolute value and using, in each case,
the formula of integration by parts, we obtain
1
1
e
dx + x ln x – dx
e
1
1/ e
Suppose f , f ' and f '' are continuous
e
on [0, e] and that f ' (e) = f (e) = f (1) = 1 and
=
1
, then find the value of
2
e
f(x)
x dx
1
f ''(x) ln x dx
1
ln 2
e ·f '(x)dx
2x
0
1
= 4 f '(ln 2) 1·3
ln 2
ln 2
2x
e
·
f(x)
2
e 2x f(x)dx
+2
0
0
6
1
= (1 – 3) + 2 f(ln 2) 2·3 = – 2 + 2 6
4
4
= – 2 + 3 + 12 = 13.
1
1 1
1
= 1 + e –(e – 1) = 2 1 .
e e
e
Example 3.
v
ln xdx + ln xdx
= x ln x 1 / e +
u
+2
= e–2x · f '(x)
Solution
1
2x
0
ln 2
lnx dx
2x
0
2x
ln 2
1
sec x d x = – .
4 2
0
2
e ·f(x)dx 3 . Find the
0
/ 4
1/ e
2
e ·f ''(x)dx .
value of
(sec2 x – 1) d x
e
I= –
f(x)
1
f ' and f ''
are continuous on [0, ln 2] and that f (0) = 0, f '(0) = 3,
tan x d x
e
e
x dx
I= 1–
/ 4
1
e
1 1 3 1
+ = .
e 2 2 e
Example 4. Suppose that f,
0
Calculate I =
Example 2.
f '(x)
dx
1
x
1 1
1 1
= 1 + = –
e 2
e 2
/4
0
u
1
1
f '(x) dx = ·f(x) +
I1 =
x
1 x
1
2
/4
v
I = 1 – I1
tan x sec x d x d x
0
1
e
tan 2 x
2
tan x sec2 x dx =
''(x) ln
x dx
f
I=
= ln x·f '(x) 1 –
v
x tan 2 x
1
–
I=
2
2
0
=
Solution
x sin x
dx
cos 3 x
I = [x tan x sec2 x d x]
–
e
x . sin x
dx
cos3 x
x tan x sec 2 x d x
/4
We have
/4
Evaluate 0
Example 1.
2
Example 5.
Solution
/2 (1 2 cos x)
Evaluate 0
(2 cos x)2
/2 (1 2 cos x)
Let I = 0
(2 cos x)2
/2 cos x(2 cos x) sin 2 x
= 0
(2 cos x)2
dx
dx
dx.
2.74
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
/2 sin 2 x dx
cos x
dx
+
0
0
(2 cos x)
(2 cos x)2
In the first integral, integrating by parts taking cos x
as second function and second integral unchanged,
we have
=
/2
/2
1
sin x
I=
(2
cos
x)
0
/2
0
/2 sin 2 x dx
sin 2 x dx
2
0
(2 cos x)
(2 cos x)2
1
1
0 .
(2 0)
2
Alternative :
=
t t
I=
2
2
Example 6.
/4
0
Evaluate
/4
0
sin 2 xdx
2mx
e (cos x msin x)2
sin 2 xdx
e 2mx (cos x m sin x)2
2
Integrating by parts taking
3
3
3
3
3
n
1/n
mx
1
[ ln (n 3 + r 3) 3 ln n ]
n
3
1 3lnn ln 1 r 3 . lnn
=
n
h
3
r
ln 1
h
=
n
sin x
as the first function,
e mx
sin x
1
1
I = – (m 2 1) e mx e mx (cos x m sin x)
0
/ 4
1
=
ln (1 + x ) d x
=
ln (1 + x) d x + ln (x –1 x + 1) dx
3
0
1
1
0
0
1
= ln (1 + x) . x 0 0
1
2
1 x 1
dx
1 x
1
x (2 x 1)
x x 1 dx
ln (1 – x + x2) . x 0 0
1
3
( (m 1)e sin x)
dx
e mx e mx {e mx (cos x m sin x)}2
/4 sin x
3
n
3
r
1
lim
1
Let S = n
ln
h
n r 1
sin 2 xdx
{e mx (cos x m sin x)}2
1
=–
(m 2 1) 0
3
1
[{ln(n3 + 1) + ln(n3 + 23) + .....
n
.... + ln(n3 + n3)} – 3ln4]
2
Show that
=
1
1
1
=–
= – 0 = .
2
t 2
2
/4
=
(m 1)e m/2
1
1
2m
(m 1) 2m(m 1)
2
Hence T r =
Let I = 0
Solution
=–
.e 3 .
Solution ln P
(2 cosec x cot x)2
Put 2 cosec x + cot x = t
(2 cosec x cot x + cosec2 x) dx = – dt
When x = 0, t = ,
dt
1
1
1 m /2
1)
(e
(m 2 1) (m 1)e m /2 2m
= 4e
0
2 dt
=–
n
/2 (cosec 2 x 2 cosec x cot x)dx
, t = 2.
2
1
. mx
dx
e (cos x msin x)
1
2mx /4
1
e
2
=–
(m 2 1) e m /2 1 m 2m 0
2
dx
(2 cos x)2
Dividing Nr and Dr by sin2 x, we get
when x =
2mx
lim (n 1) (n 2 ) (n 3n 3 ) ....... (n n )
/2 (1 2 cos x)
e
Example 7.
I= 0
I =
/4 e mx cos x sin xme mx
+ 0
2
2.75
DEFINITE INTEGRATION
2 x2 x
1
n
= ln 2 (1 ln 2) 0 x 2 x 1 dx.
=
S = ln 4 1 I1
1
x2
I1 = 0 2 2
.
dx = 2
x x 1
3
= ln 4 +
3.
3
3
0
f (x) [x] x
Hence sin n sec d
=
n
n
denotes the greatest integer function and n N.
Now,
Solution
I=
n
n
n
0
0
1
n
n
0
r 1 r 1
n
1
0
n
0
=
r 1
n
/2
0
sin (n - 2) sec d
2 sin 3 cos 5
d
cos
1
I 1
2 8
I8 =
r
1
1
f(n) f(0) f(x)dx
2
2
0
2
cos (n 1)
( n 1)
1 / 2 sin 8 sin 2
cos
2 0
I=
(r 1) f (x)dx nf(n)
r 1
1
2
0
f (x)[x]dx (xf(x)) f(x)dx 2 (f(x))
=
=
n
f (x)[x]dx xf (x)dx 2 f (x)dx
r
cos 5 sin 3
d.
cos
Consider sin n + sin (n 2)
= 2 sin (n 1) cos
sin n sec = 2 sin (n 1) sin (n 2) sec
1
dx
2
1
1
= 0 f(x)dx 2 f(0) 2 f(n) f(r) , where [.]
r0
/2
Solution
Example 8. Show that for a differentiable function
Prove that sin n sec d
evaluate 0
n
0
2 cos(n 1)
sin (n 2) sec d .
n 1
Hence or otherwise
P = 4 e 3 . e 3.
f(x),
n
=
3
3
ln P = ln 4 +
1
r 1
Example 9.
S = ln 4 1 2 +
1
f(r) 2 f(n) 2 f(0) f(x)dx .
/ 2 sin 6
2
cos 7 ] 0/ 2 0
d .
cos
7
2 2
/2
cos 5 0
7 5
/2
0
/2
=
2 2 2 cos3
0
7 5 3
sin 4
d
cos
/2
0
sin 2
d
cos
= (r 1){f(r) f(r 1)}
=
2 2 2
2 2 2 2
2 =
5
3
7 5 3 1
7
n
1
1
f(n) f(x)dx f(0)
0
2
2
= – f(1) – f(2) – ...... – f(n – 1) – f(n)
=
30 42 70 210
152
=
105
105
n
r 1
nf(n)
+
1
1
f(n) + f(0) +
2
2
n
f(x)dx
0
I=
76 105
181
1
I 1=–
=
.
105
105
2 8
2.76
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 10.
0
I=
Solution
2
2
0
Example 12.
2
x
dx
x sin x cos x
x cos x
x
·
dx
2
(x sin x cos x) cos x
2
=–
1
x
cos x x sin x cos x 0
+
2
0
sin x(xsin x cos x) x 2
Solution
I=
2
0
1 0
2
=
–0= .
2
Evaluate I =
3
= 2 tan 2
2
+
0
tan 1 x
x dx
0
2
0
sec 2 3
· d
3
u
tan
v
3 2 2
d
2 0 tan 2
3 2 2
(cosec2 1) d
=
2 0
2
2
3 2
3
( cot )d
cot 0 2
0
2
24
3
(0 ) 2
=
2
=
3
3
ln 2 .
2
16
2
0
/2
cos x·
= cos · ln sin +
/2
(cos ecx – sin x) dx
= lim {(cos – 1) ln sin } + ln2 –1
0
ln sin
ln 2 1
1
0
cos 1
[applying L' Hospital's Rule]
cos
sin
ln 2 1
= lim
1
0
·(–
sin
)
(cos 1)2
= lim
cos (cos 1)2
ln 2 1
0
sin 2
cos (1 cos )2
ln 2 1
= lim
0
1 cos2
cos (1 cos )
ln 2 1
= lim
0
1 cos
= 0 + ln 2–1 = ln 2 –1.
= lim
3
3 2 2
2
(
·cosec
)
d
= 2 0
24
u
v
=
cos x dx,
using by parts.
sin x
/2 1 sin 2 x
dx
= cos · ln sin +
sin x
0
3
Put x = tan
3
·sec2 d =
tan 3
sin x · ln sin xdx .
= cos ln sin + [– ln (cosec x + cot x) + cos x] / 2
= cos · ln sin + ln (cosec + cot ) – cos
1 cos
= cos · ln sin + ln
– cos
sin
= (cos – 1) ln sin + ln (1 + cos ) – cos
LHS = lim I
0
Example 11.
/2
sin x cos x x cos2 x 2
=
cos x(xsin x cos x) 0
2
Let I =
Solution
–
sin x x cos x
= xsin x cos x
0
sin x · ln sin xdx ln 2 – 1.
Now, I = [ln sin x.(– cos x)] / 2
cos x x sin x
·sec 2 x dx
(x sin x cos x)
cos x(xsin x cos x)
/2
0
I.
Clearly, the required integral = lim
0
1
sin x 2
x
+
=
·
cos x x sin x cos x cos x 0
=
Show that
3
2 4
Example 13.
/2
0
Prove that
xsin x cos x
dx
.
2
2
2
2
2
(a cos x b sin x)
4ab (a b)
2
2.77
DEFINITE INTEGRATION
xsin x cos x
Solution Let I = 0 (a 2 cos2 x b2 sin 2 x)2 dx
Integrating by parts taking x as the first function, we
have
/2
/2
1
I = x
2
2
2
2
2
2
2(b
a
)(a
cos
x
b
sin
x)
0
/2
1
– 0
dx
2(b2 a 2 )(a 2 cos2 x b2 sin 2 x)
/2
1
sec 2 xdx
=–
4(b2 a 2 )b 2 2(b2 a 2 ) 0 a 2 (b tan x)2
Put b tan x = t sec2 x dx =
dt
b
1
dt
=–
2
2
2
2
2
2
0
4(b a )b
2(b a ) b(a t 2 )
1
tan 1 t
4(b2 a 2 )b2 2ab(b2 a 2 )
a 0
1
0
=
4(b2 a 2 )b 2 2ab(b2 a 2 ) 2
1
=
4(b2 a 2 )b2 4ab(b2 a 2 )
(b a)
=
.
4ab2 (b2 a 2 )
Example 14. Evaluate
3 /2
0
(ln | sin x |) cos (2nx) dx, n N.
Solution
Let In =
3 /2
0
(ln | sin x |) cos (2nx) dx
Integrating by parts, taking cos 2nx as the second
function, we have
=
3 /2
3 /2 cot xsin 2nx
sin 2nx
dx
0
2n 0
2n
1 3 /2 sin 2nx cos x
=0–
dx
2n 0
sin x
1
I
...(1)
=–
2n n
In – In – 1
1 3 /2 cos x{sin 2nx sin(2n 2)x}
=–
dx
2n 0
sin x
1 3 /2 cos x2 cos(2n 1)x sin x
dx
=–
2n 0
sin x
1 3 /2
(2 cos (2n – 1) x cos x) dx
=–
2n 0
1 3 /2
(cos 2nx + cos (2n – 2) x) dx
=–
2n 0
In = ln | sin x |
3 /2
1 3 /2 sin 2nx sin(2n 2)x
2n 0
2n
2n 2
0
=–
1
{(0 + 0) – (0 + 0)} = 0.
2n
In = In – 1 = In – 2 = ......= I1
=–
Now, I1 =
3 /2 sin 2x cos x
sin x
0
=
3 /2
dx
(2 cos2 x) dx =
0
3 /2
0
(1 + cos 2x) dx
3 /2
sin 2x
3
3
= 0 (0 0) =
.
2
2 0
2
3
1
In =
. From (1), In = –
I
2
2n n
3
.
Hence In = –
4n
= x
H
1.
Applying the formula for integration by parts
calculate the following integrals :
(a)
n 2
0
xe x dx
(b)
2
x cos x dx
2
0
1
(c) cos 1 xdx
0
2.
Let f and g be differentiable on [a, b] and
suppose f and g are both continuous on [a, b],
then prove that
b
b
f (x) g(x) dx + a f(x)g(x) dx = f(b)g(b) – f(a) g(a).
a
3.
Suppose f and g are continuous and
f(a) = f(b) = 0.
Prove
b
b
f(x) g(x) dx = – f (x) G(x) dx,
a
a
x
where G(x) = g(t) dt.
a
2.78
4.
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Suppose that f(1) = 2, f(4) = 7, f'(1) = 5, f'(4) = 3
and f" is continuous, Find the value of
5.
Show that
x
(a)
e t t dt = e–x(ex – 1 – x).
0
x
x 2 x3
x
(b)
e t t 3dt = 3!e–x e 1 x 2! 3! .
0
4
xf "(x) dx
1
H
6.
Calculate the following integrals :
1
xdx
(a) 1 2
x x 1
(b)
e
(xnx) dx
2
1
x
dx
1 x
2
dx
(d) 0
(2 cos x)(3 cos x)
0 sin
(e)
(f)
7.
3
(c)
/2
0
1
Show that
2
0
sin x
2
b
dx
(iii) 1
ln(x x e x )
0
8.
Compute
(c) Similarly, show that
dx
1
0
2.10 REDUCTION FORMULA
If In =
/4
0
Solution
/4
tan n 1 x
tan n xdx
0
n 1 0
1
In 2 .
In =
n 1
/4
Now I5 =
=
/4
0
tan n 2 xdx
I n + In – 2 =
tan n xdx show that
1
and deduce the value of I5.
I n + In – 2 =
n 1
f ( 2 ) (0 ) 2
b
2
1 b (3)
f (x) (b – x) 2 dx
2 0
(d) Check that (c) is correct for any quadratic
polynomial.
(e) Use another integration by parts on the
formula in (c) to obtain the next formula.
+
f(0) = 1, f(3) = 4, and f '(3) = –2.
(2)
f(b) = f(0) + f '(0)b +
x f "(3x)dx , given that f'(0) is defined,
Example 1.
0
0
e
(iv) x (tan x)
b
f (x) dx .
f (x)(b – x)dx
0
2
0
vdx = x – b, show that f(b) = f(0) + f '(0)b +
2
[cos x(1 x) (1 x) sin x]dx
(ii) x (1 x cos x ·ln x sin x) dx
-1
1
(b) Using an integration by parts on the
derivative integral in (a), with u = f '(x) and
(x sin x) dx
1
0
(a) Explain why f(b) = f(0) +
2
xf (x)dx 3 for every function f(x)
that satisfies the following conditions : (i) f(x) is
defined for all x, (ii) f"(x) is continuous,
(iii) f (0) = f (1), (iv) f ' (1) = 3.
1
10. (a) Find an integer n such that n xf (2x)dx
Evaluate the following definite integrals by finding
antiderivatives :
(i)
1
= tf (t)dt . (b) Compute xf (2x)dx , given
0
0
that f(0) = 1, f(2) = 3, and f '(2) = 5.
11. Let f have derivatives of all orders.
sin x sin 2x sin 3x dx
9.
1
Hence proved.
n 1
1
1 1
I3 I1
4
4 2
/4
1 1
1
tan xdx ln | sec x | 0 / 4
0
4 2
4
=
1
1
1
ln 2 ln 2 .
4
2
4
2.79
DEFINITE INTEGRATION
If un =
Example 2.
/2
0
x n sin x dx, (n > 0), then
1
prove that un + n(n – 1)un–2 = n
2
Integrating by parts, un =
Solution
u n x n cos x
= n x
n 1
1
2
= n
/2
n
0
/2
sin x 0
0
(n 1)
Example 3.
/2
If un =
0
0
x
un =
/2
0
sin x dx
=
1
1 n 1
1
dx
x dx x n 1 ·
0
0
4
1 x 2
=
1
1 1
– + xn–1 tan x 0
4 n
1
0
x(sin x) n dx , n > 0,
0 + (n–1)
/2
0
/2
0
(n + 1) un + (n – 1)un – 2 =
1
n 1
cos x dx
x (sin x)n 2 (1 sin 2 x) dx
(sin x) n 1.cos x dx
1
If un = x n tan 1 x dx then prove
0
that
(n +1) un + (n – 1) un–2 =
1
.
2 n
In+2 =
0
t n dt
0
t n 1
– In.
n 1
Also evaluate
t6
dt.
0 (1 t 2 )
1
1
Solution
1
– .
n
2
If In = 1 t 2 , show that
Example 5.
1
1
un = (n –1) un–2 – (n – 1) un +
n
1
n un = (n – 1) un–2 +
n
n 1
1
u n 2 2 .
un =
n
n
Example 4.
1
–
– (n – 1)un – 2
n
2
(n + 1) un =
/2
.cos x (sin x)
– (n 1)x n 2 · tan 1 x dx
.
– x (sinx)n–1 . cos x 0
+ 0 (n 1) x (sin x)
n 1
(n + 1)un
n 1
x (sin x) n 1 . sin x dx
n 2
1
1 1 x n 1[1 x 2 1]
dx
=
4(n 1) n 1 0
1 x2
n 1
1
u n 2 2 .
n
n
/2
=
xnsin x dx
0
– n(n – 1)un–2
then prove that un =
=
/2
n 2
xn tan–1 x dx
1
x n 1 cos x dx
/2
1
0
1
x
tan 1 x· x n 1
=
–
n 1
n 1 0 1 x2
0
.
n 1
1
un + n(n – 1)un–2 = n
2
Solution
/2
n 1
un =
Solution
tn2
1
In+2 = 1 t 2 dt =
0
0
t n { (1 t 2 ) – 1}
1 t2
1
n
tn
t n 1
in result
= t – 1 t 2 dt we require
n 1
0
n 1
t
=
–I .
n 1 n
dt
Now I0 =
= tan–1 t.
1 t2
Using the reduction formula,
For n = 0, I2 = t – I0 = t – tan–1 t,
For n = 2, I4 =
t3
t3
– I2 =
– t + tan–1 t,
3
3
dt
2.80
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
t6
t5 t3
–1
–
t
–
tan
t
dt
=
2
5 3
0 (1 t )
0
1
13 – .
= 1 – 1 1 – – 0 =
15 4
4
5 3
Compute the integral
Example 6.
a
In = (a 2 x 2 )n dx , where n is a natural number..
0
Solution The integral can be computed by
expanding the integrand (a 2 – x2)n according to
Binomial theorem, but it involves cumbersome
calculations. it is simpler to deduce a formula for
reducing the integral In to the integral In–1. For this let
us expand the integral In in the following way :
a
In = (a 2 x 2 )n 1 (a 2 x 2 )dx
0
a
= a2In–1 – x(a 2 x 2 )n 1 xdx
0
and integrate the latter integral by parts.
We obtain
1 a 2
1 x (a 2 x 2 ) n a
(a x 2 ) n dx
In = a2In–1 +
0
2n
2n 0
1 I
= a2In–1 –
n
2n
2 2n
I n 1
In = a
2n 1
This formula is valid at any real n other than 0 and
1.
2
In particular, at natural n, taking into account that
2n(2n 2)(2n 4)....6.4.2
(2n 1)(2n 1)(2n 3)...5.3
Expanding the integrand by the Binomial theorem
and integrating within the limits from 0 to 1, we get :
1
In = (1 x 2 ) n dx
0
1
= 0 (1 n C1 x 2 n C2 x 4 n C3 x 6 ... ( 1)n n Cn x 2n )dx
1
n
C1 x 3 n C2 x 5 n C3 x 7
(1) n n Cn x 2n 1
dx
= x 3 5 7 ...
2n 1
0
n
C1 n C2 n C3
Cn
...... ( 1)n
.
3
5
7
2n 1
Using the result of the previous example we can easily
complete the proof.
= 1
n
Find a reduction formula for the
sin nx
sin nx dx
or 0,
and show that 0
integral
sin x
sin x
according as n is odd or even.
Solution We have
sinnx – sin(n – 2) x = 2 cos(n – 1) x sinx
sin nx
sin( n 2)x
= 2 cos(n – 1)x +
sin x
sin x
Integrating both sides we get
2 sin(n 1)x
sin(n 2)x
sin nx
sin x dx (n 1) sin x dx
Above is the required reduction formula.
Taking the limits from 0 to , we get
In = 0 + In – 2 = In–4 ... = I2 or I1
according as n is even or odd.
Example 8.
a
I0 = dx a , we get
0
In = a2n + 1
2n(2n 2)(2n 4)....6.4.2
(2n 1)(2n 1)(2n 3)...5.3
Using the result of the preceding
Example 7.
example obtain the following formula :
1
n
C1 n C2 n C3
Cn
...... ( 1)n
3
5
7
2n 1
2n(2n 2)(2n 4)....6.4.2
=
.
(2n 1)(2n 1)(2n 3)...5.3
1
Consider the integral.
In = (1 x ) dx
0
2 n
sin 2x
dx 2 cos xdx
0
sin x
sin x
dx x 0 = .
sin x
= 2sin x 0 0 .
If n is odd = In = I1 = 0
n
Solution
If n is even, In = I2 = 0
2
sin n
Prove that 0
d n ,
sin
Example 9.
n W.
Solution If the given integral is denoted by In then
In – In–1 =
sin 2 n sin 2 (n 1)
0
sin 2
d
2.81
DEFINITE INTEGRATION
=
sin(2n 1) sin
sin
0
Solution
d
2
sin(2n 1)
= 0
d
sin
[using the result of the previous question]
In – In–1 = or In = + In–1 = + ( + In –2)
or In = 2 + In–2 = 3 + In–3 .... = n + I0 = n
I0 =
sin 2 0
d
0 d 0 .
0 sin
0
Alternative :
We show that I1, I2, I3,....constitute an arithmetic
progression.
In + 1 – 2In + In – 1 = (In + 1 – In) – (In – In – 1)
/2 sin 2 (n 1)x sin 2 nx) (sin 2 nx sin 2 (n 1)x)
= 0
=
sin 2 x
/2 (sin(2n 1)xsin x sin(2n 1)xsin x)
2
sin x
0
dx
dx
sin x
/2 2 cos 2nx sin x
= 0
dx = 2
sin x
/2
0
cos 2nx dx
/2
sin 2nx
1
= (sin n – sin 0) = 0 – 0 = 0.
2n 0
n
In – 1 + In + 1 = 2In
=2.
i.e., In – 1, In, In +1 from an A.P.
Now I0 = 0 and I1 =
0
d
Hence I2 = I1 + (I1 – I0) = + – 0 = 2
Similarly, In = nI.
Example 10.
Vn
1
2
If Sn
1
2
0
sin nx
sin(2n 1)x
dx ,
sin x
2
sin x dx ,
0
n being an integer, then show that
Sn+1 = Sn =
1
, and Vn+1 – Vn = Sn+1.
2
Also obtain the value of Vn.
/2 sin(2n 1)x sin(2n 1)x
sin x
0
/2 2.cos 2nx.sin x
sin x
0
dx 2
/2
0
dx
cos 2nx dx
/2
sin 2 nx
2
0 for all integral values of n.
2 n 0
Sn+1 = Sn = Sn–1 ..... = S1.
/2 sin x
Now, S1 0
dx
sin x
Sn+1 = Sn =
/2
0
dx
.
2
.
2
Also, Vn+1 – Vn =
/2 (sin(2n 1)x sin(2n 1)x)
= 0
Sn 1 Sn
/2 sin 2 (n 1)x sin 2 nx
sin 2 x
0
dx
/2 sin(2n 1)x.sin x
dx
sin 2 x
/2 sin(2n 1)x
dx Sn 1
0
sin x
Vn – Vn–1 = Sn = , Vn–1 – Vn–2 = ,
2
2
.... , V2 – V1 = .
2
On adding, Vn – V1 = (n – 1) .
2
/2
n
dx , we haveVn =
.
Since V1
0
2
2
1 x cos nx
Example 11. If Un = 0 1 cos x dx where n
is a positive integer or zero, then show that Un + 2 + Un
0
= 2Un + 1. Hence show that
sin 2 n
n
sin 2 .
0
2
1 cos nx
dx
Un = 0
1 cos x
Un + 2 – Un + 1
Solution
=
=
(1 cos(n 2)x) (1cos(n 1)x)
0
(1 cos x)
cos(n 1)x cos(n 2)x
0
(1 cos x)
dx
dx
2.82
=
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
2sin(n 3 / 2)xsin x / 2
dx
2sin 2 x / 2
sin(n 3 / 2)x
Un + 2 – Un + 1 =
dx
0
sin x / 2
sin(n 1 / 2)
Similarly, Un + 1 – Un =
dx
0
sin x / 2
From (1) and (2), we get
(Un + 2 – Un + 1) – (Un + 1 – Un)
sin(n 3 / 2)x sin(n 1 / 2)x
dx
= 0
sin x / 2
2 cos(n 1)xsin x / 2
dx
= 0
sin x / 2
sin(n 1)x
=2
(n 1) 0
...(1)
...(2)
0
If In = ex sinn xdx n 2 N,
Example 12.
then prove that In – 2, In, In + 2 cannot be in G.P.
0
In = ex sinn x dx
Solution
= sin n x e x dx
= –n
0
sinn – 1 x cos x ex dx
0
0
= n(n – 1) (sin n 2 x (1 sin 2 x))e x dx
0
n sin n x e x dx
0
= n(n – 1) sin
0
Un – Un – 1 = U1 – U0 = – 0
Un = + Un – 1 = + Un – 2 = 2 + Un – 2
Un = n + U0
Un = n
[ U0 = 0]
/2 1 cos 2n
d .
1 cos 2
sin
Put 2 = x
dx
d =
2
/2 sin 2 n
1 1 cos nx
d
dx
Hence, 0
2
2 0 1 cos x
sin
0
x e dx n(n 1)
x
0
sin x e dx n sin x e dx
= Un – Un – 1 = .... = U1 – U0
d
n2
Similarly, Un + 2 – Un + 1 = Un + 1 – Un
2
– n sinn – 1 x cos x ex dx
= 2Un + 1
/2 sin 2 n
0
n ((n 1) sin n 2 x cos 2 x sin n 1 x sin x)e x dx
Un + 2 – Un + 1 = Un +1 – Un
Hence, 0
0
= –n [sin n 1 x cos xe x ]0
Hence, (Un + 2 – Un + 1) – (Un + 1 – Un) = 0
Un + 2 + Un
1
1
U = n from above.
2 n 2
=
0
n
x
n
x
In = n(n – 1)In – 2 – n(n – 1)In – nIn
In(1 + n2) = n(n – 1)In – 2
n(n 1)
In = 2
I
n 1 n–2
(n 1)(n 2)
I
Now In + 2 = 2
n 4n 5 n
From equation (1) and (2)
...(1)
...(2)
In
I
n(n 1)
(n 1)(n 2)
and n 2 2
2
I n 2
In
n 1
n 4n 5
Let In – 2, In and In + 2 are in G.P., then
I n2
n(n 1) (n 1)(n 2)
2
2
I n 2
In
n 1
n 4n 5
2n2 + 8n + 2 = 0 which is not possible n N.
In – 2, In and In + 2 cannot be in G.P.
In
I
1.
Derive a reduction formula and compute the
0
integral x e dx , (n is a positive integer).
n x
1
2.
e
Prove that if J m ln
1
m
x dx , then Jm = e – mJm–1,
(m is a positive integer).
3.
Evaluate 04 d
dx
4.
If In =
I5 =
/2
0
sec d dx.
x
4
1
x n sinx dx, prove that
5 4
– 152 + 120.
16
DEFINITE INTEGRATION
5.
6.
7.
a
(a x ) dx .
Evaluate
Evaluate
a
2
2 5/2
0
(i)
(ii)
x
dx
0 (1 x 2 ) 2
0
2
x (a2 – x2)5/2 dx,
4
1
Prove that
/2
0
(iii)
1
x e dx
3 x
2
0
8 .
cos4 x cos 3x dx =
35
sin 7x
dx .
sin x
1 cos 5x
5.
9. Prove that 0
1 cos x
dx
10. Prove that ( x 2 1) 4
11. Show that
dx
2n 3
I n 1
(i) I n 0
(1 x 2 )n 2n 2
dx
1.3.5.7
(ii) 0
.
(1 x 2 )5 2.4.6.8 2
12. Employing Euler's formula eix = cosx + isinx, prove
8.
Prove that 0
2 inx
that 0 e
.eimx dx
02 forfor mm n,n
13. Using Euler's formulae cosx =
1 (e ix e ix )
,
2
ix
1 ix
sin x = (e e ) calculate the integrals :
2i
(a)
/2
0
sin 2m x cos 2n xdx
2.11 EVALUATION OF LIMIT
OF SUM USING NEWTONLEIBNITZ FORMULA
From the definition of definite integral, we have :
n 1
(b a) r
ba
b
f a
= f (x) dx
1. nlim
n
a
n
r0
n
2.
ba ba
b
f a
r = f (x) dx
n r 1 n
a
n
lim
sin nx
(b)
sin x dx
(c)
cos x cos nxdx
(d)
sin x cos nxdx
14. Find
2.83
0
n
0
n
0
1
x (1 x) dx (p and q positive integers).
p
q
0
22n (n!) 2
.
(2n 1)!
1
1
n –1
16. Utilizing the equation 0 x dx , compute
n
1
15. Show that 0 (1 x ) dx
2 n
1
the integral x n –1 (ln x)k dx .
0
17. If un = cos n cosec d, prove that
2 cos(n 1)
. Hence or otherwise
n 1
/2 sin 3 sin 5
d .
find the value of 0
sin
m
1 x dx
18. Compute 0
when m is (a) even, (b) odd
1 x2
(m > 0).
1 (1 x) n
dx ,
19. Derive a formula for In =
0
x
(n is a positive integer).
1
3 n
20. If In = (1 x ) dx, prove that In = 3n In–1.
0
3n 1
n
C0 n C1 n C2
– ...
Hence, evaluate
1
4
7
u n u n 2
n 1
3.
lim
n
1
r
n f n = f (x) dx
r0
1
0
(x)
4.
b
1
r
f f (x)dx , where
n a
n n
r (x)
lim
(i) is replaced by
(ii)
sign
r
is replaced by x,
n
2.84
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
is replaced by dx,
n
(iv) To obtain the limits of integration, we use
(iii)
a = lim
n
(x)
(x)
and b = nlim
n
n
Solution
Limit = lim
n
Note: that we have to see the following
things before we can express the limit of its sum as a
definite integral.
(i) Each term of the series must be multiplied by 1/n
which 0 when n .
(ii) All the terms should be some function of r/n which
varies from term to term in A.P. with common
difference 1/n
Example 1.
Evaluate
Let L
1
1
1
1
.....
= nlim
4n 4
4n 9
3n
4n 1
n
= nlim
= lim
n
2
1 0
4 0 r
n
2
ba
(1 3x )1 2
dx =
3/ 2
1 3x
1
0
Evaluate
2
2 2i
1 1
n
n
i 1 n
n
lim
2
2 2i
1 1
n
n
i 1 n
n
Solution
Example 4.
b a
Here b = 1, a = 0 and f(x) =
dx
4x
lim
2
= sin
1
4 x2
1 x
1
20
=
Evaluate lim
n
.
n1
1
Solution Limit = lim 2 n 2 1 2 dx
n r 0 n (2r)
0 1 (2x)
.
6
=
1
n 3kn .
k 1
Evaluate
n
lim n2 2 n 2 2 n 2 ..... 2
.
2
n n
n 2
n 4
n (2 n 2 )
kn
Example 2.
3k
n
27 5 32
= 2 =
.
6 6
3
r 1
1
1 2
1
1
n f a r n
So, L = 0
k 1
dx
1
= 0
3
27
1
= 2 (1 2x) x = 2
1
6
6
2·3
0
which is of the form
lim
1
2
2
[2 – 1] = .
3
3
2
1
r 1
n
3k
n
1
n·
0
=
1
0
(1 0)
n
n
n
1
1
4n r
n
= (1 3x)
1
n
= 2 (1 2x) 2 1 dx
1
r 1
1
Example 3.
1
1
1
1
lim
.....
.
n
4n 4
4n 9
3n
4n 1
Solution
Tk =
2
1 1
1
tan 1 2x tan 1 2 .
0 2
2
2.85
DEFINITE INTEGRATION
Example 5. Evaluate
1
1
1
1
lim
.........
n
1
n
2
n
3
n
2n
1
1
m
m
m
1 1
2 ... n
n n n
n
n
lim
1
1
1 n r
n n
r 1 n
.........
Solution nlim
2n
1 n 2 n 3 n
n
= nlim
1
n
r 1
1
n
1 dx
1
= 0
r
x 1
1
n
r 1
lim
rn
n
= lim
Limit
Solution
1
xm 1
1
x m dx
m 1.
0
m
1
0
1
Evaluate
Example 9.
2
2
2
6
n
Evaluate nlim
i
i
2
2
n sin n
i 1
n
2
1
1
sin t dt = .
0
2
Evaluate
n r lim 1 (1 r / n )
n
n r 1 n 2 r 2
n
(1 r 2 / n 2 )
1
x 1
1
dx ln(x 2 1) tan 1 x
= 0 2
x 1
2
0
1
1
ln 2 .
2
4
lim
1 3
1
x3 x 4
3 4
0 0
1
x7
7
0
m
1 2 3 .... n
1
, m > –1.
m 1
m 1
n
11
3 4 7 .
1
12
7
Evaluate
n
1
n
1
....
lim
.
n (n 1)
(2n 1 (n 2) 2(2n 2)
n
2 n (n.3n )
n
Solution General term =
(n r) r(2n r)
n
Prove that
m
3
Example 10.
n
n
=
= lim
m
n
1 r 1 r
n r 1 n n r 1 n
= nlim
6
1 n r
n r 1 n
1
lim n 1 n2 2 2 .... 1
n n 2 12
n
n 2
Limit
Solution
m
r6
r 1
2
n
1 2
Put x2 = t 2x dx = dt
Example 8.
r 1
x dx x dx
= 0 1 0
6
0 x dx
0
=
n
2
2
Example 7.
3
n
The given limit is lim r 1 n
Solution
= x sin x dx
S=
3
6
r2 r3
1 r
r
r
r
·sin 2 = · sin
2
n
n
n
n
n
1 n r
r
Tr lim
sin
n
n n
n
n
r 1
r 1
1
3
2
2
Limit = lim
3
6
n
n
Tr =
6
1 2 3 ... n
1
Solution
2
lim (1 2 3 ... n ) (1 2 3 ... n )
= ln(x 1) 0 = ln 2.
Example 6.
m
1
r 1
r r r2
1 .n 2 2
n n n
limit = nlim
2.86
1
dx
1
= 0
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(1 x) (x 2x)
2
0
dx
(x 1) {(x 1) 1}
2
1
b
2b
ln a ln a
n
n
n
ln L = nlim
nb
... ln a
n
1
= sec 1 (x 1) = sec–12 – sec–11 = /3–0 = /3
0
1
n! n
Evaluate nlim
n
n
Example 11.
= nlim
1 ab
ln (z) dx
b a
which is the desired result.
1
1 n!
ln n
n n
ln L = nlim
1 1
1
n
lim n 2
n
n
1 n 1
ln n
= nlim
2 n
1
1
= (0 – 1) – lim x ln x + 0 = – 1 – 0 = –1
x 0
1
L= .
e
Example 12.
1 n
k ln k k ln n k ln n
n 2 k 1
1 n 1
1 n
k ln n n
lim
ln
n
k
ln
k
= n
2 n
n n 2 k 1
n 2 k 1
1/n
n
Show that lim a br
r 1
n
=
1
1 n 1
1 n 1
ln n 0 x ln x dx
ln n
2 n
2 n
1
= x ln x dx = –
0
1/n
a br
L = nlim
n
r 1
a b · a 2 b ... a nb
= nlim
n
n
n
Taking logarithm on both sides, we have
1 n 1
1 n k k ln n n(n 1)
ln n ln 2 ·
= nlim
2 n
n k 1 n n n
2
e
1 ab
ln (x) dx.
b a
Solution We have
n
2
= 0 ln x dx x ln x x
0
where =
1
·(11 ·2 2 ·33.......n n ) n
1 n 1
1 n
ln n 2
k ln k
ln L = nlim
2 n
n k 1
r
1
ln
n r 1 n
n
2
1 1
1
n
L = lim n 2
n
= nlim
1
·(11 ·2 2 ·33.......n n ) n .
Let
Solution
1 1
2
3
n
ln
ln ln ..... ln
n
n
n n
n
[Putting a + bx = z]
Evaluate
Example 13.
1 1. 2 . 3 ........n
= nlim
ln
n
nn
= nlim
=
n! n
Let L = nlim
n
n
Solution
n
1 ln a br 1
n 0 ln (a + bx) dx
n r 1
1
. (by integrating by parts).
4
1
L= e 4 .
1/ n
Example 14.
Evaluate
1/n
1 2 2 32 n 2
lim 1 2 1 2 1 2 .... 1 2
n
n n n n
2.87
DEFINITE INTEGRATION
If L be the required limit then ln
Solution
1
1
L = 0 ln(1 x )dx x ln(1 x ) 2x 2 tan x 0
= ln 2 +
2
4
2
ln L – ln 2 =
1
2
L = 2e(–4)/2.
Example 15.
4
2
r
lim = 0, when r = 1, lower limit = 0
n
n
r lim 2n
= n = 2, when r = 2n,
and nlim
n
n
upper limit = 2. The given limit
2
1 x
1
1 2 2x
dx =
dx
+
dx
2
2
0 1 x
2 0 1 x 2
1 x
2
= 0
2
( 2n C n )1 n .
Evaluate nlim
= tan–1x]20 +
1n
Solution
(2n)!
Let L = nlim
n!n!
1
ln(1 x 2 )
2
0
1
ln 5.
2
Example 17. Evaulate
= tan–1 2 +
1n
n!(n 1)(n 2).....(n n)
L=
n!n!
1n
n 1 · n 2 ...... n n
=
2
n
1
1 n 1
n2
nn
ln L = n ln 1 ln 2 ...... ln n
1 nr 1
1
1
=
Here, Tr = ln
n
r n r n
1 n
1
1
Sn =
n r 1 r n
1
n
lim
= nlim
1
1
1
1
1
.....
1
n
2
3
4
4n
= nlim
n . r = lim n . (r / n)
1
1
1
1
1
1
.....
n
2
3
4
4n
Let
L
Solution
4n
1
n
r 1
= ln(1 x) ln x dx
1
n2
n3
3
.........
5n
n 2 2 2 n 2 32
2n
nr
1
= lim
Solution nlim
2
2
n
n
r
r 1
r 1 n
0
x
4n
=4
n
4
2 x
= 2 (2 – 0) = 4.
0
Evaluate
n2
n2
1
.........
lim
.
3
3
n (n 1)
64n
(n 2)
Evaluate
2n
4 dx
Example 18.
= (2 ln 2 – 1 – 0) – (0 – 1)
Thus, ln L = ln 4 L = 4.
r 1
1
=0
n n
L=
= [(1 x)ln(1 x) 1 x ln x] 0
n 1
1
When r = 1
0
= (1 + x)ln (1 + x) – (1 + x) – [x ln x – x]
lim
1
and when r = 4n then x = nlim
1
n
n 2 12
4n
1
lim
1
Hence, ln L = ln 1 x dx
0
Example 16.
r
n
2
r
1
n
1
Solution
Let
n2
n2
1
.........
L = lim
3
n (n 1)3
64n
(n 2)
n2
3
n
r 1 (n r)
3n
= lim
Put 3n = m, we get
2.88
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
n
Solution Transform the given expression in the
following way :
m2 / 9
m
L = lim
r 1 m
r
3
3
3 1
n
n ......
n
n
n3
n6
n 3( n 1)
m
3
1
= lim
n
r 1 m 1 3r
m
3
dx
= 0 (1 x)3 =
3
lim
S = n r 1
n
r 3 r 4 n
Tr =
.
2
2
x (3 x 4)2
3 1
dx dt
Put 3 x 4 = t
2 x
4
2 dt 2 1
2 1 1
=
2 =
3 4t
3 t 10 3 4 10
2 6
1
.
= ·
3 40 10
Example 20. Compute
lim 3 1
n n
......
n
n
n
n3
n6
n9
n
n 3(n 1)
3
0
dx
10
3
3
1
dx (1 x) 1/ 2 dx 2 1 x
0
0
1 x
= 4 – 2 = 2.
Example 21. Compute
=
1
r
r
· n 3
4
n
n
1
on the interval [0, 3].
1 x
Therefore,
f(x) =
n
n .....
n
lim 3 1
n n
n3
n6
n 3( n 1)
1 4n
1
2
n 1
r
r
3
4 ·
n
n
4
= 0
3 1
1
1
1
.......
3
6
3(n 1)
n 1 0
1
1
1
n
n
n
The given sum is the integral sum for the function
1
15
=
.
2
2(1 x) 0 32
r 4n
S = nlim
=
Find the value of
Example 19.
Solution
3
1
1
1
.....
n 1 n 2
an
where a is a positive integer. Calculate approximately
lim
1
n n
1
1
1
1
.....
.
100 101 102
300
Solution Let
1
1
1
1
P = nlim
n 1 n 2 .... an
n
...(1)
1
1
1
1
....
= nlim
n 0
n 1 n 2
n n(a 1)
n(a 1)
n(a 1)
1
1
= nlim
(n
r)
n(1
r / n)
r 0
r 0
= nlim
dx
= [ln (1 + x) ] a0 1
(1 x)
Hence P = ln a.
Put n = 100, and a = 3 in (1), we get
(a 1)
= 0
1
1
1
1
.....
= ln 3
100 101 102
300
= 2.303 log103 = (2.303) (0.4771) = 1.1 approx.
2.89
DEFINITE INTEGRATION
1 2n
Example 22. For positive integers k = 1, 2, 3, ....., n,
let Sk denotes the area of AOBk (where 'O' is origin)
k
, OA = 1 and OBk = k. Find the
such that AOBk =
2n
= nlim
Solution
Y
2
= 0
Bk
k
k
1
k sin
2n
2
(using =
k
k
1 n
sin
Sk =
2
2
n n
2n
2n n 1
k 1
L = lim
=
1
ab sin )
2
k 1 1
1 k
x
sin
dx
= 0 x ·sin
2n n 1 n
2n 2
2
1
1 2
x
2 1
x
cos
dx
= x cos
0
2
2 0
2
1
1 2 2 x
2
0
·
sin
= 2.
= 2
2
0
Example 23.
Let
lim
1
1
1
1
.......
n
2
n
1
2
n
3
2
n
5
4
n
1
=
3
1
ln 2 = 2 ln 2
2
2
1
A
ln 2 =
ln C .
2
B
Hence, least value A + B + C = 1 + 2 + 2 = 5.
Example 24. If Hn denote the harmonic mean of
n positive integers n + 1, n + 2, n + 3, ......, n + n, then
Hn
find the value of nlim
.
n
n
1
1
1
.....
n 1 n 2
nn
n
1
1
1
H n = n 1 + n 2 + ...... + n n
Solution
Hn =
n
n
n
1
lim
nlim
=
= lim
H
n
n
n
r 1 n r
r 1
1
= 0
A + B + C.
Let given limit be L, then
1
1
ln(1 x) 0
2
=
A
ln C, where A, B, C N. Find the least value of
B
Solution
1
1
1
dx –
dx
0 2 2x
2 x
= ln 4 – ln 2 –
k
2n
1
r
n
1
H
lim n =
.
n n
ln 2
Example 25.
1
1
1
1
........
4n
2n 2 2n 4 2n 6
n 2 ·C n .
find the value of nlim
– nlim
1
n
dx
= ln 2
1 x
1
1
1
1
........
4n
2 n 1 2 n 2 2n 3
L = nlim
r 1
2
X
Sk =
r 1
= ln (2 x) 0 –
A
1
OBk = k, AOBk =
n
n
1 2n 1
1
1
= nlim
n r 1
r
n r 1 2 r
2 2
n
n
1 n
Sk .
n n 2
k 1
value of lim
1 n
n
2n r n 2n 2r
n
Let Cn =
1
n
1
n 1
tan 1 (nx)
dx then
sin 1 (nx)
2.90
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
tan 1 (nx)
1
n
Cn = 1 sin 1 (nx) dx
Solution
(put nx = t)
n 1
1 1 tan 1 (t)
dt
n
n n 1 sin 1 (t)
Cn =
tan 1 t
dt ( × 0)
1
t
n 1 sin
n
1 1
n
2
n
(n 1)
sin 1
2 1
n 1
L = lim
= · = .
n
1
4 2
2
n
Example 26. Evaluate
tan 1
1
n ·C n lim n · n
L = nlim
n
2
1 n
1
lim tan 1
n
n k 1 1 tan k n .
1
tan 1 t
dt
sin 1 t
n
= n 1
1
1 n
1
lim n tan 1 · lim
n
n
n
n k 1 1 tan k n
Solution
(applying L'Hospital's Rule)
1
n
1
= l . 0
dx
1 ln(sin1 cos1)
=
.
2
1 tan x
J
1.
Express the following limits in the form of an
integral.
(i)
1
n
cos n cos n ... cos n
lim
n n
2
2
sec sec 2 ...
(ii) nlim
6n
6n
6n
2.
1 1 1 ..... 1
(iii) nlim
na na 1 na 2
nb
4.
n
lim
n
n
(i)
5.
(n r) r(2n r)
n2 r2
r0
(b)
2n
1
n
r
n
r 1
6.
Evaluate the following limits :
(i)
lim
2
1
n n 3 13
(ii) nlim
4
n n2
(iii) lim
3.
1 2 1/2 3 1/3 n 1/n
1 1 1 .... 1
(ii) nlim
n n n
n
Prove that
(a) lim
1
(ii) nlim
2
2
2
n
.... 3
n3 23
2n
1 2 3 ......... 2 n
n n
1
2
3
n
lim tan tan
tan .....tan
n
2n
2n
2n
2n
r 1
n 1
1
1/n
4
... sec 2 ( n 1)
6n 3
Evaluate the following limits :
(i)
1
......
(iv) nlim
n n 1
n n2
n nn
Evaluate the following limits :
3
1
( n 2 1 n 2 2 2 ... n 2 n 2 ) .
4 n 2
(n 2in 2i 2 ) .
n n3
i 1
(x 2 1)dx lim
Evaluate the following limits :
(i)
lim
n
1
1
1
1
......
n n 1 n 2
4n
1
n2
n2
1
(ii) lim
.....
3
3
n n
8n
( n 1)
( n 2)
n 1
n2
n3
3
2
......
(iii) nlim
n 2 12
5n
n 2 2 n 2 32
DEFINITE INTEGRATION
7.
Evaluate the following limits :
n 1 n 2 ..... 1
(i) nlim
n 2 12
n
n 2 22
2 k 4 k 6 k .. (2n)k
(ii) lim
, k –1
n
n k 1
n
n
n
3
.....
(iii) nlim
1
n
n3
n6
n9
n
....
n 3(n 1)
n2
n2
n ( n 2 1)3 / 2
(n 2 22 )3 / 2
n2
...
3/2
n 2 (n 1)2
(iv) lim
2.12 LEIBNITZ RULE FOR
DIFFERENTIATION OF
INTEGRALS
If f is continuous on [a, b], and u(x) and v(x) are
differentiable functions of x whose values lie in [a,
b], then
d v(x)
dv
du
f(t)dt f(v(x)) f(u(x))
dx u(x)
dx
dx
T h e fol lowi n g fi gure gi ves a geom etr i c
interpretation of Leibnitz rule. It shows a carpet of
variable width f(t) that is being rolled up at the left
at the same time x as it is being unrolled at the
right. (In this interpretation time is x, not t.)
At time x, the floor is covered from u(x) to v(x). The
du
at which the carpet is being rolled up need
dx
dv
not be the same as the rate
at which the carpet
dx
rate
is being laid down.
At any given time x, the area covered by carpet is
Y
Uncovering
y=(u(x))
y=f(t) Covering
f(v(x))
1k 2 k ..... n k
for k > 0. Find
10. Find nlim
n k 1
the approximate value of 15 + 25 + ..... + 1005.
11. Prove that when a is large the sum to infinity
1
1
1
of the series a 2 a 2 12 a 2 2 2 .... is
1
/a, approximately..
2
A(x) =
v( x)
u( x)
f (t )dt .
The question is : at what rate is the covered area
changing ?
At the instant x, A(x) is increasing by the width
f(v(x)) of the unrolling carpet times the rate dv/dx
at which the carpet is being unrolled. That is, A(x)
is being increased at the rate
dv
.
dx
At the same time, A is being decreased at the rate
f(v(x))
du
,
dx
the width at the end that is being rolled up times
the rate du/dx. The net rate of change in A is
f(u(x))
dA
dv
du
= f(v(x))
– f(u(x))
.
dx
dx
dx
which is precisely the Leibnitz rule.
Proof : To prove the rule, let F be an antiderivative
of f on [a, b]. Then
v(x)
u(x)
f(t) dt = F(v(x)) – F(u(x)).
...(1)
Differentiating both sides of this equation with
respect to x gives the equation we want :
u(x)
9.
Show that for each integer m > 1,
1 1
1
1
1
...
ln m 1 ...
2 3
m
2
m 1 .
Show that for each integer m > 1,
ln 1 + ln 2 + .... + ln(m – 1) < m ln m – m + 1
< ln 2 + ln 3 + ... + ln m.
d v(x)
d
f(t)dt
[F(v(x)) – F(u(x))]
u(x)
dx
dx
O
A( x) =
8.
2.91
= F(v(x))
v(x )
u( x )
f (t )dt
v(x)
t
dv
du
– F(u(x))
dx
dx
[Chain Rule]
2.92
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
dv
du
– f(u(x)) .
dx
dx
= f(v(x))
For example, if F(x) =
x2
x
e3x
t
dt , then find first
Example 1. If F(x) = e2x
ln t
and second derivative of F(x) with respect to ln x
at x = ln 2.
dF(x)
dx
dF(x)
d ln x = dx . d (ln x)
l = xlim
ln 2x – ln x = ln 2, and
Solution
x
ln t
ln x
ln x
= lim
= 1.
x 1 ln x
1
x· ln x
x
Hence l × m = ln 2 · 1 = ln 2.
= lim
m= 1
x ln x x
2x
e3x
3x
2x e
6x
4x
3
.
e
.
2
.
e
=
3x
2x x = e – e .
ln e
ln e
d
(e6x – e4x)
Now,
2 =
d
ln
x
d ln x
Evaluate xlim
dx
d
(e6x – e4x) ×
= (6 e6x – 4 e4x) x
d
(ln
x)
dx
The first derivative of F(x) at x = ln 2 (i.e. ex = 2)
is 26 – 24 = 48.
The second derivative of F(x) at x = ln 2 (i.e. ex = 2)
is (6 . 26 – 4 . 24). ln 2 = 5 . 26 . ln 2.
Example 2. Which of the following functions
are differentiable in ( 1, 2)
(i)
(log x) dx
x
x
(iii) 0
(ii)
2x
x
sin x
dx
x
lim
Solution
x
sin x
Solution Since the functions (logx) and
x
are not well defined in ( 1, 2), therefore the
functions (i) and (ii) are not differentiable.
and g(x) = 0
1 t t2
is continuous on (1, 2)
1 t t2
1 t t2
dt is the integral function
1 t t2
of f(t), therefore g(x) is differentiable on ( 1, 2)
such that, g (x) = f(x) .
x
e
2t 2
x
2
2 . e t dt . e x
= xlim
dt
e
2t 2
dt
2
form
2
Applying L Hospital Rule
0
1. e 2x
2
2
2 . e t dt
= xlim
= xlim
0
ex
2
x
x
0
x
1 t t
dt
1 t t2
x t2
e dt
0
2
The function f(t) =
2
0
=
2
x t2
e dt
0
Example 4.
d 2 F(x)
2x
dt
and
t
x
1
m = x ln x 1 ln t dt then prove that l m = ln 2.
sin tdt , then
F(x) = 2x . sin x 2 – 1 . sin x .
Solution
2x
Let l = xlim
x
Example 3.
2
2 . ex
2
2x . e x
2
= 0.
Let f : (0, ) R be a continuous
Example 5.
x
function such that F(x) = t f (t)dt . If F (x2) = x4 + x5,
0
12
then find
f(r ) .
2
r 1
We have
Solution
F(x2) =
x2
0
t f (t) dt x 4 x 5
...(1)
2.93
DEFINITE INTEGRATION
On differentiating both the sides w.r.t. x, we get
2x (x2) f (x2) = 4x3 + 5x4
5
x
2
f (x2) = 2 +
5
5 (12)(13)
f (r 2 ) = 2 2 r = 24 + 2 2
r 1
r 1
...(2)
12
12
= 24 + (15)(13) = 24 + 195 = 216
f (r 2 ) = 219.
t2
Example 6. Let f (x) = 1 e dt and h (x) = f(1 + g(x)),
where g (x) is defined for all x, g'(x) exists for all x, and
g (x) 0 for x > 0. If h'(1) = e and g'(1) = 1, then find the
possible value of g(1).
x
t2
Given f (x) = e dt
Solution
Now
x3
dt
x
1 t2
2
3x
2x
dy
=
+
6
dx
1 x
1 x2
Slope of tangent at x = 1:
m=
3
2
2
2
=
5
2
x
We have F (x) = f (t) dt and
1
1
4 t2
x
Let H(x) = F(x) · G(x)
H'(x) = F(x) · G'(x) + G(x) · F'(x)
2
2
1
x
2
2
= (4 + x2) x 4 t dt 1 4 t dt .
(given)
(1 + g(1))2 = 1
1 + g (1) = ± 1
g (1) = 0 or g(1) = – 2.
Example 7. Find the equation of the tangent
y= 2
4 t 2 dt then compute the value of (FG)' (0).
x
x
1 g ( x ) · g ' ( x )
h'(x) = e
h'(1) = e
2
e1g (1) · g' (1) = e
Solution
4 t 2 dt and
1
G(x) = f (t) dt where f (t) =
2
1 t2
1
x
1
1
Let F (x) =
x
t2
dt
(x – 1).
Solution
Differentiating w.r.t. x,
x3
to the curve y = x 2
2
= 1 f (t) dt
e dt
h (x) =
5
y=
4 x
+ f (t) dt 4 x
1
h (x) = f(1+g(x)) , g (x) 0 for x > 0
1 g ( x )
=0
1 t2
The equation of the tangent is
1
G (x) =
r 1
x
dt
1
y=
Example 8.
12
Hence,
Also, the value of y at x = 1 is :
at x = 1.
1
0
2
2
H'(0) = 4 0 4 t dt 1 4 t dt
Now
We have x 2 a 2 dx
=
a2
x x2 a 2
ln x x 2 a 2
+
2
2
Hence,
1
0
4 t 2 dt
1
t t2 4
2 ln t t 2 4
=
2
0
5
= 2 2 ln
Also,
0
1
5 1 – 2 ln 2
5
2 ln 5 1
4 t 2 dt = 2 ln 2 –
2
= 2 ln 2 +
5
– 2 ln ( 5 1)
2
2.94
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
H'(0) = 4 [2ln ( 5 + 1) + 2ln( 5 – 1) – 4 ln 2]
= 4[2 ln (4) – 4 ln 2] = 4[4 ln (2) – 4 ln 2]
= 0.
Given a function g, continuous
Example 9.
1
everywhere such that g(1)=5 and g (t) dt = 2.
0
1 x
(x t)2 g (t) dt , then compute the value of
2 0
f (1) f (1).
If f(x) =
1
g(1) = 5 and g (t) dt = 2.
Solution
0
x
2f (x) =
x
x
x
0
0
0
= x 2 g(t) dt 2x t g(t) dt t 2 g(t) dt
Differentiating w.r.t. x,
x
2 f '(x) = x2 g(x) + g(t)dt·2x
0
2
– 2 x g(x) 0 t g(t)dt + x2g(x)
x
x
x
g(t)dt – 2 t g(t)dt
2 f '(x) = 2x
0
f " (x) = x g (x) +
h(x)
If F(x) = g(x) f(x, t) dt, then
h(x) f(x, t)
F(x) = g(x)
dt + f(x, h(x)) h(x)
– f(x, g(x)) . g(x)
x
dt
, then find f(x).
Example 10. If f(x) = ln x
xt
x
1
1
dt
Solution f(x) = ln x x t 2 + 1 . 2x
1
1
–
x (x ln x)
x
x
(x2 – 2xt + t2) g(t) dt
0
Modified Leibnitz Theorem
0
x
g(t)dt – x g (x)
0
x
= g(t)dt
0
1
1
1
=
+
–
(x t) ln x 2x x (x ln x)
1
1
1
1
=
–
+
–
2x x ln x
2x x (x ln x)
ln x 1
x 1
1
= –
=
.
x (x ln x)
x x (x ln x)
Alternative:
x
x
dt
ln(x t) (treating ‘t’ as constant)
f(x)= ln x
xt
ln x
f(x) = ln x – ln (x + ln x)
1
ln x 1
1
1 1
=
f(x) = –
.
x (x ln x) x x(x ln x)
1
Hence, f " (1) = g(t)dt = 2
0
Also, f ''' (x) = g (x)
f ''' (1) = g (1) = 5
f ''' (1) – f ''(1) = 5 – 2 = 3.
(given)
I
1.
2.
3.
If F(x) =
2 cos t dt , find F(x).
x
x
3
cos t dt, find f(x).
If f(x) =
0
Find the derivative of the function
x1 t t 2
y
dt at x = 1.
0 1 t t2
If f(x) = x 2 sin t dt then find f(x).
5.
d2y
If x = 0
and
= ky then find k
dx 2
1 4t 2
Find the derivative with respect to x of the function
x
y
dt
t
t
0
0
sin t dt , y cos t dt ;
(ii) x t ln t dt , y t ln t dt .
(i)
7.
x
t2
1
1
t2
If f(x) = eg(x) and g(x) =
value of f(2).
x2
4.
6.
y represented parametrically
ex
8.
2
x
t
1 t dt then find the
2
4
Find the second derivative with respect to z of
the function
y
z2
dx
1 x for z = 1.
0
3
2.95
DEFINITE INTEGRATION
K
9.
Find the derivative with respect to x of the function
y specified implicitly by
y
15. Find the interval in which
e dt cos t dt 0 .
0
0
((ln t) 2 ln t) dt where f(x) vanishes.
+
2
t
increasing.
10. Find the value of the function f(x) = 1 + x
x
x
1 (e 1) (2 t)dt, (x 1) is
F(x) =
x
t
x2 t 2 t
e 1 dt.
16. Find the point of maxima of f(x) =
d 2 x2 dt
11. Find 2 0
dx
1 5t 3
17. At what value of x does the function
2
12. If x =
t2
1
I(x)
1
2
z ln z dz and y = 2 z ln z dz find
t
101
13. If {F(x)}
x
(F(t))100
0
dy
.
dx
dt
, then find
1 sin t
F(x).
14. If (x) = cos x –
b
b
f (x) dx f (t)dt f (u)du
a
a
a
Property P
Order of integration : if we reverse a and b then sign
of the integral is changed.
a
b
f(x) dx = –
b
a
f(x) dx
b
a
c
An interval can be decomposed into two intervals,
Example 1.
15
Evaluate sgn ({x}) dx, where {.}
1
denotes the fractional part function.
1
1
19. Let g(x) = xce2x and let f(x) =
x
e (3t 1) dt.
2t
2
1/2
0
For a certain value of c, the limit of f(x)/g(x) as
x is finite and nonzero. Determine c and
compute the value of the limit.
Solution
We have
1, if x is not an integer
sgn ({x}) =
0, if x is an integer
15
sgn ({x}) dx
= sgn ({x}) dx + sgn ({x}) dx
= 1.dx + 15 1.dx
1
0
15
1
0
0
1
1
0
[ P–3 ]
= 1 (0 + 1) + 15 (1 – 0) =16.
Example 2.
Find the value of the definite integral
x4 x2 2
dx .
(x 2 1)2
2 1
c
f(x)dx = f(x)dx + f(x)dx
a
x
g(t) t t dt.
2 1
Property P
b
dx have an extremum ? What is it
Compute the limit of f(x)/g(x) as x .
0
Note that the definite integral is independent of what
letter denotes the variable of integration. Thus,
x2
0
2
x
Property P–1
xe
18. Let g(x) = xe x and let f(x) =
(x t) (t) dt, then find the
2.13 PROPERTIES OF
DEFINITE INTEGRAL
x
equal to?
value of (x) + (x).
b
t
0
1
2 1
Solution
2 1
I=
(x 2 1)2 (x 2 1)
dx
(x 2 1)2
2 1
(x 2 1)
1
=
dx
(x 2 1)2
2 1
2.96
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
2 1
2
2
2 1
(x 2 1)
dx
(x 2 1)2
a
put x =
I =
1a
a
1
= – a
where (a =
2 1)
1
1
dx = – 2 dt
t
t
2
4
(1 t )t
dt = –
t 4 (1 t 2 )2
cot x
dt
t
=1
2 dt + 1/e
t(1 t 2 )
1 t
tan x
L.H.S. = 1/e
Solution
Put t =
1
1
1
t2
· 2 dt
2
1 1 t
t2
1a
tan x
1/e
I1
I1 = 1 a
Prove that
Example 4.
(x 2 1)
(x 1) dx
=2–
2
(1 t )
L.H.S. =
(1 t ) dt
2 2
a
a
t2 1
t2 1
dt = – I [ P–1 ]
dt
=
–
1 a (t 2 1) 2
1
(t 2 1)2
2I1 = 0 I1 = 0 2.
dx
1
dx
Example 3. Prove that 0 1 xn 0 (1 xn )1/n (n > 1)
dx
1
Solution
Put x =
R.H.S. =
(1 x )
n 1/n
0
1
2 dt
dt
t
I=
1/n
n
1
t(t 1)1/n
1 1
n
t
Put tn – 1 = yn tn – 1 dt = yn – 1 dy
y n 1dy
Now, R.H.S. = 0
(1 y n )y
1
dy
n
y
1
y2
dy
n
0 (1 y n ) y 2
0
1
1
y
1
1
Put = z 2 dy = – dz
y
y
0 dz
dz
I=
n
0 1 zn
1 z
dx
= 0
= L.H.S.
1 xn
cot x
e
1 1
du
u u 2 cot x dt
1/e t(1 t 2 )
1 1
u2
e
e
=
dt
cot x t(1 t 2 )
cot x
e
1
dt
t(1 t 2 )
1/e
dt
2
t
2
1/e
[ P–2]
[ P–1 ]
[ P–3]
e
1
2
= ln t ln(1 t )
2
1/e
1
1
1
2
= 1 ln(1 e ) 1 ln 1 2
2 e
2
1
(e 2 1)
2
ln(1 e ) ln
2
e2
=2–
1
[ln e2] = 2 – 1 = 1 = R.H.S.
2
Example 5. Find the value of
=2–
e
dt
t(1 t 2 )
cot x
du
dt
1
2
cot x u(1 u )
t(1 t 2 )
e
=
= 1/e
1/e
t(1 t ) t 1 t dt
= I (say)
1
1
dx = 2 dt
t
t
1
cot x
1
du
u2
1a
= a
1
in the first integral
u
then dt =
1a
t
dt +
1 t2
sin 2 x
0
sin 1 t dt
cos2 x
0
cos 1 tdt
Put t = sin2 y in the first integral and
t = cos u in the second integral, we have
Solution
2
[ P–2 ]
sin 2 x
0
cos 2 x
cos 1 t dt
0
x
[ P–1 ]
sin 1 dt +
= y sin 2y dy –1 / 2 u sin 2u du
0
x
2.97
DEFINITE INTEGRATION
x
= 0 y sin 2y dy +
/2
x
y sin 2y dy
[ P–2 ]
/2
= 0 y sin 2y dy
[ P–3 ]
In the first integral on the far right side we make the
substitution u = –x. Then du = –dx and when x = –a,
u = a. Therefore
/2
cos 2 y sin 2 y
= y
2
4 0
e dt
= a , then show that
t 1
1
b
e t dt
1
Given 0 t 1 = a
Solution
b
Now let I = b 1
0
0
0
a
a
a
a
0
0
a
a
a
a
a
0
0
0
f(x)dx = f(u)du + f(x)dx = 2 f(x)dx
(iii) If f is odd, then
a
a
a
a
0
0
f(x)dx = – f(u)du + f(x)dx = 0.
e t dt
is equal to ae b.
b 1 t b 1
a
(ii) If f is even, then
t
If 0
a
f(x)dx = f( u) du + f(x)dx
= 0 (0 + 0) = .
4
4
Example 6.
a
f(x)dx = – f( u) (–du) = f( u) du
–
...(1)
e t dt
t b 1
The above property is illustrated by Figures (a) and
(b) For the case where f is even, part (ii) says that the
area under y = f(x) from – a to a is twice the area from
0 to a because of symmetry of the graph about y-axis.
Thus, part (iii) says the integral of odd function is 0
because the areas cancel.
Y
Put t = b y dt = dy
e b y
0
0
ey
dy
I = 1 b y b 1 ( dy) = e b 1 y 1
1
ey
dy
= e b 0 y 1
1
e
t
[ P–2 ]
dt
= e b 0 t 1
–a
(a ) f even ,
[ P–1 ]
a
?a
0
Y
Property P
(i)
(ii)
a
f(x)dx =
0
a
a
a
0
i.e. f(x) = f(x) .
a
(iii) a f(x) dx = 0, if f(x) is an odd function
i.e. f(x) = f(x) .
Proof : (i)
a
a X
(f(x) f( x)) dx
( b ) f od d ,
f(x) dx = 2 f(x) dx, if f(x) is an even function
a
0
–a
a
X
f( x)d x = 2 f(x)d x
ae b from (1)
a
a
0
a
f(x)dx =
0
a
a
a
0
0
f(x)dx +
= – f(x)dx + f(x)dx
a
0
-a
Since f(x) = x6 + 1 satisfies f(–x) = f(x), it is even and so
2
2
(x6 + 1) dx = 2
2
f(x)dx
a
f(x)d x = 0
2
(x6 + 1) dx
0
1 7
128 2 284
.
= 2 x x 2
7
7
0
7
Since f(x) = (tan x)/(1 + x2 + x4) satisfies f(–x) = –f(x), it
is odd and so
2.98
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
tan x
2
4 dx = 0.
1 1 x x
This property of odd functions can save a lot of
computation. For example,
1
/4
sin9 x dx = 0 by inspection.
– /4
Evaluate
Example 7.
Solution
1
= 0
1 ex
1 ex ex
= 0
1 ex
Example 10.
ex ex
dx
1 e x
[ P–4(i) ]
1
Example 8.
Evaluate
2
(x3 – 3x) dx.
The integrand, f(x) = x3 – 3x, is an an
odd function, i.e., the equation f(–x) = – f(x) is satisfied
for every x. Its graph, drawn in the figure, is therefore
symmetric about the origin. It follows that the region
above the x-axis has the same areas as the region
below it. We conclude that
Solution
2
2
(x3 – 3x) dx = 0
3
X
Example 9.
/2
Solution
4
.
3
1/2
1 x
dx = 0
Prove 1/2 (cos x) ln
1 x
Let I =
1 x
(cos x) ln
dx
1 x
1/2
1/2
f(–x) =
1 x
cos( x) ln
1 x
1 x
= (cos x) ln
1 x
1 x
= –f(x)
= (cos x) ln
1 x
Hence f is odd function
I = 0.
[ P–4 ]
7
Example 11. If f(x) =
5
3
x 3x 7 x x 1
then,
cos2 x
/4
y = x – 3x
Solution
I =
evaluate / 4 f(x) dx.
Y
/2
(Putting cos x = t 2)
1 x
Let f(x) = cos x ln
1 x
e x (e x e x )
dx
e x 1
2
cosx d x
sin x
1
(e 1 1)
e2 1
(e x e x ) dx = e – 1 +
=
=
.
0
1
e
I = 2 0
= 4 0 t 2 d t
1
1 ex ex
[P–4]
/2
0
ex e x
dx
1 1 e x
cos x cos 3 x d x
I = 2 1 2 t 2 d t
ex ex
dx
1 1 e x
/2
I = 2 0
Find the value of the integral
cos x cos x dx.
3
I=
/2
/2
cos x cos 3 x d x
Solution
f (x) =
x 7 3x 5 7 x3 x 1
cos2 x
x7 3x5 7 x3 x
=
+ sec2 x
cos 2 x
= (odd function) + (even function)
/4
/4
f (x) d x =
/4
/4
sec2 x d x
/4
x7 3x5 7 x3 x
dx
cos 2 x
/4
= 2 + 0 = 2.
+
[ P–4 ]
2.99
DEFINITE INTEGRATION
n
1
1
1
1
.......
= 2
1·2
2·4
4·6
2n(2n
2)
(–1)[x] dx, n N,
Example 12. Evaluate
n
where [x] denotes the greatest integer function less
than or equal to x.
(–1)[x] dx
1 1 1 1 1 1
= 1 2 1 2 2 3 3 4 .....
Suppose f (x) = (–1)[x]
f(–x) = (–1)[–x] = (–1)–1 – [x], x I
= – (–1)–[x]
1
1
n
n
1
Let I =
Solution
n
n
1
( 1)[x]
[x]
( 1)
( 1)2[x]
= – (–1)[x] = – f(x), x I
Note that the function is not odd, but the property
can be applied since difference at few isolated points
does not affect the integral.
=–
I=
n
The graph of an odd function is symmetric with respect
to the origin. A function may be symmetric with respect
to some other point of the x-axis.
Example 13. Find
6
x(x– 1) (x – 2) (x – 3) (x – 4) (x – 5) (x – 6) dx.
0
Solution The point of symmetry is (3, 0). To
exploit this we make the change of variable x = u + 3.
Then
x (x – 1) ... (x – 6) = g(u), where
g(u) = (u + 3) (u + 2) (u + 1) u(u – 1) (u – 2) (u – 3)
Thus,
6
0
x (x – 1) ... (x – 6) dx =
3
3
g(u) du.
x
x
x
......
dx .
In = 1 | x 1 x
2
3
2n
2
3
2n
In .
Find nlim
Proof: We have g(–x) =
g(–x) =
a
g(–x) = 0 +
x
x
x
a
f(t) dt is an
f(t) dt
a
f(t) dt +
a
x
a
f(t) dt
f(t) dt
a
[f is odd
g(–x) = –
x
x
a
a
a
f(t) dt = 0]
f(–y) dy, where t = –y
f(y) dy
a
g(–x) =
x
f(t) dt g(–x) = g(x)
a
Hence, g(x) =
x
a
We have
Solution
If f is an odd function, then g(x) =
[f is odd ]
Let
1
1.
g(–x) =
But g(– u) = – g(u), so the integral is 0.
Example 14.
Note:
even function.
(–1)[x] dx = 0.
n
1
1
3
1 2 1 n 1 = .
Hence, nlim
2
Example 15.
1
x 2 x 4 x6
x 2n
2
x
1
......
dx
In =
2
4
6
2n
0
1 (x odd )dx 0
1
[ P–4 ]
1
x2
x4
x6
x 2n 2
= 2 1·2 2· 4 4·6 ....... 2n(2n 2)
0
f(t) is even function, if f(t) is odd.
If g(x) =
x
1 t
ln 1 t dt, then find
0
whether f is even or odd.
Solution
1 t
Let f(t) = ln
1 t
1 t
1 t
= –f(t)
= – ln
f(–t) = ln
1 t
1 t
f(–t) = –f(t) i.e., f(t) is an odd function
2.100
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1 t
x
a
will be an odd function if f(t) dt = 0.
ln 1 t dt is an even function, using
g(x) =
0
0
the above property.
2.
Because if g(x) =
x
If f(t) is an even function, then g(x) = f(t) dt is
0
an odd function.
Proof: We have, g(–x) =
x
f(t) dt
0
g(–x) = –g(x)
0
f(t) dt
a
x
a
0
x
a
0
a
0
0
x
x
0
0
f(t) dt = f(y) dy – f(t) dt
2 f(t) dt dt = 0
– f(t) dt = 0
f(t) dt = 0
or g(x) = f(t) dt is an odd function when f(t) is
a
0
0
x
0
g(–x) = – g(x)
Hence, g(x) is an odd function.
3. If f(t) is an even function, then for non-zero 'a',
x
a
0
2
x
g(–x) = –
x
{Put y = –t in the second integral of LHS}
{f(–y) = f(y)}
= – f( y) dy, where t = –y
f(y) dy [ f is even]
a
f(t) dt = – f(t) dt
f(t) dt + f(t) dt = – f(t) dt – f(t) dt
x
g(–x) = –
x
x
f(t) dt is an odd function.
a
a
a
0
0
x
a
even and
f(t) dt is not necessarily an odd function. It
a
f(t) dt = 0.
0
L
1 dx
1.
ln x x 1 x ln 2
2.
Let f(x) =
3.
x
ln t
(t 1) dt if x > 0. Compute
1
1 1
f(x) + f . Also prove that f(2) + = ln2 2.
x
2 2
Prove that the function
x
1/x
1
1
dt
dt
F(x) = 0
2
0 1 t2
1 t
is constant on the interval (0, ).
Prove that
a
a
a
(x )dx 2 (x )dx, (x )x dx 0 .
2
a
5.
6.
2
0
a
2
7.
(i)
(ii)
/2
– /2
dx
1 sin x
ln
dx = 0
1 sin x
101
Prove that
3
x
x17 cos x dx = 0
tan
x
2 7
1
(1 x )
1
(i)
(ii)
(3x x sin x x 1 x )dx = 250
5
2
10
5
4
5
Evaluate the following integrals :
(i)
(ii)
Evaluate the following integrals :
7
2
sin x
2
73
10
1
4.
10
[3 7x 100x ]dx
(iii)
2
0
1/ 2
1/ 2
sec x ln
32
1
1 x
dx
1 x
| xsin x |dx
2
x
1
1 x 1
tan
tan
dx
1
x
x2 1
(iii)
(iv)
3
1/ 2
1/ 2
1/2
x 1 2 x 1 2
2
x 1 x 1
dx
DEFINITE INTEGRATION
8.
2.101
Y
Prove that one of the antiderivatives of an even
function is an odd function and ever y
antiderivative of an odd function is an even
f(a–x) f(x –a)
function.
9.
x f(x) x·f "(x) 2 dx , where f(x)
2
Evaluate
3
2
is an even differentiable function
Property P
(i)
a
f(x) dx =
a
0
0
(ii)
f(x) dx =
a
a
0
0
(i) Proof :
a
0
0
LHS
LHS
f(a x)dx
Put x = a – t dx = – dt
Also, when x = 0, t = a and when x = a, t = 0.
0
a
a
0
b
a
a
f(a t)(dt) f(a t)dt
a
f(a x)dx RHS
a
b
b
a
f (a b t) (dt) f (a b t)dt
b
f (a b x)dx RHS .
a
Graphical proof :
We can get the graph f(a + b – x) by shifting the graph
of f(x) rightward by (a + b) units to get f{x – (a + b)}
and then reflecting the resulting graph about the line
x=a +b
0
Y
x=a+b
Graphical proof :
To draw y = f(x – a), the graph of y = f(x) is shifted right
ward by 'a' units and to get y = f(–x) we draw the image
of y = f(x) in the line x = 0.
Hence, the graph of f(a – x) is the image of f(x – a) in
the line x = a ( see figure).
Finally, we observe that the graph of f(a – x) is obtained
when f(x) is inverted laterally in the region {0, a] i.e.
the graph of f(x) is reflected about the line x = a/2.
Thus, the graphs y = f(x) and y = f(a – x) form equal
areas with the x-axis in 0 x a. Hence, the formula is
established.
Y
f(x)
O
f(x)
a
f{x–(a+b)
b 2a+b
Y
a+2b X
x=a+b
f(a+b–x) f{x–(a+b)
a
b 2a+b
From the graph it is clear that
b
b
a
a
a+2b X
f (x)dx = f (a b x)dx
f(x–a)
a
X
Put x = a + b – t dx = – dt
When x = a ; t = b and when x = b; t = a
f(a + b x) dx
a
2a
b
(ii) Proof : f (x)dx f (a b x)dx
f(a x)dx
f(x) dx =
a
O
2a X
This property is the generalised form of the previous
property. In this property too, f(x) inverts itself laterally
in the region [a, b]. As a result graphs of f(x) and
f(a + b – x) form equal areas with x-axis in the interval
[a, b]
2.102
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Evaluate I =
Example 16.
We have I =
Solution
cos2 x
dx
1 a x
cos 2 ( x)
dx
1 ax
a x cos2 x dx
1 ax
Adding the above integrals, we have
2
=
3.
/2
0
g (sec x)
dx =
g(cos ecx) g(sec x)
4
g (x)
a
a
g (x) g (a x) dx = 2
4.
[ f(x) = cos2x = f(–x)]
g (cot x)
dx =
g(tan x) g(cot x)
4
g (cos ecx)
dx
g(cos ecx) g(sec x)
/2
0
/2
0
g (cos x)
dx =
g(sin x) g(cos x)
4
g (tan x)
dx
g(tan x) g(cot x)
= 0
2
2I = cos xdx = – 2 cos x dx
/2
0
I =
/2
0
2.
using b f(x)dx b f(a b x)dx
a
a
=
0
sin 2x
= (1 cos 2x)dx = x
0
2 0
This gives I = /2.
Example 17. Prove that
2
2
g (sin x)
g (sin x) g (cos x) dx
0
g (cos x)
2
= 0
Let =
2
g (sin x)
g (sin x) g (cos x) dx
0
g sin x
2
[ P–5 ]
x g cos x
g sin
2
2
2
g (cos x)
g (cos x) g (sin x) dx
=
0
Adding the above integrals, we have
g (sin x)
g (cos x)
2I = 02
dx
g (sin x) g (cos x) g (cos x) g (sin x)
dx = .
4
/2
0
Note:
1.
/2
0
g (sin x)
dx
g(sin x) g(cos x)
Evaluate
Let f(x) =
/2
0
/2
0
cos x
dx
cos x sin x
cos x
dx
cos x sin x
cos( / 2 – x )
f(/2 – x) = cos( / 2 – x ) sin( / 2 – x )
=
0
Solution
Solution
g (sin x) g (cos x) dx = 4
=
=
Example 18.
Since
/2
0
sin x
sin x cos x
f(x) dx =
/2
0
f(/2 – x) dx,
[ P–5 ]
therefore
/2
sin x
cos x
dx ...(1)
dx = 0
sin
x
cos x
cos x sin x
/2
cos x sin x dx = / 2 1.dx = /2.
2I = 0
0
cos x sin x
I =/4.
Note that we have added the two expressions for I as
given in (1) to obtain a simple expression for I which
could be easily evaluated.
Example 19. Find the value of the definite integral
/2
I = 0
dx
0
a
(1 x )(1 x ) (a > 0).
Solution
I=
2
0
2
Put x = tan
d
=
1 (tan )a
2
0
(cos )a
d
(sin )a (cos )a
DEFINITE INTEGRATION
2
0
(sin )a
d
(sin )a (cos )a
[ P–5 ]
In first integral put x = t
dx = dt
.
4
dx
e
1
dx
dx
e
1
/2
=
2
4
2
10
Then, I = 4
sin x
/2
10
0
I=
Solution
2
[(14 x) .dx
[x 2 ] [(14 x)2 ]
...(2)
I=
.
2
1 . dx = x /2 =
...(1)
[ P–5 ]
sin x
Example 23. Find I =
[x 2 ].dx
Let I = 4
[(14 x)2 ] [x 2 ]
Solution
/2
0
[x 2 ]dx
[x 28x 196] [x ] .
/2
= 0 e sin x 1 0
Example 20. If [.] stands for the greatest integer
function, then evaluate
10
dt
= / 2 e sin t 1 0
0
Adding the integrals, we get I =
2.103
sin4 t
f (z)dz
cos4 t
f (cos2t z) f (z)
sin 4 t
f (z)dz
cos4 t
f (cos 2t z) f (z)
f(cos2t z)dz
f(cos 2t z) f(z)
sin 4 t
cos4 t
[ P–5 ]
Adding (1) and (2),
10
2I =
4
I= 3
cos 4 t
2I = z | sin4 t
dx
Evaluate
tan x
6 1
3
cos t
1
I = – (sin4 t + cos4 t)
2
1 1 2
= – 1 sin 2 t
2 2
1 1
I = sin2 2t.
2 4
Solution
dx
3
1 tan x
6
Then I
3
6
2
cos x
6
cos x sin x
dx
sin x
dx
sin x cos x
b f(x) dx
a
...(1)
...(2)
b
f(a b x)dx [ P–5 ]
a
Adding (1) and (2), we get 2I = dx .
3 6 6
6
I=
Example 22.
Solution
=
3
.
12
0
dz
4
Example 21.
Let I
sin 4 t
...(2)
[ P–5 ]
Adding (1) and (2), we get
2I = 1.dx 6
...(1)
Evaluate
/2
/2
dx
.
e
1
sin x
/2
dx
Let I = / 2 e sin x 1
dx
/ 2 e sin x 1
/2
0
dx
e
1
sin x
Example 24.
=
Prove that
ln 2
8
Solution
Let I =
/4
0
/4
0
ln (1 + tan ) d
ln (1 + tan ) d
/4
= 0 ln 1 tan d
4
tan tan
/4
4
=
ln 1
d
0
1 tan tan
4
/4
1
tan
=
ln 1
d
0
1 tan
/4
2
d
= 0 ln
1 tan
...(1)
[ P–5 ]
2.104
=
/4
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
ln 2 d –
0
/4
0
ln (1 + tan ) d
/4
1.d – I
= ln 2
=
[ from (1)]
0
2I = ln 2.
4
Hence, I = ln 2.
8
ln x
dx.
ln x ln(150 x)
Evaluate 50
ln x
l n x l n (150 x) d x (1)
100
Solution I = 50
100
Also I = 50
ln (150 x)
ln (150 x) ln x) d x
[ P–5 ]
ln (150 x)
ln(150 x) ln x d x
Adding (1) and (2) we get
100
I = 50
(2)
2I =
Example 26. Suppose f is continuous and
satisfies f (x) + f (– x) = x2 then find the value of the
integral
1
f(x)dx .
I=
2I =
1
1
f(x)dx =
1
1
f( x)dx
1
1
1
1
[ P–5 ]
f(x) f(x) dx = x dx
2I = 2
1
2
find the value of
96V
.
Note: Removal of x from the integral
V=
Solution
=
2
0
x sin 2 x
0
Suppose you know the integral of f(x), then in order to
evalutate xf(x) we try to remove the factor x. This is
done by the help of property P–5, provided f(x) does
not change when x is replaced by (a – x).
i.e. f(a – x) = f(x)
...(1)
xf(x)dx
I = (a x)f (a x)dx
Let I =
0
a
0
[ P–5 ]
x sin 2 x
1
dx ,
2
1
dx
2
dt
2
Put 2x = t dx =
V=
1
1
t cos t dt = ( t) cos t dt
8 0
8 0
[ P–5 ]
2
cos t dt =
8 0
8
96
· = 12.
8
8
Example 28. Show that
2
0
Solution
2
I = 0
Let I =
2
0
x sin 2n x
dx = 2.
sin 2n x cos 2n
x sin 2n x
sin x cos 2n
2n
2
...(1)
(2 x)sin 2n (2 x)
dx [ P–5 ]
sin (2 x) cos2n (2 x)
2n
(2 x)sin 2n x
= 0
dx
(sin 2n x cos2n x)
Adding (1) and (2) we get
a
2
0
a
xf (x)dx .
1 2
1 2
x 2sin 2 x 1 dx = x cos2x dx
2 0
2 0
V=
1
2
I = x dx = .
0
3
1
0
Suppose V =
2
0
0
2V =
x dx
a
Example 27.
1
Solution
a
(x a x)f(x)dx a f(x)dx
a a
f(x)dx provided f(a – x) = f(x)
...(2)
2 0
It is to be noted that x is eliminated only when f(x)
remains unchanged when (a – x) is put in place of x.
hence before applying the above method it should be
observed that f(a – x) = f(x).
100
2 I = 50 d x 2 I = 100 – 50 I = 25.
by (1)
0
I=
100
Example 25.
a
(a x)f (x)dx
...(2)
DEFINITE INTEGRATION
2I =
0
2
2I = 2 0
I=
= 2
2
Example 30.
2 sin 2 nx
dx
sin 2n x cos2n x
2
0
sin 2n x
sin x cos x dx
0
I = 4
/2
0
[ P–6]
2n
sin 2n x
dx
sin 2n x cos 2n x
...(3)
[ P–6]
= 4
/2
0
cos 2n x
dx
(cos x sin 2n x)
...(4)
2n
Adding (3) and (4) we get 2 I = 4
/2
0
1.dx
/4
Example 29. Let I = 0 (x – 4x2) ln(1 + tan x)dx.
Solution
(x 4 x2) ln (1 + tan x) dx
I = 0
/4
I=
3ln 2
where k N, find k.
k
/4
(p x - 4 x ) [ ln2 - ln (1 + tanx) ] dx
2
sin 2 t
8
2
=0+
2.
1
2
Example 31. Evaluate
I=
/4 x 2 (sin 2x cos 2x)
0
Solution
0
[ P–5 ]
/4
2I = ln2 0
(x 4 x2)
/4
ln 2 x 2 4 3
ln 2 3 3
x =
I=
2 2
3 0
2 32 48
=
1
Hence, I = 2.
If the value of I =
[ P–5]
/2
sin 2n x
2
dx
2n
2n
sin x cos x
2
2
I = 4 0
3
ln 2.
192
sin 2n x
dx
sin 2n x cos2n x
2n
xsin 2xsin cos x
2
dx
Evaluate I =
0
2x –
Solution We have
( – x)sin 2( – x)sin cos( x)
2
I=
0
2( – x) –
( – x)sin 2xsin cos x
2
dx
=
0
– 2x
( – x)sin 2xsin cos x
2
dx
=
0
2x
Adding the above integrals, we have
2I = 0 sin 2xsin cos x dx
2
I = 0 sin x cos xsin cos x dx
2
1
=
t sin t dt [putting cos x = t]
–1
2
1
–t cos t
2
1
+ 2
cos t dt
=
1 2
1
2
sin 2n x
dx
sin x cos2n x
2n
2.105
/4
= 0
(1 sin 2x)cos2 x
dx
/4 x 2 (sin 2x cos 2x)
We have I = 0
(1 sin 2x) cos 2 x
2x 2 (sin 2x cos 2x)
dx
(1 sin 2x)(1 cos 2x)
...(1)
Also,
2
x sin 2 x – cos 2 x
/4
4
4
4
dx
I=
0
1 sin 2 x 1 cos 2 x
4
4
dx
2.106
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
2
2 x (cos 2x sin 2x)
/4 4
=
0
(1 cos 2x)(1 sin 2x)
Adding the integrals (1) and (2), we have
...(2)
2
2 x 2 – x (sin 2x – cos 2x)
/4
4
dx
2I =
0
(1 cos 2x)(1 sin 2x)
2x (sin 2x cos 2x)
/4
4
4
I= 0
(1 sin 2x)(1 cos 2x)
=
4
/4
0
1
1
dx
–
2x
4 1 cos2x 1 sin 2x
2x
/4 2x 4
4
=
dx
–
4 0 2 cos2 x
2
2 cos – x
4
[using sin 2x = cos (/2 – x)]
/4
=
x sec2 x – x sec 2 x
dx
4 0
8
8
4
= (I1 – I2).
4
Now, we have
/4
I1 = x tan x
8
0
=
and
–
/4
0
tan x dx
+ [ln |cos x|] 0 / 4 = + ln 1
8
8
2
/4
I2 = x tan x – / 4 tan x dx
8
4 0
0
4
/4
= – + ln cos x
4 0
8
Hence, we have
I=
=
= – – ln 1
8
2
ln 1 – ln 1
4 8
2 8
2
ln 1 –
=
ln 2.
2 8
2 16 2
2
Evaluate
Example 32.
=
2
0
x 2 cos x
dx
0 (1 sin x)2
Let I =
Solution
x 2 cos x
1
(1 sin x) dx
1
x {(1 + sin x)–2 cos x}dx
2
0
(applying integration by parts.)
2 (1 sin x) 1
–
= x
1
0
I = (–2 + 0) + 2
0
2 x.
x
0 1 sin x
I = – 2 + 2
( x)
0 1 sin x
(1 sin x)1
dx
1
dx
...(1)
[ P–5 ]
dx
...(2)
Adding (1) and (2), we get
.dx
2I = – 22 + 2
0 1 sin x
1 sin x
2I = – 22 + 2 0
dx
cos2 x
= – 22 + 2 (tan x – sec x) 0
= – 22 + 2 {0 – (–1 – 1)}
2I = – 22 + 4
I = – 2 + 2
Example 33.
2
x
If 0
dx = then show
1 sin x
2x 2 cos 2 (x / 2)
that 0
2x 2 cos 2 (x / 2)
Let I = 0
Solution
x 2 (1 cos x)
= 0
=
dx = + 2 – 2.
(1 sin x)2
(1 sin x)2
(1 sin x)2
dx
dx
x 2 cos x
x2
dx
dx
0 (1 sin x)2
0 (1 sin x)2
cos x
dx
(1 sin x)2
Integrating by parts taking x2 as the first function, we
get
2
=+ 0 x .
DEFINITE INTEGRATION
1
x
2
2 0
dx
I = + x
1 sin x
(1 sin x) 0
x
dx
0 1 sin x
I = – 2 + 2
...(1)
( x)
1 sin( x) dx
and I = – 2 + 2 0
[ P–5 ]
dx
0
0
2
=
or, I
f(sin 2x) sin x dx
0
f(cos2x).cos x dx .
0
2
f(sin 2x)sin x dx
...(1)
[ P–5 ]
0
2
f[sin( 2x)]sin 2 x dx
Then I
0
2
f(sin 2x)cos x dx
2
f(sin 2x)(sin x cos x)dx
1
1
2 f(sin 2x)
sin x
cos x dx
0
2
2
2
2
f(sin 2x)cos x 4 dx
0
Put x –
=t
4
4
(6 x) 3 (6 x) 4 (6 x)
2
4
2
...(2)
4
sin x sin(6 x) dx
2I =
2
= cos x cos(6 x)2
= – cos 4 + cos 2 + cos 2 – cos 4 = 2(cos 2 – cos 4)
Hence, I = cos 2 – cos 4.
4
Example 36. For any t R and f being a
continuous function,
let I1 =
1 cos2 t
1 cos2 t
sin 2 t
I2 =
sin 2 t
Solution
0
2
I=
(6 x)(x 3)(10 x)x(4 x) sin(6 x) dx
Adding (1) and (2), we get
...(1)
2
...(2)
0
2I
Let I
(x(3 x)(4 x)(6 x)(10 x) sin x)dx
2
4
Solution
Find the value of the definite integral
On adding (1) and (2), we get
Show that
Let I
0
(6 (6 x))(10 (6 x)) sin(6 x) dx [ P–5 ]
Hence, I = – 2 + 2.
2
f(cos 2x).cos x dx .
4
(sec 2 x sec x tan x) dx
= – 2 + 2
=
0
(x(3 x)(4 x)(6 x)(10 x) sin x)dx .
Now
= – 2 + {tan x – sec x} 0
= – 2 + {(0 + 1) – (0 – 1}
Example 34.
f(cos2t).cos t dt
4
=
1 sin x dx
or, I 2
4
2
(1 sin x)
= – 2 +
[ P–4 ]
4
Solution
(1 sin x)
or, I = – 2 + 0
4
2I 2 2
...(2)
1 sin x
Adding (1) and (2) we get
2I = 2 – 22 + 2
4
f sin 2 t 4 cos tdt
Example 35.
( x)dx
= – 2 + 2 0
2
2.107
1 cos 2 t
= sin 2 t
x f(x(2 x)) dx and
f(x(2 x)) dx then find
I1 =
1 cos2 t
sin 2 t
x f(x(2 x)) dx
(1 + cos 2 t + sin 2 t – x) f{(1 + cos 2 t + sin 2 t – x)
(2 (1 + cos2 t + sin 2 t x)} dx
=2
2
1 cos t
sin 2 t
f{(2 – x) x}dx –
I1
=I
I2
[ P–5 ]
2
1 cos t
sin 2 t
I1 = 2 I2 I1 2 I1 = 2 I2
I1
I2 .
x f{{2 – x)x}dx
2.108
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
/2
Suppose I1 = 0 cos( sin 2 x)dx ,
Example 37.
/2
I2 = 0 cos(2 sin 2 x)dx and
/2
I3 = 0 cos( sin x)dx , then show that
/2
I1 = 0 cos( sin 2 x)dx
/2
I1 = 0 cos( cos 2 x)dx
1
2
2
2
Hence, I =
=
.
6 3
108
n
I1 = 0 and I2 + I3 = 0.
Solution
[ P–5 ]
On adding
n = 108.
Example 39. Let f (x) be a continuous function on
[0, 4] satisfying f (x) f (4 – x) = 1. Find the value of the
4
definite integral 0
/2
2I1 = 0 cos( sin 2 x) cos( cos2 x)dx
/2
= 0 2 cos ·cos cos 2x dx = 0
2
2
I2 =
0
4
cos((1 cos 2x)dx = –
cos( cos 2x)dx
1
cos( cos t)dt
2 0
I2 = –
/2
0
cos( sin t)dt = – I
of n.
Solution
sin 1 x
dx
x2 x 1
1
1 cos
sin 1 1 x
x
dx = 2
dx
I = 0 2
0 x x 1
x x 1
On adding (1) and (2), we get
...(1)
...(2)
[ P–5 ]
1
dx
sin 1 x cos 1 x
dx
dx = 0 2
2
2 x x 1
x x 1
1
2 0
dx
2
3
1
x
2
2
2
dx
...(2)
Example 40. If f and g are continuous functions
on [0, a] satisfying f(x) = f(a – x) and g(x) + g(a – x) = 2,
then show that
a
a
0
0
f(x) g(x)dx f(x)dx .
Let I
Solution
1
f (x)
4
sin 1 x
2
0 x 2 x 1 dx n (where n N), then find the value
1
4
1
f (x)
2I = 0 dx I = 2.
3
If the value of the definite integral
I = 0
1
On adding (1) and (2), we get
1
1
...(1)
I = 0 f (x) 1 dx
I2 + I 3 = 0
Example 38.
4
I = 0 1 f (4 x) dx , put f (4 – x) =
[Put 2x = t]
2 /2
= – 0 cos( cos t)dt
2
2I = 0
1
dx
1 f(x)
[ P–5 ]
/2
0
=–
1
dx .
1 f(x)
Let I = 0
Solution
I1 = 0.
/2
2
1 1 2x 1
2I =
=
tan
2 3
3 0 3 3
2
then I
a
f(x) g(x)dx
0
a
f(a x) g(a x) dx
[ P–5 ]
0
a
f(x)[2 g(x)]dx
0
[ f(a – x) = f(x) and g(a – x) + g(x) = 2]
a
a
0
0
2
f(x)dx f(x)g(x) dx
2
f(x)dx I
a
0
2I 2
a
0
f(x)dx or, I
...(1)
a
f(x)dx .
0
DEFINITE INTEGRATION
Example 41.
3
Let f (x) = x
3x 2
1
x . Find
2
4
34
the value of 1 4 f(f(x))dx .
Solution
3
We have f (x) = x
34
34
3x 2
1
x
2
4
I = 1 4 f (f(x))dx = 1 4 f(f(1 x))dx
34
[ P–5 ]
f (1 – x) = (1 – x)3 –
f (x) + f (1 – x) = 1
Replace x by f (x)
f(f(x)) + f(1 – f(x)) = 1
...(1)
Now, I = 1 4 f(f(x))dx
Also I = 1 4
...(3)
f(f(1 x))dx
34
= a 1 4 f(1 f(x))dx
34
= 1 4 f(1 x)dx
..(4)
Adding 2I = 1 4 f(f(x)) f(1 f(x)) dx = 1 4 dx
34
1
1
2I = I = .
2
4
Alternative :
Given f (x) = x3 –
1
3x 2
+x+
4
2
1 3
(4x – 6x2 + 4x + 1)
4
1
= (4x3 – 6x2 + 4x – 1 + 2)
4
1
2
f (x) = [x4 – (1 – x)4] +
4
4
1
2
f (1 – x) = [(1 – x)4 – x4] +
4
4
2 2
f (x) + f (1 – x) = = 1
4 4
{Applying P–5 and adding}
34
34
2I = 1 4 f(x) f(1 x)}dx = 1 4 1dx
I=
1
.
4
Example 42.
1
1
1
0
cot (1 – x + x2) dx
1
0
=
cot (1 – x (1 – x)) dx
=
tan 1 x(1 x) dx
=
tan 1 x(1 x) dx
=
(tan x tan (1 x))dx ( 0 x < 1)
=
tan x dx + tan (1 – x) dx
=
tan dx + tan (1 – (1 – x) ) dx
1
0
1
1
1
1
x (1 x)
0
1
0
1
1
1
0
1
1
1
0
=2
=
1
cot (1 –x +x2) dx.
Evaluate
Let I =
Solution
[ using (1) ]
34
34
...(2)
34
34
Replacing x by f (x) we have
f {f (x)} + f [1 – f (x)] = 1
f [f (x)] = 1 – f (1 – f(x))
From equation (1), 1 – f (x) = f (1 – x) = 1 – f (1 – x)
Again, 1 – f (1 – x) = f (x)
I = 1 4 f{f(x)}dx = 1 4 f(x)dx
3
1
(1 – x)2 + 1 – x +
2
4
3
1
= 1 – x3 – 3x + 3x2 – (1 + x2 – 2x) + 1 – x +
2
4
Now,
2.109
1
0
1
1
0
1
1
0
1
[ P–5 ]
tan x dx
1
0
Integrating by parts taking unity as the second
function, we have
1
x
1
1
dx
I = 2 [tan x.x]0 0
2
1 x
1
1
2 1
= 2 2 2 [ln | 1 x |]0 = 2 ln 2
2 2
...(1)
Hence, I =
– ln 2.
2
2.110
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 43.
= k
1
Adding equations (1) and (2), we get
x4
2x
cos 1
dx
If
1 3 1 x 4
1 x2
1
3
(1 e 2 4x )dx
2I = 0
(5 2x 2x 2 )(e 2 4x 1)
x4
dx then find k.
1 x4
3
0
=
x4
2x
cos 1
dx
I=
1 3 1 x 4
1 x2
Solution
1
3
...(1)
[ P–5 ]
x4
1 2x
I = 1 3
cos
dx
1 x4
1 x4
Adding (1) and (2),
...(2)
3
1
3
3
1
3
2I = 2
0
k = .
dx
0
2
[5 2(1 x) 2(1 x) ][1 e
2
0
dx
= 0 (5 2x 2x 2 )(1 e 2 4x )
2
5 2 (x x)
1
dx
0
1
5 2 x
2
2
1
2
11 / 2 x 1
1
2
ln
=
4 11 / 2
11 / 2 x 1
2 0
= 1 ln
2 11
2 4x
...(1)
)
2 4(1 x)
1
...(2)
11 1
2
2
11 1
2
2
1 2 ln 11 1
=
2 11
11 1
=
1
ln
11
= 1 ln
11
[ P–5]
11 1
2
2 ln
11 1
2
2
=
]
1
e 2 4x dx
= 0
(5 2x 2x 2 )(e 2 4x 1)
dx
1
Also, I =
dx
1
dx
=1
2
0
2 11
1
x
4
2
1
dx
Example 44. Evaluate 0 (5 2x 2x 2 )(1 e 2 4x )
1
1
2
[ P–4]
(5 2x 2x )(1 e
2
1
x4
dx
1 x4
Solution Let I =
dx
=
x4
dx
2I =
1 3 1 x 4
1
1
0
5 2(x x)
0 1
x4
2x
cos 1
dx
I=
1 x 4
1 3 1 x 4
1
1
I=
11 1
11 1
1
ln 11 1
11
11 1
11 1
11 1
11 12
10
2
1 ln ( 11 1) .
10
2 11
J
a
1.
2.
(a) Let I =
f(x)
f(x) f(a x) dx.
b
Prove that xf(x)dx
0
Show that I = a/2.
(b) Use the result of part (a) to find
3
x
(i)
dx
0
x 3x
/2
sin x
dx
(ii
0
sin x cos x
Let f : [a, b] R be a bounded integrable
function for which x [a, b], f(a + b – x) = f(x).
a
ab b
. f(x)dx .
a
2
3.
Prove that x(sin x)dx
4.
Prove that
0
1
2
(i)
( )sin d 4 ,
(ii)
( ) cos d 0 .
0
0
(sin x)dx.
0
DEFINITE INTEGRATION
5.
If a continuous function f on [0, a] satisfies
f(x) f(a – x) = 1, a > 0 then find the value of
a
dx
.
0 1 f(x)
6.
xf (sin3 x + cos2 x) dx and
If I1 =
I2 =
2.111
0
/2
0
f(sin 3 x cos 2 x) dx then relate I1 and I2.
M
7.
Evaluate the following integrals :
1
1
ln 1 dx
(i)
x
0
9.
1 ln(1 x)
1 x
(ii)
0
(iii)
2
0
2
dx
sin 2 sin d
2
(i)
/3
(ii)
/6
3 /8
/8
sin 2x ln (tan x) dx
(iv)
dx
0 1 2 tan x
0
a
a
0
0
f(x) dx + f(2a – x) dx
2a
a
2a
0
0
a
f(x)dx f(x)(dx) f(x)dx
Put x = 2a – t in the second integral dt = – dx
Also when x = a; t = a and when x = 2a; t = 0
2a
a
0
f(x)dx f(x)(dx) f(2a t)(dt)
0
1
2x 1
1
2
0
2a
if f(2a x) f(x)
0,
a
=
2 f(x)dx, if f(2a x) f(x)
0
tan 1 x x dx
[2sin x]dx
f(x) dx =
Proof: We have
(iii)
2
Property P-6
2a
3 2
(ii)
| 2 sin x |dx dx
tan 2 x
dx
/ 4 1 ex
/4
3 2
2
4 3sin x
4 3cos x dx
ln
(iii)
0
8.
(i)
10. Evaluate the following integrals :
x
x
(i)
dx
(ii) 2
dx
0 1 sin x
0 sin x cos x
x sin x cos x
/2
dx
(iii)
sin 4 x cos4 x
0
11. Evaluate the following integrals :
1
x dx
4
(i)
0 1 cos 2x sin 2x
3 / 4 x sin x
(ii) / 4 1 sin x
x 2 sin x dx
(iii) 0
(2x ) (1 cos2 x)
x 3 cos4 x sin 2 x
dx
(iv) 0 2
3x 3x 2
2
sin x
dx
1 sin x cos x
Evaluate the following integrals :
(iv)
Evaluate the following integrals :
0
a
a
a
0
0
f (x)dx + f (2a x)dx
Thus, f (x)dx
0
a
a
0 f (x)dx 0 f (x)(dx) if f (2ax) f (x)
a
= a
f (x)dx f (x)dx if f (2ax) f (x)
0
0
if f (2a x) f (x)
0,
2a
a
f (x)dx
0
2 0 f (x) dx, if f (2a x) f (x)
Graphical proof :
First of all consider the case when f(2a – x) = – f(x).
Functions satisfying this condition are symmetric about
the point (a, 0) and such functions are obtained simply
by shifting an odd function horizontally by 'a' units.
2a
Thus, f (x)dx 0 where f(x) = – f(2a – x).
0
2.112
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Y
3
f(2a–x)
x
a
f(x)
2a X
f(x)
x=a
2a–x
Y
f(2a–x)
2a X
a
0
Evaluate
2
2
( x)dx
a cos 2 x b 2 sin 2 x
Adding and dividing by 2, we get
dx
I = 0 2
2
2 a cos x b 2 sin 2 x
/2
dx
= 2. 0
[ P–6 ]
2
2
2
a cos x b 2 sin 2 x
/2
/2
sec 2 xdx
1
1 b tan x
.
tan
I = 0
a 0
a 2 b 2 tan 2 x b a
2
0
=
2ab .
ab 2
2
Example 46.
Evaluate
3
1
x e
1/ 3
2
2/3
3
Solution
1
2
1
x
2
2
dx .
1 x
Let g(x) = x e 2 . Then, we
2
have
3
Evaluate I =
x 1 | cos x | dx
0
I = ( x) 1 | cos x | dx
[ P–5 ]
0
1
1 x
1
g x x e 6
6
3
2
2I = 2
2
1 cos x dx
0
2
[ P–6 ]
2
x
x
dx = 2 2 sin
2 0 = 2
2
Show that the value of the definite
I = 2 0 cos
Example 48.
integral
2
3 cos x
x ln
dx
3 cos x , is equal to
0
(i)
I = 0
Example 47.
1/3
xdx
2
2
0 a cos x b sin x
Solution Here f(–x) = f(x)
I=
g(x)dx 0 .
Thus, f (x)dx 2 f (x)dx when f(x) = f(2a – x).
0
2/3
0
Similarly, when f(2a – x) = f(x), the graph is symmetric
about the line x = a and such functions are obtained
by shifting an even function horizontally by 'a' units.
Example 45.
2
2I = ( 1 | cos x | dx
2a–x
2a
1
Hence, we have
Solution
x
2
1
x
and g x x e 6 = – g 1 x
3
6
3
1
/
3
2
/
3
1.
g(x) is odd symmetric about x =
2
2
2
3 cos x
ln 3 cos x dx
0
3 cos x
(ii) 2 0 ln
dx
3 cos x
3 cos x
dx
(iii) 2 0 ln
3 cos x
(iv) zero
Solution
2
3 cos x
(i) I =
x ln 3 cos x dx
2
3 cos x
I=
0
(2 x) ln 3 cos x dx
0
[ P–5 ]
2
3 cos x
dx
2I = 2 0 ln
3 cos x
2
3 cos x
I = 0 ln
dx .
3 cos x
(ii) Using P–6, we get
3 cos x
dx
I = 2 0 ln
3 cos x
3 cos x
dx
(iii) Now using P–5, I = 2 0 ln
3 cos x
On adding (ii) and (iii) we get I = 0.
2.113
DEFINITE INTEGRATION
Example 49. Without evaluating the integral
at any stage, prove that ,
I=
cos4 2 x . sin 2 4 x d x =
0
I=
Solution
/2
cos4 2 x sin 2 4 x d x
0
I=2
/2
I=8
/2
I= 4
cos6 2 x sin 2 2 x d x
0
cos6 t sin 2 t d t
0
[ P–6 ]
cos4 2 x . 4 sin 2 2 x cos2 2 x d x
0
(Putting 2 x = t )
/ 2
I = 8 cos6 t sin 2 t d t
0
I = 8
/2
sin t cos t d t
6
0
Adding, 2 I = 8
/2
/2
0
I=
2
0
2I =
0
=
/2
0
2I=
1
/2
0
sin 2 t cos2 t (1 2 sin 2 t cos2 t) d t
/2
1
1 1 sin 2 2 z
dz
4
2
Let I =
/2
0
0
/2
=
/2
=
0
0
/2
0
ln cos x dx ln 2 .
2
0
/2
0
ln sin xdx
/2
ln sin x dx
ln cos xdx
0
2
/2
=
ln sin xdx
/2
0
ln cos xdx
(ln sin x ln cos x) dx
ln sin x cos x dx
ln sin 2x dx
/2
0
/2
0
ln
sin 2x
dx
2
ln 2 dx
In the first integral put 2x = t and adjust the limits.
1
ln sin t dt x ln 2 0 / 2
2I =
2 0
1 /2
= .2 0 ln sin t dt ln 2 I ln 2
2
2
2
Taking I to the L.H.S., we get
1
I = ln 2 ln .
2
2 2
[ P–5 ]
1
(sin 4 z + cos4 z) dz
2
1
(1 2 sin 2 z cos2 z) dz
2
ln sin xdx
0
I=
1
2
cos2 z 1 cos z dz
2
Adding, 2 I =
/2
[ P–5 ]
1
2 /2
2
I=
sin 2 z 1 sin z dz
2
0
2
/2
Prove that
/2
Solution
Putting 2 t = z
0
Example 50.
1
1
2
sin 2 z 1 sin z dz
I=
0
2
2
I=
1 /2 1 1
cos2 y dz
4
2
2 0
1
1 /2
1 dy
Adding, 2 I = 0
4
2
5 /2
dy..
I=
16 0
I=
[ P–6 ]
1
2
sin 2 2 t 1 sin 2 t dt
2
/2
/2 1
1
1
sin 2 y dz
2 0
4
2
4
I=
sin 2 t cos2 t (sin 4 t + cos4 t) d t
0
I = 4
(Putting 2 z = y)
cos4 2 x sin 2 4 x d x
0
I=2
5 /2
dy.
16 0
1 1 1
sin 2 y dz
0
2
4
2
2I=
Evaluate
Example 51.
Solution
/4
– /4
ln
/4
– /4
ln (sin x + cos x) dx.
2 sin x dx
4
2.114
= , dx = d
4
Putting x +
=
/4
=
1
2
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
ln
0
2 sin d
/4
0
ln 2 d +
0
ln sin d
I=
1 ln(1 x)
1 x2
0
/4 ln(1 tan )
dx
sec2 d
sec 2
0
2
ln
dx = (ln 2) /4 –I
1 tan x
/4
0
·ln 2
8
( k = –ln 2
Evaluate I =
sin / 4 x cos( / 4 x)
ln
dx
cos( / 4 x)
I=
= – 1 ln 2.
4
Solution
=
k
.
8
I=–
then rove that
2 sin
/4
4 d
ln
= 0
cos
Solution
=
0
ln 2 d +
0
–
ln sin d
4
/4
0
ln (cos ) d
= ln 2 []0 / 4 + I1 – I2 = ln 2 + I1 – I2
8
Let us put + = – t, d = – dt in I1.
2
4
Also, when = 0, then t = /4 and
when = /4, then t = 0. Thus, we have
I1=
0
/4
ln sin (/2 – t) (– dt) =
/4
0
ln (cos t) dt = I2
Hence, we have
I = ln 2 + I2 – I2 = ln 2.
8
8
If
Example 53.
value of
/4
0
Solution
0
ln sin xdx = k, then find the
l n 1 tan x dx in terms of k.
I=
/4
0
ln 1 tan x dx
Let I =
0
3 2
ln ( 2 sin ) d.
2 0
3 ln sin d =
0
/4
ln sin d = – ln 2
Assume
Example 54.
[putting x = tan ]
/4
sin cos
d
= 0 ln
cos
/4
ln 1 tan / 4 x dx
0
/4
/4
/4
= 0
1
1
= 4 ln 2 – 2 ln 2
Example 52.
=
ln sin d
3
...(1)
0
=
( ) ln sin d
=
3
[ P–5 ]
0
(3 – 32 + 32 –3) ln sin d
0
= 3
+ 3
= 3
+ 3
0
0
0
0
ln sin d – 32
2 ln sin d –
ln sin d – 32
I2 =
=
0
=
0
0
0
0
0
0
2 ln sin d – I
2I = 3I1 – 3I2 + 3
Now I1 =
0
ln sin d
3 ln sin d
ln sin d
[From (1)]
2 ln sin d
ln sin d = – ln 2
(given)
ln sin d
( –) ln sin ( – ) d
() ln sin d
[ P–5 ]
2I2 =
ln sin d = – 2 ln 2
0
(given)
2
ln 2.
2
I2 = –
3
4
ln 2 +
I=
2
2
0
0
2 lnsin d
2 ln sin d
x(sin 2 (sin x) cos2 (cos x)) dx
I=
Solution
=
x(sin (sin x) cos (cos x))dx
2
2
0
( x)(sin (sin x) cos (cos x))dx [ P–5 ]
2
2I = 2
0
/2
/2
0
Adding, 2I =
I=
/2
0
(sin 2 (sin x) cos2 (cos x))dx [ P–6 ]
Example 57.
Solution
I
/2
0
2 dx
I=
2
x 2 sin x
dx .
8 sin 2 x
x 2 sin x
Let I
8 sin x
2 sin x
2 2 x
2
0
8 sin 2 x
Adding the above integrals, we have
2 2 (2x 2 )sin x
2I
0
8 sin 2 x
0
[ P–5 ]
2
2
xsin x
sin x
dx [ P–6 ]
2 2
2
9 cos x
8 sin 2 x
0
0
I = 2
x 3 cos x 2 2 2 3 cos x
I 2
ln
6 ln 3 cos x dx
6 3 cos x 0
0
0
4 2 2
ln 2 .
I = 3 2 ln 2 =
3
2
.
2
dx =
2
Evaluate 0
2
(sin 1) .
4
(sin 2 (cos x) cos 2 (sin x))dx
Example 58.
2
2
2
I = 0 x
2
0
[ P–5 ]
2
I = 0 sin(cos 2 x) cos(sin 2 x) dx
d(cos x)
dx
1 cos 2 x
d(cos x)
xd(tan 1 (cos x))
2
0
1 cos x
= –xtan–1 (cos x) |0 + 0 tan 1 (cos x)dx
2
= – 0 I1 I1
4
4
2I = 0 sin(cos2 x) cos(sin 2 x) dx
2I = 2 0 sin(cos 2 x) cos(sin 2 x) dx
Evaluate I = 0 x
We use integration by parts
Solution
x(sin(cos x)cos(sin x))dx
I = 0 ( x) sin(cos 2 x) cos(sin 2 x) dx
2
0
Example 56. Evaluate 0 x(sin(cos x)cos(sin x))dx
Solution
2
sin(cos2 x sin 2 x) dx
2 0
(sin 2 (sin x) cos 2 (cos x)) dx
0
Also I =
2I =
2
0
2I =
...(2)
[ P–5 ]
2
2
0
I = 0 sin(sin 2 x) cos(cos 2 x)dx
2I = 0 (sin1)dx I =
3 2
3 2
=
ln 2 d +
ln sin d
0
2
2 0
3 2
ln ( 2 sin ) d
=
2 0
Example 55. Evaluate
2.115
Adding (1) and (2)
3 4
ln 2 + 3
Now, 2I = –2 ln 2 +
2
DEFINITE INTEGRATION
[ P–6 ]
...(1)
where I1 = 0 tan 1 (cos x)dx
To find I1, we make note of the fact that the graph of
the function f(x) = tan–1(cosx) is symmetric with respect
2.116
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
to the point (/2, f(/2)) = (/2, 0). Therefore the
integrals of this function over the closed intervals
[0, /2] and [/2, ] are equal in absolute value and
opposite in sign, and hence, the sum is zero, i.e.. I1 = 0.
We can also establish this fact as follows : we divide
the integral I1 into two itnegrals over the intervals
[0, /2] and [/2, ] respectively and make a change
of variable x = – t in the second integral. We obtain
/2
I1 = 0 tan 1 (cos x)dx / 2 tan 1 (cos x)dx
/2
0
= 0 tan 1 (cos x)dx / 2 tan 1 ( cos t)( dt)
=
/2
0
tan 1 (cos x)dx
/2
0
tan 1 (cos t)dt = 0.
x
2
2010
(1 x)
dx
1
1004
(1 x 2010 )1004 dx
Consider I2 =
Solution
=2
1004
0
...(2)
[ P–5 ]
.
x
1 1
t(2 t)1004 dt
1005 0
1 1 1004
t (2 t)2004 dx
=
1005 0
Put t = 2y dt = 2dy
1 1/ 2
(2y)1004 (2 2y)1004 dt
So, I 2
1005 0
1/ 2
1
2·21004 ·21004
y1004 (1 y)1004 dy
=
0
1005
1 2009 1/ 2 1004
I2
2
y (1 y)1004 dy
0
1005
0
1
...(1)
1004
1 1
1 (1 t)2 dt
1005 0
I2
1
x
Now I1
0
x
Also I 2
dt
1
1004
2 1004
0
Thus I1 = 0 and therefore I = 2/4.
Example 59. Find the value of
1004
1
(1 t )
So, I2 =
(1 x 2010 )1004 dx
Put x1005 = t 1005 x1004 dx = dt
1004
0
1/ 2
0
...(3)
(1 x)1004 dx
x1004 (1 x)1004 dx
...(4)
From (3) and (4), we get
2010 I1
1 2010 I1
4020 .
I2
2
2
I2
1005
4
[ P–6 ]
N
1.
Let f be a continuous function. Show that
2
1
0
0
4.
f(x) dx [f(x) f(x 1)]dx .
2.
Evaluate the following integrals :
1 sin 1 x
x tan xdx
(i)
dx
(ii) 0
0
sec x tan x
x
1
ln x
x
ln(1 x 2 )
2
0
(iii)
1 x dx (iv) 0 1 x 2 dx
Prove that
1
1
ln 1 dx 0
(i)
0
x
(ii)
/2
0
sin 2x ln tan xdx 0
xdx
(iii) 0
1 cos sin x sin
x ln (sin x)dx
(iii) (2 cos x) ln(sin 2x)dx
(ii)
0
2
2
0
3.
Evaluate the following integrals :
2
x(sin x)2n
dx , n N
(i) 0
(sin x)2n (cos x)2n
3
dx
x
Evaluate the following integrals :
(iv)
5.
tan 1 x
0
x cos x sin x dx
(ii) x sin x cos xdx
xdx
(iii)
4 cos x 9sin x
(i)
0
2
4
6
4
0
0
(iv)
2
x sin xdx
0
5
2
2.117
DEFINITE INTEGRATION
6.
Let A denote the value of the integral
cos x
dx. Compute the integral
0 (x 2)2
Property P-7
If f(x) is a periodic function with period T, then
(i)
nT
a nT
0
(ii)
a
(iv)
(iii)
nT
f(x) dx = n
T
0
f(x) dx = n
0
f(x) dx, nI, a R
mT
b nT
a mT
f(x) dx, m, n I
0
f(x) dx = (n – m)
T
0
f(x) dx
b
f(x) dx, n I, a, b R
Let us prove f(x) dx = n f(x) dx, n I.
+
a
nT
T
0
0
Proof : We have
nT
0
n
f(x) dx =
n
=
rT
(r 1)T
f(x) dx
T
=n
f(u) du , [ since f(x) is periodic with period
0
r 1
T, f ((r 1)T y) f (y) ]
T
0
f(u) du = n
T
0
sin(x ) dx
0
tan =
Put x + = t
= 6
1
2
and cos =
.
3
3
–1
2 = tan ( 2 )
sin t dt
=
6 sin t dt
0
[ P–7 ]
= 2 6.
10
Evaluate 3 / 2 {2x}dx , where {.}
Example 61.
denotes the fractional part of x.
Solution f(x) = {2x}is a periodic function with
period 1/2.
20(1/ 2)
10
= 23
0
T
= 6
dx in terms of A.
Let I = 3 / 2 {2x}dx = 3(1/ 2) {2x}dx
f((r 1)T u) du,
r 1
n
=
r 1
Put x = (r – 1) T + u
x 1
T
f(x) dx = (n – m)
/2 sin x cos x
0
where sin =
f(x) dx, n I
T
f(x) dx.
From the following graph the proof is obvious.
Y
1/ 2
2x dx (as {2x} = 2x – [2x] and
0
when x [0, 1 / 2), [2x] 0)
1/2
23
= 23 x 2 =
.
0
4
Example 62.
Evaluate I =
32 / 3
– / 4
1 cos2x dx
Solution
f(x) = 1 cos2x =
period .
2 | cos x | is periodic with
Y
O
T
0
(n–1)T
2 sin x 2 cos x dx .
I=
6
1
0
3
sin x
Hence,
I=
=
We have
Solution
nT X
Find the value of the definite integral
Example 60.
2T
2
3
–
2
11 –
– / 4
3
0
2
X
| cos x | dx
2 . 11
– / 3
0
– / 4
/2
– / 3
cos x dx +
cos x dx
=
2 . 11 – / 2 cos x dx + – / 4 cos x dx
=
2 11 2 1 sin – – sin
4
3
cos x dx
2.118
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
2 22 – 3 1 .
=
2
2
(k 1)
Example 63.
(k N) and J =
| sin 2x |dx
| sin x | | cos x | ,
Let I =
k
4
0
that
dx
, then prove
sin x cos x
the curve from x = a to x = b. Also the area from x = a to
x = a + is 2 square units. Hence b – a = 4.
Similarly, from the second integralm
9
9
a+ b–0=
i.e. a + b =
.
2
2
17
a= ,b=
.
4
4
Hence,
sin 2x dx
and I = 4 – 4J.
I=2 0
sin x cos x
(k 1)
| sin 2x |dx
Solution We have I = k | sin x | | cos x |
Since the integrand is periodic with period
| sin 2x | dx
I=
[ P–7 ]
0 | sin x | | cos x |
/ 2 sin 2x dx
= 2
[ P–6 ]
0
sin x cos x
/2
= 2
sin x cos x
0
=
dx
2
2
0
0
sin x cos x dx 2
4
=4–4 0
Example 64.
Evaluate I =
f(x) = |x – 1| –
Evaluate
0
100
0
1
| f(2x) | dx
f(x)dx (ii)
0
2
f(x)dx 0 .
From the figure, we have
0
Y
1/2
–2
0
–1
cot –1 (cot x) dx
2 X
1
–1/2
(ii)
x dx +
/4
2 /3
x dx
3/ 2
.
3/ 2
Example 65.
b
ab
a
0
| sin x | dx 8 and
value of sin x dx .
| cosx | dx 9 , then find the
a
Here
b
| sin x | x dx is the area under
a
100
0
(1, –1/2)
f(x)dx 50
2
f(x)dx 0 .
0
| f(2x) | dx 2 | f(t) | dt 2 .4 2 2 2 4 .
1
1
1
2
0
0
Example 67.
Show that
Solution
=
Let I =
pq
0
=q
pq
0
pq
0
0
p
0
0
1
1
1
| cos x | dx = 2q +
| cos x | dx
q
| cos x | dx +
1
<p< .
2
2
sin p where q N and –
b
Solution
sin x dx
1
, 0 x 2, f (x + 2) = f(x) for all x R.
2
dx
sin x cos x
cot –1 (cot x)
2 33 2
= 3 8 3
If
/4
(i) Using the property of periodic functions,
3
4
2
–
3
=4
(i)
Solution
Hence,
17 /4
17
17
4
cos
cos
2 2 .
4
4
4
4
2
Example 66. Let the function f be defined by
dx
= 4 – 4J.
sin x cos x
13
4
–
3
The integrand is periodic with period .
Solution
I=
17 /4
/4
=
/ 2 (sin x cos x)2 1
x sin x dx x cos x |
| cos x | dx
| cos x | dx + | cos x | dx
2.119
DEFINITE INTEGRATION
{ period of |cos x| is }
=q
| cos x | dx cos x dx
=
0
= Area of the shaded region as shown in the figure.
/2
p
Y
0
/2
0
0
| cos x | dx
/2
cos x dx cos x dx cos xdx
=q
= q {(sin x)0 /2 (sin x) /2} (sin x)0p
= q {(1 – 0) – (0 – 1)} + sin p – sin 0
= 2q + sin p.
Example 68.
Evaluate
2n
0
Let I =
Solution
2n
0
1
0
1
[sin x + cos x] =
2
1
0
,
,
,
,
,
,
5/4 3/2 7/4
0x
+
3 /4
0
/2
3 /2
(2)dx
0.dx
7 /4
3 /2
3 /4
( 1)dx
7 /4
2n
0
[sin x cos x]dx
2
[sin x cos x]dx = – n
=n
0
Alternative :
We have
2
0
2
= 0
[sin x + cos x] dx
2 sin x 4 dx
[sin x cos x]dx = – n
[x] dx , (where [x] and
Evaluate
{x} dx
0
n
Example 69.
0
{x} are integral and fractional part of x and n N).
Solution
0.dx
3
7 3
= + 0 – +
– 3 + 2 –
+ 0 = –
2
4
4
2
Since sin x + cos x is periodic function with period 2,
so I =
0
n
We have,
1
2
n
[x] dx
0
3
n
(n 1) dx
= 0.dx 1.dx 2.dx ...
0
1
2
n 1
= 0 + 1.(2 – 1) + 2.(3 – 2) + .... + (n – 1)(n – (n – 1))
n( n 1)
= 1 + 2 + 3 + ..... + (n – 1) =
...(1)
2
(1)dx
2
2 n
Hence,
2
/2
9/4 X
5
3
= 1 + 1 × 0 (1)
4
4 4
7
7 5
( 2) 2
(1) = –
4
4
4
[sin x cos x] dx
0
1.dx
2
/4 /2 3/4
–2
[sin x cos x]dx
=
[ 2 sin x]dx
y= 2 sin x
O
–1
[sin x cos x] dx,
2
3
x
2
4
3
x
4
3
x
2
3
7
x
2
4
7
x 2
4
/4
1
where[.] is the greatest integer function.
So,
9 /4
p
/2
and
n
0
{x} dx = n
1
1
{x}dx = n. 2 .1.1
...(2)
0
n(n 1)
2
= (n – 1) [using (1) and (2)]
n
0
2
5 tan –1 (x – [x])
Example 70. Evaluate 0 1 (x – [x])2 dx
Solution We have
n
[x]dx
{x}dx
0
n
5 tan –1{x}
I=
1 {x} dx
=5
/4
tan –1 x
dx 5
u du
2
0 1 x
0
0
1
[period 1]
2
[Putting tan–1x = u]= 5
/2
u2
2
0
2
5 .
32
2.120
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 71.
2
1 sin 3 x dx =
0
Alternative :
Show that
2
1 cos3 x dx.
0
Solution In the integral on the left, replace sin x by
cos (/2 – x). Then make the substitution u = /2 – x:
2
2
3
1 sin 3 x dx = 0 1 cos – x dx
2
0
=
–3 /2
=
/2
2
1 cos3 u (–du)
/2
1 cos3 u du
–3 /2
=
2
1 2
1 2
dx +
cos (200x) dx =
2 0
2 0
The first integral on the right is . The second is zero
because cos 200 x makes a whole number of compute
cycles from 0 to 2.
Example 74. Find the value of the definite integral
[ since the integral is periodic with period 2]
The next example uses the same idea.
Example 72. Suppose n is a positive integer and
f(t) is any function defined for – 1 t 1. Show that
2
2
0
0
=
2n
0
1 cos u du.
11
cos2.
2 2
0 sin2(100x) dx
3
0
Use the identity sin2 =
max.(sin x,sin 1 (sin x)) dx (where n I).
Solution
f(sin nx)dx = f(cos nx)dx .
Solution Substitute x = /(2n) – u into the integral
on the left.
Then sin nx = cos (/2 – nx) = cos nu, dx = – du
2
0
=
f(sin nx)dx = –
( /2n
( /2n)–2
( /2n)–2
/2n
f(cos nu)du
2
f(cos nu) du
f(cos nu)du =
0
We have used two facts :
(i) f(cos nu) has period 2,
(ii)
a 2
a
Set A =
Solution
Compute
2
0
2n
f(cos nu)du,
= n 0 x dx 2 ( x) dx (sin x) dx
2
=n
0
where a = /2n – 2.
Example 73.
I = 0 max . sin x,sin 1 (sin x) dx
2
f(cos nu)du =
0
sin2 (100 x) dx.
cos2(100 x) dx and
2
sin2(100x) dx.
B=
0
By the last example, A = B. On the other hand,
A+B=
=
2
=
2
0
0
2
0
cos2(100 x) dx +
2
0
[cos2(100x) + sin2(100x)] dx
1 · dx = 2.
Hence, A = .
The period of the function is 2.
sin2 (100x) dx
2
2
8
2
=n
2
2 1 2 2
2
2 2
4
2
2 3 2
2 = n ( 8) .
2
8
4
8
Example 75.
It is known that f(x) is an odd
T T
function in the interval , and has a period
2 2
x
equal to T. Prove that
f(t) dt is also perodic with
a
period T.
Solution Given that f(–x) = –f(x) and f(x + T) = f(x)
Let F(x) =
x
f(t) dt
a
DEFINITE INTEGRATION
F(x + T) =
=
xT
=
xT
a
x
f(t) dt +
a
x
T/2
T/2
T/2
T/2
= F(x) +
Put t = u + T
= F(x) +
= F(x) +
= F(x) +
x
x
T / 2
f(t) dt +
T / 2
xT
f(t) dt
T/2
f(t) dt +
x
x
f(t) dt
f(t) dt +
x
T / 2
f(u + T) du
f(u) du
[ f(u + T) = f(u)]
f(t) dt
= F(x) + 0
{ f(t) is odd}
= F(x)
F(x + T) = F(x)
F(x) is periodic with period T.
Let
Example 76.
and
I1 =
p
0
f(t)dt I2 =
p n 10
10
xp
x
f(t)dt be independent of x
Fx
xp
x
I2=
10
p
10
...(1)
1
in (1), we get
2
...(2)
From (1) and (2), we get F(x) = F(x + 1)
F(x) is a periodic function with period 1.
1500
0
F(x)dx = 1500
1/2
1
F(x)dx
[ P–7 ]
0
1
= 1500 F(x)dx F(x)dx
0
1/2
f(t)dt
f(z)dz
F(x)dx .
1
2
1
in the second integral, we get
2
Putting x = y +
. p 10
1500
0
1
We have F(x) + F x = 3
2
1/ 2
1
1/ 2
I = 1500 0 F(x)dx 0 F y by
2
1/ 2
1
= 1500 0 F(x) F x dx
2
1/ 2
3 dx
= 1500
[ P–7 ]
0
F x + F(x + 1) = 3
0
f(z)dz
Replacing x by x +
f(t)dt and
p 10
1
= 3.
2
Solution
p
n 1
p
f(z)dz
I2
p n 1 .
I1
Find the value of
f(z)dz for some p, n N. Then
Let g(x) =
n
0
f(z)dz p n 1
Let F(x) be a non-negative
continuous function defined on R such that F(x) +
Since g(x) is independent of x, g(x) = 0,
f(x + p) – f(x) = 0
f(x) is periodic with period p.
Here, I1 =
p n 1 .p
Now I =
I2
evaluate I .
1
Solution
f(t) dt
2.121
0
[using (1)]
1
2
Hence, I = 1500(3) = 750 × 3 = 2250.
O
1.
Prove that
(i)
1000
(ii)
200
2000
(iii)
e
0
0
dx = 1000(e – 1)
1 cos x dx = 400
0
10
3
10
6
(iv)
2.
x-[x]
2
dx
= 1000
1 esin x
(sin x cos x) dx = ( 3 1) .
Evaluate the following integrals :
2
e dx
sin x dx
(ii)
(i)
{3x}
1
41 / 2
0
(iv)
(iii)
7/2
5 / 4
| cos x | dx
sin 2x
dx
sin x cos 4 x
4
2.122
3.
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Show that
(i)
a
(x 2 )dx 2
a
a
a
0
(x 2 )dx
5.
x(x )dx 0
2
a
(ii)
/2
0
1
2
=
/2
0
6.
(sin x) dx
0
m
2
2
m being an integer.
A periodic function with period 1 is integrable
over any finite interval. Also for two real numbers
a, b and for two unequal non-zero postive integers
2.14 ADDITIONAL
PROPERTIES
1. Shift property
The function y = f(x – c) is obtained by shifting
the graph of y = f(x) 'c' units rightward.
Thus, it is obvious that
b
bc
a
a c
b
b c
a
a c
f (x) dx
f (x) dx
f (x c)dx and also
f(x)
1
f(x–c)
1
2
0
1
(x 1)3 dx x 3 dx (x 1)3 dx .
2. Expansion–Contraction property
K
b/k
a/k
f(kx)dx
b
f(x)dx for every k > 0.
a
ap
a
f(x) dx =
ap
a
f(x) dx. Calculate the
f(x) dx = 0.
t
2
dt = , find
dt in
Given 0
4 2 4 2 – t
1 t
terms of .
Given an odd function f, defined everywhere,
periodic with period 2, and integrable on every
1 sin t
interval. Let g(x) =
4
sin
x
f(t)dt .
0
(a) Prove that g(2n) = 0 for every integer n.
(b) Prove that g is even and periodic with period 2.
Proof: Put kx = t in the left hand side
k dx = dt
When x = a/k ; t = a
when x = b/k ; t = b
b
b
dt
Thus, LHS = k a f(t) a f(x)dx RHS
k
Y
y(kx)
y=f(x)
A/k
.
b a+c
b+c X
O a
We can prove the above results analytically, by
substituting respectively x – c = t and x + c = t in the
two formulae.
Also, we have for example,
a
n
m
f (x c)dx
Y
0
7.
bm
b
Suppose f is continuous on (– ) with period
p. Suppose f and f are also continuous. Prove
that
(sin x)dx
0 (cos x ) dx m 0 (cos x) dx ,
(iii)
4.
(cos x)dx
a n
f(x) dx =
value of f ( x) dx.
m and n,
A
a
O a/k b/k
bX
In order to sketch y = f(kx) from y = f(x) the later is
compressed 'k' times along x-axis. Note that the graph of
y = f(kx) contracts w.r.t. y = f(x) when k > 1 and expands
when 0 < k < 1. As a result the area formed by f(x) with
x-axis in between the limits a and b contracts to the area
formed by f(kx) with x-axis in between the limits a/k and
b/k; and the latter area is (1/k) times the first area.
/3
For example, 0 sin xdx 3a sin(3x)dx
3. Reflection property
b
a
a
b
f (x)dx f ( x)dx
The curves y = f(x) and y = f(–x) are mirror images of
each other in the line x = 0 i.e. y-axis.
2.123
DEFINITE INTEGRATION
Y
integral of f.
f(–x)
f(x)
a
–a
–b
bX
1
Hence, a f (x)dx (b a)0 f[(b a)t a]dt .
5
4
e
(x 5)2
Compute the sum of two integrals
dx 3
2
2 / 3 9 x
3
e
1/ 3
2
0
1
2
2
I1 = e(x 5) dx e( t 1) dt e(t 1) dt
4
0
0
t 1
Similarly, we apply the substitution x = to the
3 3
dt
second integral. Then dx =
and
3
2
I2 =
1
0
2
1
2
0
2
Note that neither of the integrals e ( x 5) dx nor
9
a
a
We have y = f(x) dy = f (x) dx.
Also, x = f–1(y)
When x = a, y = f(a) and when x = b, y = f(b).
b
f (b)
a
f (a )
f (x)dx bf (b) – af (a)
f 1 (y)dy .
This theorem can be given in a different form as :
If g (x) is the inverse of f (x) and f (x) has domain
x [a, b] where f (a) = c and f (b) = d then the value of
b
d
a
c
f(x)dx + g(y)dy = (bd – ac).
Proof: y = f (x) x = f –1(y) = g(y)
d
b
f(x)dx + c g(y)dy
= f(x)dx + x f '(x)dx [y = f (x) f ' (x) dx = dy)]
I=
a
b
b
a
a
Evaluate
Example 2.
I=
1
0
x
e e dx 2
e e
e
ln(ln x)dx .
x
Let y = f (x) = e e .
x
Then ln y = e x ln(ln y) = x = 2 ln(ln y)
2
f–1(y) = 2 ln(ln y).
Solution
Hence, I =
b 1
a0
x
e e dx 2
de e
ce
ln(ln y)dy
If the value of the integral
Example 3.
Hence, I1 + I2 = – e(t 1) dt e(t 1) dt 0 .
0
b
where f(0) = e and f(1) = e e
I = bd – ac = 1. e e – 0.e = e e .
2 / 3 9 x 2
1
2
3 e 3 dx e ( t 1) dt
1/ 3
b
a
b
Let us tansform each of the given
integrals into an integral with limits 0 and 1.
To this end apply the substitution
x = (b – a) t + a
x = – t – 4 to the first integral. Then dx = –dt and
2
b
= x f(x) a = b f (b) – a f (a) = bd – ac.
dx .
Solution
5
f 1 (y)dy .
f(x).1dx xf(x) f '(x) x dx
For any given integral with finite limits a and b, one
can a l wa ys c h oose t h e l i n ea r subs t i t ut i on
x = pt + q (p, q constants) so as to transform this
integral into a new one with limits 0 and 1.
Let x = pt + q
Since t must equal zero at x = a and t must equal unity
at x = b, we have for p and q the following system of
equations :
a = p.0 + q,
b = p.1 + q,
p = b – a, q = a.
Example 1.
f (b)
f (a )
Proof: Using integration by parts
4. Transformation of an integral into a new
one with limits 0 and 1
b
b
a
f (x)dx bf (b) – af (a)
2
is , then find the value of
Solution
e
2
e
1
x2
dx
4
e
The functions e
n x dx.
x2
and
n x are
2
e x 3 dx can be evaluated separately in
elementary functions.
inverses of each other. Assuming f(x) = e
1
5. Integral of an inverse function
b = 2, we have a f (x)dx bf (b) – af (a) f (a ) f 1 (y)dy
If f is invertible and f ' is continuous, then a definite
integral of f–1 can be expressed in terms of a definite
1
b
2
2
x2
and a =
and
f (b)
2
e x d x = 2. e 2 –1.e –
e4
e
n y dy
2.124
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
{ x = f(y) dx = f (y)dy}
e4
= 2 e4 – e –
n x dx
e
Hence,
e4
n x d x = 2 e4 – e – .
e
f (b)
b
d [{f(y)}2]
a
b
{f(y)}2 · (–1) dy,,
= [(b – y){f(y)}2] ab –
a
(using integration by parts)
Example 4. Let f(x) be a differentiable real valued
function which is strictly monotonic and a, b two real
numbers. Show that
=
b
{f(y)}2 dy
= – {f(a)}2 dy + {f(y)}2 dy
= [{f(y)}2 – {f(a)}2]dy
= [{f(x)}2 – {f(a)}2] dx,
= {f(x) + f(a)} (f(x) – f(a)} dx.
= – (b – a) {f(a)}2 +
b
(f(x) + f(a)}{f(x) – f(a)} dx
a
–1
= 2 f (a) x{b – f (x)}dx .
a
b
b
a
a
b
a
b
As f(x) is strictly monotonic, f–1(x) exists.
Also, if x = f(y) then f–1(x) = y
and x = f(a) y = f–1 {f(a)} = a
x = f(b) y = f–1(f(b)} = b.
Solution
a
b
a
b
RHS = 2 f(y) (b – y) f (y) dy
a
P
1.
Prove that
/6
1
1
(a) sin 1 x dx = sin–1 0 sin xdx
0
2
2
1/2
(b)
2.
e2
ln xdx (2e e) e dx .
2
e
x
1
Show that for any number a > 1
a
1
ln xdx +
ln a
0
0
6.
e y dy a ln a .
y = ln x
Y
ln a
3.
2
4.
5.
1
5
10
f(x)dx g(y)dy .
the value of
1
2
Suppose f is continuous, f(0) = 0, f(1) = 1, f'(x) > 0,
1
1
and f(x)dx = . Find the value of the integral
0
3
1
1
f (y)dy
0
7.
a
1
X
8.
Use the accompanying figure to show that
/2
1
1
0 sin xdx 2 – 0 sin xdx .
Y
–1
y=sinx
2
1
0
3
7
Evaluate ( 1 x 7 1 x3 )dx
0
If g (x) is the inverse of f (x) and f (x) has domain
x [1, 5], where f (1) = 2 and f (5) = 10 then find
y=sinx
1
9.
1
2
Show that 0 e x dx 0 nydy by interpreting
the integrals as areas.
Let f be an increasing function with f(0) = 0, and
assume that it has an elementary antiderivative.
Then f–1 is an increasing function, and f–1(0) = 0.
Prove that if f–1 is elementary, then it also has an
elementary antiderivative.
Let a > 0, b > 0, and f a continuous strictly
increasing function with f(0) = 0. Prove that
a
b
ab f(x)dx + f 1 (x)dx .
0
0
Prove, moreover, that equality occurs if and
only if b = f(a).
2.15 ESTIMATION OF
DEFINITE INTEGRALS
2
X
Not all integrals can be evaluated using the methods
discussed so far. So we try to find the approximate value
of such integrals using some estimation techniques.
2.125
DEFINITE INTEGRATION
1. Domination Law
If f(x) g(x) for a x b, then
b
b
a
b
f(x) dx
b
a
f(x) dx g(x) dx unless f(x) = g(x) for
all x, in which case f(x) dx g(x) dx .
dx
1
.
Example 1. Show that
3
1 x
Actually
a
b
a
4
1
1 1
, it follows from
1 x4 x4
domination law that
3
3 dx
3 dx
1 1 x 4 1 x 4 – 3x1 3 13 – 811 13 .
1
This example illustrates an important technique : replacing
the integrand f(x) by a slightly larger one, g(x), which can
be easily integrated, gives the upper bound.
Note:
(i) If f(x) is an integrable function and f(x) 0 for a
x b, then actually
b
f(x) dx 0 unless f(x)
a
= 0 for all x.
(ii) If at every point x of an interval [a, b], the
inequalities (x) f(x) (x) are fulfilled then
b
a
(x)dx
b
a
f (x)dx
b
a
(x)dx, a < b.
This means that an inequality between functions
implies an inequality of the same sense between
their definite integrals, or, briefly speaking, that it
is allowable to integrate inequalities termwise.
On the other hand, simple geometrical
considerations indicate that the differentiation of
an inequality may lead to a senseless result. For
instance, the inequality f(x) < C (C is a constant)
does not, of course, imply that f'(x) < 0.
(iii) Let f and g be integrable on [a, b]. Define M and m
by M(x) = max {f(x), g(x)} and m(x) = min {f(x),
g(x)}. Then M(x) and m(x) are integrable and max
b
a
f(x)dx,
b
a
g(x)dx
b
a
max{f (x),g(x)}dx,
min{f (x),g(x)}dx min f(x)dx, g(x)dx
b
b
b
a
a
a
Example 2.
1
0
2
Estimate the value of
e x dx by using
Solution
2
ex
a
3
Solution
ex
b
a
Since
Y
e
g(x) dx .
1
0
e x dx
For x (0, 1), e x < ex
2
1
1
1
1
X
1dx < 0 e dx < e dx
1 < e dx < e – 1.
1
Example 3. If I = 3 ln x dx , then prove
x2
x
0
0
1
x2
0
4
3
that 0.92 < I < 1.
Solution For x > e , we know that
1/3
x e
1 < ln x <
e x
4
< 3
1
<1
log x
4
e1/3 x 1/3 dx < I < 3 dx
3 1/3 2/3 2/3
e (4 3 ) < I < 1
2
3
0.92 < I < 1.
Example 4.
Consider the integrals,
1
x
cos2 x dx,
I1 = 0 e x cos2 x dx , I2 = e
1
2
0
1
1
x / 2
dx and I4 = e
dx
I3 = e
0
0
Find the greatest integral.
Solution For 0 < x < 1 , x2 < x2 < x
2
x2 > x e x > e x
x2
1
2
1
2
2
x
2
e x cos x dx > 0 e cos x dx
2
and cos x 1
0
1
0
1
2
2
2
e x cos x dx 0 e x dx
1
1 x2
< 0 e 2 dx = 1
Hence, I4 is the greatest integral .
1
1
Example 5. Show that 0
2
dx
2
4x x
Let
the
given
integral
be I.
Solution
For x (0, 1), 4 x 2 x 5 > 4 x 2
1
1
1
1
0
<
2
5
0
4x x
4 x2
1
1
1 x
I < sin 2 = sin–1 = .
2 6
0
5
6.
2.126
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
>
2
5
4 (x x )
4
Also,
1
1
0
4 (x 2 x 5 )
>
– sin2x – sin3x – sin4x
1 – sin2x 1 – sin3x 1 – sin4x
So, we get 1 sin 2 x 1 sin 3 x 1 sin 4 x
11
2 dx
0
and,
Prove that
Example 6.
2
dx
1
(4 3x x 3 )
0.573
2
dx
(4 3x x3 )
Put x = 1 + u, then
1
du
I=
0
(2 3u u 3 )
Since, 2 + 3u2 < 2 + 3u2 + u3 < 2 + 4u2, (0 < u < 1)
du
1
2
(2 4u )
0
1
du
0
(2 3u 2 u3 )
Solving the two integrals, the inequality is established.
dx
Show that
0 1 x 2 2x 5
4
3 3
1
dx
1
2
0
1
dx
0
2
1
5
0
3 3
1
0
dx
,
1 x 2 2x 5 4
which is the desired result.
Example 8.
1
2
0
Show that
1
1 sin 3 x dx ( 2 ln(1 2 )) .
2
2
0
1 sin 3 x dx
Example 9.
If In =
/4
0
1
n 1
(i) In + In – 2 =
1
1 t 2 dt
1
( 2 ln(1 2 )) .
2
tann x dx, prove that
(ii) In – 1 + In + 1 =
1
n
1
1
< 2In <
, where n > 1 is a natural
n 1
n 1
number.
Solution
=
/4
=
/4
0
0
In =
/4
tann x dx
0
tann – 2 x(sec2 x – 1) dx
tann – 2 x sec2 x dx –
/4
0
tann – 2 x dx
/4
tan n 1 x
In =
(n 1) 0
2
1
cos x 1 sin 2 x dx =
0
1
dx
1
tan –1 3x
1
dx
tan –1 x
0
2
5
0
1 x 2x
3 0
(1 sin 2 x) cos 2 x dx
1
( 2 ln(1 2 ))
2
=
We have 1 + x2 + 2x5 > 1 + x2
2
and 1 + x + 2x5 < 1 + x2 + 2x2 = 1 + 3x2
[ x5 < x2 on (0, 1)]
1
1
1
Hence, we have
1 3x 2 1 x 2 2 x 5 1 x 2
1 3x 1 x 2x 1 x
2
0
2
cos x dx = [sin x]0 2 = 1,
1
Solution
2
0
1 sin 4 x dx =
0
(iii)
.
(2 3u )
Example 7.
2
1 2
2
= t t 1 ln( t t 1)
0
2
2
1 sin 2 x dx =
0
du
1
0
=
0.595 .
1
Let I =
Solution
2
0
1
.
2
I>
sin4x sin3x sin2x
Solution
1
– In – 2
1
n 1
Replace n by (n + 1)
1
Then In + 1 + In – 1 =
n
In the interval 0, , tann x < tann – 2 x
4
In + I n – 2 =
/4
0
tann x dx <
/4
0
tann – 2 x dx.
...(1)
...(2)
2.127
DEFINITE INTEGRATION
I n < In – 2
1
In <
– I (from (1))
n 1 n
1
2In <
(n 1)
And similarly, In + 2 < In
Y
f(a)
f(b)
...(3)
x=a
x=b X
Estimate the value of the integrals :
1
< 2In
...(4)
n 1
1
1
< 2In <
.
From (1) and (2) we get
(n 1)
(n 1)
Example 11.
x dx
144
or
0 x 3 16
160
less by breaking up the integration into two intervals
[0, 1] and [1, 10] and using appropriate approximations
for the integrand.
sin x
x
x cos x sin x (cos x)(x tan x)
f(x) =
=
<0
x2
x2
f(x) is a strictly decreasing function.
f(0) is not defined, so we evaluate
1
– In < In
n 1
Show that
Example 10.
10
10
x dx
1
x dx
10
x dx
0
3
0
3
1
3
x 16 x 16 x 16.
Solution
02
(i)
sin x
dx
x
(ii)
/3
/4
sin x
dx .
x
Solution
(i) Let f(x) =
lim f(x) = lim sin x = 1.
x 0
x
x 0
1
xdx
1 xdx
0
3
0
1
10
xdx
10 xdx
1
3
1
x 16 16 32 ; x 16
Thus,
10
x dx
0
3
1
9
x
2
= .
2
9
10
f
149
x 16 32 10 160 .
2. Max-Min Inequality
If m f(x) M for a x b, then
m (b – a)
3
Hence,
1< 2
0
b
f(x) dx M(b – a).
b
(b – a) f(a) < f (x) dx < (b – a) f(b)
a
sin x
dx < .
2
x
(ii) f(x) =
a
Further,
(i) For an increasing function in (a, b)
2
sin x
. 0 < 2
dx < 1 . 0
0
2
2
x
sin x
decreases on the interval
x
,
4 3 .
Hence, the least value of the function is
3 3
, and
m = f
3 2
Y
f(b)
2 2
.
its greatest value is M = f
4
Therefore,
f(a)
/ 3 sin x
3 3
2 2
dx
/ 4
2 3 4
x
3 4
x=a
x=b X
(ii) For a decreasing function in (a, b)
b
(b – a) . f(b) < f (x) dx < (b – a) f(a)
a
/ 3 sin x
3
2
dx
.
/
4
8
x
6
2.128
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 12.
Prove that
dx
3
.
0
10 5
x 10x 9sin x 5 5
Solution Let f(x) = x3 + 10 x + 9 sin x + 5
Thus, the absolute value of the definite integral does
not exceed the integral of the absolute value of the
integrand function.
3
f(x) = 3x2 + 10 + 9 cos x > 0 x R
f(x) is strictly increasing
1
is strictly decreasing in (0, )
f(x)
Absolute maximum of f(x) in [0, ] is
Note: Let K be a number such that |f(x)| K
x [a, b], then
1
and
5
–K f(x) K
–K m f(x) M K
It then follows from (1) that
–K(b – a) m(b – a)
dx
3
.
So, 3
0 x 10x 9sin x 5
5
10 5
6
5x
Example 13. Prove that 1 < 0 9 x 2 dx 5 .
5
x
,
Solution Let f(x) =
(9 x 2 )
We first find the greatest and the least values of the
integrand f(x) in the interval [0, 2].
( x 9)( x 1)
f(x) = –
(9 x 2 ) 2
For f(x) = 0, we have x = 1 as x [0, 2].
Now f(0) = 5/9, f(1) = 1/2, f(2) = 3/5.
The greatest and the least values of the integrand
in the interval [0, 2] are, respectively, equal to f(2) = 3/5
and f(1) = 1/2.
2
5x
3
1
dx < (2 – 0) .
Hence, (2 – 0) < 0
2
9 x2
5
2
or, l <
3.
b
2
5x
9 x dx < 65 .
2
0
b
a f (x)dx a | f (x) | dx
Proof For any x we have, –|f(x)| f(x) |f(x)|.
(If f(x) > 0 the right inequality turns into an equality
and the left inequality is obvious; if f(x) < 0 these
inequalities are proved similarly) It follows that
b
b
b
a
a
a
| f (x) | dx f (x)dx | f (x) | dx
b
b
that is, a f (x)dx a | f (x) | dx
a
Proof If a = b, the result is trivial.
Let m, M be the bounds of f on [a, b].
Let b > a. Then for all x in [a, b], we have f(x)| K
1
absolute minimum is 3
.
10 5
b
f(x)dx K b a .
...(1)
b
a
f(x) dx
M(b – a) K(b – a)
b
f(x)dx K(b a)
...(2)
a
If b < a, then a > b.
Hence we get from (2) that
b
f(x)dx K(a b)
a
b
f(x)dx K(a b)
a
b
f(x)dx K(a b)
...(3)
a
From (2) and (3), we get
b
f(x)dx K b a .
a
Example 14. Estimate the absolute value of the
integral
19 sin x
10 1 x8 dx.
19 sin x
19 sin x
Solution I = 10 1 x8 dx 10 1 x8 dx ...(1)
Since |sin x| 1 for x 10, the inequality
sin x
1
holds
...(2)
1 x8 | 1 x 6 |
8
8
Also, since 10 x 19, 1 + x > 10
1
1
1
8
< 10–8
...(3)
8
1 x
10
| 1 x8 |
2.129
DEFINITE INTEGRATION
From (2) and (3), we have
sin x
< 10–8
1 x8
19 sin x
19
dx <
10–8 dx
10 1 x8
10
19 sin x
dx < (19 – 10).10–8 < 10–7.
10 1 x8
1
a
a
2
2
f (x)dx g (x)dx
a
Proof Consider the function F(x) = [f(x) – g(x)]2
where is any real number. Since F(x) 0, then
b
[f (x) g(x)] dx 0 ,
1
n
0
1
1
(1 x )dx
b
a
Proof We have
b
b
a
a
a
2 g 2 (x)dx 2 f (x)g(x)dx f 2 (x)dx 0
The expression in the left side of the latter inequality
is a quadratic trinomial with respect to . It follows
from the inequality that at any this trinomial is nonnegative. Hence, its discriminant is non-positive, i.e.
b
f (x)g(x)dx
a
2
b
b
f (x)dx g (x)dx 0
2
a
b
Hence f (x)g(x)dx
a
2
a
b
b
f (x)dx g (x)dx ,
2
a
2
a
which completes the proof.
Example 15. Prove that
1
b
b
a
a
b
b
a
a
b
a
b
b
b
a
a
g(x)dx f(x)g(x)dx M g(x)dx .
m
a
(1 x)(1 x ) dx cannot exceed 15 / 8 .
0
Solution
(1 x)(1 x ) dx
3
0
(1 x)dx (1 x )dx
x2
x4
x 2 x 4
0
0
1
1
0
0
1
3
1
3 5
15
.
.
2 4
8
(n 0) .
1
1
dx dx 1.
Solution
0 1 xn
0
Using Schwartz-Bunyakovsky inequality,
2 sin xdx
0
1 x
2
.
2
We regard the integral as
2
f(x) g(x) dx, f(x) = sin x, g(x) = 1 1x 2 .
0
Since |f(x)| 1 and g(x) > 0,
2
2
xdx
0 sin
0 1 dxx2 = tan–1 x 0 = tan–1 2.
1 x2
2
But tan–1 x < /2 for all values of x, hence
2 sin xdx
.
0
2
1 x2
1
b
f(x) dx – 2
a
b
a
(x – a) (b – x) f(x) dx.
Further, if |f(x)| M, a x b, then prove that
b
M
3
a f(x) dx 12 (b – a) .
Solution We derive this result by two
applications of integration by parts :
b
[(x – a) (b – x)] f(x) dx
a
Example 16. Prove that
1
1
n 1
1
dx
0 1 xn
n2
1
Solution
Show that
Example 18. If f(a) = 0 and f(b) = 0, then prove that
3
1
n 1
n2.
m.g(x)dx f(x)g(x)dx M.g(x)dx
Example 17.
b
n
g(x)dx f(x)g(x)dx M g(x)dx .
m
2
a
0
5. If m f(x) M and g(x) > 0 on [a, b], then
For any functions f(x) and g(x), integrable on the
interval (a, b), the following inequality holds :
b
0
0
4. Schwartz-Bunyakovsky inequality
b
2
1
n
1 x dx
or,
f (x)g(x) dx
1
n
0
b
1
1
1 x dx . (1 x )dx dx 1
=–
=
b
a
b
[(b – x) – (x – a)] f(x) dx
(– 2) f(x) dx = – 2 a f(x) dx.
a
b
Now, suppose |f(x)| M, a x b.
From the preceding result,
2.130
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1 b
(x – a)(b – x)f (x) dx
a
2 a
but (x – a) (b – x) 0 between a and b,
b
f(x) dx
b
f(x) dx
Assume f and g are continuous on [a, b]. If g never
changes sign in [a, b] then, for some c in [a, b], we
have
a
f(x)g(x)dx f(c)
b
a
If
b
a
b
b
a
a
for some c in [a, b]. Therefore, (2) becomes
b
f(x) g(x)dx = f(b)G(b) – G(c)[f(b) – f(a)]
a
and G(b) – G(c) =
Example 19.
a
a
a
Solution
g(x)dx = 0, this inequality shows that
b
f(x)g(x)dx = 0.
a
In this case, the theorem holds trivially for any
choice of c since both members are zero. Otherwise,
the integral of g is positive, and we may divide by
this integral and apply the Intermediate Value
Theorem to complete the proof. If g is nonpositive,
we apply the same argument to –g.
The weighted mean value theorem sometimes leads
to a useful estimate for the integral of a product of
two functions, especially if the integral of one of
the factors is easy to compute.
Generalized Mean Value Theorem for
integrals
Assume g is continuous on [a, b], and assume f
has a derivative which is continuous and never
changes sign in [a, b]. Then, for some c in [a, b], we
have
b
c
b
a
a
c
f(x)g(x)dx f(a) g(x)dx f(b) g(x)dx ...(1)
g(t)dt . Since g is continuous,
a
1
we have 0
a
b
g(x) dx.
c
sin x
dx .
1 x2
By the generalized mean value theorem
b
g(x)dx f(x)g(x)dx M g(x)dx .
c
g(x) dx
This proves (1) since G(c) =
b
x
...(2)
a
f (x) G(x) dx = G(c) f (x) dx = G(c)[f(b) – f(a)]
b
Proof Let G(x) =
f (x) G(x) dx,
= f(a)G(c) + f(b)[G(b) – G(c)].
g(x)dx .
Proof Since g never changes sign in [a, b], g is
always nonnegative or always nonpositive on
[a, b]. Let us assume that g is nonnegative on
[a, b]. T hen we in t egr a t e t he i nequal i t ies
mg(x) f(x) g(x) Mg(x) to obtain,
m
a
b
since G(a) = 0. By the weighted Mean Value
Theorem, we have
Weighted Mean Value Theorem for
integrals
b
b
= f(b)G(b) –
b
f(x)g(x)dx = f(x) G(x)dx
a
b
1
M (x – a)(b – x) dx
a
a
2
b
M
f(x) dx (b – a)3 .
a
12
we have G(x) = g(x). Therefore, integration by parts
gives us
1
Estimate the integral I = 0
sin x
dx
1 x2
1
= sin 0
dx
1
sin tan 1 x 0 sin , (0 1) .
2
4
1 x
Since the function sinx increases on the interval
[0, 1] then sin < sin1. Thus, we get an upper estimate
of the integral :
1
sin x
0 1 x 2 dx 4 sin1 0.64 .
It is possible to get a better estimation if we apply the
same theorem in the form
1
sin x
0
2
1
1
1
1 x dx 1 sin xdx 1 (1 cos1)
2
2
0
< 1 – cos1 0.46.
6. Concavity
By making the use of concavity of the graph we can
make more appropriate approximation integrals as
shown in the illustration below.
b
Let us estimate the integral f (x)dx , where
a
f(x) = px2 + qx + r.
2.131
DEFINITE INTEGRATION
Y
Y
D
B
B'
A
C
B0
C0
a
a+b
0
b
X
2
It is clear that this area is less than the area of a
trapezoid with bases A0A and B0B. Let a midline C0C
be drawn. If we draw a tangent to the curve at point C,
then this tangent will intersect the vertical lines at the
points A' and B' and will form a trapezoid A'B'B0A0.
The area of this trapezoid is obviously less than the area
under the curve. Thus, in the case of the above function
f(x) with p > 0
A'
B
f(b)
y= px2+qx+r
A0
f(a)
b X
0 a
(b) This can be proved as above.
Example 21.
f (a) f (b)
2
(b) If the function f(x) increases and has a
concave down graph in the interval [a, b], then
b
(b – a) f(a) < a f (x)dx (b a)
(b – a)
b
f (a ) f ( b )
< f (x)dx < (b – a)f(b).
a
2
Solution
1
2
Estimate the value of e x dx .
0
Solution :
Y
C
e
b
a b
f (a) f (b)
(b a)f
f (x)dx (b a)
.
2 a
2
Example 20. Using geometry, prove that :
(a) If the function f(x) increases and has a
concave up graph in the interval [a, b], then
C
A
B
E
1D
O
1
A
X
From the figure,
1
2
Area of rectangle OAED < e x dx < Area of trapezium
0
OABD
1
2
1 < e x dx <
0
1
2
1 < e x dx <
0
1
.1. (e + 1)
2
e 1
.
2
1
(a) Without loss of generality we may assume
f(x) > 0. If the graph of a function is concave up, it
means that the curve lies below the chord through
the points A(a, f(a)) and B(b, f(b)) (see figure).
Therefore, the area of trapezoid aABb is greater
than that of the curvilinear trapezoid bounded
above by the graph of the function i.e.,
b
f (a ) f ( b)
a f (x)dx < Area of aABb = (b – a). 2
b
The inequality (b – a) f(a) < f (x)dx is obvious
a
because the area of rectangle aACb < the area
under the curve.
4
Example 22. Estimate the integral 0 1 x dx
using
(a) The mean value theorem for definite integral,
(b) Concavity and geometry,
4
4
x
(c) The inequality 1 x 1
,
2
(d) Schwartz-Bunyakovsky inequality.
Solution
(a) By the mean value theorem
I=
1
0
1 x 4 dx 1 4 , where 0 x 1.
But 1 < 1 4 <
2.
1 < I < 2 1.414.
2.132
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(b) The function f(x) =
the interval [0, 1], since
1 x 4 is concave on
7.
2 x 2 (x 4 3)
0 , 0 x 1.
f"(x) =
(1 x 4 )3 / 2
Hence, 1 <
4
0
1
(c) 1 < 0
x4
1
1 x dx 1 dx 1
1.1
0
2
10
1
4
(d) Put f(x) = 1 x 4 , g(x) = 1 and taking
advantage of the Schwartz-Bunyakovsky
inequality
1
1
1
1
0 1 x dx 0 1 x dx 0 (1 x )dx.0 1 dx
4
4
4
2
= 1.2 1.095.
t
0
0
t
t
sin x 0 x 0
t
t
0
0
sin xdx x dx
t
x2
cos x 0
2
t
0
1 – cost
...(1)
t2
.
2
Writing 1 – cos t t2 / 2 as
cosx 1 – x2/2,
we continue this process :
logarithmic, and arithmetic mean
We knows that the graph of ex is concave up over
every interval of x-values.
x
y=e
F
E
M
B
A
D
ln a ln a + ln b ln b
2
t
t
0
0
t
t
0
0
cosx 1 – x2/2 cos xdx (1 x 2 / 2)dx
sint t – t3 /3!
...(3)
sinx x – x3/3! sin xdx (x x 3 / 3!)dx
...(4)
Example 23. Let f (x) be a continuous function
with continuous first derivative on (a, b), where b > a,
and let lim f(x) , lim f(x) – and f '(x) + f 2 (x)
xb
x a
X
With reference to the above figure
we conclude that if 0 < a < b then
ln b
...(2)
1 – cost t2/2! – t4/4!
Clearly, this process can be continued.
C
– 1, for all x in (a, b) then show that the minimum value
of (b – a) equals
Solution f(x) + f 2(x) – 1
f 2(x) + 1 – f ' (x) in (a, b)
e dx
x
1–
ln a
e ln a e ln b
. (ln b – ln a).
2
From here we find that
<
ab
t
sin t t
Reverting to the x notation, we have
sin x x
Note: Compar i sion of ge ome tr i c ,
e(ln a + ln b)/2 . (ln b – ln a) <
We can begin with the inequality and generate
other inequalities by successive integrations.
For example :
cos x 1 cos xdx 1 dx
1 2
1.207 .
1 x dx
2
1
mean, which is turn is less than their arithmetic
mean.
ba
ab
.
ln b ln a
2
This inequality says that the geometric mean of
two positive numbers is less than their logarithmic
f '(x)
1 f 2 (x)
in (a, b)
b
b
f '(x)
0
a
2
dx – 1 f (x) dx
2 2
b – a – (tan 1 (f(x)) ab = –
(b – a) .
2.133
DEFINITE INTEGRATION
Example 24. Suppose f is a differentiable real
function such that f (x) + f '(x) 1 for all x, and f (0) = 0,
then find the largest possible value of f (1)
Solution Given f ' (x) + f (x) 1
Multiplying by ex
f(x) ex + f (x) ex ex
d x
(e · f(x)) ex
dx
Integrating between 0 and 1
1 d
1
x
0 dx (e f (x))dx 0 ex dx
1
e x f(x)
e
0
x 1
0
e · f (1) – e0 · f (0) e – 1
e 1
.
f (1)
e
K
1.
b
b
b
a
a
7.
It is known that f (x)dx g(x)dx . Does it
4.
a
/2
(ii) 1 <
/2
sin n 1 x dx <
0
sin x dx <
0
(iii)
/2
1
e 4<
1
e
x2 x
0
sin 2 x dx, n > 1
2
(iii)
6.
100
1
8.
Show that
2
f (x) sin 2xdx f (x) dx .
0
5
4
Estimate the integral (1 sin 2 x) dx .
4
1
11. Show that 0.78
4
1 x
1 dx
ln 5
e–x sin2x dx < 1(iv) 0
4
4 x3
(a) Show that 1
d
(0 < k2 < 1).
2
2
1 k sin 2 1 k 2
dx
10. If I = 0
, prove that, n 2 I .
1 x3 / 2
4
dx < 1
x dx
1
(ii)
3 <
16 x
6
3x 3 1 dx < 10 15 – 8 6 /5
2
9.
Prove that
2
5
0
3
1 x cos x
1
1
dx < .
(iv) – 0
2
2
2 x2
(i) 0 < 0
1
dx
4
1 sin 2 x
/2
(iv) 2 0
is continuous on [a, b]. Prove that f(x) = 0 for all
x in [a, b].
Prove that
0
5.
4
(iii) 2 < 0
f ( x)g( x) dx = 0 for every function g that
(i) 0 <
dx
2
b
2
1 x 3 3x 1 5
(ii) 3 23 <
Assume f is continuous on [a, b]. Assume also
that
1 x 3 dx 1.25.
Prove that
(i)
follow that f(x) g(x) x [a, b] ? Give examples.
3.
1
0
a
that f(x) 0 x [a, b]? Give examples.
2.
(b) Show that 1
It is known that f (x)dx 0 . Does it follow
1
dx
0
4
1 x 3 1 + x for x 0
3
dx
1
1 x 0.93 .
4
0
12. Prove that, if n > 1
/2
/2
sin xdx < sin xdx,
/4
/4
(ii) 0 < tan n 1 xdx < tan n xdx .
0
0
dx
(iii) 0.5 <
(1 x ) 0.524.
(i) 0 <
n 1
0
n
0
1/2
0
2n
Q
13. Prove the inequalities :
(i)
3
1
26
x 4 1 dx
3
(ii)
/2
2
0 x sin xdx 8
2.134
(iii)
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
2
1
1
7
dx
.
17 1 1 x 4
24
14. Prove that
2
2
dx
2
(i) 13 0 10 3 cos x 7
/4
2
x tan x
(ii) 0
0
32
/2 sin x
1
1
dx
(iii)
/4
2
x
2
4 sin x
3
(iv)
dx .
1
x
2
15. Integrating by parts, prove that
200 cos x
1
0
dx
.
100
x
100
16. Determine the signs of the integrals without
evaluating them :
2
3
0
0
0
5
1 x dx
.
2
3
2.16 DETERMINATION OF
FUNCTION
Example 1.
0
22. If and are positive acute angles then prove
that
dx
.
< 0
2
2
(1 sin sin x)
(1 sin 2 sin 2 )
If = = 1/6 , then prove that the integral lies
between 0.523 and 0.541.
23. Let p be a polynomial of degree atmost 4 such
that p(–1) = p(1) = 0 and p(0) = 1. If p(x) 1 for
1
x [–1, 1], find the largest value of p(x)dx .
(b a)
17. Prove that
100 tan 1 x
99
99
dx
(i)
2
1
400
200
x
5
609(ln 2) 2
609(ln 5)2
3
(ii)
< x (ln x)2 dx <
2
4
4
10 1 e – x
dx ln10
(iii) (1 – e–1) ln 10 <
1
x
1
1
x9
1
dx .
(iv)
0
10
10 2
1 x
1
dx
2
lies between
and
18. Prove that 0
2 x x2
3
1
. Also find the exact value of the integral.
2
19. Using the Schwartz-Bunyakovsky inequality,
prove that
1
0.692 x x dx 1 are valid.
24. Proceeding from geometrical reasoning prove that:
(i) if the function f(x) increases in the interval
[a, b] and has a concave down graph, then
1
2
1
1 , g2(x) = 1 + x2, show that
1 x2
1
3
1
0 1 x 2 dx 4 .
21. Show that the inequalities
f 2 (x)
1
x dx
sin x
dx
(b)
x
(c) x cos x dx .
(a)
20. Using Schwartz-Bunyakovsky inequality with
If f is continuous function such
f(a) f(b)
2
b
f(x) dx (b a) f(b).
a
(ii) Estimate the integral
3 x 2 dx
1 x using the
2
2
above results.
25. Let f be twice continuously differentiable in
[0, 2] and concave up. Prove that
2
f(x) cos x dx 0.
0
26. Find the greatest and least values of the function
2t 1
x
t 2t 2 dt on the interval [–1, 1].
I(x)
0
2
27. Given that f satisfies | f(u) – f(v) | < |u – v | for u
and v in [a, b] then prove that
(i) f is continuous in [a, b] and
(ii)
b
f (x) dx (b a) f (a) <
a
(b a)2
.
2
that f(x + y) = f(x) + f(y) for all real numbers x and y,
then prove that f(x) = kx for some constant k.
...(1)
Solution f(x + y) = f(x) + f(y)
Let us substitute t for y in (1) and then integrate
DEFINITE INTEGRATION
from t = 0 to t = y with x held constant :
y
y
t0
t 0
y
f (x t)dt = f (x)dt + t 0 f(t)dt .
Then substituting u = x + t, du = dt on the left hand
side yields the equation
xy
ux
f(u)du y . f(x)
x y
t 0
f(t)dt –
y
f(t)dt
x
y
t 0
t 0
f(t)dt – f(t)dt .
f (x) f (y)
y . f(x) = x . f(y), so that
,
x
y
for all x and y. Because x and y are independent, if
follows that the function f(x)/x must be constantvalued, and therefore f(x) = kx for some constant k.
Example 2. If f is a positive-valued continuous functin such that f(x + y) = f(x) . f(y) for all
real numbers x andy, then prove that f(x) = ekx for
some constant k.
...(1)
Solution f(x + y) = f(x) . f(y)
Let g(x) = ln(f(x)). Taking the natural logarithm of
both sides in (1) gives
g(x + y) = ln (f(x + y)) = ln (f(x) . f(y))
= ln (f(x)) + ln(f(y)) = g(x) + g(y).
The application of the previous example to g now
yields ln(f(x)) = kx,
so that f(x) = ekx.
Find the real number a such that
f (t)dt
=2 x.
t2
Solution Differentiating both sides of the given
x
6 + a
f(x)
1
2·
f (x) = x3/2.
x2
2 x
Substituting this value in the given relation, we get,
expression, we get
x
6 + a
dt
=2 x
t
x
A function f (x) satisfies
Example 4.
x
f (x) = sin x + f '(t) (2 sin t – sin2t) dt then find f (x).
Differentiating both sides w.r.t. x, we get
f(x) = cos x + f '(x)(2 sin x – sin2x)
(1 + sin2x – 2 sin x) f '(x) = cos x
cos x
cos x
f(x) =
=
2
2
1 sin x 2sin x (1 sin x)
Solution
t 0
The right hand side is symmetric in the variables x
and y, so interchanging them gives
Example 3.
a= 9
Hence, f (x) = x3/2 and a = 9.
0
from which we find that
y . f(x) =
2.135
Integrating, f(x) =
cos x dx
(1 sin x ) 2
Put 1 – sin x = t
f(x) = –
dt
1
1
t 2 = t = 1 sin x + C.
Also , f (0) = 0, hence C = –1
sin x
1 1 sin x
1
f(x) =
–1=
=
.
1 sin x
1 sin x
1 sin x
Example 5. Let f(x) be a continuous function
x
such that f(x) > 0 for all x 0 and (f(x))101 = 1 + f (t) dt .
0
Solution
Given (f(x)) 101 = 1 +
x
f (t) dt
0
Differentiating,
101 · (f(x))100 · f '(x) = f (x)
101 · (f(x))99 · f '(x) = 1 (as f (x) > 0)
Integrating,
(101)(f(x))100
=x+C
100
but f (0) = 1
101
100
101
101
(f(x))100 = x +
.
100
100
C=
Putting x = 101,
101
101 (101)(101)
(f(101))100 = 101 +
=
100
100
100
6+ 2 t a = 2 x
(f(101))100 = 101.
6 + 2[ x a ] = 2 x
Example 6.
2 a =6
and f(2) = 2, find the value of f(t) dt .
xy
If f(t) dt is independent of x
x
x
1
2.136
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
xy
We have f(t) dt .
Solution
x
Differentiating both sides with respect to x, treating y
as constant, we get
d xy
f(t) dt = 0
dx x
yf(xy) – f(x) = 0
...(1)
1
in equation (1), we have
Putting y =
x
1
f(1) – f(x) = 0
x
f(x) = f(1) · 1
x
x1
1
1
...(2)
Let f(x) be a continuous function
0
Put x = 0 in the given equation
0
Differentiating the given equation, we get
/2
/ 2
(sin x t f (t)) dt .
/2
f (x) = ( + 1) sinx + A
0
Integrating, we get
f(x) = ex – 2e2x
1
1
Now, k = (e t 2e2t )e t dt = (1 2e t ) dt = 3 – 2e
0
0
1
1
c= =
.
k 3 2e
1
0
...(1)
t ( 1) sin t A dt
t sin t dt
A 2 ( 1)
Hence, f (x) = ( + 1) sinx + 2( + 1)
fmax. = 3( + 1) and fmin. = ( + 1).
Example 8. Let f (x) is a derivable function satisfying
f(x) = e t sin(x t) dt and g (x) = f '' (x) – f (x).
x
f (t) dt = ex – e2x,
Example 10. If f(x) = ex + (e x te x ) f(t) dt, find f(x).
/2
= sin x + sin x + / 2 t f (t) dt
Find the range of g (x).
...(1)
1
1
Let c =
where k = f (t)e t dt .
0
k
Solution We have f (x) = sin x + / 2 (sin x t f (t)) dt
0
x
1
Find the maximum and minimum values of the
function f.
x
t
0
0 = 1 – c · f (t)e t dt
Example 7. Consider a real valued continuous
/2
x
f ' (x) = sin x + f (x)
f '' (x) = cos x + f ' (x)
= cos x + sin x + f (x) [using (1)]
f '' (x) – f (x) = sin x + cos x
g (x) = sin x + cos x
The range of g (x) is [ 2 , 2 ] .
Solution
1
function f such that f (x) = sinx +
e sin t dt e
0
1
and x = 2 in equation (1),
2
x
0
f ' (x) = ex · e–x sin x +
1
Hence, we have f(x) f(x) dx = 4 ln x.
= 2( + 1)
f (x) = ex e t sin t dt
c e2x f (t)e t dt then find f(x) and the value of c.
1
f(1) – f(2) = 0
2
f(1) = 2f(2) = 4.
/ 2
x
x
we have
Now, A =
[ P–5 ]
0
and c is a constant satisfying f (t) dt = e x –
f(x) dx = f(1) x dx = f(1). ln x
/2
x
= e x t sin(t) dt
Example 9.
x
Now, putting y =
0
0
x
f (x) = e t sin(x t) dt
Solution
We can write f(x) = Aex + Be–x, where
Solution
A=1+
A=1+
1
1
0
0
f(t) dt and B = t f(t) dt.
1
(Ae Be )dt = 1 + (Ae Be )
t
t
t
0
A = 1 + A(e1 – 1) – B(e–1 – 1)
(2 – e) A + (e–1 – 1) B = 1
Now, B =
1
t(Ae Be )dt = A (tet – et) 0t
t
0
+ B( te t e t ) 1
0
t
t
t
0
...(1)
2.137
DEFINITE INTEGRATION
From (1),
B = A + B(1 – 2e–1)
A – 2e–1 B = 0
From (1) and (2), we get
2(e 1)
e 1
,B=
.
A=
4e 2 e 2
4 2e
Hence, f(x) = 2(e 1) . ex + e 1 . e–x.
4 2e
4 2e 2
If f(x) = x +
Example 11.
...(2)
f(x) =
Example 12.
Solution
1
1
0
1
1
= x 1 0 y f(y)dy x
2
0
yf(y) dy
1
0
f(x) is a quadratic expression.
Let f(x) = ax + bx2
then f(y) = ay + by2,
...(1)
1
0
y2(ay + by2) dy
1
1
0
4
y f(y) dy =
1
a y3 b y
=
4 0
3
a b
b= 3 4
12 b = 4 a + 3 b
9b–4a=0
From (2) and (3),
180
80
a=
and b =
119
119
1
0
1
1
t3
B 2
(1
A)
t
Now A = t t(1 A) B dt = 3
+ 2
0
0
0
1 A B
A=
6A = 2(1 + A) + 3B
3
2
4A – 3B = 2
...(1)
1
1
t 4 (1 A) Bt 3
Again B = t t(1 A) B dt =
4
3 0
0
2
=
0
1
0
A = t f (t) dt and B = t 2 f (t) dt
1 A B
4
3
12B = 3 + 3A + 4B
8B – 3A = 3
(1) × 3 gives 12A – 9B = 6
(2) × 4 gives – 12A + 32B = 12
Adding , we get 23B = 18
1
ay 4 by 5
= 1 +
5 0
4
a b
a=1+
4 5
20 a = 20 + 5 a + 4 b
15 a – 4b = 20.
Also, b =
1
0
f (x) = x (1 + A) + B where
1
y2f(y) dy
where a = 1 +
=1+
1
1
0
f (x) = x + x t f (t) dt + t 2 f (t) dt
Solution
0
yf(y)dy
2
0
0
(xy2 + x2y) (f(y)) dy,,
y2f(y) dy + x2
1
If f (x) = x + t(x t) f (t) dt , then
find the value of the definite integral f (x) dx .
find f(x).
f(x) = x + x
80x 2 180x
.
119
...(2)
1
0
y (ay + by2) dy
B=
18
23
54 46 100
18
4A = (3) + 2 4A =
=
23
23
23
A=
...(3)
...(2)
25
.
23
1
1
0
0
25
18
f (x) dx = 1 23 x 23 dx
48
18
25 1 18
+
=
23 2 23 (23) ( 2) 23
24 18 42
=
=
.
23 23 23
= 1
R
1.
If f is a cont inuous function such th at
f(xy) = f(x) . f(y) for all positive real numbers x
and y, then prove that f(x) = x k for some
constant k.
2.138
2.
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
If F is a continuous function and
F(x)
x
0
9.
F(t)dt , show that F(x) = 0 for every x.
3.
A function f, defined for all positive real
numbers, satisfies the equation f(x2) = x3 for
every x > 0. Determine f(4).
4.
Find f(x) if [f(x)]2 = 2
5.
Find a function f and a value of the constant c
x
0
f(t) dt.
1
for all real x.
2
Find a function f and a value of the constant c
x
such that c f(t)dt cos x
6.
such that
7.
c
all real x.
In each case, compute f(2) if f is continuous
and satisfies the given formula for all x 0.
(a)
x
f(t)dt = x (1 + x).
x2
f(t)dt = x (1 + x).
(c)
f (x)
x 2 (1 x)
2
0
0
0
t 2 dt = x2(1 + x).
f(t)dt x .
A function f is defined for all real x by the
formula f(x) = 3 +
x 1 sin t
2 t dt.
2
0
Without attempting to evaluate this integral,
find a quadratic polynomial p(x) = a + bx + cx2
such that p(0) = f(0), p(0) = f(0), and p(0) =
f(0).
2.17 WALLIS’ FORMULA
The formula to evaluate the integral
/2
sin m x . cos n x dx , where m, n are any
m,n =
0
positive integers, is called Wallis' Formula.
First of all, we find a formula to evaluate a special case
of the above integral i.e. n =
/2
0
To evaluate n =
/2
0
2+
x
f(t)dt e .
3x
a
10. If each case, give an example of a continuous
function f satisfying the conditions stated for
all real x, or else explain why there is no such
function :
x
f(t)dt = e
(b) f(t)dt = 1 – 2 x .
(c) f(t)dt f (x) – 1.
(a)
x
0
x2
2
0
x
2
0
11. Let F(x) =
x
f(t)dt . Determine a formula for
0
computing F(x) for all real x if f is defined as
follows :
(a) f(t) = (t + |t|)2
1 t 2 if | t | 1,
(b) f(t) =
1 | t | if | t | 1.
(c) f(t) = e–|t|.
(d) f(t) = the maximum of 1 and t2.
2
0
(b)
(d)
8.
1
x
tf(t)dt = sin x – x cos x – 2 x2 for
Find a function f and a number a such that
sin n x dx.
sin n x dx , we need to find a
12. There is a function f, defined and continuous
for all real x, which satisfies an equation of the
form
x
1
x16 x18
f(t)dt t 2 f(t)dt
+ c.
0
0
8
9
Find the function f and the value of the constant c.
13. A function f, continuous on the positive real
axis, has the property that
xy
x
y
1
1
1
f(t)dt = y f(t)dt + x f(t)dt
for all x > 0 and all y > 0. If f(1) = 3,
compute f(x) for each x > 0.
reduction formula. We have
n =
/2
0
sin n x dx
= [ sin n 1 x cos x]02 +
/2
0
(n 1) sin n 2 x . cos 2 x dx
DEFINITE INTEGRATION
= (n – 1)
/2
= (n – 1)
/2
0
0
/2
Put 2 sin = sin t
cos t
cos d =
dt
2
When 0 then t 0
When /4 then t /2
sin n 2 x . (1 sin 2 x) dx
sin n 2 x dx (n 1)
sin n x dx
0
n + (n – 1) n = (n – 1) n–2
L.H.S. =
n 1
.
n =
n n–2
Using the above formula repeatedly, we obtain
m,n =
0
/2
0
sin n x dx =
/2
0
cos n x dx
/2
0
[ P–5 ]
n
sin xdx
or
cos n xdx we start with (n–1) in the
numerator and go on diminishing by 2 till we get
either 2 or 1. Similarly we start with n in the
denominator and go on diminishing by 2 till we
in
get either 2 or 1. Further, we multiply by
2
case n is even.
3
/4
3/2
Example 1. Prove that 0 (cos2) cosd= 16 2 .
L.H.S. =
Solution
=
/4
0
1 3 1
. . .
2 4 2 2
m 1
,n
m n m–2
sin m 1 x (sin x cos n x) dx
sin m 1 x . cos n 1 x 2
=
n 1
0
/2
+ 0
cos n 1 x
(m – 1) sinm–2 x cos x dx
n 1
m 1 /2
sin m 2 x . cos n x . cos 2 x dx
=
n 1 0
In order to evaluate
/2
dt =
sin m x . cos n x dx .
We have m,n =
Note:
2.
2
0
Then show that m,n =
n 1 n 3 n 5 .... 1 . if n is even
n
n 2 n 4 2 2
=
n 1 n 3 n 5
.... 2 . 1 if n is odd
n n 2 n 4 3
/2
/2
0
Hence, n
0
/2 cos 4 t
3
= R.H.S.
16 2
Now, we develop a reduction formula for
We have0 = , 1 = 1.
2
=
n 1 n 3 n 5
..... 0 or 1
n =
n n2 n4
according as n is even or odd.
1.
2.139
/4
0
(cos2)3/2 cos d
(1 – 2sin2 )3/2 cos d
m 1 /2
(sin m 2 x . cos n x sin m x . cos n x) dx
=
n 1 0
m 1
m 1
–
=
n 1 m–2,n n 1 m,n
m 1
m 1
m,n =
1
n 1
n 1 m–2,n
m 1
m,n =
m n m–2,n
Using the above formula repeatedly, we obtain :
Im–2, n = m 3 Im–4,n,
mn2
Im–4,n = m 5 Im–6, n,
mn4
.................................
I2,n = 2 . I1,n, if m is odd
3n
2.140
I2,n =
Thus,
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
The above formulae (7), (8) and (9) can be combined
to have the following simple formula covering all the
cases :
1 , I , if m is even
2 n 0,n
m–5
2
m 1 m 3
.
.
. ...
. I1,n ,
Im,n =
3 n
m n m n 2 m n 4
if m is even
...(1)
Im,n =
m 1 m 3
m–5
1
.
.
. ...
. I 0,n ,
mn mn2 mn4
2 n
if m is even
...(2)
Now I1,n =
I0, n =
/2
0
/2
0
/2
– cos n 1
1
sin x cosn xdx
n 1
n 1 0
...(3)
sin 0 x cos n xdx
2
n 1 n 3 n 5
n . n 2 . n 4 . .... 3
=
n 1 . n 3 . n 5 . .... 1
n n 2 n 4
2
/2
0
Im,n =
I 0,1 if n is odd
I 0,0 if n is even
(m 1)(m 3)....(n 1)(n 3)....
K,
(m n)(m n 2).....
the three sets of factors starting with m – 1, n – 1 and
m + n and diminishing by 2 at a time, end up with either
1 or 2 according as the first factor of the set is odd or
even, and K = /2 if m and n are both even, and K = 1
if atleast one of m and n is odd.
/2
Thus, 0
I0,1 =
/2
sin 0 x cos1 x dx sin x 0 1
sinmx . cosnx dx
(m 1) (m 3)..1 or 2 (n 1) (n 3) (n 5)..1 or 2
K
(m n) (m n 2) (m n 4)....1 or 2
if both m and n are even (m, n N);
2
= 1 otherwise.
The above formula is called Wallis formula.
where K =
Note: Wallis formula is applicable only when
...(4)
/2
.
2
...(5)
limits are 0 to
0
0
I0,0 = 0 sin x cos x dx x 0 / 2. ...(6)
From (1) and (3) we find that if m is odd,
Im,n= m 1 . m 3 . m 5 ..... 2 . 1 , ...(7)
mn mn2 mn4
3 n 1 n
whether n is odd or even.
From (2), (4) and (5) we find that if m is even and n is
odd, then
m 1 m 3
m5
.
.
.....
Im,n =
mn mn2 mn4
1 n –1 1
....
.
·
,
...(8)
2 n n 1 n
From (2), (4), and (6) we find that if m is even and n is
also even, then
Im,n = m 1 . m 3 .....
mn mn2
Example 2.
0
/2
/2
.... 1 . n – 1· n 3 ... 1 ·
2n n n2 2 2
...(9)
sinm x cosn x dx
=
=
cos n xdx
/2
0
(a)
/2
0
Solution
Evaluate
7
sin 5 x cos x dx (b)
I5,7 =
/2
0
/2
0
sin 6 x cos8x dx.
sin 5 x cos7 x dx
4.2.6.4.2
K
12.10.8.6.4.2
K ,
=
where K is 1 since the exponents are not
120
both even.
/2
5.3.1.7.5.3.1 K,
I6, 8 =
sin 6 x cos8 x dx =
14.12.10.8.6.4.2
0
=
5K ,
where K is /2 since the exponents are
2048
both even = 5 .
4096
=
Example 3.
2
0
sin4x cos3x dx.
Solution Since sin4 (2 – x) cos3 (2 – x) = sin4x cos3x,
2
0
sin4x cos3x dx = 2
0
sin4x cos3x dx. ...(1)
2.141
DEFINITE INTEGRATION
Again, since sin4( – x) cos3( – x) = – sin4x cos3x,
sin4x cos3x dx = 0
0
...(2)
From (1) and (2), we find that
2
0
Example 4.
/2
0
0
sin
4
sin 4 x cos6x dx = 2
sin x cos6x dx ...(3)
4
0
Again, since sin4( – x) cos6( – x) = sin4x cos6x,
therefore,
/2
sin x cos6x dx = 2 0 sin4x cos6x dx
4
0
/2
/2
0
...(4)
sin 4 x cos6x dx = 2.2 3 = 3 .
512 128
Example 5.
/2
/2
/ 2
/2
sin 3 x cos 2 x dx +
=0+ 2
/2
0
sin 2 x cos3 x dx
( sin3x cos2x is odd and sin2x cos3x is even)
1. 2
4
= 2.
=
.
5 . 3 . 1 15
Example 6.
Evaluate x sin 5 x cos6 x dx .
0
Let = x sin 5 x cos6 x dx
Solution
= ( x) sin ( x) cos6 ( x) dx [ P–5 ]
5
0
= sin 5 x . cos6 x dx –
0
x sin x . cos x dx
0
5
6
0
Here if we replace x by – x in sin6x
Solution
cos2x, it does not change.
I = ( x) sin 6 x cos 2 xdx
[ P–5 ]
2 I= 0 (x x) sin 6 x cos2 xdx 0 sin 6 x cos 2 xdx
/2
2I = .2
0
I = .
sin 6 x cos 2 xdx
1
Evaluate x 3 (1 x)5 dx
0
Put x = sin2 dx = 2 sin cos d
Solution
x= 0
=0; x=1
.
2
=
1
[ P–6 ]
5.3.1.1 52
.
.
8.6.4.2 2 256
x (1 x) dx
sin (cos ) 2 . sin . cos d
=
sin cos d
=2.
3
5
0
/2
6
2
5
0
/2
7
11
0
6 . 4 . 2 . 10 . 8 . 6 . 4 . 2
18 . 16 . 14 . 12 . 10 . 8 . 6 . 4 . 2
6 . 4 . 2 . 10 . 8 . 6 . 4 . 2
=
.
18 . 16 . 14 . 12 . 10 . 8 . 6 . 4 . 2
=2.
0
Evaluate x sin 6 x cos 2 xdx .
Example 7.
sin 2 x cos3 x dx
/ 2
8
.
693
Example 8.
The given integral
Solution
=
Evaluate
sin 2 x cos 2 x(sin x cos x) dx .
/ 2
4 . 2 . 5. 3.1
11. 9 . 7 . 5 . 3 . 1
=
sin 4 x cos6x dx =
0
sin 5 x . cos6 x dx
0
3.1.5.3.1 · 3 .
...(5)
10.8.6.4.2 2 512
From (3), (4) and (5), we have
Also,
/2
0
=
x cos6x dx
sin4(2 – x) sin6(2 – x) = sin4x cos6x.
Solution
/2
sin4x cos3x dx = 0.
2 = . 2
Example 9.
Evaluate 100C30. I if
1
I = x 70 (1 x)30 dx .
0
Put x = sin2
Solution
I= 2
/2
0
sin141 · cos61 d
2.142
I=
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
2(140·138......2) (60·58.......2)
202·200........·2
For all x of the interval 0, the inequalities
2
sin2m–1x > sin2mx > sin2m+1x hold.
2·2 70 (70·69......1) 230 (30·29.....1)
I=
2101 (101·100.......1)
Integrating from 0 to
I2m–1 I2m I2m+1
70! 30!
1
1
= 101 100! =
· 100
101
C30
1
.
101
Finally, let us find the Wallis product, which expresses
I · 100C30 =
in the form of an infinite product.
2
Recall the formulae
the number
/2
I2m = 0 sin
2m
2m 1 2m 3 5 3 1
xdx
·
... · · · ...(1)
2m
2m 2 6 4 2 2
/2
I2m+1 = 0 sin 2m 1 xdx
2m
2m 2 6 4 2
·
... · ·
...(2)
2m 1 2m 1 7 5 3
We find, by means of term wise division,
2
I 2m
2·4·6...2m
1
=
2 3.5...(2m 1) 2m 1 I 2 m 1
I 2m
1
We shall now prove that mlim
I
2 m 1
...(3)
, we get
2
I 2m –1
I
2m 1
I 2m 1 I 2m 1
...(4)
I 2m –1
2m 1
=
2m
I 2 m 1
I
Hence, lim 2m 1 lim 2m 1 1
m I 2 m 1
m 2 m
We have
From inequality (4) we have lim
I 2m
m I 2 m 1
1.
Passing to the limit in formula (3), we get the Wallis
product for
2
lim 2.4.6...2m
1
2 m 3.5...(2m 1) 2m 1
This formula may be written in the form
lim 2 . 2 . 4 . 4 . 6 .... 2m 2 . 2m . 2m
.
2 m 1 3 3 5 5 2m 1 2m 1 2m 1
S
1.
Evaluate the following integrals :
(i)
2.
/2
0
5
sin x dx
(ii)
Evaluate the following integrals :
(i)
/2
0
2
(iii)
1
2 cos6 x dx
0
sin 5 x cos 4 x dx
0
/2
7
4.
6 x
8 x
(iv) 0 sin cos dx
2
2
Evaluate the following integrals :
7
0 x a x 2 dx
2
(ii) x 3/ 2 2 x dx
0
(i)
a
0
2
2
5.
3 2
(i)
(ii)
x sin x dx
(iii)
x (1 x) dx
(iv)
x (1 x) dx
5
0
3.
2 5/2
Evaluate the following integrals :
4
3
3
0
2a
(iv) x 5 (2ax x 2 )dx
sin x cos xdx
(iii) sin x cos x dx
(ii)
1
x (1 x ) dx
cos 4 3x ·sin 2 6x dx
0
1
6
1
0
1
3
9/2
0
1
4
1/4
0
Evaluate the following integrals :
(i)
1
(1 x ) dx
0
2 n
2.143
DEFINITE INTEGRATION
1
x 2n dx
0
2
1 x
(ii)
the correct value of the integral
0
x 4 dx
(a x 2 ) 2
Prove that
Y
1
6.
/2
0
7.
2
cos m x sin m xdx 2 – m
/2
0
cos m xdx .
One of the numbers , /2, 35/128, 1 – is
2.18 LIMIT UNDER THE SIGN
OF INTEGRAL
The value of limit of a definite integral can be
determined by finding the limit of the integrand with
respect to a quantity of which the limits of integration
are independent.
b
if a and b are independent of n.
2
n
Example 1. Evaluate nlim
0
lim n
Put n =
n
2
0
1 sin x dx .
n
1 sin x dx
1
t
n
1 (sin x) t
lim
dx
t
t 0
2
= – ln(sin x) dx = ln 2.
0
2
2 | sin(x t) sin x |
dx
Example 2. Evaluate lim
t 0 0
|t|
2
= 0
Solution
sin(x t) sin x
dx
t
2
sin(x t) sin x
dx
= 0 lim
t0
t
(since modulus function is continuous)
2
= 0 lim
t 0
= | cos x |dx = 4.
t3
lim 1 (1 form)
n
n
Solution
=
e
t3
lim n 1 1
n
n
et
3
n
3 a1 3 t 3 2
n (a )1 3 1 · t dt
2
n
lim
n
=
3 a1 3
t3 2
lim
1 · t dt
2 (a )1 3 n
n
a1 3
2 1 (sin x)
= lim 0
dx
t
t0
0
n
3 a1 3 t 3 2
lim
Example 3. If n 2 (a)1 3 1 n ·t dt 2 ,
where n N, then find the value of 'a'.
n
t
2
X
b
f (x, n) dx lim f (x, n) dx ,
For example, nlim
a n
a
Solution
8
0
Use the graph of y = sin8 x and a logical process
of elimination to find the correct value.
2a
(iii) x 9 / 2 (2a x) 1/ 2 dx
(iv) 0
sin xdx .
3
= 1 3 e t · t 2 dt .
a
Put e
t3
3
= y – t2 · e t dt =
dy
3
1 ea
1
dy = (ea – e–a).
3
3 e a
1 a –a
2 2
It is given that (e – e ) =
3
3
ea – e–a = 2 2 e2a – 2 2 ea – 1 = 0
=
ea =
a = ln
2 3 .
Example 4.
Solution
( 2 3 rejected)
2 3
a
Evaluate lim 0
n
e x dx
0 1 xn
lim
n
a
e x dx
.
1 xn
2.144
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
= lim 0
n
x
a e dx
e x dx
lim
n
n 1 1 x n
1 x
x
k
n
x
2
n
x
lim
C
.
k dx
= 0 n
n
k 0
x
a
e dx
e dx
lim
n
1
n
1 x
1 xn
n and x (0, 1), xn 0
and n and x (1, a), xn
1
= 0 lim
a
I = e dx 0dx = e – 1.
x
0
1
n
n
n
n
n
n
= nlim
n
k 0
n
Ck
1
n (k 3) = lim k 3 nCk .
k
k 0
1
Ck . k
n
1
x
n
k 2
0
dx
1
0
k 0
x
k2
1
dx
k3
1
1
x . ex dx
2
0
n
x
lim 1 e x
n
n
1
2x . ex dx
= (x2 . ex) 10 –
k 0
lim
Solution
C
Show that nlim
n k (k k 3) = e – 2
Example 5.
n
n
1 x dx
2
x
lim
= 0 n n =
k 0
n
1
1
0
=e–2
1
nk
1
x 1
x
( x e )0 0 e dx
= e – 2 {e – e + 1}
= e – 2.
T
1
dx 0 .
0 1 x 4
2. If f is continuous in [0, 1], show that
1 nf (x)
lim
dx f (0) .
n 0 1 n2x2
2
3. (a) Make a conjecture about the value of the
limit
1.
Prove that lim
lim
b
k 0 1
t
4.
Examine lim
5.
A definite integral can be differentiated with respect
to a quantity of which the limits of integration are
independent.
Let a function f(x, ) be continuous for a x b and
c d. Then for any [c, d] if
b
I() = f (x, )dx , then
a
b
dI( )
d
f (x, ) dx .
a
d
d
Prove that
2 2
0 if x b,
e k x
lim b 2 2
e k x dx if x b
a
( > 0, k > 0, b > a > 0).
dt (b > 0)
2.19 DIFFERENTIATION
UNDER THE SIGN OF
INTEGRAL
b
f (x)sin cx dx .
c a
k 1
(b) Check your conjecture by evaluating the
integral and finding the limit. [Hint : Interpret
the limit as the definition of the derivative of
an exponential function]
Let f have a continuous derivative for x in [a, b].
6.
Show that
x
– rx
sin ax dx equals a/(a2 + r2),
0
where r > 0 and a are constant.
For example, if I() =
ln(1 sin cos x)
dx
cos x
dI
cos dx
then we have
.
0 1 sin cos x
d
With the help of the above result firstly, new integrals
can be deduced from certain known standard integrals.
Secondly, the value of a given integral can be found
by first differentiating the integral then evaluating the
new integral thus obtained and finally integrating the
result with respect to the same quantity with respect
to which the integral was first differentiated.
1 xk 1
dx, (k 0).
Example 1. Evaluate I(k) = 0
ln x
0
2.145
DEFINITE INTEGRATION
I(k) =
Solution
d
(I(k)) =
dk
=
1 xk 1
ln x dx
1 d
0
xk 1
dk ln x dx
0
1 x k ln x
ln x dx
0
...(2)
...(3)
tan 1 (ax)
2
2
ln(1 a 2 x 2 )
0
1 b2 x 2
0
1 b2 x 2
/2
= 0
/2
0
I(a) =
=
dx
ab
ln
.
b
b
/2
0
sec . tan–1 (a cos ) d.
/2
0
sec . tan–1( cos ) d
sec .
1
. cos d
1 a 2 cos 2
1
d
1 a cos 2
2
/2
sec 2 d
, put tan = t
1 tan 2 a 2
dt
=
0 t 2 (a 2 1)
0
.
a 1 2
1
t
1
tan
2
2
a 1
a 1 0
1
2
I(a) =
dx
Then, I(a) =
=
Evaluate
ln(1 a 2 x 2 )
0
x(1 x ) dx 2 ln(1 a) .
Example 3.
Now when a = 0, I = 0 A = – ln b.
b
=
tan ax
ln(a + b) + A.
b
Solution The given definite integral is a function
of 'a'. Let its value be I(a).
x(1 x ) dx
a
1
tan 1 x
tan 1 ax
2
2
1 a
1 a
0
1 2 a 2
.
2 1 a
1 a 2(1 a )
da
I
ln(1 a) A .
2 1 a 2
When a = 0, I = 0 A = 0.
0
2a 1 1 .
.
b 2 a 2 a b 2 b(a b)
Example 4. Evaluate
Hence,
Hence
1
2a 1 tan 1 ax 1 tan 1 bx
b
b 2 a 2 a
0
Therefore I =
Differentiating w.r.t. a, we have
dI
1
dx
2
0
da
(1 x ) (1 a 2 x 2 )
1
a2
dx
2
2
2
2
2
0 (1 a ) (1 x )
(1 a ) (1 a x )
...(1)
0
2ax 2
dx
0 (1 b 2 x 2 ) (1 a 2 x 2 )
dx
Let I
Then dI
da
1 b2 x 2
0
1
1
2a 2 2
2 2
dx
2 2
2 2
(b a )(1 a x ) (b a )(1 b x )
0
k
Solution
ln(1 a 2 x 2 )
Let I
1
x k 1
= (x )dx =
0
k 1 0
1
1
=
[1 – 0] =
.
k 1
k 1
d
1
(I(k)) =
.
dx
k 1
Integrating both sides w.r.t. 'k'. we get,
I(k) = ln (k + 1) + c
1 xk 1
Given I(k) =
dx
0 ln x
I(0) = 0 (when k = 0)
also from (1), I(0) = ln(1) + c
I(0) = c.
From (2) and (3), c = 0
I(k) = ln (k + 1).
tan 1 (ax)
Example 2. Evaluate 0 x(1 x 2 ) dx
1
Solution
,
2 a2 1
Integrating both sides w.r.t. 'a', we get
2.146
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
ln |a + a 2 1 | + C
2
Since I(0) = 0, i.e. 0 + C = 0, we have C = 0.
I(a) =
1 a2
I = sin–1a, since at a = 0, I = 0.
ln |a + 1 a 2 |.
2
I(a) =
tan 1 ax tan 1 x
dx .
Evaluate 0
x
tan 1 ax tan 1 x
dx
Let I = 0
x
0
/2
Solution I = 0
1 a sin x dx
ln 1 a sin x
, (| a | < 1).
sin x
1 a sin x dx
ln
1 a sin x sin x
(| a | < 1)
dI
db
2sec2 x
dx
1 tan 2 x a 2 tan 2 x
/2
2 sec 2 x
dx (put tanx = t)
(1 a 2 ) tan 2 x 1
e ax sin bx
tan 1 t 1 a 2 =
2
0
1 a
ax
0
0
cos bx dx
a
a b 2 , a > 0.
Now, integrating with respect to b,
2
db
1
b
a tan 1 C
a
a
a 2 b2
e ax sin bx
b
...(2)
x
a
Assuming I a continuous function of a, we deduce
from (2), when a 0,
2
dt
2 0
2
(1 a )
1
=
t2
1 a 2
dx .
b
C
...(1)
a
where C is the constant of integration.
From the given integral, we see that when b = 0, I = 0.
From (1), we deduce C = 0.
2 dt
= 0
(1 a 2 )t 2 1
2
x
tan 1
=
e
I a
/ 2
= 0
sin bx
0
dx, a 0 .
x
Using differentiation under the integral sign, we have
/2
dI
2sin x
dx
.
2
2
0
da
(1 a sin x) sin x
= 0
Evaluate the integral
Let I
Solution
ln a.
2
Example 8.
Example 6.
dI
a
e – ax cos mxdx 2
.
0
dm
a m2
Solution
m
I = tan –1
a
da
I = ln a + C.
2 a
2
When a = 1, I = 0 C = 0
/2
0
e –ax sin mx
dx.
x
dm tan –1 m + C
a
a2 m2
Since the integral vanishes when m = 0, C = 0.
1 dx
1
1
2
= a 0 x 2 1 = 2 a tan x =
.
a
0
2a
a2
Hence, I =
I=a
1· x
dI
dx
= 0 (1 a 2 x 2 )x dx = 0
1 a2x2
da
I=
Evaluate I =
Example 7.
Example 5.
Solution
da
Now, dI =
0
sin bx
0
1 a2
x
dx tan 1
dx
, or
2
2
according as b > or < 0.
.
sin x
When b = 1 we have 0
x
dx
.
2
2.147
DEFINITE INTEGRATION
Application of integration under the sign of
integral
Find the value of
Example 9.
e
– x2
0
dx.
Denoting the proposed integral by k,
and substituting ax for x, we get
Solution
e
–a 2 x 2
0
adx k
–2
Multiplying both sides by ea , we get
adx ke
e
adadx k e da
Hence e
...(1)
1 1
ada
Since, e
, we have from (1)
2 1 x
1
dx
k
2 1 x
–a 2 (1 x 2 )
a –2
0
0
0
–a 2 (1 x 2 )
0
– a 2 (1 x 2 )
2
0
2
1
1
Solution
We have
0 a0
a1 x a
a –1
0 a0
x dadx
a–1
a1
1
x dadx a da
a1 da
a1
= a 0 a ln a
0
a1
We have
e – ax da cos mx dx
=
0 a0
Simplifying both sides, we get
a1
e –a1x – e –a 0 x
x
0
ada
a1
a m
cos mx dx =
a0
1 x a1 1 x a 0 –1
0
m
, prove that
a 2 m2
Solution We get
a–1
1
Using the equation 0 x dx
a
e –a1x – e –a 0 x
a0
dz ln .
0
z
a1
prove that
1
Solution
2
a2
ln x
0
a1
0
a0
e
– ax
sin mxdadx
sin mx
x
0
2
1 a12 m 2
ln
2 a 02 m 2
Example 10.
Also,
Example 12. Using the equation 0 e
1
e – x dx k
.
Hence,
0
2
a
–ax
,
Example 11. Given 0 e cos mxdx 2
a m2
prove that
e –a1x – e –a 0 x
1 a 2 m2
cos mx dx = ln 12
0
2 a 0 m 2
x
2
k2
4
0
2
0
–a 2
1 x a1 1 x a 0 –1
a
dx ln 1
ln x
a0
Now, if we put x = e–z in this equation, we get
e –a1x – e –a 0 x
a
dz ln 0
0
z
a1
Hence
–ax
dx
a1
mda
a0
2
a m
sin mxdx
.
2
2
e –a 0 x – e –a1x
a
a
sin mxdx tan –1 1 – tan –1 0 .
x
m
m
If we make a0 = 0 and a1 = in the latter result, we
obtain
0
dx
sin mx
0
dx
x
.
2
U
3
1.
If F(t) = sin(x t 2 )dx , find F(t).
2.
Show that
3.
Evaluate ln(1 b cos x) dx
4.
2
0
n (1 a cos x)
d x = sin–1a, (| a | < 1)
cos x
5.
Evaluate
/2
6.
Evaluate
a
0
ln(1 cos cos x)
x
a 2 x 2 cos –1 dx.
a
0
0
1
tan 1 a x
0
x 1 x2
Evaluate
dx
7.
dx
cos x
1
n (1 a 2 x 2 )
0
x2
Show that
(1 x 2 )
= [ 1 a 2 1] , (a2 < 1) .
dx
2.148
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
2.20 INTEGRATION OF
INFINITE SERIES
The integral of the sum of a finite number of terms is
equal to the sum of the integral of these terms. Now,
the question arises whether this principle can
be extended to the case when the number of terms is
not finite.
In other words, is it always permissible to integrate an
infinite series term by term ? It is beyond the scope of
this book to investigate the conditions under which
an infinite series can properly be integrated term by
term. We should merely state the theorem that applies
to most of the series that are ordinarily met with in
elementary mathematics. For a complete discussion,
students may consult any textbook on Mathematical
Analysis.
Theorem. A power series can be integrated term by
term throughout any interval of convergence, but not
necessarily extending to the end points of the interval.
Thus, if f(x) can be expanded in a convergent infinite
power series for all values of x in certain continuous
range, viz.,
f(x) = a0 + a1x + a2x2 + ..... to ,
then
b
b
a
a
f(x) dx (a a x a x .....)dx
Also,
b
1
2
a
a
f(x)dx (a a x a x .....)dx
a
1
1
1
... 2 ln 2 1
1.2 2.3 3.4
We can also apply the processes of differentiation
to functions expressed by power-series.
Assuming that f(x) = a0 + a1x + a2x2 + ..... to is a
continuous and differentiable function of x, then
f(x) = a1 + 2a2x + 3a3x2 + ... + nanxn–1 + ...
Here we mention some important expansions
which are useful in certain problems :
2
0
1
1
(ii)
1 1
1
1
2
.....
6
12 2 2 32 42
(iii)
1 1
1
1
2
.....
12
12 2 2 32 42
(iv)
1 1
1
1
2
2 2 2 .....
2
8
1 3
5
7
(v)
1
1
1
1
2
.....
24
2 2 4 2 6 2 82
2
a r x r dx ,
provided the intervals (a, b) and (a, x) lie within the
interval of convergence of the power series.
For example, If |x| < 1, we have
1
1
1
1. 3 4 1. 3. 5 2
1 x2 x2
x
x ...
2
2
2
2.4
2. 4.6
(1 x )
Hence, integrating term by term between the limits 0
and x,
2
7
1x
1. 3
1. 3 . 5 x
...
2 3 2. 4. 5 2 . 4. 5 7
Note that this series is due to Newton.
1
If we put x = we get
2
sin 1 x x
1
1. 3
1
6
... ,
2
2
2 . 4 . 5. 2
2 2 . 3. 2
1 1 1 1
..... n2
2 3 4 5
(i)
r
x
x2 x2 x2
...
1. 2 2 . 3 3 . 4
Assuming that the function on the right-hand is
continuous up to the limit x = 1. We infer that
(1 + x) ln(1 + x) – x =
r
x
x
x2 x2
...
2
3
Integrating this between the limits 0 and x,
Also, if |x| < 1, ln(1 x) x
2
0
a x dx ,
a
from which can be calculated without much trouble.
1
Example 1. Prove that 0
1
1
2
ln(1 x 4 )dx
.
x
48
1
1
1
4
8
12
16
Solution I = 0 x x x x 3 x 4 x .... dx
1
1
1 7 1 11 1 15
3
= 0 x x x x ...... dx
2
3
4
1
x 4 x 8 x12 x16
= 4 16 36 64 .....
0
2.149
DEFINITE INTEGRATION
=
1
1
1
1
2 2 2 ....
2
2
4
6
8
=
1
1 1
1
1
1
1
= 2 2 2 2 ... 2 2 2 2 ...
2
4
6
8
4
8
12
=
2
1 1 1
1
2. 2 2 2 2 ......
24
4 1 2
3
=
2 1 . 2 22 2 2
.
24 8 6
48
48
I=
1 3 1 5 1 7
x + x – x + .... , –1 < x < 1.
3
5
7
1
1
.
Example 3. Prove that 3n 1 3n 2 =
3
3
n 0
1
Solution
(t t
3n
3n 1
1
) dt =
n 0 0
(t (1 t) dt
n 0 0
1
3n
= 0 (1 t) t dt
n0
1
= (1 t)·{1 t 3 t 6 t 8 ......}dt
/2
3n
=
1 01
ln(1 t 2 )dt
2 1 t
1 01 2 1 4 1 6 1 8
t t t t .... dt
2 1 t
2
3
4
1 0
1 3 1 5 1 7
t t t t ....
1
2
2
3
4
Integrating and putting the limits, we get
=
1 1 1 1 1 .....
I=
2 2 2.4 3.6 4.8
1 1 1 1 1
= 1
2 2 2 2 4
11 1 1 1 1
...
33 6 4 4 8
1
1
1
1
= 1 2 2 2 ....
2
2
3
4
1
1
1
1
1 2 2 2 ......
2
2
3
4
2
2
2
= 1 1. .
2 6 2 6
24
=
1
1
Example 5.
1
dt
= –
0
t 1 3
2 2
x
2
1
1
2
t (1 2)
tan 1
=
3
3
0
2
2
1
tan 1 3 tan 1
=
3
3
2
.
12
dx = –
Let I =
Solution
ln x
(1 x) dx
Show that 0
1 ln (1 x)
0
2
tan x ln sin xdx
0
0
1
(1 t)
dt
dt =
= 0
0 1 t2 t
1 t3
/2
0
sin x
ln (1 cos 2 x)dx
cos x
Now put cosx = t and adjust the limits for t.
tan–1 x = x –
Evaluate I =
Example 4.
Solution
Example 2. Find by integration the expansion
series for tan–1 x.
1
Solution We have 1 x 2 = 1 – x2 + x4 – x6 +.......+ .... ,
if x2 < 1.
Now, integrating both sides between the limits 0 and x
x
x
dx
(1 x 2 x 4 ....)dx .
0 1 x2
0
2
=
.
3
6
3
3 3
ln x
(1 x) dx
0
Integrating by parts taking ln x as the first function,
we have
I = [ln x . ln (1 + x)] 10
1 ln (1 x)
0
1 ln (1 x)
I = 0 – 0
x
dx
x
dx
2.150
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
1
2z 1
1
1
2
tan 1
= ln(z 1) ln(z z 1) +
6
3
3
3 0
x 2 x3 x 4
x
...
1
2
3
4
dx
=– 0
x
1
1
1
tan 1
= ln 2 0
3
3
3
1
x x 2 x3
= – 0 1 ... dx
4
2 3
1
1
tan 1
– 0 0
3
3
1
x 2 x3 x 4
= 1 2 3 2 ...
2
3
4
0
=
Show that sum of infinite series
1 1 1 1 ....
can be
a a b a 2b a 3b
1
expressed in the form 0
z a 1
dz and hence
1 zb
prove that
1 1 1 1 1 1
1
.....
ln 2 .
1 4 7 10 13 16
3 3
Solution
1
Let I = z a 1 (1 z b ) 1 dz
1
I = z a 1 (1 z b z 2b z3b z 4b ....)dz
0
1
= (z a 1 z a b 1 z a 2b 1 z a 3b 1 .....) dz
0
1
z a z a b z a 2 b z a 3b
.....
=
a a b a 2 b a 3b
0
1
1 1 1 .....
=
a a b a 2b a 3b
Again if we put a = 1 and b = 3, we get
1
= 0
1
2x
4x3
8x 7
....
1 x 1 x 2 1 x 4 1 x8
Solution Let
n
1
2x
4x3
8x 7
2 n x 2 1
....
n
1 x 1 x 2 1 x 4 1 x8
1 x2
Integrating both sides
S=
S dx = ln (1 + x) + ln (1 x2) + ln(1 + x4)
n
+ ln (1 + x8) + .... + ln (1 x 2 ) + c
n
= ln {(1 + x) (1 + x2) (1 + x4) (1 + x8) ..... (1 x 2 ) } + c
(1 x)(1 x)(1 x 2 )(1 x 4 )(1x x8 )...(1 x 2n )
+c
=l n
(1 x)
0
Expanding by binomial theorem,
I
If |x| < 1 then find the sum of the series
Example 7.
1
1 1
2
= 1 2 3 2 ... =
.
3 4
2
12
Example 6.
1
1 1
ln 2
ln 2 .
3
3 6 6 3 3
z0
1 1 1 1
1
dz ...
1 4 7 10 13
(1 z 3 )
L.H.S. on integrating using partial fractions
(1 x 2 )(1 x 2 )(1 x 4 )(1x x8 )...(1 x 2 )
+c
= ln
(1 x)
n
(1 x 4 )(1 x 4 )(1 x8 )....(1 x 2 )
+c
= ln
(1 x)
n
.........................................................
n 1
(1 x 2 )(1 x 2 )
1 x 2
+c
+ c = ln
= ln
(1 x)
(1 x)
n
n
When n S dx = ln
+ c = – ln (1 – x) + c
1 0
1 x
Differentiating both sides, we get S =
1
.
(1 x)
2.151
DEFINITE INTEGRATION
Polynomial approximation to logarithm
Prove that
x u2
x2
– ln(1 – x) = x +
du.
0 1 u
2
Solution We have ln(1 – x)
Example 8.
1 x dt
= 1
, which is valid if x < 1.
t
The change of variable t = 1 – u converts this to
the form –ln(1 – x)
du
, valid for x < 1.
1 u
From the algebraic identity 1 – u2 = (1 – u) (1 + u),
we obtain the formula
x
= 0
1
u2
1 u
, valid for any real u 1.
1 u
1 u
Integrating this from 0 to x, where x < 1, we have
– ln(1 – x) = x +
x u2
x2
du.
0 1 u
2
V
1.
2.
Find the sum of the series
x n 1
x 2 x3 x 4
n 1
... , | x | < 1.
–... ( 1)
n (n 1)
1.2 2.3 3.4
If x < 1 then find the sum of the series
1
2x
4x 3
8x 7
...... .
1 x 1 x 2 1 x 4 1 x8
5.
6.
7.
3.
Starting from
1
x 2n
1 x x 2 ... x 2n 1
,
1 x
1 x
show that
1
1
n
Prove that 0 x ln x dx (n 1) 2 ,
Prove that if |x| < 1
x3
x5 x7
1
1
... (1 x 2 ) tan 1 x x .
1 . 3 3.5 5.7
2
2
Prove that
x m 1
1
(ii)
x sin x
t2 t2
t 2n
t 2 t3
t 2n 1
...
ln(1 t) t
2 3
2n
2 3
2n 1
for t 0.
(iv)
8.
1
1
1
dx x
x3
x5
...
3.3! 5.5!
b
b 2 a 2 b3 a 3
e
..
dx ln (b a)
a
2.2!
3.3!
x
x
1 tan 1 x
0
x
dx
1
(1) (2n 1) .
0
n
2
Evaluate
n3
n2
9.
x
0
(iii) a
t
Evaluate the following integrals :
x dx
(i)
x 4 1
1 ln(1 x)
dx
(ii)
0
x
dx
(iii) 0
(x 1) (x 2)
x 2 dx
(iv) 0 2 2
, a, b > 0.
(x a ) (x 2 b2 )
1
(i) 0 1 x n dx m m n m 2n m 3n ..
b
4.
n 1.
f (x)dx, where f(x) = ex + 2e2x + 3e3x + ..
Let Pn denote the polynomial of degree n given
n
xn
xk
x 2 x3
by Pn(x) = x +
+ ....+
.
n k 1 k
2
3
Then, for every x < 1 and every n 1, prove that
x un
du.
–ln(1 – x) = Pn(x) +
0 1 u
2.21 APPROXIMATION OF
DEFINITE INTEGRALS
for wanting to make such an approximation. First, the
integrand f(x) may not have an elementary
antiderivative; thus the fundamental theorem of
This section presents some ways of approximating a
calculus could not be used (for example,
definite integral
b
f(x) dx. There are several reasons
a
1
e dx ).
0
x2
Second, even though the antiderivative is elementary,
2.152
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
it may be tedious to compute [for example,
1
1 / (1 x ) dx]. Third, the values of the integrand
5
0
f(x) may be known only at a few values of x.
The definite integral
b
f(x) dx is, by definition, a limit
a
n
of sums of the form
f(c )(x – x )
1
i
i 1
Any such sum consequently provides an estimate of
b
a
x0 = a x1 x2 x3 x4 = b X
...(1)
i–1
Example 1.
Use th e t r a pez oi da l met h od
dx
.
1 x2
Solution In this case, a = 0, b = 1, and n = 4,
so h = (1 – 0)/4 = 1/4. The trapezoidal estimate is
1
f(x) dx. However, the two methods described in
this section, the trapezoidal method and Simpson's
method, generally provide much better estimates for
the same amount of arithmetic.
h f (0) 2f 1 2f 2 2f 3 f (1)
.
2
4
4
4
The trapezoidal sum is therefore approximately
1
1
= (1 + 1.882 + 1.6 + 1.28 + 0.5) = (6.262) 0.783.
8
8
1 dx
0.783 .
Thus,
0 1 x2
The integral can be evaluated by the Fundamental
Theorem of Calculus. It equals tan–1 1 – tan–1 0 =/4
0.782.
=
Trapezoidal method
The sum (1) can be thought of as a sum of areas of
rectangles. In the trapezoidal method, trapezoids are
used instead of rectangles. Recall that the area of a
trapezoid of height h and bases b1 and b2 is (b1 + b2) h/2.
Let n be a positive integer. Divide the interval [a, b]
into n sections of equal length h = (b – a)/n with
x0 = a, x1 = a + h, x2 = a + 2h, ... xn = b
The sum
f(x 0 ) f(x1 )
f(x ) f(x 2 )
·h 1
· h + .....
2
2
f(x n -1 ) f(x n )
·h
+
2
is the trapezoidal estimate of
wit h n = 4 to estimate 0
b
f(x) dx.
a
It is usually written
h
[f(x0) + 2f(x1) + 2f(x2) + ...+ 2f(xn – 1) + f(xn)] ...(2)
2
Note that f(x0) and f(xn) have coefficient 1, while all the
other f(x)'s have coefficient 2. This is due to the double
counting of the edges common to two trapezoids.
The diagram illustrates the trapezoidal approximation
for the case n = 4. Note that if f is concave down, the
trapezoidal approximation underestimates
b
f(x) dx.
a
If f is a linear function, the trapezoidal method, of
course, gives the integral exactly.
Simpson's method
In the trapezoidal method a curve is approximated by
lines. In Simpson's method a curve is approximated by
parabolas (see figure). Simpson's method is exact if
f(x) is a polynomial of degree atmost 3.
Y
a
bX
The dashed lines are
parts of parabolas
In Simpson's method the interval [a, b] is divided into an
even number of sections.
Divide the interval [a,b] into n sections of equal length
h = (b – a)/n with
x0 = a, x1 = a + h, x2 = a + 2h, ..., xn = b.
h
[f (x 0 ) 4f (x1 ) 2f (x 2 ) 4f (x 3 )
The sum 3
.... 2f (x n – 2 ) 4f (x n –1 ) f (x n )]
DEFINITE INTEGRATION
Show that the approximation
Example 3.
b
is the Simpson's estimate of f(x) dx.
2.153
a
Example 2.
Use Simpson's method with n = 4 to
dx
.
1 x2
Solution Here h = 1/4. Simpson's formula takes
1
1
2
3
4
the form f(0) 4f 2f 4f f(1)
3
4
4
4
1
estimate 0
The Simpson's approximation of
1
dx
0 1 x2
ba
a b
F(a) 4F
F(b),
6
2
is exact for F(x) a cubic polynomial.
Solution Let F(x) = A + 2Bx + 3Cx2 + 4Dx3.
It then follows that
b
F(x) dx
a
b
b
F(x) dx (Ax Bx 2 Cx3 Dx 4 )
a
a
2
2
= (b – a) (A + B(b + a) + C(b + ba + a )
is therefore
+ D(b3 + b2a + ba2 + a3))
1
(1 + 3.765 + 1.6 + 2.56 + 0.5)
=
12
1
(9.425) 0.785.
=
12
1 dx
0.785.
Thus
0 1 x2
Simpson's method usually provides a much more
accurate estimate of an integral than the trapezoidal
estimate for the same amount of arithmetic.
ba
(A + 2Ba + 3Ca2 + 4Da3)
6
=
+ (A + 2Bb + 3Cb2 + 4Db3)
a b
a b 4D a b
4 A 2B
3C
.
2
2
2
2
=
3
ba
a b
F(a) 4F
F(b) .
6
2
W
1.
2.
Let f(x) = Ax2 + Bx + C. Show that
h
h
– h f(x) dx 3 [f(–h) 4f(0) f(h)] .
Let f be a function. Show that there is a parabola
y = Ax2 + Bx + C that passes through the three
points (–h, f(–h)), (0, f(0)), and (h, f(h)).
1
5.
1
1
f(x) = 1, x, x2 and x3.
(b) Let a and b be two numbers, – 1 a < b 1,
1
such that f(x) dx f(a) f(b) for f(x) = 1,
Q
y = f(x)
P
–1
f(x) dx f 3 f 3 for
(a) Show that
1
3 and b = 1/ 3 .
(c) Show that the approximation
R
x, x2, and x3. Show that a = – 1/
h X
f(x) dx f(–1 / 3) f(1 / 3) has no error
1
‒h
3.
4.
0
Let f(x) = Ax2 + Bx + C. Shows that
c h
h
ch f(x) dx 3 [f(c h) 4f(c) f(c h)] .
Show that if f(x) = x3,
h
h
h f(x) dx 3 [f( h) 4f(0) f(h)]
1
when f is a polynomial of degree atmost 3.
6.
E st i m a t e
3
f(x) dx i f i t i s kn own t h a t
0
f (0) = 10,f (0.5) = 13, f (1) = 14, f (1.5) = 16, f (2) = 18,
f (2.5) = 10, f (3) = 6 by (a) the trapezoidal method.
(b) Simpson's method.
2.154
Problem 1.
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Find a mistake in the following
evaluation:
0
3
dx
1
2x
tan 1
1 x2 2
1 x2 0
3
1
tan–1 ( 3 ) – tan–10] = – ,
2
6
d 1 1 2x
1 (x 1)
.
where dx 2 tan
=
1 x2 1 x2
Solution The result is wrong since the integral of
a function positive everywhere cannot be negative.
The mistake is due to the fact that the function
1 tan 1 2 x
has a discontinuity at the point
2
1 x2
x =1:
1
2x
lim tan 1
;
2
x1 2
4
1 x
1
1 2x
lim tan
2
x1 2
4
1 x
The correct value of the integral under consideration is
equal to
3 dx
3
1
1
1
0 1 x 2 tan x 0 tan 3 tan 0 3 .
Here the Newton-Leibnitz formula is applicable, since
the function F(x) = tan–1 x is continuous on the interval
=
0,
3 and the equality F'(x) = f(x) is fulfilled on the
whole interval.
Problem 2.
4x2
Solution There are two natural ways to go at this.
One would be to evaluate the integral, (by substitution
t = s2), then take the derivative, then work out what
values of x give a derivative of zero. The other, which
we choose here, would be to find the derivative
without doing the integral, with the rest of the plan
being the same.
d 4x2
f(t)dt 8xf(4x 2 ) 2xf(x 2 ) .
Thus, f(x)
dx x2
Since here, f(x) = sin x , we have f'(x) = 8x sin(2x) – 2x
sin(x).
Clearly this is zero at x = 0, , and 2. But from the
graph, there must be other points as well.
With the trigonometric identity sin 2x = 2 sin x cosx,
we have F'(x) = 2x sin x(8 cosx – 1). Now the other two
critical points come into focus :
they are the places where cosx = 1/8, and those are
cos–1(1/8) and 2 – cos–1(1/8).
Problem 3. Let a be a positive real number. Find
the value of a such that the definite integral
a2
dx
a x
x
achieves its smallest possible value.
Solution Let F(a) denote the given definite
integral. Then
d a 2 dx
1
1
.
F '(a)
2a .
2
2
a
da
x x
a a
a a
Setting F'(a) = 0, we find that
2a + 2 a = a + a or ( a + 1)2 = 2.
a = 2 – 1, and because
Let F(x) 2 sin( t )dt .
We find
y = F(x)
the integral (a 3 4x a 5 x 2 )eax dx .
x
a > 0, a = ( 2 – 1)2 = 3 – 2 2 .
Problem 4. For a > 0, find the minimum value of
Y
20
1a
0
10
Let (a 4x a x )e
ax
2
= e [Ax +Bx+C]+D
Differentiating both sides (a3 + 4x – a5x2)eax
= eax [2Ax + B] + [Ax2 + Bx + C]eax a
A = – a4 ,
a 4 4
4 2a 4
, C=
B=
a2
a
Solution
1
2
3
4
5
6
–10
Find all critical points of F between 0 and 2.
X
3
5 2
ax
DEFINITE INTEGRATION
1
4 2a 4
a 4 4 a
x 2
I eax a 4 x 2
a
a
0
4
4
2 4 2a
a 4 a 4 4
2 2
= e a
a2
a a
4
4
2 a a 4
= e a 2 2
a
a
2
4
2
I = a2 + 2 = a 4 .
a
a
At a = 2 , I has a minimum value of 4.
If f (x) = a | cos x | + b | sin x | (a, b R)
and satisfies
has a local minimum at x = –
3
2
2
f (x) dx = 2. Find the values of a and b and
Problem 5.
2
hence find b2/a2.
a cos x bsin x if 0 x
2
Solution f(x) =
a cos x bsin x if x 0
2
For – /2 < x < 0
...(1)
f(x) = – a sin x – b cos x
...(2)
and f(x) = – a cos x + b sin
Since f (x) has a minima at x = – /3
f(– /3) = 0 and f ''(– /3) > 0
3 b
– = 0 3 a – b = 0.
2
2
a
1
3
and f(– /3) = – – b ·
= [a + b 3 ]
2
2
2
Now f(– /3) = a ·
= – 2a > 0.
Hence, a < 0 and b < 0
2
Now, I = 2 f (x) dx
2
0
2
2
= 2 f (x) dx 0 f (x) dx
0
= 2 (a cos x 2ab sin x cos x b sin x) dx
+
2
0
2
2
2
2
(a 2 cos 2 x 2ab sin x cos x b 2 sin 2 x) dx
2
2
a
b
+
+ 2ab = 2 (given)
2
2
2 ( 3 ) a 2 = 2
On solving, I =
a=–
1
3
Problem 6. For a positive constant t, let , be
the roots of the quadratic equation x2 + t2x – 2t = 0.
If the minimum value of
2
1
1 1
a
1 x 2 x 2 dx is b c ,
where a, b, c N, then find the least value of (a + b + c).
Solution If and are the roots of x2 + t2x – 2t = 0,
then we have + = – t2 and = – 2t.
2 2 ( )2 2 t 2 1
So
4 t
()2
()2
1
1
1
1
and ()2 4t 2 2t .
2
1
1 1
dx
Now I = 1 x 2 x 2
2
1
t 2 1
1
2
x
= 1
4 t x 4t 2 2t dx
3t 2
3
2 3.
8
4t
Differentiating I w.r.t. t,
dI 3t
3
0.
we get
dt 4 2t 2
=
So we get t = 4 2 , and since t is taken to be
positive then
3 2
18
9
4
3
3
3
Imin = I 2
4
16
8
a
c.
b
The least value of a + b + c = 20.
Problem 7. Consider the function
x
2
and b = –
3
3
.
2.155
f(x) = f(1 + t3)–1/2 dt. If g is the inverse of f, then
0
g(y)
.
find the value of 2
g (y)
x
Solution We have f(x) = f(1 + t3)–1/2 dt
0
i.e. f{g(x)} =
g(x)
0
i.e. x =
g(x)
0
(1 + t3)–1/2 dt
(1 + t3)–1/2 dt
[ g is inverse of f f{g(x)} = x]
Differentiating w.r.t. x, we have
1 = (1 + g3)–1/2 . g
i.e. (g)2 = 1 + g3
Differentiating again w.r.t. x, we have
2gg = 3g2g
2.156
g
3.
2
g2
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
1
1
(1 cos 4) .
= – cos 4 =
16 16
16
Let a + b = 4 where a < 2 and let g(x) be
Problem 8.
dg
> 0 for all x, prove that
a differentiable function of x. If
dx
b
a
0 g(x) + dx + 0 g(x) dx increases as (b – a) increases.
a
Let y = g(x) dx +
Solution
0
b
g(x) dx
0
...(1)
z = b – a = (4 – a) – a = 4 – 2a
...(2)
dy
> 0.
We have to prove that
dx
dy
db
Differentiating (1) w.r.t a,
= g(a) + g(b) ·
da
da
dy
d
or
= g(a) + g(b)
(4 – a) = g(a) – g(b).
da
da
dz
Differentiating (2) w.r.t. a,
= – 2.
da
dy
dy da g(a ) – g( b) g( 4 a ) g(a )
dz dz
...(3)
–2
2
da
dg
0 for all x, implies that g
As a < 2, 4 – a > a. Also
dx
is an increasing function.
g(4 – a) > g(a).
Hence (3)
dy
0 , i.e., y is an increasing function of z.
dz
a
g(x) dx +
0
b
g(x) dx increases as b – a increases.
0
n
Problem 9.
nk
4k
.
n
n
k 1
n
1 k
k
1 cos 4
We have S =
n k 1 n
n
Evaluate lim
n
2
Solution
1
1 x cos
4x
dx
= 0
u
n
lim
n
If a, b, c R+ then show that
n
(k an)(k bn) is equal to
k 1
(i)
1
a(b 1)
ln
if a b
a b b(a 1)
(ii)
1
if a = b.
a(1 a)
Solution
(i) We have lim
n
n
n
k
k
k 1 n 2 a
b
n
n
1 n
1
n n
k
k
k 1 a
b
n
n
= lim
1
= 0
1
dx
(a x)(b x)
=
1 1 (a x) (b x)
dx
a b 0 (a x)(b x)
=
1 1 1
1
dx
0
ab
b x a x
=
1
[ln(b x) ln(a x)] 10
ab
1
1 (b x)
= a b ln (a x)
0
=
1
(b 1)a
ln
, if a b.
a b (a 1)b
(ii) Now if a = b then the given limit
n
v
1
= (1 x)
cos
Example 10.
sin 4x
1 1
sin 4x dx
4 0 4 0
=
1
1
1 1
= 0 + 0 sin 4x dx = 16 cos 4x
0
4
=
1
1
1
dx
1
2
0
(a x) 2 = 1
k
k 1 n
a
a x 0
n
lim
n
1
1
1
=
.
a a 1 a(a 1)
2.157
DEFINITE INTEGRATION
Evaluate
Problem 11.
n
6
4
2
1 n 2 2 n 32 n
n2
lim 1 2 1 2 1 2 ...1 2
n
n n n
n
2n
2
2
2
Let A denote the given expression, then
Solution
ln A
n
2
2
1 n r
r2
2 ln 1 2
n n
n
r 1 n
lim ln A lim
n
r2
2r
n ln 1 n
r 1
2
1
2x ln(1 x )dx
2
1
negative. Show that the integral f (x ) dx ln 2 .
0
2
ln z dz ,
1
0
putting 1 + x2 = z
1
4
e
2
4
Since ln lim A = lim ln A = ln ,
n
n
e
The required limit = lim A =
n
Let f(x), x 0, be a non-negative
x
f(t) dt, x 0.
0
If for some c > 0, f(x) c F(x) for all x 0, then show
that f(x) = 0 for all x 0.
Given that, for x 0, F(x) =
Solution
F(0) =
x
f(t) dt
0
0
f(t) dt = 0
0
As f(x) c F(x) x 0, we get
f(0) c F(0) f(0) 0.
Since f(x) 0 x 0, we get
f(0) 0
f(0) = 0
Since, f is continuous on [0, ], F is differentiable
on [0, ) and F(x) = f(x) x 0.
Since, f(x) cF (x) 0 x 0, multiplying both
sides by e–cx (the integrating factor) we get
e–cx F(x) – ce–cx F(x) 0 [ e–cx > 0 x]
d
[e–cx F(x)] 0
dx
So, g(x) = e–cxF(x) is a decreasing function on [0, ]
i.e. g(x) g (0) for each x 0.
1/2
1/3
1/4
4dx
1/3
1/4
4dx
1/4
5dx ....
1/5
1 1 1 1 .... ln 2 .
2 3 4
Problem 14. Evaluate the definite integral
4
.
e
continuous function, and let F(x) =
1
Solution We have 0 f(x)dx 1/2 2dx 1/3 3dx
= z ln z z 1 2ln 2 1 ln .
Problem 12.
But we know that g(0) = e–c(0) F(0) = 0
g(x) 0 x 0
e–cx F(x) 0 x 0
F(x) 0 x 0
So, f(x) c F(x) 0 x 0.
But it is given that f(x) 0 x 0.
hence, f(x) = 0 x 0.
Problem 13. Let f(x) be defined in the interval
0 < x 1 as follows :
1
1
1
f(x) = 2, < x 1, f(x) = –3, < x
2
3
2
1
1
1
1
f(x) = 4, < x , f(x) = –5, < x ,
4
3
5
4
and so on, the values being alternatively positive and
1 2u 332 u 998 4u1664 sin u 691
1 u 666
1
du .
4 u1664 sin u 691
is odd in u,
1 u 666
so its integral is 0. Now make the substitution
1 –332/333
v
dv to find that
u = v1/333 du =
333
332
998
1 2u
u
1 1 2 v 2
du
dv
1 1 u 666
333 1 1 v 2
1 1
1
1
dv
1
333
1 v 2
2 1
1
1
dv
333 0
1 v 2
1
2
1
2
dv
(1 tan 1 1)
1 0
333
1 v 2 333
2
1 .
333
4
Solution
The term is
dx
Problem 15. Show that 0 (a cos x)
2
(a 1)
.
2.158
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Hence or otherwise evaluate 0
dx
Let I =
Solution
dx
.
( 5 cos x)3
(a cos x)
...(1)
0
dx
=
a cos( x)
=
(a cos x)
dx
...(2)
0
= 2a . 2
I = 2a
/2
0
/2
= 2a 0
(a cos x)
2
0
2
dt
2
(a 1)
=
=
{tan–1 – tan–1 0}
dx
or, 0 (a cos x)
Hence, I =
2
2
(a 1)
(a 1)
Differentiating both sides w.r.t. ‘a’, we get
=
2
2
2
1
dx
(x cos ) (x 1)
2
1
1
1 cos
dt
(1 t cos )2 t 2
0
2
1
cos 1
t t
1
1 cos
dt
2
3/2
Again differentiating both sides w.r.t. ‘a’ we get
=
2
( t sin 2t cos 1)
0
1
1
1
1 cos
| sin | 0
a
dx
(a cos x) (a 1)
0
1
1 cos 1
dx
(x cos ) (x 1) ,
1
2 dt
t
0
1
1 cos
= | sin | 0
–
Evaluate
Let I =
2
0 .
(a 1) 2
2
I=
t
tan 1
2
(a 2 1)
(a 1) 0
2
11
16 .
1
1
dx = – 2
t
t
1
When x = 1 then t =
and
1 cos
when x then t = 0
2
2
Put x – cos =
sec 2 dx
a 2 (1 tan 2 x) 1
2
2
3/2
=
0
3/2
dx
(a cos2 x)
( a 1) t
=
dx
( 5 cos x)
Solution
/2
0
0
0 < < 2
2
=2
3
Problem 16.
dx
(a cos2 x)
sec2 x dx
0
(a 2 1) (a tan x)2
Put a tan x = t a sec2 x dx = dt When x = 0, t = 0 ;
x = /2 t =
= 2a
( 5 cos x) (4)
or,
(11)
dx
2
2
0
/2
2a dx
Adding (1) and (2), 2I =
dx
(2a 2 1)
0 (a cos x)3
(a 2 1)3/2
Put a = 5 on both sides, we get
[ P–5 ]
0
2
dt
2t cos
1
t 2
sin 2 sin 2
dt
cos cos2
1
t 2
2
4
sin sin
sin
2
1
1
dt
1 cos
2
2
0
| sin |
1 t cos
2
sin sin 2
2.159
DEFINITE INTEGRATION
1
=
1 cos
1
sin 1 (t sin 2 cos )
| sin |
0
1
2
2
1 1 sin 2
1
= | sin | sin 1 cos cos sin (0 cos )
=
=
cos ( cos ) cos cos( ) | |
| sin |
| sin |
| sin |
,0
sin
=
, 2
(sin )
.
sin
Solution
dx
(x 1 x )
2 n
0
/2
0
1
(x 1 x2 )n
/2 sec 2 cos n
(1 sin )n
/2
0
d
/2
0
/2
0
n 2
sin
d
(1 cos )n
2sin cos
/2
2
2
n
0
2 cos2
2
Prove that
cos n 2
d
(1 sin )n
2 (a b)
dx
2
2
2
2
2
dx
=
1
(x3 b3 ) (x3 a 3 )
dx
(a 3 b3 ) (x 2 ax a 2 )(x 2 bx b2 )
1
xb
xa
dx
dx
3 2
2
2
(a b )
x ax a
x bx b2
3
1
2 (2x a) (a / 2 b)
1
dx
= (a 3 b3 )
x 2 ax a 2
1
(2x b) (a b / 2)
2
dx
x 2 bx b2
1
1 (2x a)dx
= 3 3
(a b ) 2 (x 2 ax a 2 )
(a / 2 b)
dx
(a 3 b3 ) (x a / 2)2 (a 3 / 2)2
–
1
1 (2x a)dx
3
(a b ) 2 (x 2 bx b2 )
(a b / 2)
dx
(a 3 b3 ) (x b / 2)2 (b 3 / 2)2
3
x 2 ax a 2
1
1
= 2 . (a 3 b3 ) ln x 2 bx b 2
n 2
d
2
Solution Let I = (x 2 ax a 2 )(x 2 bx b2 )
cos n 2
d
(1 sin )
cos n 2
/2
2
d
n
0
1
sin
2
dx .
sec 2
d
(tan sec )n
[Putting x = tan]
= 0
1 z n 1 z n 1
1 1
1
n
2 .
2 n 1 n 1 0 2 n 1 n 1 n 1
(x ax a )(x bx b ) 3ab(a ab b ) .
=
2
Problem 18.
1
Problem 17. If n > 1, evaluate 0
1 /2 n 2 4
tan
sec d
4 0
2
2
1
=
Finally, I =
cos
2 d
n2
1
sin 1 ( cos )
| sin | 2
1
0
n 2
1 1 n 2
2
[Putting z = tan ] 0 z (1 z )dz
2
2
1
{sin–1 (1) – sin–1 (– cos )}
| sin |
/2 sin
(a / 2 b) 1
(a 3 b3 ) a 3 / 2 2 2
2.160
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(a b / 2) 1
3
3
(a b ) b 3 / 2 2 2
+
n
=
(a 2b)
3
3
(a b )a 3
ab 2a 2 ab 2b2
=0+
2
2
ab
(a b)(a ab b ) 3
2 (a b)
= R.H.S.
3ab(a 2 ab b 2 )
0
Let I =
Solution
=
1
0
0
1
C k ( n 1)
(tx 1 x) dx
0
We have
1
0
1
xm (nC0 – nC1x + nC2 x2 – ...... + (–1)n nCn xn) dx
0
1
xn (mC0 – mC1 x + mC2 x2 – .... + (–1)m mCm xm) dx
0
1
(nC0 xm – nC1xm + 1 + nC2 xm + 2 – ...
0
1
(mC0xn – mC1 xn + 1 + mC2xn + 2 – ....
0
n
+ nC2(1 – x)n – 2(tx)2 + ... + nCr(1 – x)n – r(tx)r
+ .....+ nCn (tx)n} dx
= 0 C r (1 x)
r 0
m
m
Ck
Ck
(1)k
(k m 1) k 0
k n 1
+ (–1)m mCm xm + n) dx
0
nr
Prove the identity
n
1
=
...(1)
r
(tx) dx
n
n
C0
C1
C2
(1)n n Cn
...
(m 1) m 2 (m 3)
(m n 1)
m
=
{(1 – x) + tx}n dx
n
( 1) k
1
.
C k ( n 1)
+ (–1)n nCnxm + n) dx
{nC0(1 – x)n + nC1(1 – x)n – 1(tx)
1 n
x r dx = n
0
=
n
nr
xm (1 – x)n dx = xn(1 – x)m dx [ P–5 ]
m
C0 m C1 m C2
Cm
... ( 1)m
n 1 n 2 n 3
(m n 1)
n
1
=
1
0
k 0
n
1
=
(tn + tn – 2 + ... + t + 1}
n 1
0
(( t 1)x 1) n 1
=
( n 1)( t 1) 0
r
1
1 (1 x) n r x r dx
=
k
0
n 1
n
1
Again, I =
nr
0
(1 x)
((t – 1) x + 1)n dx
1
1
r
Problem 20.
(tx 1 x) dx , n N
1
r
Solution
and is independent of x. Hence show that
n
nC
x k (1 – x)n – k dx = n
...(2)
1
{tn + tn – 1 + tn – 2 + ... + t + 1}
n 1
Equating the coefficient of tk on both sides,
b 2a a 2b
+ 3
3
a
(a b ) 3 b
1
r
=
(a 3 b3 )b 3
Evaluate
nr
0
r 0
Problem 19.
1
r
r 0
n
(b 2a)
1
r
C t (1 x) x dx
1
1
=
[ln 1 – ln 1]
2 (a 3 b3 )
=
n
From (1) and (2)
a a 2
1
1
x x2
= 1
ln
2 (a 3 b3 )
b b2
1
x x 2
C t (1 x) x dx
n
Ck
m
m
Ck
(1) (k m 1) (1) (k n 1)
k
k 0
k
k 0
Hence proved.
1
Problem 21. Evaluate I = 0 2 sin (t) sin (t) dt if
(a) tan and tan
(b) tan , tan and
DEFINITE INTEGRATION
Solution
I=
1
2 sin t sin t dt
0
(a) Integrating by parts, taking sin t as the second
function, we have
1
cos t 1 2 cos t cos t
dt
0 0
I = 2sin t
2
2 1
cos sin +
cos t cos t dt
0
I=–
=–
2
cos sin
2
sin t 1
+ cos t 0 sin t sin t dt
0
2
2
2
cos sin +
sin cos + 2 I
I=–
2 2
2
I 1 2
sin cos – cos sin
2
2
= cos cos tan tan
= 0, given tan and tan
(b) Given
I=
1
1 2t 1 2t
te e 0
= lim
t
4
4
2
t 1
1
= – lim
2t
2 t e 4
1 2t
[ We have lim
e = 0 an d using
t 4
L’Hospital’s rule]
1
1 1
1
= – 2 lim
4 = .
t 2e 2t
4
Prove that
Problem 23.
1
1
1
0
0
1
x
n–1
0
1.2.3...(m – 1)
(1 – x)m–1dx = n( n 1)...( n m – 1)
when m and n are positive, and m is an integer.
Solution Integrating by parts, we have
n
(1 – x)m–1dx = x (1 – x)m–1
n
m 1 xn
+
(1 – x)m–2dx
n
Moreover, since n and m – 1 are positive, the term
xn(1 – x)m–1 vanishes for both limits.
1
m –1 1 n
n–1
m–1
x (1 – x)m–2 dx.
0 x (1 – x) dx
0
n
The repeated application of this formula reduces the
x
n 1
2 sin2 t dt = (1 – cos 2t) dt
integral to depend on
1
sin 2t 1 sin 2
= t
2 0
2
=1–
=1–
0
dx , the value of which
1.2.3...(m – 1)
n(n 1)....(m m – 1)
This shows that when either m or n is an integer the
1
x (1 x)
n–1
m–1
dx
0
definite integral x n –1 (1 x) m –1 dx can be easily
xe–2x dx.
2x
x e
dx
xe–2x dx = lim
t
0 u
t
x
t 1
2x
2x
lim
e
e
dx
= t 2
0 0 2
t
xe 2x e 2x
= lim
t
4
2
0
0
evaluated.
m n–2
0
1
Evaluate
x
1
.
m n –1
Hence we have
0
1
2
=
.
1 2 1 2
Solution
1
is
2 tan
2
=1–
2
2(1 tan )
2 (1 2 )
Problem 22.
2.161
dv
If the value of the definite integral
Problem 24.
1
1
cot 1 x · cot
1
1
2
1
dx
1 (x )
x
2 |x|
2 ( a b)
, where a, b, c N in their lowest
c
form, then find the value of (a + b + c).
=
2.162
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Let
Solution
1
I = 1 cot
1
2
x
1
· cot
dx ...(1)
1 x2
1 (x 2 )| x |
1
[ P–5 ]
1 1
x
1
dx ...(2)
cot
I = 1 cot
2
2 |x|
1 x
1 (x )
1
On adding
1
1
dt
2
0 2 t2
2
I = 2
1
t
2
2 2
tan 1
=
2
2
2 0
2
2
= 2
2 ( 2 1) ( a b)
=
2
c
a = 2, b = 1 and c = 4
a + b + c = 2 + 1 + 4 = 7.
Problem 25. Let <n> be a sequence of positive
=
1
x
cot
1 x
1 (x 2 )|x|
1
2
dx
1 (x 2 ) |x|
x
1
+ – cot
real numbers, such that lim n = 0. Find
n
1
1
2I = 1 cot
dx
1 x2
n
1
lim 1 ln k n .
n
n n k 1
1
= tan 1 1 x 2 dx
1 x is even function)
1
= 2 tan 1 1 x 2 dx
0
...(3)
1
1 tan 1 ( 1 x 2 ) dx
I = 0
...(4)
u
Integrating by parts
1
1
0
0
I = tan1( 1 x2 )·x
x
1
= 0 + 0
x
(x)
dx
2
(11 x ) 1 x2
2
dx
(2 x ) 1 x 2
Put x = sin dx = cos d
2
I = 0
2
sin
d
(2 sin 2 )
2
= – 0
2
2 sin 2
d
2 sin 2
2
I = 2 0
2
I = 2 0
2
d
2
2 sin 2 2
sec2 d
2
2 2 tan 2 tan 2 2
1
2
v
It is well known that
Solution
1
(As tan
sec2 d 2
Put tan = t
2 tan 2 2
2
2I = 1 cot
1
= 2 0
1
n
ln xdx lim
n
n
k
ln n
k 1
n
1 n k
1
k
Then, n ln n n n ln n
k 1
k 1
Given > 0 there exists n0 such that 0 < n for all
n n0.
n
n
k 1
k 1
k
1
k
1
Then n ln n n n ln n
n
1 ln k 1 ln( x )dx
0
n n k 1 n
Since lim
1
ln xdx
we obtain the
result when goes to 0 and so
1 n k
lim
ln n 1
n n
n
k 1
Evaluate the limit
Problem 26.
lim
x 0
2x sin
x
Solution
t
m
n
t
dt (m, n N).
We use the fact that
in the interval (0, ) and lim
t0
sin t
is decreasing
t
sin t
1.
t
2.163
DEFINITE INTEGRATION
For all x 0, and t [x, 2x] we have
2
sin 2 x sin t
1 , thus
2x
t
m
m
m
m
sin 2x 2x t 2x sin t dt 2x t dt
,
x
x tn
tn
2x x t n
2x t m
2
dt x m n 1 u m n du .
n
x t
1
m
The factor sin 2 x tends to 1. If m – n + 1 < 0, the
2x
limit of xm–n+1 is infinity ; if m – n + 1 > 0 then 0. If
2
m – n + 1 = 0 then xm–n+1 u m n du ln 2 . Hence,
1
lim
x 0
2x sin m t
tn
x
mn
0,
dt ln 2, n m 1
.
, n m 1
0
–2
1
1
2
2
0
1
1
2
0
1 1
1
f(t)dt dx
1
1 1 x2
2
This completes the proof.
0 x
Problem 28.
defined by F(x)
0
2
dx or,
1
1
1
tf(t)dt 3 .
0
Let F : (1, ) R be the function
x 2 dx
ln t . Show that F is one-to-one
x
(i.e. injective) and find the range of F.
Solution From the definition we have
x 1
, x > 1.
ln x
Therefore F'(x) > 0 for x (1, ). Thus F is strictly
in creasing and hence one-to-one. Since
F' ( x)
ln x
2005
f(x)dx ?
a 6
a
g(x)dx
a 3
a
a 3
a
g(x)dx
a 6
a 3
g(x)dx
(g(x) g(x 3))dx 0
2005
g(x)
333
6n 7
6n 1
n 0
g(x)dx 0 ,
so that
1
f (x) 2 xf(x)dx x dx 2 xf(x)dx .
0
0
0
0
3
From the hypotheses, we have
2
2 ln x
Solution Let g(x) = f(x) – x. Then g(x – 1) + x – 1 +
g(x + 1) + x + 1 x + g(x) + x, or, g(x – 1) + g(x + 1) g(x).
But now, g(x + 3) g(x + 2) – g(x + 1) –g(x).
Therefore
1
We get
1
2 ln x 1
dv x 2 ln 2
v
and similarly F(x) > x ln 2. Thus F(1+) = ln 2.
Hence the range of F(x) is (ln2 , ).
Problem 29. If f is a continuous real function
such that f(x – 1) + f(x + 1) x + f(x) for all x, what is the
Hence, F(x) e
xf(x)dx x dx
0
dv .
v
ln x
It follows that
(f(x) x) dx f (x)dx
0
2 ln x e v
1
1
1
1 x2
2
f(t)dt
. Show that 0 f (t)dt .
x
3
2
From
the
inequality
Solution
F(x)
minimum possible value of
Problem 27. Let f be a continuous function on
[0, 1] such that for every x [0, 1] we have
1
2
1
x x
F(x) (x 2 x).min :x t x 2
2
ln
t
ln x
as x , it follows that the range of F is (F(1+), ). In
order to determine F(1+) we substitute t = ev in the
definition of F and we get
2005
1
f(x)dx
2005
1
(g(x) x)dx
2005
1
x dx
2005
20052 1
2
x
2010012
2
2 1
The equality holds for f(x) = x.
Problem 30.
For 0, , find the value of
2
ln 1 tan tan x dx .
0
Solution
Let I = ln 1 tan tan x dx
0
I = ln 1 tan tan( x) dx
0
tan (tan tan x)
I = 0 ln 1 1 tan tan x dx
[ P–5 ]
2.164
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1 tan 2
= 0 ln 1 tan tan x dx
0
0
1 (b a)
ab
( k = b/a)
= ln
= ln
2
2a
I = ln (1 tan 2 ) dx ln 1 tan tan x dx
I = 2 ln sec – I
2I = 2 ln sec
I = ln sec .
Problem 31.
/2
Solution I = 0
=
/2
0
/2
0
ln(a2 cos2 x + b2 sin2 x) dx.
ln(a cos x + b sin x) dx (a > 0 , b > 0)
2
2
2
2
ln a 2 (cos 2 x k 2 sin 2 x) dx
/2
ln(cos x k sin x)dx
0
= ln a +
2
2
2
I1
I1 =
/2
0
ln(cos 2 x k 2 sin 2 x)dx
/2
dI1
2k sin 2 x
dx
2
0
dk
cos x k 2 sin 2 x
tan 2 x sec 2 x
2k
=
0 (1 k 2 tan 2 x) (1 tan 2 x) dx
(put tanx = t)
/2
= 2k 0
t2
dx
(1 k 2 t 2 ) (1 t 2 )
(1 k t ) (1 t )
2k
dx
= 2
0
k 1
(1 k 2 t 2 ) (1 t 2 )
2 2
=
2
2k
dt
dt
k 2 1 0 1 t 2 0 1 k 2 t 2
2k 1
1
tan t k
= 2
k 1 2 k
0
2k
= 2
k 1 2 2k
dI1
2 k k 1
=
.
dk k 2 1 2 k
k 1
I1 = ln (1 + k) + C
If k = 1, I1 = 0
C = – ln2
1 k
I1 = ln
2
ab
I = lna + ln
2a
a.(a b) ln a b
=
.
= ln
2a
2
Problem 32. Using definite integral as a limit of
sum for the integral
1000
0
x10 dx , determine an
approximate value for the sum 110 + 210 + 310 + ... +
100010.
100011
0
11
The graph of y = x10 is increasing. Hence, Left end
estimation < I < Right end estimation Dividing the
interval [0, 1000] into [0, 1], [1, 2], [2, 3], ... , [999, 1000],
we get using left end estimation
Solution
Let I =
999
1000
r
10
0
r0
1000
x10 dx =
x10 dx
100011
11
11
1000
S <100010 +
11
Using right end estimation
S –100010 <
1000
r
10
r 1
1000
0
x10 dx
100011
11
100011
100011
S 100010
Hence,
.
11
11
Problem 33. Find a step function S(x) on [1, 2.5]
S>
such that 1/x S(x) and
2.5
1
S(x) dx < 1. Also conclude
that e > 2.5.
Solution Define S(x) by 1 <
5 6 7 8 9 10
4 4 4 4 4 4
= 2.5 with the values 1, 4 , 4 , ... , 4 .
5 6
9
Then 1/x S(x) and
2.165
DEFINITE INTEGRATION
Thus, Rolle's Theorem is applicable to F (x).
There exists some c (a, b) such that F ' (c) = 0.
1 4 4 4 4 4
a S dx = 4 1 5 6 7 8 9
= 1 1 ... 1 = 2509 / 25820 < 1.
4 5
9
b
Therefore
2.5
1
0
16
I = f (t) dt
0
g (1) g (0)
= g ' ()
1 0
...(1)
g (2) g (1)
= g ' ()
2 1
...(2)
Similarly LMVT in [1, 2] gives some (1, 2) such
zero
3
4
But g ' (x) = f (x4) · 4x3 4 f( ) f( )
0
Let f be a continuous function on
f (t) dt f (t) dt (2x (a b) then
x
b
a
x
prove that there exist some c (a, b) such that
c
f (t) dt – f (t) dt = f (c)(a + b – 2c).
c
Given F (x)
Solution
x
b
...(1)
Since, f is continuous, F (x) is also continuous.
Also, put x = a :
b
b
F (a) = a f (t) dt (a b) (b a) a f (t) dt
and put x = b :
F (b) =
f (t) dt (b a)
b
a
Hence, F (a) = F (b).
0
Let k 0,
t2
I = | k t | dt = (k t)dt = kt
0
0
2
1
1
1
0
1
2
1
t2
1
I = – (k t)dt = – kt 2 = – k
0
2
0
1
,
1
= a f (t) dt x f (t) dt 2x (a b)
1
D = 4 + 4k – 4 k = 2 > 0.
2
Let – 1 < k < 0
b
a
1
3
= 4 + 8k + 2 = 8k + 6 > 0 = 4 2k .
2
Let k – 1,
= f (t) dt .
[a, b]. If F(x) =
D = 4 + 4 k 0 | k t | dt
1
Hence, D = 4 + 4k + 4 k
2
x4
Problem 35.
1
= 4 + 4k + 4 | k t | dt .
=k+
Addind (1) and (2).
( 0) ;
g '() + g ' () = g (2) – g
4
Comment upon the nature of roots
1
0
LMVT for g in [0, 1] gives, some (0, 1) such that
3
b
c
Problem 36.
Solution
Consider g (x) = f (t) dt g (0) = 0
c
a
of the quadratic equation x2 + 2x = k + | t k | dt
0
depending on the value of k R.
x4
that
f (t) dt f (t) dt f (c)[(a b) 2c] .
F(c) =
f (t) dt = 4(3f(4) + 3 f(4))
Solution
b
= + (2x – (a + b)) [f(x) + f(x)] = 0.
dx / x 1, ln 2.5 < 1, 2.5 < e.
Problem 34. If the function f : [0, 16] R is
differentiable. If 0 < < 1 and 1 < < 2, then prove that
16
x
Now, F ' (x) = 2 a f (t) dt x f (t) dt
I = | k t | dt
0
Let k = – y 0 < y < 1
y
1
1
I = | t y |dt = 0 (y t) dt y (t y) dt
0
y
1
t2
t2
yt
ty
=
2 0 2
y
2
2 y2 1
y y2
y
y
=
2 2
2
2.166
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
1
= k2 + k +
2
2
1
D = 4 + 4k + 4(k2 + k + )
2
1
3
2
2
= 4 1 k k k = 4 k 2k
2
2
3
2
= 4 ( k 1) > 0.
2
0 ln y dy
ln y dy
I2 = y 2 y 1 = – 0 y 2 y 1 = – I2
= y2 – y +
Hence, D > 0 k R roots are real and distinct.
For a > 0, b > 0 verify that
Problem 37.
ln x dx
0
2
ax bx a reduces to zero by a substitution
x = 1/t. Using this or otherwise, evaluate
ln x dx
0 x 2 2x 4 .dx
Now, I1 =
ln 2 ln t
= 2 0 4(t 2 t 1) dt
I2
1
ln t dt
1
put t =
dt = – 2 dy
y
t t 1
y
2
ln y·(1)
1 1 2
= 1 y dy..
y2 y
1
=
x2
and du = cos ax dx,
1
2
x2
1 2xe x
2
0
sin ax dx
a
e sin a 2 1 x2
–
xe sin ax dx .
a
a 0
Hence, – e =
r 1
(b a) r
ba
f a
.
= nlim
n
n
r0
1
1
0
x2
0
1
0
xe sin ax dx e
x
2
0
esin a – 1 1 2xe x2 sin ax dx 0
and lim
.
a
a
a 0
Thus, the limit of
1
0
x2
0
n 1
f (x) dx = nlim
(x)
3.
1
e cos ax dx as a is 0.
If a = 0, b = 1, then
ba
n f a b n a r
1
–e dx xe sin ax dx e dx e .
Consequently,
a
Letting u = e
x
e cos ax dx = e sin ax –
0
a
0
2.
b
f (x) dx
n 1
x2
2
0
= nlim
1
e cos ax dx .
a 0
Now, for x in [0, 1], – e xe x sin ax e.
ln 2 dt
1 ln t dt
2
2
0
0 t t 1
2
t
t 1
= 2
n
Find lim
Problem 38.
we have
1.
2
ln x dx
I = 0 2
dx (put x = 2t to make
x 2x 4
coefficient of x2 and constant term same dx = 2dt)
t 1 3
2 2
2
2
(t (1 2))2
tan 1
=
3
3
0
2
2
=
=
.
3 2 6 3 3
ln 2 2
2
ln 2
·
=
=
.
I=
2 3 3
3 3
3 3
I2 = 0
dt
0
Solution
Solution
I1
1
r
n f n
r0
b
1
r
f f(x)dx , where
n n
r (x) n
a
lim
a = nlim
( x)
(x)
and b = nlim
n
n
2.167
DEFINITE INTEGRATION
4.
Rules of definite integration
(i)
(ii)
(iii)
(iv)
(v)
f(x) dx 0
b
b
a
f(x)dx f(x)dx
b
b
a
a
a
cf(x)dx c
b
f(x)dx
b
f (x)dx lim
a
b
b
a
a
a
5.
a
f(x)dx
a
f(x)dx
b
c
b
a
b
f(x)dx = f(c) (b – a)
a
Average value of a function : If f is integrable on
the interval [a, b], the average value of f on this
interval is given by the integral
6
b
1
f(x)dx .
ba a
If a function f is integrable on a closed interval
[a, b], then the function g(x) =
7.
x
f(t)dt is
a
continuous at any point x [a, b].
First Fundamental Theorem of Calculus:
If f is continuous on [a, b], then the function g
definedby g(x) =
x
f(t)dt a x b is continuous
a
on [a, b] and differentiable on (a, b), and g(x) = f(x).
8.
The Newton-Leibnitz Formula : If f is continuous
on [a, b] and F is any antiderivative of f on the
interval [a, b], that is a function F exists such that
F(x) = f(x), then
9.
b
f(x)dx = F(b) – F(a).
a
Improper integrals
(i)
(ii)
b
a
b a
b
f (x)dx lim f (x)dx
c
b
a
a
a
10.
b
u(x)v(x)dx
a
u(x) v(x)dx
u '(x) v(x)dx dx
b
b
a
a
11. Leibnitz rule for differentiation of integrals : If
f is continuous on [a, b], and u(x) and v(x) are
differentiable functions of x whose values lie
in [a, b], then
d v(x)
dv
du
f(t)dt f(v(x)) f(u(x))
dx u(x)
dx
dx
h (x )
12. Modified Leibnitz Rule : If F(x) = g(x ) f (x, t) dt,
then
h (x )
F(x) = g(x )
f (x, t)
dt + f(x, h(x)l) h(x) – f(x, g(x)) . g(x)
x
b
b
13. Property P f (x)dx f (t)dt
a
a
b
= f (u)du
a
Property P
Property P
a
b
f(x)dx f(x)dx
b
a
b
c
b
f(x)dx = f(x)dx + f(x)dx
a
a
c
Property P
f (x)dx lim f (x)dx
b
b
f (x)dx f (x)dx f (x)dx
f(x)dx
Mean Value Theorem for Integrals : If f is
continuous on the interval [a, b], there is atleast
one number c between a and b such that
favg =
f (x)dx, where > 0.
(vi) If the function f(x) has an infinite
discontinuity at an intermediate point x = c
of the interval [a, b] (i.e. a < c < b) then, by
definition
[f(x) g(x)]dx f(x)dx g(x)dx
b
b
0 a
a
b
c
f (x)dx where > 0.
(v) If f(x) has an infinite discontinuity only at
the right end point x = b of the interval [a, b],
(vi) a [c1f(x) c 2 g(x)]dx c1 a f(x)dx c 2 a g(x)dx
(vii)
b
0 a
a
b
b
0
f (x)dx lim
f(x)dx f(t)dt
b
0
(iv) If f(x) has an infinite discontinuity only at
the left end point x = a of the interval [a, b],
a
a
(iii) f (x)dx f (x)dx f (x)dx
a
a a
a
a
a
0
(i)
f (x) dx = (f (x) f ( x)) dx
(ii)
f(x) dx
a
a
a
= 2 f(x) dx, if f(x) is an
0
2.168
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
even function
(iii) f(x) dx = 0, if f(x) is an odd function
Property P-5
a
a
0
0
m (b – a)
f (x)dx = f (a x)dx
b
b
a
a
a
a
0
0
0
nT
T
0
a nT
T
0
(iii)
nT
(iv)
b nT
mT
T
f (x) dx = (n – m) f (x) dx, m, n I
a mT
0
b
bc
a
a c
14. Shift property : f (x)dx
f (x c)dx
15. Expansion–Contraction property
b/k
a/k
[a, b], we have
a
a
b
b
1
22.
a
0
f (x)dx (b a) f[(b a)t a]dt
b
f (b)
a
f (a )
f (x)dx bf (b) – af (a)
f 1 (y)dy
19. Estimation of definite integral :
(i) If f(x) 0 for a x b, then
b
f(x)dx 0
a
(ii) If f(x) g(x) for for every x in [a, b], then
b
a
a
b
c
b
a
a
c
f(x)g(x)dx f(a) g(x)dx f(b) g(x)dx
a
b
b
f(x)g(x)dx f(c) g(x)dx
21. Generalized Mean Value Theorem for integrals:
Assume g is continuous on [a, b], and assume
f has a derivative which is continuous and
never changes sign in [a, b]. Then, for some c
in [a, b], we have
b
f (x)dx f ( x)dx
18.
20. Weighted Mean Value Theorem for integrals :
Assume f and g are continuous on [a, b]. If g
never changes sign in [a, b] then, for some c in
f (kx)dx f (x)dx for every k > 0.
16. Reflection property
17.
b
f (a ) f ( b )
< f (x)dx < (b – a)f(b).
a
2
T
a
k
(b – a)
0
+ f (x) dx, n I, a, b R
2
f (a) f (b)
2
(b) If the function f(x) increases and has a
concave down graph in the interval [a, b],
then
f (x) dx = (n – m) f (x) dx
b
b
b
f (x) dx = n f (x) dx, n I
f (x) = n f (x) dx, n I, a R
2
(b – a) f(a) < a f (x)dx (b a)
Property P
If f(x) is a periodic function with period T, then
0
b
(vi) (a) if the function f(x) increases and has a
concave up graph in the interval [a, b], then
if f (2a x) f (x)
0,
a
= 2 f (x) dx, if f (2a x) f (x)
0
a
a
a f (x)g(x)dx a f (x)dx a g (x)dx
f(x) dx = f(x) dx + f(2a - x) dx
b
a
b
2a
(ii)
a
b
(v) Schwartz-Bunyakovsky inequality
Property P
(i)
b
f(x)dx M(b – a)
f (x)dx | f (x) | dx
(iv)
f(x) dx = f(a + b - x) dx
(ii)
b
a
(iii) If m f(x) M for a x b, then
a
(i)
b
a
f(x)dx g(x)dx
a
/2
0
sin n x dx =
/2
0
cos n x dx
n 1 n 3 n 5 ... 1 . if
n n 2
n4 2 2
=
n 1 n 3 n 5
... 2 . 1 if
n n 2 n 4 3
23. Wallis formula :
/2
0
n is even
n is odd
sinmx . cosnx dx =
(m 1) (m 3)....1 or 2 (n 1) (n 3) (n 5)....1 or 2
(m n) (m n 2) (m n 4)....1 or 2
K
DEFINITE INTEGRATION
where K
if both m and n are even (m, n N);
2
=
25. Let a function f(x, ) be continuous for a x b
and c d. Then for any [c, d] if
b
I() = f (x, )dx , then
= 1 otherwise.
b
2.169
a
b
f (x, n) dx lim f (x, n) dx , if a and b
24. nlim
a n
a
b
dI( )
d
f (x, ) dx .
a d
d
are independent of n.
SINGLE CORRECT ANSWER TYPE
1.
3
5
3
x 1 x 4 dx is equal to
5
(1 x 4/3 )6/5 C
8
5
2/3 6/5
(B) (1 x ) C
8
5
6/5 4/3
(C) (1 x ) C
8
(D) None of these
5.
equal to
(A)
2.
x
1
3.
(A) x = b is a point of local minimum
(B) x = b is a point of local maximum
(C) x = a is a point of local minimum
(D) x = a is a point of local maximum
If f : [0, ] R is continuous and
0
0
number of roots of f(x) in (0, ) is
(A) zero
(B) exactly one
(C) exactly two
(D) atleast two
1
If I log log x
dx , then I is equal
(log
x) 2
(A) loglog – loglog
(B)
(C) –1
a
b
b
a
(D)
1
2
x
x (a, b); if g(x) = f (t)dt , f(x) being continuous
0
and
differentiable
in
(a,
b)
then
b
f (x) g(x)dx 0 implies
a
(A) g(x) = 0 has atmost one root in (a, b)
(B) g(x) = 0 has atleast one root in (a, b)
(C) g(x) = 0 has exactly one root in (a, b)
(D) g(x) = 0 x (a,b)
7.
Given
1
(x x x ) 4x 5x 10dx .19 19
5
4
2
3
2
0
then is equal to
1
1
(A)
(B)
30
20
8.
/2
If 1 0
(C)
1
15
(D)
1
10
1
1 tan
x
x
dx and 2
dx , then
0
sin x
x
1
value of is
2
1 1
log log log log
log log log log
(D) None of these
(C)
(B) 0
6. Given f (x)dx f (x) dx and f'(x) 0 at any
(t – a)2n ((t – b)2m+1 dt, a b, then
f (x) sin x dx = f (x) cos x dx = 0, then the
4.
(A) 1
If m and n are positive integers and
f(x) =
tan 1 (nx)
dx , then lim (n 2 I n ) is
n
sin 1 (nx)
1/ n
Let In = 1/(n 1)
(A) 1
(B) 2
(C)
1
2
(D) 3
2[x]
3x [x]
The value of the integral 10 2[x] dx, where
3x [x]
0
9.
[.] denotes the greatest integer function is,
2.170
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(A) 0
(C) 10
(B) –10
(D) None of these
16. If
=k
function f(x). Then the value of the definite
integral h '(x).sin xdx is
(A) equal to zero
(B) equal to 1
(C) equal to –1
(D) non existent
11. If f : R R is a continuous and differentiable
x
0
x
–1
x
1
17.
function such that f(t)dt f (3) dt t 3dt
x
3
0
x
(B) 48 – 8f(1) – f(2)
(D) None of these
12. If f(y) = ey, g(y) = y, y > 0 and
F(t) =
1
f(t – y) g(y) dt, then
0
(A) F(t) = et – (1 + t)
(C) F(t) = te–1
(B) F(t) = tet
(D) F(t) = 1 – et (1 + t)
13. Suppose that F(x) is an antiderivative of f(x) =
3 sin 2x
sin x ,
where x > 0, then
dx can be
1
x
x
expressed as
1
(A) F(6) – F(2)
(B) (F(6) – F(2))
2
1
(C)
(F(3) – F(1))
(D) 2(F(6) – F(2))
2
14. The value of
k
(1)
(A) –2(k – 1)
(C) k
sgn[x]
k
X
15. If f(x) = 0
(B) – 2k
(D) none of these
1 tan 2 x sin 2 t
(A) f(0+) = –
for 0 < x <
, then
2
2
(B) f
4 8
(C) f is continuous and differentiable in 0,
2
(D) f is continuous but not differentiable
in 0,
2
1
2
1
2
(B)
1
(C) 2 2
1
(D) – 2
0
– / 3
2
6
(B)
2
3
(C)
1
2
8
(D)
3 2
8
1 x – 1 x ln x dx is equal to
0
2
(A) –
ln
2
(C)
2
2
(B) 0
ln 2
2
(D) None of these
(y2 4y 5)sin(y – 2)dy
19. 0
is equal to
2y2 8y 1
4
(A) 0
(C) – 2
(B) 2
(D) None
1
20. If f(x) = 0
dt
1
, then f is equal to
2
1 | x t |
1
2
(D) None
(A) 0
(B)
(C) 1
dx is
t sin t dt
18.
sec x dx, then the value of k is
–1
–1
2
1
cot 2 cos x – 1 cot cos x – 2 dx
is equal to
(A)
– f(1) t 2 dt f (2) t dt , then the value of f(4) is
(A) 48 – 8f(1) + f(2)
(C) 48 + 8f(1) + f(2)
/2
– / 2
(A)
/4
x
– / 4
10. Let a function h(x) be defined as h(x) = 0, for all
x 0. Also h(x).f (x)dx f (0) for every
e / 4 dx
(e e ) (sin x cos x )
3 / 4
21. The value of 0 (x r)
r 1
1
n
1
k 1 x k dx
equals
(A) n
(B) n!
(C) (n + 1)!
(D) n · n!
22. f(x) is a continuous for all real values of x and
satisfies
n 1
n
f(x) dx =
n2
n I, then
2
5
f(| x |) dx is equal to
3
(A) 19/2
(C) 17/2
(B) 35/2
(D) None
2.171
DEFINITE INTEGRATION
23. Let f be integrable over [0, a] for any real values
of a. If I1 =
/2
cos f (sin + cos2 ) d and
0
I2 =
/2
sin 2 f (sin + cos2 ) d, then
0
(A) I1 = – 2I2
(B) I1 = I2
(C) 2I1 = I2
(D) I1 = – I2
24. If are the roots of g(x) = ax2 + bx + c = 0
and f(x) is an even function, then
f
g(x)
e x dx
e
g(x)
f
x
(A)
e
g(x)
f
x
b
2a
b 4ac
| 2a |
b
(D) None of these
a
25. If the function f : [0, 8] R is differentiable, then
8
f(t)dt is equal to
0
(A) 3[3f(2) + 2f(2)]
(B) 3[3f() + 2f()]
(C) 3[2f(3) + 2f(3)]
(D) 3[2f(2) + 2f(2)]
26. Let g(x) be a continuous and differentiable
function such that
2
[2x 3]dx . g(x) dx = 0, then g(x) = 0
0 2
when x (0, 2) has (where [.] denotes the greatest
integer function)
(A) exactly one real root
(B) atleast one real root
(C) no real root
(D) none of these
27. The values of x satisfying
5/2
2
x
dx
[x + 14]dx, lie in the interval
0
2
(where [.] and {.} denotes the greatest integer
and fractional part of x)
(A) [–14, 13)
(B) (0, 1)
(C) (–15, –14]
(D) none of these
2[x 14]
0
28. The value of
equal to
{x}
2
dx
0
2
(17 8x 4x )(e
1
2 21
(B) – 8 21 ln 21 2
1 2 21
2 21
(C) – 8 21 ln 21 2 ln 21 2
(D) none of these
10
1
sgn (x – [x]) dx, is equal to
(where [.] denotes the greatest integer function)
(A) 9
(B) 10
(C) 11
(D) none of these
2
(C)
for 0 < a, b < 2,
2 21
29. The value of
is equal to
(B)
1
(A) – 8 21 ln 2 21
6(1 x )
1) , is
30. The function
1
dt , x (0, 2)
2
strictly increases in the interval
f(x) =
x
log
0
|sin t| sin t
5
(A) ,
6 6
5
(B) ,2
6
7
(C) ,
6 6
5 7
(D) ,
6 6
31. Let f : (0, ) R and F(x) =
12
2
4
5
F(x ) = x + x , then
(A) 216
(C) 221
x
t f(t) dt. If
0
f (r ) , is equal to
2
r 1
(B) 219
(D) 223
32. Let f(x) be a continuous functions for all x, such
that f(x))2 =
x
0
f(t).
2sec 2 t
dt and f(0) = 0, then
4 tan t
5
3
(A) f = ln (B) f =
4
4 4
4
(C) f = 2
2
33. The value of
(D) none of these
1 3
2
n
2 sin3
... n sin3 , is
sin
4n
4n
4n
equal to
lim
n n 2
2.172
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(A)
2
(52 – 15n)
9 2
(B)
2
(52 – 15n)
92
1
32
4
(C)
32
(A)
/2
0
0
2
2
2
ln (sin + k cos ) d, is equal to
1
(A) 1
(B) e
(C) e + 1
(D) e – 1
36. If f : R R is continuous and differentiable
x
0
1
x
function such that f(t)dt + f(3) dt
x
3
1
x
3
0
1 e x dx
x
1
0
is equal to
(A) 3/e
(C) 3e
(B) e/3
(D) 1/3e
38. If g(x) =
I1
x 2 dx
3 , then I
e (2 x )
2
x3
n
(| sin t | | cos t)dt , then g x 2
0
is equal to, where n N
(A) g(x) + g()
n
(B) g(x) + g
2
(C) g(x) + g (D) none of these
2
1
39. Let f(x) = minimum | x |,1 | x |, , x R, then
4
1
the value of f(x)dx is equal to
1
5
1
x, y R and f(0) = 0. If I1 = f(x)dx ,
I2 = f ( x)dx and I3 =
1
(A) I1 = I2 > I3
(C) I1 = I2 < I3
2
f (x)dx , then
1/ 2
(B) I1 > I2 > I3
(D) I1 < I2 < I3
1
e
0
2x [2x]
d(x [x]) . (where [.]
denotes the greatest integer function) is
(A) e + 1
(B) e
(C) e – 1
(D) none of these
44. The value of
x
3
(0 < x 4), then the number of points where f(x)
assumes local maximum value is
(A) one
(B) two
(C) three
(D) none of these
42. Let f : R R such that f(x + 2y) = f(x) + f(2y) + 4xy
43. The value of
and I2 = 0
1 x
t
0
0
t dt + f(2) t .dt, then the
2
x
(sin t cot t) (e – 2) (t – 1) (t – 2) dt
0
value of f(4) is
(A) 48 – 8f(1) + f(2)
(B) 48 – 8 f(1) – f(2)
(C) 48 + 8 f(1) + f(2)
(D) none of these
37. Let I1 =
has atleast two roots of opposite signs in (–1, 1) is
(A) a (0, 1)
(B) a (0, 3)
(C) a (–, 1) (D) a (3, )
41. Let f(x) =
n
= t dt – f(1)
0
x
(ln x) dx then In + nIn – 1 is
35. If In =
x
f(t) dt 5 as |x| 1, then the
value of 'a' so that the equation 2x + f(t)dt = a
(A) ln (1 + k) – ln 2
(B) ln 2 – ln (1 + k)
(C) ln (1 + k) – ln 2
(D) none of these
e
3
8
(D) none of these
40. If the function
1
(15n – 15)
(C)
92
(D) none of these
34. The value of
(B)
(A) 0
(C) 2
45. I1 =
/2
0
I3 =
/4
0
cot( /4 x)
ln cos x
4
(B) 1
(D) not defined
dx is
/ 2 sin 2 x
cos 2 x
dx,
I
=
dx,
2
0
1 cos 2 x
1 sin 2 x
/2 1 2 cos 2 x.sin 2 x
dx, then
4 2 cos2 xsin 2 x
(A) I1 = I2 > I3
(B) I3 > I1 = I
(D) none of these
(C) I1 = I2 = I3
0
46. The value of
9 /4 (| sin x cos x |)dx 5 { x}dx
0
is (where
1
[.] and {.} represents greatest integer and
fractional part).
2.173
DEFINITE INTEGRATION
(A) 3
(C) 2
(B) –4
(D) 4
54. Let f be integrable
x 1
47. The value of 2 x 1 cos
dx, where
2
[.] denotes greatest integer function, is
(A) 1
(B) 1/2
(C) 2
(D) none of these
0
2 | sin x |
| sin x |
dx , n
48. If m
dx
2
x
1
0 x
1
2
2
2
(where [ ] = G.I.F.), then
(A) m = n
(B) m = –n
(C) m = 2n
(D) None of these
1
49. If [.] is G.I.F. then
(A) 0
(C) 3
2
[x ]dx
10
[x 28x 196] [x ] is
2
4
2
(B) 1
(D) 4
0
e
/2
0
(B) e + 1/e
(D) e
n
x cos x dx , then the value of
28(I8 + 56I6) is
(A) 8
(C) 5
53. Given I m
(B) 8
(D) 5
e
((ln x) dx
I m I m2
e , then the value of K and L are
K
L
(A) 1 – m, 1/m
(B) (1/1 – m), m
1 m(m 2)
,
(C)
1 m m 1
(D)
m
,m 2
m 1
(B) I1 + I2 = 0
(D) none of these
a
ka 4
(B) ka2
24
(C) a3/3
(D) none of these
56. If x satisfies the equation
dt
1
3 t 2 sin 2t
x2
20
x
2
2
0 t 2t cos 1 3 t 1
(0 < < ), then x is
57.
(A) 2
sin
(B) 4
(C) 2
cos
(D) none of these
sin
sin(x a) cos(x a)
sin(b x) cos(b x) dx
a
b
a
sin(b x) cos(b x)
dx .
sin(x a) cos(x a)
Then the value of is
(A) 1
(B) 1/2
(C) 2
(D) a function of a and b
m
1
If
sin 2 . f(sin cos2 )d , then
a
b
equals
(A) 0
(C) –e – 1/e
/2
0
(A)
sec3x(sin2x + cosx + sinx + sin x cos x)dx
52. If I n
cos f(sin cos2 )d and
x2
x3
f(x)dx xf(x)
f '(x)
f "(x) is
0
2!
3
0
b
sec x
I2
0
55. If f"(x) = k in [0, a], then
51. The value of
(A) I1 = I2
(C) I1 = 2I2
xn
dx 6 then
a x n (16 x)
(A) a = 4, b = 12, n R
(B) a = 14, n R
(C) a = –4, b = 20, n R
(D) a = 2, b = 8, n R
50. If
/2
I1
58. If F(x)
x3
0
f(x)g(t)dt, f(x)
x dt
t
1
1
and g( t ) 1 t 2 sin 2 t , and F'(x) is
x3
3x 2 ln x
g(t)dt then (x) is
(x)
0
1 x 6 sin 2 x3
(A) x
(B) x2
(C) 1/x2
(D) 1/x
2.174
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
2
n
1
...
59. f ( x) nlim
1.3.5 ..(2 n 1)
1.3 1.3.5
(A)
(B) 1
(D) None
60. If f(x)
x
1
2
1
(f(t)) dt and (f(t)) dt 6 ,
2
0
0
3
2
then f(9) is equal to
(A) 2
(B) 0
(C) 3
(D) None
n 1 n
. x , then the value of
61. If fn ( x)
( n 1)!
1
fn (x) dx is
0
n 1
(A) 2
(B) 3
(C) 4
(D) None
n
e
62. If
x(a n)
a n
n
a n
(D) None of these
63. For 0 x ,
cos x.d(cos x) is
2 1/ 2
(A) 1
(B) 2
(C) 3
(D) None of these
64. If f(x + y) = f(x) + f(y) for all x and y and
1/2
0
(A) 0
(C) 2
(B) a
(C) 2a
1
f(x)d (x [x]) ([ . ] is G.I.F.) is
then
a
2
dx , then the value of
xe x(a n)dx , a 2n, is
3
a
(x 1) f(x 1)dx and
b
(x 1) f(x 1)dx , then a is equal to
2
1
1
2
3
(A) b
(C) 3b
(B) 2b
(D) None of these
65. The value of
/2
/4
sin 2x tan 1 (sin x)dx is equal to
(A)
2
2
(B)
2
2
(C)
2
2
(D)
2
2
MULTIPLE CORRECT ANSWER TYPE FOR JEE ADVANCED
1
2
66. If f(x) = [x] + x + x , then
3
3
([ .] denotes the greatest integer function)
(A) f(x) is discontinuous at x = 1, 10, 15
(B) f(x) is continuous at x = n/3, where n is any
integer
2/3
(C) 0
f (x) dx
1
3
(D) xlim
23 f(x) = 2
67. Consider a real valued continuous function f (x)
defined on the interval [a, b]. Which of the
following statements does not hold(s) good?
(A) If f (x) 0 on [a, b] then
b
b
a
a
f(x)dx f (x)dx .
(B)
2
If f (x) is increasing on [a, b], then f 2(x) is
increasing on [a, b].
(C) If f (x) is increasing on [a, b], then f (x) 0 on
(a, b).
(D) If f (x) attains a minimum at x = c where
a < c < b, then f ' (c) = 0.
68. Let f (x) be a non constant twice derivable function
defined on R such that f (2 + x) = f (2 – x) and
1
f ' = 0 = f ' (1). Then which of the following
2
alternative(s) is/are correct?
(A) f (– 4) = f (8).
(B) Minimum number of roots of the equation
f '' (x) = 0 in (0, 4) are 4.
/4
(C) / 4 f (2 x)sin x dx = 0.
2
4
0
2
(D) f (t) 5cos t dt f (4 t) 5cos t dt .
69. Let [ . ] represent the greatest integer function.
x
[x]
0
0
[t]dt = t dt , if
(A) x = 5/2
(B) x = 6
7
2
(D) x =
(C) x = –
101
2
x
70. Let f(x) = | t 1| dt, then
2
(A) f(x) is continuous in [–1, 1]
DEFINITE INTEGRATION
(B) f(x) is differentiable in [–1, 1]
(C) f’(x) is continuous in [–1, 1]
(D) f’(x) is differentiable in [–1, 1]
71. Which of the following function(s) is/are even?
x
(A)
2
(A) f (x) = 0 ln t 1 t dt
(C)
(2 t 1)t
dt
(B) g (x) = 0 t
2 1
x
x
(C) h (x) = 0
1 t t 1 t t dt
2
2
x
1 t dt
(D) l(x) = 0 ln
1 t
et
dt – ex, then
t
(A) f(x) is an increasing function
x
72. Let f : [1, ) R and f(x) = x 1
(B) lim f ( x )
x
(C) f(x) has a maxima at x = e
(D) f(x) is a decreasing function
73. If x satisfies the equation
2
1
dt
x 3 t sin 2t dt 2 0
x 2
3 t 2 1
0 t 2t cos 1
2
(0 < < ), thet the value of x is
sin
(A) 2
(C)
sin
sin
(B) 2
(D) none of these
3k 2
n
74. If nlim
3
n 2 n = f (x) dx then
b
0
k 1
(B) f(x) = (9x2 + 2)
(A) b = 1
n
3k
2
3
(C) nlim
n 2 n = 15
(D) nlim
3k 2
3
2 = 5
k 1 n
n
k 1
n
x
m
m
m
n
Cx 1
75. xlim
1
n
n
n
e
m x m
.e
x!
(B)
mx m
.e
x!
0
(D)
m x1
me m x!
2.175
n x
equals to
1/ n
n!
; n N is equals to (a–1eb), then
76. nlim
( 2 n ) n
(A) a = 2
(B) a = 1
(C) b = –1
(D) b = 1
77. Let f(x) be a periodic function with period 3 and
x
2
f 7 and g(x) = 0 f (t n)dt where n = 3k,
3
k N, then
(A) g'(–2/3) = 7 (B) g'(–2/3) = –7
16
(D) g' = 7
3
78. The value of the definite integral
(C) g'(7/3) = 7
2
3 cos x
x ln 3 cos x dx , is
0
2
3 cos x dx
(A) 0 ln
3 cos x
3 cos x
(B) 2 0 ln
dx
3 cos x
(C) zero
3 cos x dx
(D) 2 0 ln
3 cos x
79. The function f (x) is defined for x 0 and has its
inverse g (x) which is differentiable. If f (x) satisfies
g(x)
0
f (t) dt = x2 and g (0) = 0 then
(A) f (x) is an odd linear polynomial
(B) f (x) is some quadratic polynomial
(C) f (2) = 1
(D) g (2) = 4
2.176
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
80. Let f(x) be a quadratic function with positive
integral coefficients such that for every
R, > ,
f (x) dx > 0. Let g(t) = f(t). f(t) and
g (0) = 12, then
(A) 16 such quadratic functions are possible
(B) f(x) = 0 has either no real or distinct roots
(C) Minimum value of f(1) is 6
(D) Maximum value of f(1) is 11.
81.
lim
2x sin m t
(m, n, N) equals
tn
(A) 0 if m n
(C) if n – m > 1
x0
x
82. If f(x) =
(B) ln 2 if n – m = 1
(D) None of these
/ 2 ln(1 x sin 2 )
sin 2
0
d, x 0 then
(A) f(t) = ( t 1 1)
2 t 1
(C) f(x) cannot be determined
(D) None of these
(B) f(t) =
83. Let In=
dx
3
In = I
1 x (n = 1, 2, 3 ... ) and Lim
0
n
n
0
(say), then which of the following statement(s) is/are
correct? (Given : e = 2.71828)
(B) I2 < I0
(A) I1 > I0
(C) I0 + I1 + I2 > 3
(D) I0 + I1 > 2
84. Let J =
2
1 1
cot x cot x dx and
1
1
sin x
dx.
| sin x |
Then which of the following alternative(s) is/are
correct ?
(A) 2J + 3K = 8
(B) 4J2 + K2 = 26 2
7
K = 2
(C) 2J – K = 3
J 2
(D)
K 5
85. If f : R R be a continuous function such that
f (x) =
x
2t f(t)dt , then which of the following
1
does not hold(s) good?
(A) f () = e
(C) f (0) = 1
2
(B) f (1) = e
(D) f (2) = 2
Comprehension - 1
A continuous function f satisfies f(2x) = 3f(x) for all x.
86.
1
2
0
1
f(x)dx = 1 and let f(x)dx = S.
Moreover
1
2 n 1 f(x)dx
1
2n
is equal to
(A) 3n S
(C) 6n S
87.
(B) 2n S
(D) None of these
1
f(x)dx is equal to
1
8
13
S
27
43
(C)
S
216
88. The value of S is
(A) 5
(C) 1
(A)
(B)
7
S
8
(D) None of these
(B) 2
(D) None of these
Comprehension - 2
Let f be a continuous real function such that
f(x – 1) + f(x + 1) x + f(x) for all x.
Let g(x) = f(x) – x. Then g(x – 1) + x – 1 + g(x + 1) + (x
+ 1) x + g(x) + x
g(x + 1) + g(x – 1) g(x).
89. g(x) satisfies the inequality
(A) g(x + 2) g(x + 3) + g(x + 1)
(B) g(x + 1) g(x + 2) – g(x)
(C) g(x + 3) + g(x)0
(D) None of these
90.
a 6
g(x) dx is equal to
a
(A)
a 3
(B)
a 3
a 3
(C)
a
a
a
(g(x) g(x 3))dx
(g(x) g(x 3))dx
g(x 3)dx
(D) None of these
25
91. The minimum value of f(x)dx is
1
(A) 12
(C) 1248
(B) 312
(D) None of these
Comprehension - 3
Let f : R R be a differentiable function such
that f(x) = x2 +
x
e f(x t)dt .
0
t
2.177
DEFINITE INTEGRATION
92. f(x) increases for
(A) x > 1
(B) x < – 2
(C) x > 2
(D) None of these
93. y = f(x) is
(A) Injective but not surjective
(B) Surjective but not injective
(C) Bijective
(D) Neither injective nor surjective
94. The value of
1
0
f(x) dx is
(A) 1/4
(C) 5/12
(B) – 1/12
(D) 12/7
Comprehension - 4
Let f : R R be a continuous function with
1
0
f(x) f(x) dx = 0 and
1
0
97.
e 2 e 2
2
(D) 2
Assertion (a) and Reason (R)
(A) Both A and R are true and R is the correct
explanation of A.
(B) Both A and R are true but R is not the correct
explanation of A.
(C) A is true, R is false.
(D) A is false, R is true.
101. Assertion (a) : f : [0, ] R is continuous and
f(x)sin xdx 0 then equation f(x) = 0 has at
0
f(x) f(x) dx is equal to
0
486
5
243
(C)
5
486
25
243
(D)
35
(B)
I=
Comprehension - 5
dx
sec 2 xdx
I
=
0 cos2 x 3sin 2 x
1 3 tan 2 x
(B) 10 3
(A) 9 3
(C) 11
(D) 12
100. If f is a real valued derivable function satisfying
x f(x)
f
with f(1) = 2. Then the value of the
y f(y)
a
b
f(x) d (n x) is equal to
1 tan 1 3 tan x
I=
0 = 0
3
2
Let C be a curve defined by y = e a bx . The curve
C passes through the point (P(1, 1) and the slope
of the tangent at P is (–2). Also C1 and C2 are the
circles (x – a)2 + (y – b)2 = 3, (x – 6)2 + (y – 11)2
= 27 respectively.
98. The value of a2 + b2 is equal to
(A) 2
(B) 8
(C) 18
(D) 32
99. The length of the shortest line segment AB which
is tangent to C1 at A and to C2 at B is
integral
e 2 e 2
2
(B)
least one root between (0, ).
Reason (R) : f(x) is continuous and f()f() < 0,
< then f(x) = 0 has atleast one root between
(, ).
dx
102. Assertion (a) : I = 0
can be
2
cos x 3sin 2 x
evaluated by substitution cot x = t.
Reason (R) : by substitution tanx = t,
4
(A)
(C)
f(x)2 f(x) dx = 18.
95. The value of f(1)3 – f(0)3 is
(A) 0
(B) 6
(C) 18
(D) 54
96. The value of f(0) is
(A) 2
(B) 3
(C) –3
(D) 0
1
(A) 0
103. Assertion (a) :
1
tan x
x dx 1 (where [.]
0
denotes greatest integer function)
Reason (R) : For x > 0 and wherever tan x is
defined, tanx > x.
104. Let f(x) is continuous and positive for x [a,b],
g(x) is continuous for x [a,b] and
b
b
a
a
| g(x) | dx g(x)dx, then
Assertion (a) : The value of
b
a
f(x) g(x) dx can
be zero.
Reason (R) : Equation g(x) = 0 has atleast one
root in x (a, b).
105. Assertion (a) : The value of
1
+ sin(x 2 12x 33) zero.
2
5
sin(x 3)dx
4
2
2.178
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Reason (R):
a
f(x)dx 0 if f(x) is an odd
a
function.
106. Assertion (a) : The value of
0 x 1
2
3 sec 1 x dx
2
is
4 5
– ln |2 3 – 3|
3 3 3
Reason (R) : The function F(x) = x sec –1 x
– ln |x +
x 2 1 | + c is the indefinite integral of
sec–1 x for all |x| > 1 and
2
2
3 sec 1 xdx
=F
2
3
– F(–2).
d2f
dx 2
= a(x) then
a(t)dt dx = f(1) – f(0)
x
0
0
b
a
f (x)dx = f(b) – f(a).
108. Assertion (a) : lim (1 – t)
x 1
tn
0
0
0
x
f(t)dt . Integrating the last
f(x) M f(x)
0
f(x)dx f (x) dx – f (0) f(x)dx
M f(x)dx
2
1
–M
n 1
0
n 1
3
2
0
0
2
0
1 sin x x 2
3 | x |
1
Reason (R) : Since
n ln t
1 sin x x 2
that 1
3 | x |
Column - I
e x cos x 1 x
(A) The function f (x) =
is not defined at x = 0.
sin x 2
The value of f (0) so that f is continuous at x = 0 is
1
dx
(B) The value of the definite integral 0
x3x
equals a + b ln 2 where a, b N then (a + b) equals
Given e
(D) Let an =
n
n sec 2 tan
0
1/ n
1/ n 1
e
d = 1 then the value of tan (n) is equal to
tan 1 (nx)dx and bn =
an
then nlim
has the value equal to
b
n
1
dx is same as
0
1
1 e
1
1 2x 2
(C)
2
Reason (R) : Since – M f(x) f(x) . f(x) Mf(x), x
[0, 1] by integration and then multiplying by
x
1
1
f(x) we get –Mf(x) f(t)dt f3(x) – f2(0)
0
2
2
=
sin x
is an odd function so
3 | x |
1 2x 2
3 | x | dx
0
MATCH THE COLUMNS FOR JEE ADVANCED
111.
1
3 | x | dx
n
dx
= – ln (e–x + 1) | 0 = ln 2
0 1 ex
1
2
110. Assertion (a) :
1 t = ln 2
Reason (R) : xlim
(– ln t)
1
3
1
Reason (R) : The fundamental theorem of integral
calculus that
1
f (x) dx f (0) f (x) dx M f (x)dx
inequality on [0, 1], it follows that
107. Assertion (a) : If
1
109. Assertion (a) : Let f : [0, 1] [0, ) be a
continuously differentiable function with
M = max | f(x) | then
1/ n
1/ n 1
sin 1 (nx) dx
Column-II
(P) – 1
(Q) 0
(R) 1/2
(S) 1
DEFINITE INTEGRATION
112. Column-I
Column-II
n
t
2 1
n 1
dt
0
(A) nlim
n 1
is equal to
(B) Let f(x) be a function satisfying f (x) = f(x) with
f(0) = 1 and g be the function satisfying f(x) + g(x).= x2
Then the value of the integral
1
(C)
(D)
lim 1
0
(Q) e 2
1
0
x
k
(P) e – 1 e2 – 3
2
2
f(x) g(x) dx is
e e (1 + xex) is equal to
(R) e2 – 1
(1 + sin2x)1/x dx is equal to
k 0 k 0
(S) ee
113. Column-I
Column-II
(A) If f(x) is an integrable function for x , and
6 3
I1 =
/3
/6
sec2 f(2sin2)d and I2 =
/3
/6
(P) 3
2
cosec2
f(2sin2)d then I1/I2
(B)
If f(x + 1) = f(3 + x) for x, and the value of
ab
a
f(x) dx
(Q) 1
is indepedent of a then the value of b can be
(C)
2.179
tan 1 [x 2 ]
(where [·] denotes
tan 1 [x 2 ] tan 1 [25 x 2 10x]
the greatest integer function) is
4
The value 1
(D) If I =
2
0
x x x ... dx (where x > 0) then [I] is equal to
(where [·] denotes the greatest integer function)
114. Column-I
(A) If f(x) and f'(x) are continuous functions on (a, b)
lim f (x) , lim f (x) an d f'(x ) + f 2 (x) –1
x a
(R) 2
(S) 4
Column-II
(P) –2
x b
(B)
ba
x (a, b), then minimum value of
If f(x) is differentiable function such that f(x) + f'(x) 1, x R
1
and f(0) = 0, then the largest possible value of f(1) + is
e
(C)
f(x) =
x
e dt , h(x) = f(1 + g(x)), g(x) 0, x > 0,
t2
1
(Q) –1
(R) 0
h'(1) = e and g'(1) = 1,then the possible value g(1) can take is
(D) If g : [0, ) and is defined as g(x) =
Then the value of g(g(x)) is
x
e dt and f(x) g(x), x 0.
1
t2
(S) 1
2.180
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
115. Column I
lim
(A)
x
Column II
ln x x dt
is
x e3 ln t
4
(P) 0
2
lim (e x 1 e x 1 ) is
(B)
(Q)
x
lim ( 1)n sin( n 2 0.5n 1)sin
(C)
n
(n 1)
is where n N
4n
(R) 1
x
tan
1
x 1
dx
2
(D) The value of the integral 0
is
1 1 2x 2x
tan
2
1
1.
Prove that
(i)
(ii)
2.
1/ 2
1/ 2
1/ 2
lim n3 / 2 n 3 / 2 n 3 / 2 ......
n n
(n 3)
(n 6)
n
lim
n
1
(x x
k 1
n
C=
S=
4.
k 1
)dx
1
2
1
(x x
)dx
(ii) nlim
0
2
k 1
When C and S are defined by
3.
k
0
1
0
2
2
0
2
2
2
2
prove that C = a (a b) , S = b(a b) (a, b > 0).
Prove that
(i)
/2
0
sin 2 cos
d
(sin cos )2
dx
n –1
– a n –1
2 (a b )
5.
3
( 2 15)
4
x
x dx
f (sec tan )
0
2
2 sin x
2 3
Prove that
dx
f (cosecx) sin x .
=
6.
0
Prove, by the substitution (1 + x4) = (1 + x2) cos
, or otherwise, that
1 1 x2
dx
1 x (1 x ) 4 2 .
2
0
sin 2 d
(x a)2 (x b)2
(iv) 0 ( x x ) cos 2x dx
2
a cos b sin
= b
a cos b sin ,
2
a x n–1[(n – 2)x 2 (n – 1)(a b) x nab]
cos2 d
0
cos2 sin
d
(sin cos )2
(iii) b
1/(k 1)
1/ k
/2
x sin 3 x
1
dx 2
0 1 cos2 x
2
(ii)
n1 / 2
1
{n 3(n 1)}3 / 2 3
Verify the following and give a geometric
interpretation using the concept of area bounded
by graphs.
(i)
(S) Non existent
=
1
4
r2
1 1
......
ln 2
lim 3
.....
3
3
3
n n
2n 3
8 n
r n
1
2
7.
(i)
4
Prove that
cos d
a cos b sin 0
0
2
2
2
2
(ii) Prove that, when a and b are positive,
sin d
is equal to
2
2
0 a cos b 2 sin 2
2
b (a 2 b 2 )
tan 1
(a 2 b 2 )
,
b
2.181
DEFINITE INTEGRATION
1
n
b (b 2 a 2 )
b (b 2 a 2 ) b (b 2 a 2 )
according as a > b, a < b.
Prove that, when b = a, the integral is a
continuous function of at = 1.
8.
dx
Prove that 0 2
and by differentiation
2
4a
a x
a
dx
2
w.r.t. a, prove also that 0 2
.
(a x 2 )2
8a 3
Show that , when a > 0,
9.
ln x
0
2
a
ln a
(i)
a x dx 2 a
(ii)
(a x ) dx 16a (3 ln a 4)
2
0
10. Calculate
ln x
2
2 3
5
b
xdx, where 0 < a < b, by dividing
a
(a,b) into n parts by the points of division a, ar,
ar2, ...., arn–1, arn, where rn = b/a. Apply the same
method to find the more general integral
b
x dx .
m
a
11. Prove that
1 2 cos x
1
(i)
0 (2 cos x)2 2
a
dx
1
(ii) 0
2
2
4
x a x
(1 x 2 )
dx
3
2
0 1 x sin
1
(iii)
4 cos2
2
1
16(x 1)
(iv)
dx =
0 x 4 2x 3 4x 4
12. Find a function g(x) continuous in (0, ) and
positive in (0, ) satisfying g(0) = 0 and
1
2
2 x
g(t) dt .
0
x 0
13. Consider the function
x
g 2 (t)dt
1
x [x] , if x I
2
f(x) =
, where
0,
if x I
[x] denotes the greatest integer function and I is
the set of integers. If g(x) = max {x2, f(x)| x |},
– 10 x 10, then find the value of
14. Let f : R R be defined by f(x) =
3
2
g(x)dx .
2
1
1 | t x | 1
then find the value of
dt,
3
f(t)dt .
1
15. Let f : R+ R be a differentiable function with
f(1) = 3 and satisfying
xy
x
y
1
1
1
f(t)dt = y f(t)dt + x f(t)dt , x, y R+.
Find f(x).
16. Given a positive integer p. A step function s is
defined on the interval [0, p] as follows :
s(x) = (–1)nn if x lies in the interval n x < n + 1,
where n = 0, 1, 2, ..., p – 1; s(p) = 0.
Let f(p) =
p
s(x)dx .
0
(A) Calculate f(3), f(4), and f(f(3)).
(B) For what value (or values) of p is |f(p)| = 7 ?
b
c
1
t
f(x)dx =
f(x)dx
17. Prove that
ba a
ca a
1 t b
f(x)dx , a < c < b, 0 < t < 1.
+
b c c
18. Given two functions f and g, integrable on
every interval and having the following
properties :
f is odd, g is even, f(5) = 7, f(0) = 0, g(x) = f(x + 5),
x
g(t)dt for all x. Prove that
f(x) =
0
(A) f(x – 5) = –g(x) for all x;
5
(B)
f(t)dt = 7;
(C)
f(t)dt = g(0) – g(x).
0
x
0
19. Prove that, as n ,
1 n cos x
(i)
0 (1 nx)2 dx 1
1 n cos x
dx 0 .
(ii)
0 1 n2x
20. (A) If
I n (x)
=
x
t (t a )
n
2 1/2
2
0
,
use
integration by parts to show that nIn(x)
= xn–1 x 2 a 2 – (n – 1)a2 In – 2(x) if n 2.
(B)
Use part (a) to show that
= 168/5 – 40 5 /3.
2
x (x2 + 5)–1 2 dt,
0
5
2.182
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
21. A function f, continuous on the positive real
axis, has the property that for all choices of
xy
f(t)dt is
x > 0 and y > 0, the integral
x
independent of x (and therefore depends only
on y). If f(2) = 2, compute the value of the
x
f(t)dt for all x > 0.
integral A(x) =
1
1
et
(t 1) dt. Express the values of the
22. Let A =
et
dt.
a 1 t a 1
2
te t
(B)
dt.
0 t2 1
a
1
et
1
(D) e t ln (1 + t) dt. .
0 (t 1)2 dx .
0
23. Find the number a, 0 a 2 that maximizes the
(C)
1
function f(a) =
2
sin x sin(x a) dx
dx
0
2
dx
2
n 1
1
sin dx
2
what values of x is the integral a discontinuous
function of ?
27. Prove that if m 1 and
/2
0
Jm,n =
/2
0
sin m x cos nxdx,
3
1
34. Find the average value of the function
cos( / x)
(i) f(x) =
over the interval [1, 3]
x2
2
(ii) f ( x) x
on the interval [0, 2].
e 1
35. Determine a pair of numbers a and b for which
1
(ax b) (x2 + 3x + 2)–2 dx = 3/2.
0
sin m x sin nxdx,
then (m + n) Im, n = sin n/2 – mJm–1, n–1 and express
Im, n in terms of Im–2, n–2 when m 2.
28. Show that the value of the integral
2
a
1
1 2x cos x . For
1
1
f (x) cos nxdx (n = 0, 1, 2, ....., N)
x 1 x dx < 4 is satisfied.
a 2
1 e –ax
dx .
25. Evaluate the integral 0
xe x
Im,n =
a0 N
(a n cos nx b n sin nx) on R
2 n 1
Show that
f(x) =
1
f (x) sin nxdx (n = 1,2, ....., N)
33. Find all the values of a for which the inequality
26. Evaluate the integral
is equal to 2 if
bn =
.
1
(1 – 2x 2 )
– 1 1 and to 2/ if || > 1.
32. Suppose that
(x a)
dx
1
31. Prove that 1
an
x a 2 a , evaluate the integral
2.4.6...2n
.
3.5.7...(2n 1)
30. If fave [a, b] denotes the average value of f on the
interval [a, b] and a < c < b, show that
ca
bc
f [a, c] +
f [c, b]
favg[a, b] =
b a avg
b a avg
0
24. Proceeding from the equation
0
n
n(n 1) n(n 1)(n 2)
...
1.3
1.3.5
1.3.5.7
0
following integrals in terms of A :
(A)
1
375x (x2 + 1)–4 dx is 2n for some integer n.
5
0
36. Prove that
dx is equal to or to 0
sin x
according as n is odd or even.
37. Prove that
0
0
prove that if n be a positive integer,
dx
(i)
1
29. By considering the value of (1 x 2 )n dx ,
sin nx
(ii)
x
0
4
1x
3
0
2
a x
sin 1 x
1 x
2
2
dx
2 2
,
3a 4
7
9,
DEFINITE INTEGRATION
ln x
1
x dx (n – 1) , n 1 ,
(iii)
n
1
2
dx
1
(1 ln 2) .
1 (x 1) 2 (x 2 1)
4
38. Prove that
(iv)
In =
/2
0
=
x sin n x dx
149
.
225
40. Show that if n is a positive integer, then
2 cos(n 1)x cos nx
dx 2
0
1 cos x
2
nx
2 sin
2 dx 2n .
and deduce that
x
0
sin
2
x sin nx
dx,
sin x
2
then In = (n – )2 [(–1)n–1 – 1] + In–2.
1
Deduce that I5 =
1036
2
.
, and I6 =
225
2
42. The function f(x) = x is continuous on
[0, 4] and therefore integrable on this interval.
Evaluate
4
0
xdx using subin terva ls of
unequal length given by the partitions
0 < 4(1)2/n 2 < 4(2)2/n 2 < ... < 4(n – 1)2/n 2 < 4
and the right endpoint of the kth subinterval.
43. Show that for any positive integer n
(A)
(B)
/6
0
/2
/
3
b
b
a
a
dx
f (x)dx f (x)
reaches the least value only if f(x) is constant on
this interval.
45. If |x| < 1, prove that
1 1 1 1
2 3 4 5
Hence show that 1 ... 0.438...
46. Assuming that the function g defined by
g(x) = 2x is integrable over the interval [0, 2], use
1
39. Evaluate 0 (1 x 2 ) n 1/ 2 dx
41. Show that if In = 0
44. Let the function f(x) be positive on the interval
[a, b]. Prove that the expression
x3 x 4
x6
1
... x tan 1 x ln(1 x 2 ) .
1. 2 3 . 4 5 . 6
2
1
n 1 /2
x sin n 2 xdx
2
n
n 0
and hence deduce that I5 =
2.183
the
partition 0,
1
3
, 1, , 2 to show that
2
2
2
3 2 x dx 5 .
0
47. Find the limit, when n tends to infinity of
1
1
1
1 .
...
n n 1 n 2
4n
48. Let f be a continuous function such that f(0) = 2
and f(x) 3 as x . Find the limit of
(1/b) 0b f(x)dx as (a) b 0, (b) b .
49. Let f(x) = x – [x] –
1
, if x is not an integer, and
2
let f(x) = 0, if x is an integer. ([x] denotes the
greatest integer x.) Define a new function P
as follows :
P(x) =
x
f (t)dt for every real x.
0
(A) Draw the graph of f over the interval
[–3, 3] and prove that f is periodic with
period 1.
1 2
(x – x), if 0 x 1
2
sin n xdx
3. 2 n 1
(B)
Prove that P(x) =
cos n xdx
3. 2 n 1
(C)
and that P is periodic with period 1.
Express P(x) in terms of [x].
2.184
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(D) Determ ine a consta nt c such th at
1
(P(t) c)dt 0 .
50. Given an even function f, defined everywhere,
periodic with period 2, and integrable on every
(E)
For the constant c of part(d), let Q(x)
(P(t) c)dt . Prove that Q is periodic with
0
period 1 and that Q(x) =
x3 x 2 x
, if 0 x 1.
6
4 12
1.
Evaluate the following limits :
n
n
(i) nlim
(n 1) (2n 1)
(n 2) 2(2n 2)
n
..... up to n terms
(n 3) 3(2n 3)
2
3
1 3
2sin3
3sin3
....
sin
(ii) nlim
2
4n
4n
4n
n
n
.... n sin 3
4n
2
3
n
3
3
3
3
sin 4n 2sin 4n 3sin 4n ... n sin 4n
2. Prove that, when a and b are positive,
/2
b a cos
0 b2 2ab cos a 2 d is equal to
1 tan 1 a
1 1 b
, or tan
b
a
2b b
b
according as b is greater than or less than a
3. Show how it follows from the equality
x dx
x ln x that the sum of n terms of the
1
1 1
harmonic series 1 + + + ......... lies between
2 3
4.
ln(n + 1) and 1 + ln n.
Prove that
5.
sin k sin(k 1 / 2)
, where [.]
sin(1 / 2)
denotes the greatest integer function.
When a > 0 and b > 0, prove that
k
2x
sin dx 2 ·
0
ln(1 b2 x 2 )
ab
0 1 a 2 x 2 dx a ln a .
0
(A) Prove that g is odd and that
g(x + 2) – g(x) = g(2).
(B) Compute g(2) and g(5) in terms of A.
(C) For what value of A will g be periodic
with period 2 ?
x
=
x
f(t)dt , and let A = g(1).
interval. Let g(x) =
0
2
6.
Prove that, if I = e x cos 2xydx
7.
dI
then
= –2yI, and assuming the formula
dy
1
y 2
1
x2
0 e dx 2 prove that I = 2 e .
Prove that, if y1 > 0, y1 y y2, and
0
sin xy
(y) = 0 x(a 2 x 2 ) dx
1
then "(y) – a2(y) = –
2
Show further that, when y > 0 and a > 0,
ay
2
1
(y) = (1 e ) / a
2
n
8.
If nh = 1 always, then show that lim [1 +
n
1
dx.
(rh)2k]1/r = e where =
2k 0
x
Show that for a differentiable function f(x)
9.
r 1
1 ln(1 x)
n
1
f (x) [x] x 2 dx
0
n
1
1
f (r) ,
f(0) + f(n) –
2
2
0
r 0
where [.] denotes the greatest integer function
and n N.
1 ln x
ln x
dx –
dx
10. Prove that
0 1 x2
1 1 x2
ln x
dx 0 and deduce that if a > 0
and
1 1 x2
ln x
dx ln a .
then
0 a2 x2
2a
=
n
f(x)dx +
2.185
DEFINITE INTEGRATION
11. A particle moves along a straight line. Its
position at time t is f(t). When 0 t 1, the
position is given by the integral
t 1 2 sin x cos x
f(t) = 0
dx.
1 x2
For t 1, the particle moves with constant
acceleration (the acceleration it acquires at time
t = 1). Compute the following :
(A)
Its acceleration at time t = 2;
(B)
Its velocity when t = 1
(C)
Its velocity when t > 1
(D)
The difference f(t) – f(1) when t > 1.
12. Assume that the function f is defined for all x
and has a continuous derivative. Assume that
f(0) = 0 and that 0 < f '(x) 1.
2
1
1
Prove that f(x) dx [f(x)]3 dx
0
0
13. Let p(x) be a polynomial of degree at most 3.
(A) Show that there is a number c between 0
1
and 1 such that p(x)dx p(c) p( c) .
1
(B) Show that there is number c such that
1/2
1
1/2 p(x)dx 3 [p(c) p(0) p(c)] .
dx
14. (A) Let G(a) = 0 (1 x a )(1 x 2 ) . Evaluate
G(0), G(1), G(2).
(B) Show, using the substitution x = 1/y, that
x a dx
.
G(a) = 0
(1 xa )(1 x 2 )
(C) From (b), show that G(a) = /4, independent of a.
15. If f1(x) =
x
x
0
0
f(t)dx, f2(x) = f (t)dt, ....,
1
17. A function F is defined by the indefinite
f (t)dt, then prove that
0
k 1
(B) Prove that
3
1
3
3 1/2
dt , given
that (4 t 3 )1/2 dt = 11.35. Leave the answer
1
in terms of
3 and
31 .
1
x et
x e at
x
t dt, t dt, e dt .
2
1
1
1/t
1
18. The strength of an alternating current values
according to the law
2 t
i = i0sin
T
where i0 is the amplitude, t is time, T is the period
and is the initial phase, Find the mean value of
the square of the current strength :
(A) on the interval of time [0, T],
(B) on the interval [0, T/2] (the period of the
function is i2(t))
(C) on the arbitrary interval [0, t0] and the limit
of that mean value as t0
19. Evaluate the following integrals :
(i)
(iii)
tan 1 (x 1)dx
(x 1)
1
3
x2 2
1
3
(ii)
4
1
1 x x3 dx
1
2
ln 1 x 1 x
x x 1 dx .
2
20. (A) Show that 0 <
1 t 4 (1 t) 4
1 t2
0
and that
1 t 4 (1 t) 4
1 t2
0
1
dt
dt
22
.
7
t (1 t) dt then apply the
4
4
0
results of (a) to conclude that
16. Evaluate the integral t (4 t )
3
et
x
t a dt = e–a[F(x) + a) – F(1 + a)].
(C) In a similar way, express the following
integrals in terms of F :
(B) Evaluate
x
1
f(t) (x – t)k–1 dt.
0
(k 1)!
fk(x) =
1
(A) For what values of x is it true that ln x F(x)?
x
fk(x) =
x et
t dt, if x > 0.
integral F(x) =
22
1
22
1
.
7 630
7 1260
6
21. Use the identity 1 + x = (1 + x2) (1 – x2 + x4) to
prove that for a > 0, we have
a dx
1
a3 a 5
a3 a 5
a
a
3 5 0 1 x2
3 5
1 a6
2.186
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Taking a = 1/10, calculate the approximate
value of the integral.
x
2
22. Prove that y = 8 t 2 e 2t sin 2(x t)dt is a
0
solution of the differential equation
d2y
2
4 y 16x 2 e 2 x .
2
dx
Given that every solution of the differential
equation must be of the form
e–2x2 + A cos2x + B sin 2x, where A and B are
constants, prove that
x
2
2 2 t
t e sin 2(x t)dt
0
1 2 x2
(e
cos 2 x).
8
1 1
(1 t 2 ) n cos tdt
n!2n 0
and n is a positive integrer, prove that
(i) 2In+1 = (2n + 1)In – In–1,
dI n
I n 1
(ii)
d
and deduce that In() is a solution of the
23. When In () =
differential equation
d2y
dy
2n 2
y0.
d
d
2
24. Prove that
cos 2 d
3 ,
2
2
2
2
2
(a cos b sin )
4a b
/2
0
/2
sin d
2
2
2
(a cos b sin )
4ab3
2
2
0
/2
0
2
d
1
1
2 2
2
2
2
2
2
4ab a
(a cos b sin )
b
25. Prove that , when 0 x < 1,
/2
0
1 x
2 x x3
tan 1
tan d
....
1 x
8 1 32
26. Prove, when a, b, are constant, that
x
x
1
1
f (t)ea (x t ) dt
f (t)eb(x t ) dt
ab
ba
is a solution of the differential equation
y=
d2y
dx
2
(a b )
dy
aby f (x )
dx
27. (A)Give a geometric argument to show that
x 1 1
1
1
dt , x > 0
x
x 1
t
x
(B) Use the result in part (a) to prove that
1
1
1
ln1 , x > 0
x 1
x x
(C) Use the result in part (b) to prove that
x
1
ex/(x + 1) < 1 < e, x > 0
x
x
1
and hence that xlim
1 x = e.
(D)Use the inequality in part (c) to prove that
x
1 1 e 1 1
x
x
x 1
,x>0
28. Prove that, for any fixed value of ,
dx
1
1
0 x 2 a 2 a tan a
dx
1
(x a ) 2a tan
0
2
2 2
3
1
2 2
a 2a ( a 2 )
dx
and deduce that 0 (x 2 a 2 ) 2 4a 3 .
29. Show from graphical considerations that if f(x)
steadily diminishes, as x inereases from 0 to ,
the series f(1) + f(2) + f(3) + ...... is convergent,
and that its sum lies between I and
I + f(1), provided the integral I f ( x) dx , be
1
finite.
Apply this to the series
1
1
1
.... .
(n 1)2
(n 2)2
(n 3)2
30. Show that
(A)cosx – (1 – x2)/2! + x4/4! – x6/6! x8/8!
(B) sinx – (x – x3)/3! + x5/5! – x7/7! x9/9!
31. Assume that the function f defined by f(x) = x2 + 1
is integrable over the interval [0, 1]. Using the
1 2 3 4
partition 0, 5 , 5 , 5 , 5 , 1 , show that
2.187
DEFINITE INTEGRATION
31
25
32.
36
1
f(x) dx 25 .
3
Using the inequality sinx x – x (x 0) and
6
the Schwartz-Bunyakovsky inequality, show that
1.096 <
/2
0
1 1 dx
n 0 1 x 2
Deduce that In as n
37. Given that f(x) is continuous in (a, b), prove that
In
0
x sin xdx < 1.111
x
1
nf (x)
nf (x)
dx , I2 =
dx
I1 = 0
1/ n 1 n 2 x 2
1 n2 x2
Use the mean value theorem for integrals to prove
1/ n
I1 = f( n)tan –1 n , I2 = f( n){tan –1n – tan –1 n }.
Deduce that, as n , I2 and
nf (x)
1
dx f (0)
I = 0
2 2
2
1 n x
34. When f(x) is bounded and is strictly decreasing
1
in (0, 1) prove that
11/ n
0
f (x)dx
1 2
n 1
n–1 f f ...... f
n
n n
1
f (x)dx .
1/ n
4
(b a)2
b
a
f(x)dx.
Verify the theorem for the function sin2x in (0,).
36. When In =
a
(ln x)
x (ln x) x ln ln x
,
,
ln ln x
(ln x)
,
x ln ln ln x
(ln ln x)
according to their order of magnitude for large x.
39.
Show that
dt
x
1
1
1 t x – 5 x 9 x – ... if – 1
5
9
4
0
x 1. Also deduce that
1 – 1 1 ... =
5 9
40. If (x) =
2ln( 2 1)
4 2
.
x
(1 + sec t)ln sec t dt
3
0
lnsec x [x + ln(sec x + tan x)]
then
(x) is even,
(i)
(ii) (x)
3
when x through values less
2
2
2
41. (A)
If is continuous and nonzero on
[a, b] , and if there is a constant m > 0 such
that (t) m for all t in [a, b], prove that
than
35. The function f(x) is differentiable in a x b, and
f(a) = f(b) = 0; show, by dividing the range (a,b)
into two equal parts and applying the mean-value
theorem to each part that there is at least one
point in (a,b) for which
|f()| >
b
f (x) cos nxdx 0 .
n
38. Arrange the functions
33. The function f(x) is continuous in (0, 1) and
that for some n in (1/ n , 1).
1
as n , In
1 1 sin nx
dx ,
n 0 1 x 2
1 1 sin nx
dx
n 0 1 x 2
and since sinnx 1, we have
By property, I1
b
4
sin (t)dt m .
a
2
for all x > a.
a
42. A sequence of polynomia ls (called the
Bernoulli polynomials) is defined inductively
as follows :
x
(B) If a > 0, show that sin(t )dt
2
a
1
P0(x) = 1; Pn(x) = nPn – 1(x) and Pn ( x) dx = 0,
0
if n 1,
(A) Determine explicit formulas for P1(x),
P2(x),...., P5(x).
(B) Prove, by induction, that P n (x) is a
polynomial in x of degree n.
2.188
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
(C) Prove taht Pn(0) = Pn(1) if n 2.
(D) Prove that Pn(x + 1) – Pn(x) = nxn – 1 if n 1.
(E) Prove that for n 2 we have
k 1
Pn 1 (k) Pn 1 (0)
. r 0 Pn (x)dx
n 1
r 1
(F) Prove that Pn(1 – x) = (–1)n Pn (x) if n 1.
n
44. Let f(x) =
x
(1 t )
3 1/2
0
2n
0
0
n
xdx 22 1 .
ln x e t
A.
t
b
b
p (x) p (x)dx p (x) p (x)dx
3
1
1
a
48. Given two functions f and g whose derivatives
f and g satisfy the equations
f(x) = g(x), g(x) = –f(x), f(0) = 0, g(0) = 1
for every x in some open interval J containing
0. For example, these equations are satisfies
when f(x) = sin x and g(x) = cos x.
(A) Prove that f2(x) + g2(x) = 1 for every x in J.
(B) Let F and G be another pair of functions
satisfying the given conditions. Prove that
F(x) = f(x) and G(x) = g(x) for every x in J.
49. Since sin x < x for x > 0, it follows that
x2
cos x 1.
0
0
2!
In a similar fashion show that x – (x3/3!) sinx x.
50. Prove by repeated differentiation of the identity
1
= 1 + x + x2 + x2 + ... , where |x| < 1, that, if m
1 x
be a positive integer (1–x)–m = 1 + mx
+
Fill in the blanks :
x
sin t dt
Then
sec x
cos x
sec 2 x cot x cosec x
2
2
cosec 2 x
cos2 x
.
2.
x
or, 1
t dt
m(m 1) 2 m(m 1)(m 2) 3
x
x ...
1.2
1.2.3
/2
0
1.
2
p 4 (x) p1 (x)dx .
dt = Li(x) and find the value of b.
f(x) = cos x cos x
1
cos2 x
a
b
46. A function, called the integral logarithm and
denoted by Li, is defined as follows :
x dt
Li(x) =
if x 2.
2 ln t
x dt
x
2
(A)
Prove that Li(x) =
.
2
2
ln x
ln t ln 2
(B)
Show that there is a constant b such
that b
e2x
x 2 3x 2
47. Assume that integrable functions p1(x), p2(x),
p3(x), p4(x) are given on the interval [a, b], the
function p1(x) is non-negative, and the function
p2(x), p3(x) satisfy the inequality p3(x) p2(x)
p4(x). Prove that
x > 3. Show that f(x) =
a
(A)Show that f is strictly increasing on the
nonnegative real axis.
(B) Let g denote the inverse of f. Show that
the second derivative of g is proportional to
g2 [that is, g(y) = cg2(y) for each y in the
dom ain of g] and fi nd th e con stant of
proportionality.
45. Let n be a fixed integer. Let f : R R be given
x if x 0
by g(x) = n
n
n 2
n 1
n 1
2 if 2 2 x 2 2
2n
1
ln 2.
2
Let f(x) = e4 Li(e2x – 4) – e2 Li(e2x – 2) if
(D)
dt if x 0.
Prove that f (x)dx
e2t
dt in terms of the
( t 1)
integral logarithm, where c = 1 +
k
1
(G) Prove that P2n+ 1(0) = 0 and P2n–1 = 0 if n 1.
2
n
1
n
(1) k
43. Prove that
k k m 1
k 0
n
m
1
( 1) k
=
k
n 1
k
k 0
x
Express 0
(C)
f (x) dx ........
The integral
1.5
0
[IIT - 1987]
2
[x ] dx, where [ ] denotes the
greatest integer function, equals.... [IIT - 1988]
2.189
DEFINITE INTEGRATION
2
3.
The value of 2 |1 – x | dx is......... [IIT - 1989]
4.
3 / 4
The value of / 4 1 sin d is... [IIT - 1993]
5.
The value of 2
6.
x
3
dx is.. [IIT - 1994]
5x x
If for nonzero x, af(x) + bf(1/x) = 1/x – 5 where
2
a b, then f (x)dx ...............
2
xsin 2n x
dx =............
sin x cos2n x
[IIT - 1996]
For n> 0,
8.
The value of
0
2n
e37 sin( ln x)
x
1
dx
cot x tan x
is
[IIT - 1983]
(A) /4
(B) /2
(C)
(D) None of these
14. For any integer n the integral
e
T he va lue of in tegra l
[g(x) – g(–x)] dx is
(A)
(C) –1
2
4 2e sin x
d
e sin x
F(x) =
, x > 0. If
dx
1
dx
x
x
= F(k) – F(1) then one of the possible values of k
is...............
[IIT - 1997]
10. The value of the integral
f (x)
[{f (x) f (2a x)}] dx is equal to a.
17.
0
1
0
(1 + cos x) (ax + bx + c) dx =
8
2
2
2
0
8
(1+ cos x)
(ax + bx + c) dx. Then the quadratic equation
2
ax + bx + c = 0 has
[IIT - 1981]
(A) No root in (0, 2)
(B) At least one root in (0, 2)
[f(x) f( x)]
[IIT - 1990]
5
2
2
2
(A) 0
(C) 2 - 3
19.
[IIT - 1992]
(B) – 3/3
(D) 7/2 – 23
1
| (1 x) | dx equals
[IIT - 1992]
1
(A) – 2
(C) 2
20. Integral
(A) 0
(C) /4
[IIT - 1992]
(B) 0
(D) 2
(1 x ) sin x cos x dx is
2
0
1
/2
(cosax sin bx) dx where a and b are
18. The value of
11. The value of the definite integral (1 e x ) dx is
[IIT - 1981]
(A) –1
(B) 2
(C) 1+ e–1
(D) None
12. Let a, b, c be non-zero real numbers such that
/2
(B) 1
(D) 0
integers is equal to
(A) –
(C)
[IIT - 1988]
C. Multiple Choice Questions with ONE
correct answer :
x2
dx is
[IIT - 1992]
3 x2 4
(A) 2 – loge (15/7)
(B) 2 + loge (15/7)
(C) 2 + 4 loge 3 – 4 loge 7 + 4 loge 5
(D) 2 – tan–1 (15/7)
16. The value of
B. True/False :
2
cos3 (2n + 1)x dx has the value
[IIT - 1985]
(A)
(B) 1
(C) 0
(D) None
15. f : R R, g : R R are continuous functions
dx is...............
Let
cos2 x
0
[IIT - 1997]
9.
cot x
/2
13. The value of the integral 0
[IIT - 1996]
1
7.
(C) A double root in (0, 2)
(D) Two imaginary roots
2
(B) 0
(D) 4
1
| sin 2x | dx is equal to [IIT - 1992]
0
(B) 3/
(D) 2/
2.190
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
/2
21. The value of 0
(A) 0
(C) /2
22. The value of
dx
is
1 tan 3 x
(B) 1
(D) /4
1 sin x x 2
3 | x |
1
[IIT - 1993]
dx is
(A) 2
(B) –2
(C) 1/2
(D) –1/2
30. If for a real number y, [y] is the greatest integer
less than or equal to y, then the value of the
integral
[IIT - 1994]
sin x
dx
3 | x |
1 sin x x 2
(D) 2
dx
0 3 | x |
(A) –
(C) –/2
1
(B) 2 0
(A) 0
x2
0 3 | x |
(C) 2
1
1
=2
2
23. If f(x) = A sin (x/2) + B, f’
0
[IIT - 1995]
(B) 2/ and 3/
(D) 8/ and 0
24. The value of
2
[2sin x] dx where [ ] represents
0
k
1 k
x.f [x(1 – x)] dx & I2 =
where (2k – 1) > 0, then I1/I2 is
(A) 2
(B) k
(C) 1/2
(D) 1
26. If g(x) =
x
k
1 k
f[x(1 – x)] dx,
[IIT - 1997]
1
0
x
f(t) dt = x + tf(t) dt, then the value of
27. If
f(1) is
[IIT - 1998]
(A) 1/2
(B) 0
(C) 1
(D) –1/2
28. Let f(x) x – [x], for every real number x, where [x]
is the integral part of x. Then
(A) 1
(C) 0
29.
3 /4
/4
1
1
(B) 2
(D) 1/2
dx
is equal to
1 cos x
f(x) dx is
[IIT - 1998]
[IIT - 1999]
e cos x sin x for | x | 2
,
otherwise
2
3
(B) 1
(D) 3
33. The value of the integral
(A) 3/2
(C) 3
34.
th en
[IIT - 2000]
(A) 0
(C) 2
0
x
0
2
cos t dt, then g(x + ) equals
(A) g(x) + g()
(C) g(x)g()
x
f(t) dt , where f is such that
f(x) dx
4
[IIT - 1997]
(B) g(x) – g()
(D) g(x)/g()
[IIT - 1999]
(B) 0
(D) /2
32. If f(x) =
the greatest integer function is
[IIT - 1995]
(A) – 5/3
(B) –
(C) 5/3
(D) – 2
25. Let f be a positive function, It
I1 =
[2sin x] dx is
1
1
f(t) 1 for t [0, 1] and 0 f(t)
for
2
2
t (1, 2]. Then g(2) satisfies the inequality
[IIT - 2000]
(A) –3/2 g(2) < 1/2
(B) 0 g(2) < 2
(C) 3/2 < g(2) 5/2
(D) 2 < g(2) < 4
2 and
f(x) dx = , then the constant A and B are
(A) /2 and /2
(C) 0 and – 4/
3 /2
/2
31. Let g(x) =
2A
1
cos2 x
x
log e x
dx is
e
x
[IIT - 2000]
(B) 5/2
(D) 5
e2
1
1 a dx, a > 0
(A)
(C) /2
[IIT - 2001]
(B) a
(D) 2
35. Let f : (0, ) R and F(x) =
2
= x (1 + x), then f(4) equals
(A) 5/40
(B) 7
(C) 4
(D) 2
36. Let f(x) =
x
1
x
f(t)dt . If F(x )
2
0
[IIT - 2001]
2 t 2 dt. Then the real roots of
2
[IIT - 2002]
the equation x – f’(x) = 0 are
(A) ± 1
(B) 1 / 2
(C) ± 1/2
(D) 0 and 1
DEFINITE INTEGRATION
37. Let T > 0 be a fixed real number. Suppose f
is a continuous function such that for all
x R, f(x + T) = f(x). If I =
value of
33T
3
T
f(x) dx , then the
0
f(2x) dx is
(A) (3/2) I
(C) 3I
[IIT - 2002]
(B) 2I
(D) 6I
1/2
1 x
38. The integral 1/2 [x] n 1 x dx equals
[IIT - 2002]
(A) –1/2
(B) 0
(C) 1
(D) 2 n (1/2)
1
39. If I(m, n) = t m (1 + t) dt, then the expression
n
0
for I(m, n) in terms of I(m + 1, n – 1) is [IIT - 2003]
(A)
n
2n
–
I(m + 1, n – 1)
m 1 m 1
(B)
n
I(m + 1, n – 1)
m 1
(C)
n
2n
+
I(m + 1, n – 1)
m 1 m 1
(D)
m
I (m + 1, n – 1)
n 1
x 2 1
40. If f(x) = 2 e
x
t2
then f (4/25) equals
(A) 2/5
(C) 1
42. The value of the integral
0
3
f(0) = 0, then
[IIT - 2009]
1 1
1 1
(A) f and f
2 2
3 3
1 1
1 1
(B) f and f
2 2
3 3
1 1
1 1
(C) f and f
2 2
3 3
1
1
1 1
(D) f and f
2 2
3 3
45. Let f be a real-valued function defined on the
–x
interval (–1, 1) such that e f(x) = 2 +
x
–1
t 4 1 dt, for all x (–1, 1) and let f be the
–1
inverse function of f. Then (f )' (2) is equal to
[IIT - 2010]
1
(A) 1
(B)
3
1
1
(C)
(D)
2
e
2 5
t
5
[IIT - 2004]
(B) –5/2
(D) 5/2
1
0
(B) /2 – 1
(D) 1
[IIT - 2004]
2
[IIT - 2005]
Ú sin x + sin(ln 6 - x ) dx is
2
[IIT - 2011]
1 3
ln
4 2
3
(C) ln
2
1 3
ln
2 2
1 3
(D) ln
6 2
(A)
(B)
D. One or More than ONE correct :
1 x
1 x dx is
x sin x 2
2
ln 2
47. Let Sn
{x + 3x + 3x + 3 + (x + 1) cos (x + 1) } dx is
equal to
0
ln 3
0
2
x
1 [f '(t)]2 dt f (t)dt , 0 x 1 and
46. The value of I =
t2
x
0
0
41. If f(x) is differentiable and x f(x) dx =
43.
If
[IIT - 2001]
(B) no value of x
(D) (–, 0)
(A) /2 + 1
(C) –1
(A) –4
(B) 0
(C) 4
(D) 6
44. Let f be a non-negative function defined on the
interval [0, 1].
dt, then f(x) increases in
(A) (–2, 2)
(C) (0, )
2.191
Tn
n
n
n kn k and
k 1
2
n 1
2
n
n kn k for n = 1, 2, 3, ..... Then
k 0
2
(A) Sn
3 3
(C) Tn
3 3
2
[IIT - 2008]
(B) Sn
3 3
(D) Tn
3 3
2.192
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
48. Let f(x) be a non-constant twice differentiable
function defined on (–, ) such that f(x) = f(1 – x)
1
and f ' 0 . Then
[IIT - 2008]
4
(A) f”(x) vanishes at least twice on [0, 1]
1
(B) f ' 0
2
1
(C) 1/ 2 f x sin x dx 0
2
1/ 2
(D)
1/ 2
0
1
f (t)esin t dt f (1 t)esin t dt
1/ 2
sin nx
dx , n = 0, 1, 2, .... , then
49. If x
(1
) sin x
[IIT - 2009]
10
(A) In = In+2
10
(C)
I
(B)
2m 0
I
2m 1 10
m 1
(D) In = In+1
m 1
/4
55. Evaluate : 0
3/ 2
|x sin x| dx.[IIT - 1982]
1
sin x cos x
dx
9 16sin 2x
[IIT - 1983]
x sin 1 x
1/ 2
56. Evaluate the following 0
1 x2
dx
[IIT - 1984]
57. Given a function f(x) such that
(i) It is integrable over every interval on the real
line and
(ii) f(t + x) = f(x), for every x and a real t, then
show that the integral
indepedent of a.
58. Evaluate the following:
59. Evaluate 0
at
a
/2
0
f(x) dx is
[IIT - 1984]
x sin x cos x
dx
cos 4 x sin 4 x
[IIT - 1985]
x dx
, < < [IIT - 1986]
1 cos sin x
1
60. Evaluate log[ 1 x 1 x ] dx [IIT - 1988]
0
x 4 (1 x) 4
50. The value(s) of 0
dx is (are)
1 x2
1
[IIT - 2010]
(A)
54. Find the value of
22
7
(B)
2
105
E. Subjective Problems:
1
a
0
0
62. Show that
/2
0
f(sin 2x) sin x dx =
f(cos 2x) cos x dx
Hence prove that
/2
n
–1
1
0
k
x (1 – x)
n– k
dx
= [ Ck (n + 1)] for k = 0, 1,......, n. [IIT - 1981]
53. Show that
0
xf (sin x) dx = 0 f(sin x) dx.
2
[IIT - 1982]
sin 2kx
sin x
2
[IIT - 1990]
integer and t is a parameter independent of x.
Hence show that
0
sin 2kx cot x dx =
(tx + 1 – x) dx, where n is a positive
0
/4
= 2[cos x + cos 3x + ........+cos (2k – 1) x ]
0
n
2
[IIT - 1990]
63. Prove that for any positive integer k,
1 1 .... 1
= log 6
51. Show that : nlim
n 1
n2
6n
[IIT - 1981]
52. Evaluate
a
then show that f (x)g(x)dx = f (x) dx
[IIT - 1989]
71 3
(D)
15
2
(C) 0
61. If f and g are continuous function on
[0, a]satisfying f(x) = f(a – x) and g(x) + g(a – x) = 2
64. If f is a continous function with
x
0
f(t) dt
as |x| , then show that every line y = mx
x
intersects the curve y + f(t) dt = 2
2
0
[IIT - 1991]
2.193
DEFINITE INTEGRATION
xsin (2x)sin ( / 2.cos x)
65. Evaluate
2x –
0
dx
[IIT - 1991]
66. Determine a positive integer n 5, such that
1
0
x
n
e (x – 1) dx = 16 – 6e.
67. Evaluate
[IIT - 1992]
3 (2x 5 x 4 – 2x 3 2x 2 1)
(x 2 1)(x 4 – 1)
2
68. Show that
n v
0
dx
x
function and let F(x) = f (t) dt , x 0 . If for
0
some c > 0, f(x) cF(x) for all x 0, show that f(x)
= 0 for all x 0.
[IIT - 2001]
78. Let f(x) be an even function then prove that
/2
cos v
where n is a positive integer and 0 v.
[IIT - 1994]
1 cos mx
dx. Use mathematical
1 cos x
induction to prove that m = m, m = 0, 1, 2,.........
[IIT -1995]
4
1/ 3
x
–1 2x
dx
70. Evaluate
cos
–(1/ 3 ) 1 – x 4
1 x2
[IIT - 1995]
71. Let a + b = 4, where a < 2 and let g (x) be a
differentiable function, If dg/dx > 0 for all x. Prove
/4
2 0 f (sin 2x) cos x dx.
[IIT - 2003]
79. Let f(x) be a differentiable function defined as
f[0, 4] R show that
(i) 8f’ (a) f(b) = {f(4)2 – f(0)}2; when a, b (0, 4).
f (cos 2x) cos x dx =
0
[IIT - 1993]
|sin x| dx = 2n + 1 -
f(x) + f(1/x) and show that f(e) + f(1/e) = 1/2.
[IIT - 2000]
Here lnt = loget.
77. Let f(x), x 0 be a non-negative continuous
4
f (x) dx= 2 [f( ) + f ( )] 0 < a , < 2.
(ii)
2
2
0
[IIT - 2004]
69. Let m = 0
that
a
0
g(x)dx +
b
g(x)dx increases as (b – a)
0
increases
72. Evaluate
/4
0
n (1 + tan x) dx
73. Determine the value of –
1
1
dx = 2
74. Prove that tan
0
1 x x2
e cos x
75. Integrate
dx
0 e cos x e cos x
[IIT - 1999]
x
76. For x > 0, let f(x) = 1
1
dx
1
(P)
2
1
dx
0
1 x
3
dx
2
Column - II
1
2
log
2
3
(Q)
2
2 log
3
(C)
1 x
(R)
3
(D)
x x 1 (S)
2
1
[IIT - 1998]
2
1
G. Match the Columns
(B)
0
4x
dx
2 cos | x |
3
[IIT - 2004]
[IIT - 2005]
1 x
tan x dx.
/ 3
2 sin 1 cos x 3 cos 1 cos x
sin x dx
2
2
(A)
tan (1 – x + x ) dx.
0
|cos x|
0
[IIT - 1997]
Hence or otherwise, evaluate the integral
1
e
82. Column - I
1
/3
81. Evalutate
[IIT - 1997]
2x(1 sin x)
dx
1 cos 2 x
[IIT - 1997]
1
80. Find the value of
3
2
2
2
dx
1
2
H. Comprehension Based Questions.
Let the definite integral be defined by the
ln t
dt . Find the function
1 t
ba
(f(a) + f(b)). For more
2
accurate result for c (a, b), we can use
b
formula f(x) dx =
a
2.194
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
f(b) f(a)
ba
2f(b) f(a)
(C)
2b a
b
(A)
f(x) dx = f(x) dx + f(x) dx = F(c) so
b
c
a
a
c
ab
, we get
that for c =
2
ba
(f(a) + f(b) + 2f(c)).
4
83.
/2
0
sin x dx
(1 2 )
8
(C)
8 2
84. If
[IIT -2006]
(1 x ) dx
86. The value of 5050
is
(1 x ) dx
1
(1 2 )
4
(D)
4 2
x a
a
(x a)
3
0,
then f(x) is of maximum degree
[IIT - 2006]
(A) 4
(B) 3
(C) 2
(D) 1
85. If f’’(x) < 0 x (a, b) and c is a point such that
a < c < b, and (c, f(c)) is the point lying on the
curve for which F(c) is maximum, then f’(c) is equal
to
[IIT - 2006]
CONCEPT PROBLEMS—A
1.
3.
5.
2. square unit
4.5
e – 1.
(i) 13.5
(iii) 18
(v) 1
(D) 0
I. Integer Answer Type:
a
f (x)dx 2 (f (x) f (a)
lim
x a
2(f(b) f(a))
ba
f(x) dx
(B)
(A)
x
b
(B)
(ii) 12
(iv) e2 / 2
(vi) cos a – cos b
0
1
50 100
50 101
0
[IIT - 2006]
87. For any real number, let [x] denote the largest
integer less than or equal to x. Let f be a real
valued function defined on the interval [–10, 10]
x [x] if [x] is odd,
by f(x)
1 [x] x if [x] is even
2 10
f (x) cos x dx is
Then the value of
10 10
[IIT - 2010]
88. Let f : [1, ) [2, ) be a differentiable function
such that f(1) = 2. if
[IIT - 2011]
x
6 f (t)dt 3x f (x) x 3
1
for all x 1, then the value of f(2) is
b3 a 3
3
2 3/2 3/2
(iii)
(b – a ).
3
15. (i)
(ii) ln b
a
CONCEPT PROBLEMS—B
PRACTICE PROBLEMS—A
1.
3
(A) (1 + )/2
(B) – 4
(C) – 8
25
8.
2
10. The region under the graph of y = tanx from 0 to
/4.
2.
f (x) dx
3. –1
5.
(i) 4
(ii) – 2
(iii) 4.5 – 2
(A)
(B)
(C)
(i) First
(iii) First
(ii) Second
(iv) Second
7.
n
11.
lim
2 4i / n
n i 1 1 (2 4 i / n ) 5
.4
n
12.
218/3
6.
7.
5
1
DEFINITE INTEGRATION
8. (A) True
11. Yes
(B) False
(B) True, g is continuous because it is differentiable.
(C) True, since g(1) = f(1) = 0
(D) False, since g(1) = f(1) > 0.
(E) True, since g(1) = 0 and g(1) = f(1) > 0.
(F) False, g(x) = f(x) > 0, so g never changes
sign
(G) True, since g(1) = f(1) = 0 and g(x) = f(x) is an
increasing function of x (because f(x) > 0).
14. (A) x = 3
(B) Increasing on [3, ), decreasing on (–, 3]
PRACTICE PROBLEMS—B
13. 122
14. (A) True
(B) True
(C) False
15. (A) 4
(C) –3
(B) 10
(D) 2
6
9
17.
4
4
18. both the integrals are positive
16. 3 +
19. 4,
Practice Problems—C
7 23
,
.
4 4
23. 3 2
15.
3
2
17
20. (i) 1
(ii) 1
(iii) 4
22. (A) 1/4
23. e2x(1 + 2x)/(1 – e–x)
x 2dx 3 29
0
27. Yes
CONCEPT PROBLEMS—C
2.
4.
2x – 2
(i) 0
3x 2
1 x
x2
(vi) 2
t
5.
y'
7.
–
16. –1
18.
1
20
(b) 3 12
24. /2
n
, n W
2 4
(ii) {n/2 ± (1/4) cos–1(2a – 1)/3, n I} for a
(–1, 2), for a [1, 2] the function has no critical
points.
26. (A) 0, 2, 5, 7, 3
(B) (0, 3)
(C) x = 3
25. (i)
(iii) 2x 1 x 4
(ii) – sin(a2)
(iv)
1 sin 4 x . cos x
17.
3
24. 3/2
28. Yes
2.195
6
2x
1 x
4
xdx
(x 2 t 4 )3 / 2
3(1 3x )3
1 (1 3x )2
(v)
3x 2
1 x
6
2x
1 x4
3x 2
x 2 x12
6.
257
8.
x2
(1 + 3x)–3/2 (6 sin 2x + 15x sin 2x + 4x cos 2x
2
+ 12x2 cos 2x)
f(a) = 0.
9.
F(x) = t 2 sin(t 2 )dt
x
1
(D)
10. (b) sec2 x.
11. (A) (0, ) (b) x = 1
12. (a) (–3, 3) (b) x = 1
13. (A) True. Since f is continuous, g is differentiable
by First Fundamental Theorem of Calculus.
27. (A) local max. at 1 and 5; local min. at 3 and 7
(B) 9
(C) (1/2, 2), (4, 6), (8, 9)
2.196
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
F
a=0
1
(D)
2
4
6
8
0
–1
–2
x
11. (A) 0,
0
29. 1/2
30. [–1, 2]
1
33. (b b a a ) b a e 1
32. 1/2.
34. 2/3
(B) 0
1
(4 2 1) .
3
15. The curves coincide for x R.
16. The increase in the child's weight in kg between
the ages of 5 and 10.
13. K = 3, L = 1, M = 5
1
31. f(x) = x or f(x) = 0
2
14.
PRACTICE PROBLEMS—D
CONCEPT PROBLEMS—D
1.
(A) 5
2.
3.
F(x) = 3x + x – 4, x = 1
29.
4.
F(x) = y0 + f (t)dt
9.
x
a=4
3
2
12. P(x) = 4x + 8x2 + 3x3
28. f = c, f' = b, f ( t )dt a
8.
0
10. –2
–4
x
6.
a=8
(B)
4
+
+2
2
1024
2
2
(ii)
–
+ 2 loge
3
3
3
18
y
(iii) 4
(iv) 3 – 1
17. (i)
26
3
x
x0
14
(A)
3
(C) The statement is true. 7. (B)
(A) 2/3, 103/2/15, 1003/2/150.
(B) 10
(C) 0
4
(A)
3
2
(B)
3
Y
4
9
4 log e 2
19. (A) – F(0) = 0, F(3) = 0, F(5) = 6, F(7) = 6, F(10) = 3
3
37
(B) increasing on ,6 and ,10 ,
2
2
3
37
decreasing on 0, and 6,
2
4
15
9
3
(C) maximum
at x = 6, minimum – at x =
2
4
2
Y F(x)
6
18. 5 +
No
(D)
(c)
22
3
X
10 X
–2
2
(1 x ) / 2, x 0
20. F(x) =
2
(1 x ) / 2, x 0
21. y = 2 – x; y = x – 3
DEFINITE INTEGRATION
n
, n N
22. ,
6 2 6
3
1
max F(x) F – ,
23. x[–1/2,
1/2]
8
2
5
1
F(x) F – .
x[–1/2, 1/2]
2
8
min
24. f(p) = – /2.
26. f(x) = x3/2, a = 9.
28. [4, )
25. 3 3 2 2 1
27. 1
29. a = 1
– 1 8 1
31. A = 7, B = – 6, C = 3
2
32. – , – , 0
3
1
1
33. 3 x + sin x + /6 if 0 x 2/3; 2 3 x –
2
2
sin x + 5/6 if 2/3 x
34. 0, ± 2
30.
2 ,
35. (A) Displacement = –
(B) Displacement =
36. (v0 + 2gh)/2
1
1
; distance =
2
2
3
; distance = 2
2
263
37.
4
38. 1 m.
41. (i) x = 2
(ii) x = ln 4
(iii) {–1, 4}
42. Number of litres of oil leaked in the first 2 hours.
43. Number of bees in the first 15 weeks.
2
44. 46 kg
3
CONCEPT PROBLEMS—E
1.
2.
3.
4.
5.
6.
8.
(i) Yes
(ii) No
(i) Yes
(ii) Yes
(i) Yes
(ii) Yes
(iii) No
(A) Yes
(B) Yes
(C) No
(D) Yes
a, b, d are defined.
(i) Yes
(ii) No
1 if x rational
Consider f
.
1 if x irrational
(i) Does not exist
(ii) Does not exist
(iii)
10. No
11. (A) 16
(B) 4
2.197
9.
13. (i) 3
(ii) 22
(C) 10
9
(iii)
(iv) 6
2
PRACTICE PROBLEMS—E
16. a, b, c are integrable.
21. (A) Yes
(B) No
(C) No
(D) Yes
(E) Yes
(F) No
22. No, Yes, No, No
23. No, consider the example
1 if x is a rational number,
f(x) = – 1 if x is an irrational number
24. Make sure that the function f(x) is continuous
both inside the interval (0, 1) and at the end-points
[ lim f (x) = f(0) and lim f (x) = f(1)].
x 0
x 1
25. No. Consider the function
1 if x is rational
f(x)
1 if x is irrational, on the interval [0,1]
26. (A) 2/, 0, 0, (2/)cos–1a
(B) – 1, – 1/3, 1/2, 0, 1
(C) Not necessarily because of discontinuity.
27. 5/6
29. (B) 2 k 1 k ( k 1 k ) = 2 (21 – 3 2 – 3
8
–
5 –
6 –
7)
5
2
31. No, f is discontinuous.
30. (C) x = 1, x =
CONCEPT PROBLEMS—F
1.
2.
3.
4.
(B), (C), (D)
(A) Infinite interval
(B) Infinite discontinuity
(C) Infinite discontinuity
(D) Infinite interval
The integral is improper; the integrand is
undefined at 0.
both limits are ¥
2.198
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
11. The function x = ± t 5 is double-valued. Too
obtain the correct result it is necessary to divide
the initial interval of integration into two parts :
Y
1
y
5.
1
1 x2
X
7.
8.
(i) 1 – ln 2
(iii) 1
(ii) 1/2
(iv)
2
e
(ii) /2
(i) –
33
(iv)
2
3
Error when we move from step (3) to step(4)
(iii)
9.
2 5
2
x 2 dx
0 5
2
x 2 dx
x = t 5 in 0 < x < 2.
12. It is impossible, since sec t > 1 and the interval of
integration is [0, 1]
13. Yes, it is possible
14. The antiderivative F1(x) will lead to the correct
result and F2(x) to the wrong one, since this
function is discontinuous in the interval [0, ]
PRACTICE PROBLEMS—G
15. (i) ln tan = –lntan ( 1 – )
2
(ii) Use 2sin2 = 1 – cos2, quote (i) and integrate
cos2 lntan by parts. Note sin2 lntan 0 as
1
0,
2
1
1
17. (i)
(ii)
2
3
18. t = tan gives I, on using partial fractions, as
15. (i) 2 [cos2–cos3]
1
b2
1
dt
2 0 2
2 2
b a
a
b
t
1
t2
22. 8
CONCEPT PROBLEMS—G
1.
(A) 1/6
(C) 2 – /2
2.
(i)
3.
6.
8.
9.
94 2
ln
7
2
(D) 2/4
1
3
1
log 6 +
tan–1 5
5
5
4 2
3
(iv)
3
(ii)
(iii) – 2
1
ln(2 3 )
2
1
(iii) n2
3
16. (i)
17. (i)
43 2
4
(iii)
2
18. ln 11.
(B)
(iii) 0
Yes in each case ; /4 5. –4/5
2
Avni is right.
The limits of integration cannot be covered by
sint.
x 2 dx and apply
the substitutions x = – t 5 in –2 < x < 0 and
PRACTICE PROBLEMS—F
2
25
0
2
19. The substitution x = tan
the function tan
21.
83 3
3
2
(iv)
8
(ii)
(ii)
2
2
(iv) 2 – ln 2 –
3
3 3
(ii)
3
32
(iv)
33
2
x
is not applicable, since
2
x
is discontinuous at x = .
2
ln 2
8
23. 2009
22.
1
4
CONCEPT PROBLEMS—H
1.
3.
4.
(A) 0.5 ln(e/2)
(B) 4
(C) 1
Apply integration by parts to f and G. Note that G
= g.
2
DEFINITE INTEGRATION
PRACTICE PROBLEMS—H
6.
(A) 0.5 ln 3 –
3
2
(B)
5e 2
27
1
1
(D) 2
3 2 2
4
3
3
(C)
3
3
(E) 1/6 (f)
6 4
7.
(i) 2
2
(ii)
4
(iii) ln 2
1
1 n 2
(iv)
4 4
2
8. –1.
10. (A) n = 4
(B) 2
11. (E) f(b) = f(0) + f(1)(0)b +
+
3.
6.
10.
(B)
19. 2
20.
f ( 2 ) (0 )b 2 f (3) (0)b 3
2
6
5.
(ii) 5/4 – 3/8;
x
5x
5x
5
tan 1 x
3
2
2
2
6( x 1) 24( x 1) 16( x 1) 16
(i)
1
cos xdx 0
0
(ii)
(2m)!(2n)!
2
m! n!(m n)!
(B) 0, if n is even, , in n is odd
(C) /2n
(D)
3.
(i)
17.
71
105
16. (–1)k
0
sec 2 xdx
3
3
(ii)
2
1
ln 2
3
(ii)
16
3
b
a
(iv) 2
2 1
2
12
e
4.
(i) 1
(ii)
6.
(i) ln 4]
(ii) 3/8
(iii) tan –1 2 +
1
ln 5
2
1
+ n2
4 2
(ii)
2k
k 1
7.
(i)
8.
(iii) 2
(iv) 1 / 2
Consider upper and lower sums for the integral
m1
1 x dx obtained by dividing the interval [1, m]
n
sin
n
2
2
p! q!
.
( p q 1)!
/6
(iii) ln
2 m 2 n 1
14.
1
3
(iii) ln 3
2
13. (A)
Put x = sin .
3 n · n!
1 · 4 · 7 · ...· (3n 1)
(i)
1
1
1 1
... 1 .
( 1) n n! 1
1!
e n! ( n 1)!
(5/32)a 6
(m 1) (m 3) ... 4 2
. Put x = sin .
m(m 2) ... 3 1
2 n(2 n 2) ... 4 2
.
(2 n 1) (2 n 1) ... 3 1
2.
1 b ( 4)
f ( x)(b – x)3 dx
6 0
4
3
(i) a4/32;
(iii) 1/2.
(m 1) (m 3) ... 3 1
m(m 2) ... 4 2 2
PRACTICE PROBLEMS—J
1.
PRACTICE PROBLEMS—I
1.
18. (A)
2.199
k!
.
n k 1
into m – 1 equal parts.
1
10.
; 1.67 1011 .
k 1
CONCEPT PROBLEMS—I
1.
cos(ex) ex – cos(x2)2x
2. 3x2 cos x3
2.200
3.
4.
5.
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
.
3
x2 (2x sin x2 – sin x) + 2(cos x – cos x2) x
4
6.
(i)
7.
2
17
–2.
8.
dy
cot t
dx
8.
dy
t2
(ii)
dx
9.
y'
10. (i)
cos x
.
ey
8x 2
40x6 4
11.
1 5x6 (1 – 5x6 )3/2
12. – t2
10. 1 +
x2
dt
0
1 5t 3
1
(tan x sec x 1)
101
14. – cos x
16. x = 0
(iii)
2
e
11. (i)
(iii)
13.
18.
1
2
5.
15. [0, 2]
17. Minimum at x = 0.
19. c = 1; limit =
1
f(x) + f(1/x) = (ln x)2
2
(i) 0
(ii) 0
(iii) 60
7.
(i) 0
9.
(iii)
8
(B) (i)
6.
2l1 = l2
(i) 0
(ii)
2 2
ln 2
16
(ii) [
2
4
(iv)
ln (1 + 2 )
– ( 2 – 1)]
4
2
32
(ii)
4.
(i) 2
(ii) –
1
3
2
5.
6.
(ii) · ln 2
8
4
( 2)
2
(iv) ln 2
(i)
3 1
2
(iv) 4 ln 4/3
3
(ii)
2
ln2
2
(iii) ln 2
2.
(ii)
PRACTICE PROBLEMS—M
7.
(ii) – /2
2
16
(iii) –
CONCEPT PROBLEMS—J
1.
(iv) /2
PRACTICE PROBLEMS—N
PRACTICE PROBLEMS—L
2.
(iv)
(iii) 1
4
(i) 4
(iii) 0
PRACTICE PROBLEMS—K
9.
3 3
(ii) 0
4
(i) 0
(iii)
ln 2
2
(iv)
3
3
ln 2
2
16
(i)
2
32
(ii)
32
512
(iii)
2
12
(iv)
8
15
11
1
A
22 2
PRACTICE PROBLEMS—O
2.
4.
6.
2
·ln 2
2
(i) 3 (e – 1)
(ii) 1
(iii) 5
(iv)
0
–
4
DEFINITE INTEGRATION
PRACTICE PROBLEMS—P
3.
tan –1t +
5.
48
6.
2
3
t
1
1 t 2 2t
(C)
4. 0
11. (A)
x±1
2
2 2
x (x + |x|)
3
1 3
1
1|x|
x if |x| 1; x – x|x| +
if |x| > 1
3
2
6 x
(C) 1 – e–x if x 0; ex – 1 if x < 0
(B) x –
CONCEPT PROBLEMS—K
1. No
2. No
9. < I < 2.
11. To estimate the integral from below use the
inequality 1 + x4 < (1 + x2)2, the SchwartzBunyakovsky being used for estimating it from
above.
PRACTICE PROBLEMS—Q
16. (A) Plus
(C) Plus
18. 2 sin–1
2.201
(B) Minus
(D) x if |x| 1;
12. f(x) = 2x15; c = –
(i)
8
15
(ii)
5
32
2.
(i)
8
315
(ii)
16
3.5.7.11
(iii)
1
24
(iv)
5
211
(i)
a9
9
(ii)
2
(iii)
2
63
(iv)
33 7
a
16
(i)
5
192
(ii)
16
–
245
14
(iii)
25
3.7.11.13
(iv)
213
5.7.913.17
(i)
2.4.6...(2 n )
3.5.7....(2 n 1)
(ii)
1.3.5...(2 n – 1)
.
2.4.6....2 n 2
(iii)
63 5
a
8
(iv)
32a 3
3.
1
1
value m = e e 0.692 at x = and the greatest
1
26. I(1) 1.66 the greatest value ; I –0.11
2
the least value.
PRACTICE PROBLEMS—R
3.
4.
6.
3
x or 0.
f(t) = sin t – 1; c = 0
5. f(t) = – sin t; c = /3
7.
(A) 16
(C) (36)1/3
(B) 1 + 3/ 2
(D) 1/5
1
1
1
8. p(x) = 3 + x + x2
9. f(x) = 3e3x, a = ln 2
2
4
3
10. (A) No such function
(B) – 2x ln 2
13. 3 + 3 ln x
1.
continuous on the interval, it reaches the least
e
value M = 1 at x = 0 and at x = 1.
23 8/5
24. (ii) 0.85 < I < 0.90.
1
9
PRACTICE PROBLEMS—S
1
3
x x at 0 x 1
21. The function f(x) =
is
1 at x 0
1 3 2 |x|
x +
if |x| > 1
3
3 x
4.
5.
7.
35
128
PRACTICE PROBLEMS—T
4.
5.
0
Change the variable by the formula z = k2x2, and
then apply L'Hospital's rule.
2.202
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
PRACTICE PROBLEMS—U
3
1.
2
cos(x t )2t dx
5.
n a a 2 1
2
7.
2a 2 a 2
16
4
2
1 1 b2
4. ln
2
6.
1 2 – 2
2 4
PRACTICE PROBLEMS—V
1
1 x
3. Integrating both sides between 0 and t yields
1.
(x + 1) ln (1+ x) – x
2.
2
2n
t 2n
ln(1 t ) t t ... t x dx .
2n 0 1 x
2
Now use
2n 1
t 2n
t
0 x dx x 2 n dx t
01 x
0
2n 1 .
2
4. (i) 0
(ii)
6
(iii) ln 2
(iv) 2(a b)
1
8.
2
OBJECTIVE EXERCISES
43. C
44. D
45. C
46. C
47. C
48. B
49. C
50. A
51. C
52. A
53. A
54. C
55. A
56. A
57. A
58. D
59. A
60. C
61. D
62. A
63. D
64. D
65. C
66. AC
67. ABCD
68. ABCD
69. ACD
70. ABCD
71. ACD
72. AB
73. AB
74. AC
75. AD
76. AC
77. ACD
78. ABCD
79. ACD
80. ABCD
81. ABC
82. AB
83. ACD
84. AB
85. ABCD
86. C
87. C
88. A
89. C
90. A
91. B
92. B
93. B
94. C
95. D
96. C
97. A
98. A
99. C
100. B
101. A
102. C
103. C
104. A
105. B
106. C
107. D
108. A
109. A
110. A
111. (A)–(R) ; (B)–(P) ; (C)–(S) ; (D)–(R)
112. (A)–(R) ; (B)–(P) ; (C)–(S) ; (D)–(Q)
113. (A)–(Q) ; (B)–(RS) ; (C)–(QS) ; (D)–(S)
114. (A)–(S) ; (B)–(S) ; (C)–(PR) ; (D)–(R)
115. (A)–(R) ; (B)–(S) ; (C)–(Q) ; (D)–(Q)
REVIEW EXERCISES for JEE ADVANCED
3.
1. A
4. D
7. A
10. C
13. A
16. C
19. A
22. B
25. C
28. C
31. B
34. A
37. C
40. B
2. A
5. D
8. B
11. B
14. A
17. A
20. B
23. B
26. B
29. C
32. A
35. B
38. B
41. C
3. D
6. C
9. D
12. A
15. C
18. A
21. D
24. B
27. A
30 D
33. A
36. B
39. B
42. D
Having worked C one finds S by obsering that
a2C + b2S = .
12. g(x) = cx2+1, c R
14. 6 ln 3 – 4
21. 4 ln x
22. (A) – Ae–a
(C) A + 1 –
275
48
15. 3ln(ex)
13.
(B)
e
2
23. a = 0
25. ln (1 + ), ( > – 1).
A
2
(D) e ln 2 – A
24.
1.3.5...(2 n 1)
2
2 n n!
DEFINITE INTEGRATION
26. The value of integral is /2 if 2n < < (2n + 1),
and – /2 if (2n –1) < < 2n, n being any
integer; and 0 if is a multiple of
33. (0, 4)
34. (i)
2
(ii) 2 ln 2 .
e 1
3
4
35. a = 9, b =
27
2
sin n 1 x
2 = 1 + 2 cos x + 2
36. Use the identity
sin 1 x
2
cos 2x +...+ 2cos nx to prove that sin nx 2 cos
(n – 1) x + 2 cos (n – 3) x + ...,
where the last term is 1 or 2 cos x.
sin x
(2 n 2)(2 n 4)....4.2
16
42.
(2 n 1)(2 n 3).....5.3.1
3
44. Apply the Schwartz-Bunyakovsky inequality in
the form
11. (A) –
(C)
2
1
1
dx f (x)dx
dx
a f (x).
a
a f (x)
f
(x)
47. ln 4
49. (C) P(x) =
b
1
12
50. (B) g(2) = 2A, g(5) = 5A
(C) A = 0
1
1
1
+ ( – ) (t – 1) (D) (t – 1)
2
2
2
16.
2
(3 31 +
3
17. (A) x 1(c) F(ax)– F(a); F(x) –
1
ex
+ e; xe1/x –e–F
x
x
18. (A) (i 0 ) 2 / 2
(B)
(i 0 ) 2 (i 0 ) 2 T
4t 0
sin(
2) sin 2
2
8t 0
T
(C) (i 0 ) 2 / 2
32 3
3
ln 2
4
2
19. (i)
5
. Put x = cos and integrate by parts
3
(iii) 0
21. 0.099
32. Integrate the inequality
(ii)
/2
TARGET EXERCISES for JEE ADVANCED
(ii)
3 – 11.35)
4
2
x sin x x 2 x x 1 x on 0 x
6
6
6
and write Schwartz-Bunyakovskyn equality
(D)
3
1
2
1
) (t – 1)2/2
2
14. (A) G(0) = G(1) = G(2) = /4
1
1
1
(x – [x])2 – (x – [x])
2
2
(B)
+ ( –
39.
b
1
2
2.203
2
(52 – 15)
92
1.
(i)
2.
1
The integral reduces, with t = tan , to
2
1 1 dt
b2 a 2 1
dt
2
b 0 1 t
b 0 (a b) 2 (a b) 2 t 2
The difference between a > b and a < b comes in
the answer to the second of these integrals.
/2
x sin x
6
/2
xdx sin xdx
0
0
2
8
2 2
33. In I2 note that f(n) is bounded , f(x) being
continuous.
37. f(x) M, say, since f(x) is continuous; and so
In M(b – a)/ n
1
1
; P2(x) = x2 – x + ; P3(x) = x3 –
2
6
3 2 1
1
x + x; P4(x) = x4 – 2x3 + x2 –
; P (x) = x5
30 5
2
2
42. P1(x) = x –
–
5 4 5 3 1
x + x – x
6
2
3
2.204
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
43. 1/(k + m + 1) =
t
44. (B) constant =
3
2
0
k m
dt
46. (B) b = log 2
(C) e2Li(e2x – 2)
48. Consider h(x) = [F(x) –f(x)]2 + [G(x) –g(x)]2
PREVIOUS YEAR'S QUESTIONS
(JEE ADVANCED)
1.
15 32
–
60
3.
5.
4
1/2
2. 2 –
4. ( 2 – 1)
1
7b
a(log 2 5)
2
2
a b
8. 2
9. 16
11. D
12. B
14. C
15. D
17. D
18. A
20. D
21. D
23. D
24. B
26. A
27. A
29. A
30. C
32. C
33. B
35. C
36. A
38. A
39. A
41. A
42. B
6.
2
2
7.
2
10. T
13. A
16. B
19. C
22. C
25. C
28. A
31. B
34. C
37. C
40. D
43. C
44. C
47. AD
50. A
t n 1 1
52.
(t 1)(n 1)
1
55.
log 3
20
59.
sin
8
65. 2
67.
3
1
log 2
2
10
70.
1 n
2 6 2
72.
n 2
8
45. B
48. ABCD
46. A
49. ABC
3 1
2
6 3
2
56.
58.
16
12
1
log 2 1
60.
2
2
54.
66. n = 3
68. 2n + 1 – cos v
3 1 2
3 1
3
73.
2
74. log 2
75.
2
78. 2
80.
4 1
tan
4
3
24
1
1
1
e cos esin 1
5
2 2
2
82. (A)–(S) ; (B)–(S) ; (C)–(P) ; (D)–(R)
83. A]
84. D
86. 5051
87. 4
81.
85. A
88. 6
C H A P T E R
3
AREA UNDER
THE CURVE
3.1
CURVE SKETCHING
One of the applications of definite integral is the
calculation of areas bounded by curves. In the previous
chapter we dealt with definite integration as the limit of
a sum. The concept of integration arose in connection
with determination of area. We shall now use definite
integral to find the area of plane regions bounded by
curves.
Let us first focus on the techniques of tracing the
curves bounding the plane regions. The following
procedure is to be applied in sketching the graph of a
function y = f (x) which in turn will be extremely useful
to evaluate the area under the curves.
values of y for which x is imaginary. The curve does
not exist for these values of x and y.
For example, the values of y obtained from y2 = 4ax are
imaginary for negative values of x. So, the curve does
not exist on the left side of y-axis.
Similarly, the curve a2y2 = x2(a – x) does not exist for
x > a as the values of y are imaginary for x > a.
In the curve, a2y2 = x2 (x – a) (2a – x), we find that for
0 < x < a, y2 is negative, i.e., y is imaginary. Therefore,
the curve does not exist in the region bounded by the
lines x = 0 and x = a. For a < x < 2a, y2 is positive i.e., y
is real. Therefore, the curve exist in the region bounded
by the lines x = a and x = 2a.
1. Extent
2. Intercepts
Find the domain of definition of the function and the
values of the function at the points of discontinuity (if
possible) and the endpoints of the domain.
We should find out if there is any region of the plane
such that no part of the curve lies in it. Such a region is
easily obtained on solving the equation for one variable
in terms of the other.
The curve will not exist for those values of one variable
which make the other imaginary. For this, find the value
of y in terms of x from the equation of the curve and
find the value of x for which y is imaginary. Similarly,
find the value of x in terms of y and determine the
When we put x = 0 in the equation of the curve, we get
the y-intercept and this tells us where the curve
intersects the y-axis. To find the x-intercepts, we set
y = 0 and solve for x. (We can omit this step if the
equation is difficult to solve.)
We should also see whether the curve passes through
the origin or not. If the point (0, 0) satisfies the equation
of the curve, it passes through the origin.
3. Sign Scheme of f(x)
When graphing a function f, the zeros of the function,
i.e. solutions of the equation f(x) = 0 and the points of
discontinuity of the function divide its domain of
3.2
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
definition into intervals where the function is of
constant sign. The sign scheme of the function helps
in locating parts of the graph which lie above/below
the x-axis.
i.e. F(x, y) = 0 implies F(– x, –y) = 0.
If F represents a function, then it is said to be odd.
For example, the graph of y = x3 is symmetric with
respect to origin.
4. Symmetry
The aim of symmetry is to make the calculations as
short as possible. Indeed, if a function is even or odd,
then we may consider the part of the domain which
belongs to the positive x-axis, rather than the whole
domain. On this part of the domain we must carry out
complete investigation of the behaviour of the function
and construct its graph, then, resorting to symmetry,
complete the construction on the whole of the domain.
Consider the graph of an equation F(x, y) = 0 in the
x-y plane.
(i) The graph of F(x, y) = 0 is symmetric about the
y-axis if on replacing x by – x, the equation of the
curve does not change. i.e. F(x, y) = 0 implies
F(– x, y) = 0.
If F represents a function, then it is said to be
even.
For example, the parabola y = x2 is symmetric with
respect to the y-axis.
(ii) The graph of F(x, y) = 0 is symmetric about the
x-axis if on replacing y by – y, the equation of the
curve does not change. i.e. F(x, y) = 0 implies
F(x, – y) = 0.
For example, the parabola y2 = x is symmetric with
respect to the x-axis.
(iii) The graph of F(x, y) = 0 is symmetric about the
origin if on replacing x by – x and y by – y, the
equation of the curve does not change.
Further, the circle x2 + y2 = r2, ellipse
x2 y2
= 1,
a 2 b2
x2 y2
–
= 1 are symmetric with
a 2 b2
respect to the y-axis, the x-axis and the origin.
(iv) The graph of F(x, y) = 0 is symmetric about the
line y = x if on interchanging x and y, the equation
of the curve does not change. i.e. F(x, y) = 0 implies
F(y, x) = 0.
For example, the graph of xy = c2 is symmetric
with respect to the line y = x.
and hyperbola
Also, the graph of x3 + y3 = 3 a x y, a > 0 is symmetric
with respect to the line y = x.
(v) The graph of F(x, y) = 0 is symmetric about the
line y = –x if on replacing x by – y and y by – x, the
equation of the curve does not change.
i.e. F(x, y) = 0 implies F(– y, – x) = 0.
For example, the graph of xy = – c2 is symmetric
with respect to the line y = – x.
AREA UNDER THE CURVE
3.3
strictly increasing interval
f (x) 0 : equality holding for sub-intervals
non-decreasing or increasing interval
f (x) 0 : equality holding for distinct points only
strictly decreasing interval
f (x) 0 : equality holding for subintervals
non-increasing or decreasing interval.
Note that at the points where f '(x) = 0, the curve
has horizontal tangents.
Note: For graphs of algebraic equations, the
7. Local Maximum and Minimum Values
symmetry is judged as follows :
(a) If all the powers of y in the equation are even, the
curve is symmetric about the x- axis.
(b) If all the powers of x are even, the curve is
symmetric about the y-axis.
(c) If all powers of x and y are even, the curve is
symmetric about the x-axis as well as y-axis.
Thus, the curve x2 + y2 = 4ax is symmetrical about
x-axis, The curve a2x2 = y3 (2a – y) is not symmetrical
about x-axis, for besides containing a term in y4, the
equation also contains a term in y3 so that even powers
as well as odd powers of y occur in the equation.
Find the critical points of f and use the tests for
discrimination of local maximum and minimum.
The First Derivative Test
Suppose that x = a is a critical point of a continuous
function y = f(x).
(i) If f(x) changes from positive to negative at x = a,
then f has a local maximum at x = a.
(ii) If f(x) changes from negative to positive at x = a,
then f has a local minimum at x = a.
5. Periodicity
Find out whether the function is periodic or not. If
f(x + T) = f(x) for all x in the domain, where T is a
positive constant, then f is said to be a periodic
function and the smallest such number T is called the
fundamental period. If we know what the graph looks
like in an interval of length T, then we can repeat the
same segment of the graph to sketch the entire graph.
6. Monotonicity
The steps for determining intervals of monotonicity are
as follows :
(i) Determine the derivative of the function f'(x) and
find the critical points i.e. points where the
derivative is zero or does not exist.
(ii) Determine the sign of f'(x) in different intervals
formed by the critical points.
(iii) Determine monotonic nature of function in
accordance with following categorization :
f (x) 0 : equality holding for distinct points
only
(iii) If f(x) does not change sign at x = a (that is, if f(x)
is positive on both sides of x = a or negative on
both sides), then f has no local maximum or minimum
at x = a.
The Second Derivative Test
Let x = a be a stationary point of a function f (i.e. f(a) = 0).
The function f has a local maximum at a if f(a) is
negative, and a local minimum if f(a) is positive.
8. Concavity and Points of Inflection
If the second derivative f(x) is everywhere positive
within an interval, the arc of the curve y = f(x)
corresponding to that interval is concave up. If the
second derivative f(x) is everywhere negative in an
interval, the corresponding arc of the curve y = f(x) is
concave down.
The concavity of the graph of f will change only at
points where f"(x) = 0 or f''(x) does not exist. Let x = c be
such a point and the signs of f(c – h) and f(c + h) be
opposite, then the point x = c is called a point of
inflection.
The following table contrasts the interpretations of
the signs of f, f, and f. (It is assumed that f, f, and f
are continuous.)
3.4
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Where the Where the
ordinate f(x) slope f '(x)
Where f "(x)
Is positive The graph is The graph
The graph is
above the x slopes upward concave
Upward
axis
Is negative The graph is The graph
below the x slopes
downward
axis
Changes
sign
The graph
crosses
the x axis.
1
lim y = lim x = –
x
x
There is no horizontal asymptote.
x
The graph is
concave
downward
The graph has a The graph
has
horizontal
tangent and a an inflection
point
relative
maximum or
minimum.
Further, xlim
y
1 1
= 1.
= xlim
x2
x
1
lim (y – x) = lim x x = lim 1 = 0.
x
x
x x
y = x + 0 i. e. y = x is an oblique asymptote.
A rough sketch of the curve is as follows :
x
9. Asymptotes
(i) If either xlim
f(x) = L or xlim
f(x) = L, then the line
y = L is a horizontal asymptote of the curve
y = f(x).
(ii) The line x = a is a vertical asymptote if at least one
of the following statements is true :
lim f(x) =
lim f(x) =
x a
x a
lim f(x) = –
lim f(x) = –
xa
xa
(iii) If there are limits
f(x)
lim
m1 and lim [f(x) m1x] c1 ,
x
x x
then the straight line y = m1x + c1 will be an
asymptote (a right inclined asymptote or, when
m1 = 0, a right horizontal asymptote).
If there are limits
f(x)
lim
m 2 and lim [f(x) m 2 x] c 2 ,
x x
x
then the straight line y = m2x + c2 is an asymptote
(a left inclined asymptote or, when m2 = 0, a left
horizontal asymptote).
1
Example 1. Find the asymptotes of y = x +
x
and sketch the curve.
x 1
=
y = xlim
Solution We have xlim
0
0
x
1
lim y = lim x = –
x 0
x0
x
x = 0 is a vertical asymptote.
x 1
=
Now, xlim
y = xlim
x
Example 2. Analyse the equation (y – x)2 = x3
and trace its graph.
Solution Solving for y, we have
...(1)
y = x ± x3/2.
(i) Intercepts. The given equation shows that the curve
contains the points (0, 0) and (1, 0) on the coordinate
axes.
(ii) Symmetry. There is no symmetry with respect to
the axes or the origin.
(iii) Extent. Equation (1) shows that we must exclude
x < 0. It also shows that for each value of x (except x = 0),
there are two values of y, and that the curve
consists of two branches which start at the origin
and lie on opposite sides of the line y = x.
For the upper branch, lim y = and for the lower
x
branch lim y = – . Hence, the curve extends
x
indefinitely up and down.
(iv) Asymptotes. Equation (1) shows that there is no
vertical asymptote and the discussion of the extent
in y shows that there is no horizontal asymptote.
(v) Maximum, minimum, and inflectional points.
Taking the positive sign in (1) and differentiating,
we get y = 1 + 3/2x1/2, y = 3/4 x–1/2.
Since y 0, the upper branch has no extreme
AREA UNDER THE CURVE
point except the origin, which is a minimum point
since it is a left end point and the slope there is 1.
Since y > 0 for x > 0, the upper branch is concave
up and has no point of inflection.
For the lower branch we have
y = 1 – 3/2x1/2, y = – 3/4 x–1/2.
Hence again the origin is a minimum point as in the
case of the upper branch. Setting y = 0, we get x = 4/9
and y = 4/27. This point is a maximum point since
y < 0 for x > 0. The last statement also tells us that the
lower branch is concave down and has no point of
inflection.
3.5
The points x = 2 and x = 5 are the critical points.
On the intervals (– 2), (2, 3) and (5, ), f(x) is
negative and, consequently, the function f(x)
decreases. On the interval (3, 5) the derivative is
positive and the function f(x) increases.
We tabulate the various information about the
function as follows:
x: (– 2) (2 3)
(3 5)
(5 )
y:
to 0
0 to – – to –
y
–
–
+
27 – 27
to –
4
4
–
The graph of the function f(x) is shown below.
Y
O
2
3
5
X
Example 3. Applying the derivative, construct
the graphs of the following functions:
(i) f(x) =
(2 x)3
(x 3)2
(iii) f(x) =
6sin x
2 cos x
(ii) f(x) = x + e–x
Solution
(2 x)3
is defined for all x 3.
(x 3)2
It is neither even nor odd, nor periodic. Its graph
cuts the x-axis at the point x = 2 and the y- axis at the
point y = 8/9. Since f(x) – as x 3, the
straight line x = 3 is a vertical asymptote to the
graph of the function f(x). Note that
8 3x
8 12x 6x 2 x 3
f(x) =
=–x+ 2
,
2
x – 6x 9
x – 6x 9
As x and as x – , f(x) – x.
Therefore, the line y = – x is an inclined asymptote
to the graph of the function. Let us find the
derivative
3(2 x)2 (–1)(x 3)2 (2 x)3 2(x 3)
f(x) =
(x 3)4
(2 x)2 (5 x)
=
.
(x 3)3
(i) The function f(x) =
– 27
4
(ii) The function f(x) = x + e–x is defined for all x R.
It is neither even nor odd. It is not periodic.
The graph cuts the y-axis at y = 1. It does not
intersect the x-axis.
Since lim e
x
–x
0, the line y = x is an inclined
asymptote as x .
As x – there is no inclined asymptote
f (x)
– as x – .
because
x
Let us find the derivative
f(x) = 1 – e–x.
The point x = 0 is a critical point.
It is easy to see that f(x) < 0 for x < 0 and f(x) > 0
for x > 0. Therefore, the function f(x) decreases on
the infinite interval (– 0) and increases on (0, ).
It follows that at the point x = 0 the function has a
relative minimum. The graph of the function f(x) is
shown below.
3.6
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Y
1
X
O
6 sin x
is defined for all
2 cos x
x R. It is an odd function, and periodic with a
period 2. It is, therefore, sufficient to construct
its graph on the interval [0, ]. Since 2 + cos x 1,
the function possesses no vertical asymptotes.
We find the derivative
cos x(2 cos x ) – sin x( sin x )
6
f(x) = 6
(2 cos x )2
= 2 cos x 12 .
(2 cos x)
2
The points x = 2n ± , n I, are critical.
3
On the interval (0, 2/3) the derivative is positive
and, consequently, the function f(x) increases.
On the interval (2/3, ) the derivative is negative
and the function f(x) decreases.
The point x = 2/3 is a point of maximum. In fact at
2 ,
n I, the function f(x)
the points x = 2n +
3
possesses maxima and at the points x = 2n –
2 , n I, it possesses minima.
3
The graph of the function f(x) is shown below.
(iii) The function f(x) =
Y
2 3
2
3
O
2
3
X
–2 3
Example 4. Which one of the following functions
resembles the graph of a rational function as shown?
2
5( x 1)( x 3)
( x 2 4)( x 2)
5( x 2 1)( x 3)
(C)
( x 2 4)
(A)
2
( x 1)( x 3)
( x 2 4)( x 2)
5( x 2 4)( x 2)
(D)
( x 2 1)( x 3)
(B)
Solution The function seems to have a zero (i.e.
crosses the x-axis) at – 3, – 1 and 1. Of all the choices
given only (A), (B) and (C) satisfy this.
The function in (C) would change sign around the
vertical asymptote x = – 2 and the graph does not, so
(C) is not the correct answer either.
The horizontal asymptote is the limit of the function at
positive or negative infinity, and from the graph, this
limit should be 5. Answer (B) does not satisfy this, so
(A) must be the correct answer, and indeed, it satisfies
all other requirements.
x2 1
f
(x)
Construct
the
graph
of
Example 5.
x2 1
and find the area bounded by y = f(x) and x-axis.
2
x 1
x2 1
(i) The function f(x) is well defined for all real x.
Domain of f(x) is R.
(ii) f (–x) = f(x), so it is an even function and hence
graph is symmetrical about y-axis.
(iii) Obviously the function is non-periodic.
(iv) f(x) 1– for x (we are considering x > 0 only
as curve is symmetrical about y-axis). Hence y = 1
is an asymptote of the curves. It may be observed
that f(x) < 1 for any x R and consequently its
graph lies below the line y = 1 which is the
asymptote to the graph of the given function.
Solution
Here, f (x)
(v) Again
2
2
x 1
x2 1 1
2
decreases in (0, ) , thus f(x)
increases in (0, ).
(vii) The greatest value 1 for x and the least
value is –1 for x = 0. Thus, its graph is as shown in
figure.
AREA UNDER THE CURVE
3.7
(iv) f(x) 0– as x – . Hence y = 0 is an
asymptote of the curve.
(v) f (x) = (x + 1) ex f(x) increases for x 1
and decreases for x 1 .
Hence, x = –1 is the point of absolute minima.
Minimum value = f(–1) = 1 .
e
Example 6. Construct the graph of f(x) = xex.
Find the area bounded by y = f(x) and its asymptote.
(vi) f(x) = (x + 2)ex f(x) is concave up for x > –2 and
concave down for x < –2 and hence x = –2 is a
point of inflection.
Solution
(i) The function is well defined for all real x
domain of f(x) is R.
(ii) There is no symmetry in the graph.
(iii) Obviously function is non-periodic.
A
1.
2.
3.
4.
5.
Plot the following curves :
(i) y = ± x 2 1
(ii) y = 2 ± (x 1)2 1
(iii) y = ± x 2 1
(iv) 4y2 + 4y – x2 = 0
Construct the graph of the following functions :
(i) y = 1 + x2 – 0.5 x4 (ii) y = (x + 1) (x – 2)2.
Construct the graph of the following functions :
(ii) y = x4 (1 + x)–3.
(i) y = (1 – x2)–1.
4
–4
(iii) y = (1 + x) (1 – x) .
Construct the graph of the following functions :
(ii) y = 2x – 1 + (x + 1)–1
(i) y = x (1 – x2)–2
Construct the graph of the following functions :
(i) y = 0.5 ( x 2 x 1 – x 2 – x 1)
3.2
AREA OF A CURVILINEAR
TRAPEZOID
It is known that the definite integral of a non-negative
function is the area of the corresponding curvilinear
trapezoid. This is the geometrical meaning of the
definite integral, which is the basis of its application to
computing the areas of plane figures.
Consider the curvilinear trapezoid aABb bounded by
the graph of a nonnegative continuous function
6.
7.
8.
(ii) y = x 2 1 – x 2 – 1
(iii) y = (x + 2)2/3 – (x – 2)2/3.
Construct the graph of the following curves :
(ii) y2 = (x – 1) (x + 1)–1 .
(i) y2 = 8x2 – x4
Plot the graph of the following functions :
cos x
x 2 2x 3
(ii) y = 2
(i) y =
cos 2x
x 2x 8
Construct the following curves :
(i) x = cost, y = sin2t
(ii) x = cos3t, y = sin3t
(iii) x = cos(5t +1), y = sin(5t +1)
(iv) x = cost, y = cos t
4
y = f(x), x [a, b], the interval [a, b] of the x-axis, and
the line segments x = a and x = b (a < b).
3.8
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
If f(x) 0 for x [a, b], then the area bounded by
b
curve y = f(x), x-axis, x = a and x = b is A = f(x) dx
a
Example 1. Compute the area of the figure
bounded by the curves f(x) = x2 – 2x + 2, x = –1, x = 2,
and the segment [–1, 2] of the x-axis.
The region described is a curvilinear
trapezoid lying above the x-axis, therefore its area is
computed as
2
3 2
2
2
A = (x 2 2x 2)dx x
x 2 2x 1 6 .
1
3
1
1
Solution
Example 2. Find the area bounded by the curve
y = n x + tan–1x , x-axis and ordinates x = 1 and x = 2.
Solution Let f(x) = n x + tan–1x
1
1
>0
f(x) = +
x 1 x2
f(x) is a strictly increasing function.
> 0, f(x) is positive for all x [1, 2].
4
A rough sketch is as follows
Since f(1) =
The required area =
2
x sin x dx (x sin x) dx
0
x ( cos x) 0 ( cos x) dx
0
2
2
(x( cos x)) ( cos x) dx
sin x 0 (2 ) sin x
2
4 .
Example 4. If a function f(x) is defined as
f(x) = max {4 – x2, |x – 2|, (x – 2)1/3} for x [–2, 4], then
find the area bounded by the curve and x-axis
y = 4 – x2 is a parabola EMA. y = |x – 2| is
a pair of straight lines ABF and ACD y = (x – 2)1/3 is the
curve IHABG. Thus, from the graph
f(x) = max {4 – x2, |x – 2|, (x – 2)1/3} is equivalent to
Solution
2
1
The required area = (n x tan x) dx
1
2
1
1
2
= x n x x x tan x n (1 x )
2
1
1
= 2 n 2 – 2 + 2 tan–12 – n 5 – 0 + 1
2
1
–1
– tan 1 + n 2
2
5
1
= n 2 – n 5 + 2 tan–12 – – 1.
2
4
2
Example 3. Find the area bounded by y = x |sinx|
and x-axis between x = 0 and x = 2
xsin x, if sin x 0, i.e., 0 x
Solution y
xsin x,if sin x 0, i.e., x 2
2x
2
4x
1/3
f(x) =
x 2
x 2
if
2 x 1
if
1 x 2
if
2x3
if
3 x4
AREA UNDER THE CURVE
Now, the area bounded by the curve and the x-axis
1
2
3
4
2
1
2
3
2
1/3
= (2 x)dx (4 x )dx (x 2) dx (x 2)dx
2
x
1
3
x
2
3
3
x
2
4
4/3
= 2x 2 4x 3 4 x 2 2 2x
2
2
1
3
7
3 3 59
= 9
.
2
4 2 4
Example 5. Consider the function, f(x) = cos 1
(cos x) sin 1 (sin x) in [0, 2 ]. Find the area bounded
by the graph of the function and the x axis .
0x
x
Solution cos 1 (cos x) = 2 x x 2 ,
x
x
sin 1 (sin x) =
x 2
3.9
Further, if y = f(x) does not change sign in [a, b], then
area bounded by y = f(x), x-axis and the ordinates
b
f(x) dx .
x = a, x = b is
a
Example 6.
Find area bounded by y = log 1 x
2
and x-axis between x = 1 and x = 2.
Solution A rough sketch of y = log 1 x is as follows
2
0 x 2
x 3
2
2
3
2 x 2
2
The required area = –
1
2
0
if x 0 ,
2
2 x
if x ,
2
if x , 3
Hence, f (x) =
2
3
4 2 x if x 2 , 2
=–
log x dx
1
2
log x . log e dx
e
1
1
2
= – log 1 e . [x log e x x]12
2
The graph of f (x) is :
The required area = area of the trapezium
= – log 1 e . (2 loge2 – 2 – 0 + 1)
2
= – log 1 e . (2 loge 2 – 1).
2
Example 7. Find the area of the region bounded
by the curve y = e2x – 3ex + 2 and the x-axis.
Solution We find the points of intersection of the
curve and the x-axis : e2x – 3ex + 2 = 0. The curve
intersects the x-axis at x = 0 and x = ln 2. The graph of
f(x) = e2x – 3ex + 2 is shown in the figure.
1 3
= 2.
2 2 2
=
Note: If f(x) 0 for x [a, b], then the area
bounded by the curve y = f(x), x-axis, x = a and x = b is
b
A=–
f(x) dx
a
We see that on the interval [0, ln 2], f(x) 0. So the area
of the given region is
3.10
A=
ln 2
0
=
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
x
ln 2
(e
e 2x
3e 2)dx
3e x 2x
2
0
2x
x
3
– 2 ln 2
2
Example 8. The figure shows two regions in the
first quadrant. A(t) is the area under the curve y = sin x2
from 0 to t and B(t) is the area of the triangle with vertices
A(t)
.
O, P and M(t, 0). Find lim
t 0 B(t)
Y
P(t,sin t2 )
y = sinx2
We have f 3 (x) = t f 2 ( t ) dt
Solution
0
On differentiating w.r.t. x, we get
3 f 2 (x) f ' (x) = x f 2 (x)
f 2 (x) [3 f ' (x) – x] = 0
As f (x) 0
x
x2
f ' (x) = f (x) =
+C
3
6
x2
,
But f (0) = 0 C = 0 f (x) =
6
which is a non-negative function.
Y
A(t)
t
O
Y
O
X
3
3.3
B(t)
t
t
Solution We have A(t) = 0
2
A(t)
= lim
lim
t 0 B(t)
t0
= lim
2
t
t
X
sin x 2 dx , B(t) =
t sin t 2
2
sin x dx
2
0
t sin t 2
sin x dx
2
0
sin t 2
t
t2
t0
3
t
sin x dx
= lim
2
t 0
Hence, lim
t 0
X
2
P(t,sin t )
O
3
3
3
x3
1 2
x dx =
A=
= .
18 0 2
60
2
0
t3
AREA BOUNDED BY A
FUNCTION WHICH
CHANGES SIGN
If on the interval [a, b] the function f(x) 0, then, as we
know from the area of a curvilinear trapezoid bounded
by the curve y = f(x), the x-axis and the straight lines
x = a and x = b is
b
A = f (x) dx
a
If f(x) changes sign on the interval [a, b] a finite number
of times, then we break up the integral throughout
[a, b] into the sum of integrals over the subintervals.
The integral will be positive on those subintervals
where f(x) 0, and negative where f(x) 0.
A(t)
2
2sin t 2
= lim
= .
2
t 0
B(t)
3
3t
Example 9. Let f : [0, ) R be a continuous and strictly increasing function such that
x
f 3(x) = t f 2 (t)dt , x > 0. Find the area enclosed by
0
y = f (x), the x-axis and the ordinate at x = 3.
The integral over the entire interval will yield the
difference of the areas above and below the x-axis. To
find the sum of the areas in the ordinary sense, one has
to find the sum of the absolute values of the integrals
over the above indicated subintervals or compute
the integral
AREA UNDER THE CURVE
b
a
For example, if f(x) > 0 for x [a,c] and f(x) < 0 for
x [c,b] (a < c < b) then the area bounded by curve
y = f(x) and x–axis between x = a and x = b is
b
c
b
a
c
A = | f(x) |dx = | f(x) |dx + | f(x) |dx
a
c
b
a
c
Compute the area of the plane figure
Example 3.
A = | f ( x ) | dx
3.11
5
bounded by the interval , of the x-axis, the
6
graph of the function f(x) = cosx, and segments of the
5
straight lines x =
and x =
6
Solution The graph of f(x) = cosx is shown below
Y
= f(x) dx f(x) dx .
Example 1. Find the area bounded by y = x 3
and x–axis between the ordinates x = – 1 and x = 1.
Solution The graph of y = x3 is shown below :
:
–
5
6
–
2
0
–1
X
2
Solving the equation cos x = 0, we find that the graph
The required area =
0
0
1
x3dx x3dx
1
3 1
0
0
sin xdx = – cos x
0
2
sin xdx
= – (cos – cos 0)
= – (– 1 – 1) = 2.
sin xdx – cos x
0
2
/2
5 /6
/2
/2
/2
0
/2
= cos xdx cos xdx cos xdx
2
| sin x | dx sin xdx
cos x dx
5 / 6
Example 2. Compute the area bounded by the
curve y = sin x and the x-axis, for 0 x 2.
Solution Since sin x 0 when 0 x and sin x 0
when < x 2, we have
2
=
x
x4
= +
4 0
4 1
1
1 1
= 0 – + – 0 = .
4 4
2
A=
5
of the function y = cos x on the interval ,
6
intersects the x-axis at the points x1 = , x2 = .
2
2
Consequently, the require area
–(cos 2 cos ) –2 .
/2
= sin x 5 /6 sin x /2 sin x /2
7
.
2
Example 4. Compute the area of the figure bounded
by the curve (y x 2)2 = 9x and the co-ordinate axes.
Solution The given equation represents a parabola
touching y axis at (0, 2) and cutting x axis at (1, 0)
and (4, 0).
Solving (y x 2)2 = 9x for y, we get y = x + 2 + 3 x
and y = x + 2 3 x . The curve below the line y = x + 2
is given by y = x + 2 3 x .
y = x+2+3 x
Y
Consequently, A = 2 + |–2| = 4.
(0,2)
(1,0)
(4,0) X
3.12
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
The required area
1
4
= x 2 3 x dx x 2 3 x dx = 1.
0
1
The area bounded by y = x2 + 2 and
p
where p and q are
y = 2| x | – cosx is of the form of
q
relatively prime, then find p – q.
Solution Let us solve for points of intersection
Example 5.
x2 + 2 = 2| x | – cosx
(|x| – 1)2 + 1 = –cosx
x = ± 1.
For –1 < x < 1, x2 + 2 > 2 |x| – cos x
The required area
1
= x 2 2 | x | cos x dx =
2
1
8
.
3
Thus, p – q = 8 – 3 = 5.
A
1.
2.
Find the area enclosed between y = sin x and
3
.
x-axis as x varies from 0 to
2
Under what condition does the value of the integral
b
f (x)dx coincide with the value of the
a
3.
area of the curvilinear trapezoid bounded by the
curves y = f(x), x = a, x = b, y = 0?
Which of the following statements are true?
(i) If C > 0 is a constant, the region under the line
y = C on the interval [a, b] has area
A = C(b – a).
(ii) If C > 0 is a constant and b > a 0, the region
under the line y = Cx on the interval [a, b] has
1
C(b – a)
2
(iii) The region under the parabola y = x2 on the interval
1
[a, b] has area less than (b2 + a2) (b – a).
2
area A =
4.
5.
6.
7.
(iv) The region under the curve y 1 x 2 on
the interval [–1, 1] has area A
.
2
(v) Let f be a function that satisfies f(x) 0 for x
in the interval [a, b]. Then the area under the
curve y = f2(x) on the interval [a, b] must
8.
9.
always be greater than the area under y = f(x)
on the same interval.
(vi) If f is even and f(x) 0 throughout the interval
[–a, a], then the area under the curve y = f(x)
on this interval is twice the area under y = f(x)
on [0, a].
Express with the aid of an integral the area of a
figure bounded by :
(i) The coordinate axes, the straight line x = 3
and the parabola y = x2 + 1.
(ii) The x-axis, the straight lines x = a, x = b and
the curve y = ex + 2 (b > a).
Prove that the whole area (when finite) included
1 x
between the axis of x and the curve y is
a a
independent of the value of a, assuming (x) 0.
Compute the area of the figure bounded by the
parabola y = x2 – 4x + 5, the x-axis and the straight
lines x= 3, x = 5.
Find the area between curve y = x2 – 3x + 2 and x–axis
(i) bounded between x = 1 and x = 2.
(ii)
bound between x = 0 and x = 2.
Find the area bounded by the curve y = x(x – 1)
(x – 2) and the x-axis.
After four seconds of motion the speed, which is
proportional to the time, is equal to 1 cm/s. What
is the distance travelled in the first ten seconds ?
B
10. Compute the area of the curvilinear trapezoid
bounded by the x-axis and the curve y = x – x2
x.
11. Find the area of the figure bounded by the curve
y = sin3x + cos3 x and the segment of the x-axis
joining two successive points of intersection of
the curve with the x-axis.
AREA UNDER THE CURVE
3.13
12. Find the area of the region bounded by the graphs
2x
, y = 0, x = 0, and x = 4.
of y
x2 9
1
13. Show that the area under y =
on the interval
x
[1, a] equals the area under the same curve on
[k, ka] for any number k > 0.
14. Compute the area of the curvilinear trapezoid
bounded by the curve y = e–x (x2 + 3x + 1) + e2, the
x-axis and two straight lines parallel to the y-axis
drawn through the points of extremum of the
function y.
15. A rectangle with edges parallel to the coordinate axes
has one vertex at the origin and the diagonally
opposite vertex on the curve y = kxm at the point where
x = b (b > 0, k > 0, and m 0). Show that the fraction of
the area of the rectangle that lies between the curve and
the x-axis depends on m but not on k or b.
16. (a) Find the area A(b) under the curve
y = e–x cos2x in the interval [0, b].
(b) Find blim
A(b).
3.4
Thus, if f(x) > g(x) for x [a,b] then the bounded area
AREA OF A REGION
BETWEEN TWO NONINTERSECTING GRAPHS
If f and g are continuous functions on the interval
[a, b], and if f(x) g(x) for all x in [a, b], then the area of
the region bounded above by y = f(x), below by
y = g(x), on the left by the line x = a, and on the right by
the line x = b is
b
A = (f(x) g(x))dx
a
We explain this using vertical strips :
Note that the vertical strip has height f(x) – g(x) if
y = f(x) is above y = g(x) and the area of the vertical
strip is (f(x) – g(x))dx
17. Prove that the area between the curve
2/3
x y 1 and the segment (– a, a) of the axis
b
a
4
of x is ab.
5
=
b
f(x) g(x) dx .
a
Remark : It is not necessary to make an extremely
accurate sketch; the only purpose of the sketch is the
determine which curve is the upper boundary and
which is the lower boundary.
Example 1. Find the area of the region enclosed
by the parabolas y = x2 and y = 2x – x2.
Solution We first find the points of intersection
of the parabolas by solving their equations
simultaneously. This gives x2 = 2x – x2, or, 2x2 –2x = 0.
Thus 2x (x – 1) = 0, so x = 0 or 1. The points of
intersection are (0, 0) and (1, 1).
We see from the figure that the top and bottom
boundaries are y1 = 2x – x2 and y2 = x2 respectively. The
area of a typical rectangle is (y1 – y2)dx = (2x – x2 – x2)
dx and the region lies between x = 0 and x = 1. So the
total area is
1
A = (2x – 2x2) dx = 2
0
2
3 1
1
(x – x2) dx
0
x
x
1 1 1
=2 =2 .
2
3
2 3 3
0
3.14
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 2. Find the area of the region between
the curves y = x – 2 and y = 2x – x2.
Solution
Y
(2, 0)
X
R
We need to know which curve is the top
curve in [0, 1]. Solving x3 = x2 – x, or x(x2 – x + 1) = 0.
The only real root is x = 0. To see which curve is on
top, takes some representative value, such as x = 0.5,
and note that because (0.5)3 > 0.52 – 0.5, the curve y = x3
must be above y = x2 – x. Thus, the cubic curve is on
top throughout the interval [0, 1].
The region is shown in the figure.
Solution
(–1, –3)
The graphs intersect in those points whose coordinates
are the simultaneous solutions of the given equations.
Eliminating y between these equations, we have
x – 2 = 2x – x2 or x2 – x – 2 = (x – 2) (x + 1) = 0. Thus
x = 2 or x = –1, and the common points of the two
graphs are (2, 0) and (–1, –3). The graph of y = x – 2 is
a straight line. The graph of y = 2x – x2 is a parabola
that is concave down and has its vertex at (1, 1). The
region whose area is sought is shaded in the figure.
This area is given by
2
1
[(2x – x2) – (x – 2)] dx =
2
2
(–x2 + x + 2) dx
1
9
1 3 1 2
= x x 2x .
2
3
1 2
Thus, the required area is given by
1
5
1 4 1 3 1 2
2
[ x3 (x
x)]dx
x
x
x
4
3
2
12
0
Top
Bottom
curve
Example 3. Compute the area of the region
between the graphs of f and g over the interval [0, 2] if
f(x) = x(x – 2) and g(x) = x/2.
Solution The two graphs are shown in the figure.
curve
Example 5. Sketch the curves y = 2 x2 ex and
y = x3 ex and compute the area of the finite portion of
the figure bounded by these curves.
y=–x3 ex
Solution
Y
–3
27e
–2
8e
g(x)= x
2
–3
2
0
X
2
f(x)=x – 2x
The shaded portion represents the required region.
Since f g over the interval [0, 2], the required area
=
=
2
2
0
0
5
[g(x) – f(x)] dx = 2 x x dx
2
5 22 23 7
.
2 2 3 3
Example 4. Find the area of the region between
the curves y = x3 and y = x2 – x on the interval [0, 1].
2 x
y=2xe
–2
Solving the two curves 2 x2 ex = x3 ex
x = 0 or 2.
dy
Consider the curve y = 2 x2 ex
= 2 x ex (x + 2)
dx
the curve has a horizontal tangent at x = 0 , 2
Also it increases in x < 2 and x > 0, and decreases in
( 2, 0) . Similarly for y = x3 ex
dy
we have
= x2 ex (x + 3)
dx
It increases in (, 3) and decreases in ( 3, ).
0
Hence, A =
(2 x e + x e ) d x .
2
2
x
3
x
AREA UNDER THE CURVE
Integrating by parts we get the result
A = x3 x 2 2x 2 e x 2 18e 2 2 .
0
Example 6. Find the area of smaller portion of
the circle x2 + y2 = 4 cut off by the line x = 1.
Solution Equation of the circle is x2 + y2 = 4 and
equation of the line is x = 1.
3.15
h
sin 1 =
a
2
3
h
1
1
sin
=
= sin 1
2
a
6
a
h=
2
a
For this value of h =
,
2
bh
a2 h2
a
b a
3a
3 ab
=
. .
=
a 2
2
4
which agrees with the second term.
Let
Example 8. For what value of the parameter
a > 0 is the area of the figure bounded by the curves
The required area = area ABCA
2
=
2
2 ydx 2
1
4 x dx
2
1
=
=
=
2
x 2 x2 22
x
2
sin 1
2
2
2
1
3
1
0 2 sin 1 (1) 2
2sin 1
2
2
2
4 3 3
.
3
x2 y2
Example 7. The ellipse 2 2 = 1 is divided
a
b
into two parts by a line parallel to the y axis . Find the
equation of the line if the area of the smaller part is a b
3
.
3
4
y = a x and y = 1 x and the y axis equal to the
number b? Also state the value of b for which the
problem has a solution .
1
a
Solution Solving the curves : P a 2 1 , 2
a 1
1
a2 1
Hence, b =
0
1 x a x dx
1
2
2
a 2 1
b = (1 x)3/2 ax3/2
3
3
0
2 3b
a=
.
3b 4 3b
The problem has a solution for 0 < b < 4/3.
Solution Let the line be x = h parallel to the
y axis . The smaller area
a
b
a 2 x2 d x
a
h
a
2
b x a 2 x 2 a sin 1 x
= 2
2
a h
a 2
2
h
a2
h
2b a
2
2
a
h
sin 1
=
2
4
2
a
a
1
h
2b 2
b
a sin 1
Hence,
h a 2 h2
a a
a 4 2
3
= ab
a b (given)
3
4
= 2
Example 9. Sketch thecurve y2 = xe2x andcompute
the area of the figure bounded by the curve. You may
x2
dx is
.
assume that the value of the integral e
2
0
Solution
3.16
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Solution
A = 2 y dx
0
A = 2
x e
x
dx
a
2
0
Consider
ex
x e x
dx
+
0
0 2 x
I=
0
x=
Solving ex = ea – x, we get e2x = ea
x e
x
dx =
2
= 0 + e t dt
where x = t2
0
Hence A =
.
a 2
Example 10. Find the area included between the
parabola x2 = 4ay and the curve y = 8a3/(x2 + 4a2).
Solution The curve (x2 + 4a3) = 8a3 is symmetrical
about y-axis. Equating to zero the coefficient of the
highest power of x in the curve parallel to x-axis. Also
this curve cuts the y-axis at (0, 2a). Solving the two
given equations x 2 = 4ay and y = 8a3/(x2 +4a2) we
get their points of intersection as (±2a, a).
Also both the curves are symmetrical about y-axis.
(e · e
Area S =
a
x
a 2
e x ) dx = (ea · e x e x ) 0
0
=
(ea + 1) – (ea/2 + ea/2) = ea – 2ea/2 + 1 = (ea/2 – 1)2
S
ea 2 1
1 ea 2 1
=
2 =
4 a 2
a
a
2
lim
a 0
2
S
1
= .
a2
4
Example 12. F i n d t h e a r e a e n c l os e d by
curve y = x2 + x + 1 and its tangent at (1,3) and between
the ordinates x = – 1 and x = 1.
dy
dy
Solution dx = 2x + 1. The slope of tangent is dx = 3
at x = 1.
The equation of tangent is y – 3 = 3 (x – 1) y = 3x
Y
Now the required area OPAQO
= 2 × area OPA (by symmetry)
= 2 × [area OAPM – area OPM]
0
2a
8a 3
ydx, for x 2 4ay
2
(x 4a 2 ) 0
2a
8a 3 dx 2 x 2 dx
4a
2
2
0
0 x 4a
=2
2a
= 16a3 .
x=1 X
1
2a
= 2 ydx, for y
x=–
2a
y=3x
1
The required area = (x x 1 3x) dx
2
2a
1 1 x
1 x3
4a 2
2
tan
2
a
2a
2a 0 2a 3 0
3
4 2
= 2 – 2 a .
3
Example 11. Let 'a' be a positive constant.
Consider two curves C1: y = ex, C2 : y =ea – x. Let S be the
area of the part surrounding by C1, C2 and the y-axis,
S
then find lim
.
a0 a 2
1
1
=
2
(x 2x 1) dx
1
1
1
1
x3
x2 x
3
1
= 3 1 1 – 3 1 1 =
2
8
+2= .
3
3
Example 13. Suppose g(x) = 2x + 1 and h (x) = 4x2
+ 4x + 5 and h (x) = (fog)(x). Find the area enclosed by
the graph of the function y = f (x) and the pair of tangents
drawn to it from the origin
AREA UNDER THE CURVE
Given
g (x) = 2x + 1, h (x) = (2x + 1)2 + 4
h (x) = f [ g (x) ]
= (2x + 1)2 + 4 = f (2x + 1)
Let 2x + 1 = t
f (t) = t2 + 4
f (x) = x2 + 4
Solving, y = mx and y = x2 + 4, we get
x2 – mx + 4 = 0
For tangency, put D = 0
m2 = 16
m=±4
3.17
The tangents are y = 4x and y = – 4x
Solution
...(1)
2
2
The required area = 2 [(x 4) 4x]dx
0
2
2
2
2
16
3
= 2 [(x 2) dx = 3 (x 2) =
.
0
3
0
C
1.
2.
3.
4.
5.
6.
7.
8.
Find the area included between curves y = 2x – x2
and y + 3 = 0.
Find area between curves y = x2 and y = 3x – 2 from
x = 0 to x = 2.
Calculate the areas of the figures bounded by
3y = – x2 + 8x – 7, y + 1 = 4/(x – 3)
At what values of the parameter a > 0 is the area of
the figure bounded by the curves x = a, y = 2x,
y = 4x larger or equal to the area bounded by the
curves y = 2x, y= 0, x = 0, x = a?
Find the area of the figure bounded by the straight
line y = – 8x – 46 and the parabola y = 4x2 + ax + 2,
if it is known that the tangent to the parabola at
the point x = – 5 makes an angle – tan–1 20 with
the positive x-axis.
Find the area of the figure bounded by the parabola
y = ax2 + 12x – 14 and the straight line y = 9x – 32,
if the tangent drawn to the parabola at the point
x = 3 is known to make the angle – tan–16 with
the x-axis.
Find the area of the figure bounded by the curve
y – 15 = e2x and the curve y = 7 e x dx passing
through the point A(0, 10).
Find the area of the region in the first quadrant
below y = – 7x + 29 and above the portion of y = 8/
(x2 – 8) that lies in the first quadrant.
3.5
AREA OF A REGION
BETWEEN TWO
INTERSECTING GRAPHS
10. Find the approximate area of the region that lies
below the curve y = sin x and above the line
y = 0.2x, where x 0.
11. The accompanying figure shows velocity versus
time curves for two cars that move along a straight
track, accelerating from rest at a common starting
line.
(a) How far apart are the cars after 60 seconds ?
(b) How far apart are the cars after T seconds,
where 0 T 60 ?
v(ft/s)
180
v1(t)=3t
2
v2(t)=t /20
60 t(x)
12.
Prove that the curves y2 = 4x and x2 = 4y divide
the square bounded by x = 0, x = 4, y = 0, y = 4 into
three equal areas.
13. Find the area of the region bounded by the
parabola (y – 2)2 = (x – 1) and the tangent to it at
ordinate y = 3 and x–axis.
14. Find the area of the region bounded by the curves
2
2
.
y = x and y =
1 x2
If we are asked to find the area between the curves
y = f(x) and y = g(x) where f(x) g(x) for some values of
x but g(x) f(x) for other values of x, then we split the
given region S into several regions S1, S2, ...... with
3.18
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
areas A1, A2, ..... as shown in the figure. We then define
the area of the region S to be the sum of the areas of
the smaller regions S1, S2, ...., that is, A = A1 + A2 + .....
The area between the curves y = f(x) and y = g(x) and
between x = a and x = b is
=–
1 (1)4 ( 1)2 2 2 12 4 23
.
4 4
2
2
44 16
b
|f(x) – g(x)| dx
A=
a
When evaluating the integral, however, we must split
it into integrals corresponding to A1, A2, ....
f(x) g(x) when f(x) g(x)
|f(x) – g(x)| =
g(x) f(x) when g(x) f(x)
We explain this using vertical strips :
Area by vertical strips
Note that the vertical strip has height f(x) – g(x) if
y = f(x) is above y = g(x), and height g(x) – f(x) if
y = g(x) is above y = f(x). In either case, the height can
be represented by |f(x) – g(x)|, and the area of the vertical
strip is dA = |f(x) – g(x)|dx
Thus, we have A =
b
a
|f(x) – g(x)| dx.
Example 1. Compute the area of the region
between the graphs of f and g over the interval [–1, 2]
if f(x) = x and g(x) = x3/4.
Solution The region is shown in the figure.
Here we do not have f g throughout the interval
[–1, 2]. However, we do have f g over the subinterval
[–1, 0] and g f over the subinterval [0, 2].
0
2
1
0
A = [g(x) – f(x)] dx + [f(x) – g(x)] dx
2
x3
x dx 0 x dx
1 4
4
=
0 x2
Example 2. Find the area of the region bounded
by the line y = 3x and the curve y = x3 + 2x2.
Solution The region between the curve and the
line is the shaded portion of the figure.
Part of the process of graphing these curves is to find
which is the top curve and which is the bottom. To do
this we need to find where the curves intersect :
x3 + 2x2 = 3x or x(x + 3) (x – 1) = 0
The points of intersection occur at x = –3, 0 and 1. In
the subinterval [–3, 0], labeled A1 in figure, the curve
y = x3 + 2x2 is on top (test a typical point in the
subinterval, such as x = –1), and on [0, 1], the region
labeled A2, curve y = 3x is on top. The representative
vertical strips are shown in the figure and the area
between the curve and the line is given by the sum
A=
0
1
[(x3 + 2x2) – (3x)] dx + [(3x) – (x3 + 2x2)]dx
3
0
0
1
1 4 2 3 3 2
3 2 1 4 2 3
= x x x x x x
3
2 3 2
4
3 0
4
81
54
27
3
1
2
– 0 = 71 .
=0–
3
2 2 4 3
4
6
AREA UNDER THE CURVE
Example 3. Find the area of the region bounded
by the curves y = sin x, y = cos x, x = 0, and x = /2.
Solution The points of intersection occur when
sin x = cos x, that is, when x =/4 (since 0 x /2).
The region is sketched in the figure.
3.19
5 4
(sin x cos x)dx
Using strip-1, A =
4
4
Using strip-2, the same area A =
(cos x sin x) dx .
3 4
Hence, we get the desired result.
Example 5. Find the area of the figure bounded
by the curves y = |x – 1|, y = 3 – |x|
Solution The curves meet at two points (see
figure). Solving the equation 3 – |x| = |x – 1|, we find the
abscissa of these points : x = –1, x = 2.
y
Observe that cos x sin x when 0 x /4 but
sin x cos x when /4 x . Therefore , the
required area is
Therefore, A =
0
/4
0
(cos x – sin x) dx +
/2
/4
(sin x – cos x) dx
/2
1 1 0 1 0 1 1 1
=
2
2
2
2
= 2 2 – 2.
/4
(cos x – sin x) dx.
0
Using the concept of area prove
4
5 4
that
4
We divide the integral into three integrals over the closed
intervals [–1, 0], [0,1], [1, 2] respectively. We obtain
0
A =
1
1
[(3 x) (1 x)]dx [(3 x) (1 x)]dx
0
2
+ [(3 x) (x 1)]dx = 1 + 2 + 1 = 4.
1
Note: In this particular example we could
have saved some work by noticing that the region is
symmetric about x = /4 and so
Example 4.
(3 | x | | x 1| dx
1
= [sin x + cos x] 0 / 4 + [– cos x – sin ] / 4
A = 2A1 = 2
x
2
2
/2
|cosx – sinx|dx = A1 A2
A=
=
–1 0
(sin x cos x) dx =
(cos x sin x) dx .
3 4
Solution The curves y = sinx and y = cos x intersect
each other at several points enclosing regions of equal
areas. We compute the area of one such equal region.
Example 6. Find the area of the region enclosed
by x = y2 and y = x – 2.
Solution To make an accurate sketch of the region,
we need to know where the curves x = y2 and y = x – 2
intersect. We find intersection points by equating the
expressions for y.
x = y2 and x = y + 2
...(1)
This yields
y2 = y + 2 or y2 – y – 2 = 0 or (y + 1) (y – 2) = 0
from which we obtain y = –1, y = 2. Substituting these
values in either equation in (1) we see that the
corresponding x-values are x = 1 and x = 4, respectively,
so the points of intersection are (1, –1) and (4, 2). (Fig. a)
Y
2
(4,2)
y=x–2
(x=y+2)
x=y2
A
–1
(1,–1)
Fig: (a)
4
X
3.20
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Y
2
(4,2)
y=x–2
y= x
A
A1
4
(1,–1)
–1
X
y=- x
Fig: (b)
The equations of the boundaries must be written so
that y is expressed explicitly as a function of x. The
x (rewrite
x = y2 as y = ± x and choose the + for the upper
portion of the curve). The lower portion of the
upper boundary can be written as y =
boundary consists of two parts : y = – x for 0 x
1 and y = x – 2 for 1 x 4 (Figure b).
Because of this change in the formula for the lower
boundary, it is necessary to divide the region into two
parts and find the area of the each part separately.
With f(x) =
x , g(x) = – x , a = 0 and b = 1, we obtain
1
A1 = [ x ( x )] dx
0
=2
1
0
With f(x) =
A2 =
1
2
4
4
x dx 2 x3 / 2 = – 0 =
3
3
0 3
x , g(x) = x – 2, a = 1, and b = 4, we obtain
4
4
[ x (x 2)]dx = 1 ( x x 2)dx
1
4
2 3/2 1 2
= x x 2x
2
3
1
16
2
1
19
= 3 8 8 – 3 2 2 6
Thus, the area of the entire region is
4 19 9
.
A = A1 + A2 =
3 6 2
Example 7. The line y = mx bisects the area
3
enclosed by the lines x = 0, y = 0, x = and the curve
2
y = 1 + 4x – x2. Find m.
Solution Here, the curve is x2 – 4x = 1 – y,,
i.e., (x – 2)2 = – (y – 5).
It is a parabola whose vertex is (2, 5), axis is x – 2 = 0.
3
The area enclosed by the lines x = 0, y = 0, x = and
2
the curve
=
3/2
3/2
0
0
ydx (1 4x – x ) dx
2
3/2
x3
2
x
2x
–
3 0
3
9 1 27 39
2· – ·
.
=
2
4 3 8
8
The area bounded by the lines y = mx, y = 0 and x = 3/2
=
3
x2 2 9
ydx
mx dx = m m.
=
0
0
2 0 8
According to the question,
9
1 39
m= ·
8
2 8
13
m= .
6
3/2
3/2
B
1.
Supose that f and g are integrable on [a, b], but
neither f(x) g(x) nor g(x) f(x) holds for all x in [a, b]
[i.e., the curves y = f(x) and y = g(x) are intertwined].
(a) What is the geometric significance of the
integral
3.
b
[f(x) g(x)]dx ?
a
(b) What is the geometric significance of the
integral
2.
b
| f(x) g(x) | dx ?
a
Find the area of the region R lying between the
lines x = –1 and x = 2 and between the curves
y = x2 and y = x3.
4.
Find the area of the closed figure bounded by the
following curves.
2
(i) y = 3x + 18 – x , y = 0
2
2
(ii) y = x – 2x + 2, y = 2 + 4x – x
3
2
(iii) y = x – 3x – 9x + 1, x = 0, y = 6 (x < 0)
6x 2 x 4
(iv) y =
, y= 1
9
Find the area of the region in the first quadrant
bounded on the left by the y-axis, below by the
x
line y = , above left by the curve y = 1+ x and
4
above right by the curve y = 2
x
AREA UNDER THE CURVE
5.
Find the area included between the curves
–1
–1
y = sin x, y = cos x and the x-axis.
6.
3.21
Find the area bounded by f(x) = max{sin x, cos x},
x = 0, x = 2and the x-axis.
D
7.
Find the area of the closed figure bounded by the
following curves.
2
(i) y = sin x , y = x
2
4
2
(ii) y = – 2x + 5x + 3, y + 1 =
x 1
x 2 3x 1
(iii) y =
, x = 1, x = 2, y = 0
x
(iv) y = x , y = 4 3x , y = 0
Find the area of the closed figure bounded by the
3
curves y = 2 – | 2 – x | and y =
|x|
9. Compute the area of the figure bounded by the
ln x
curves y =
and y = x ln x.
4x
10. Find the area
of the figure lying between the curve
2
13. For what values of m do the line y = mx and the
curve y = x/(x2 + 1) enclose a region ? Find the area
of the region.
14. The figure shows a horizontal line y = c intersecting
the curve y = 8x – 27x3. Find the number c such that
the areas of shaded regions are equal.
8.
x
y = x e 2 and its asymptote.
11. Find the area included between curves y =
and 5y = 3|x| – 6.
12. Find the area bounded by the curve |y| +
3.6
4 x2
4 x2
1
= e–|x|.
2
AREA BY HORIZONTAL
STRIPS
For many regions it is easier to form horizontal strips
rather than vertical strips. The procedure for horizontal
strips duplicates the procedure for vertical strips. If we
want to find the area between the curve x = g(y) and
the y-axis on the interval [c, d], we form horizontal
strips. Such a region is shown in the figure, together
with a typical horizontal approximating rectangle of
widthdy, which we refer to as horizontal strip.
15. A circle with radius 1 and centre on the y-axis
touches the curve y = 2x twice. Find the area of
the region that lies between the two curves.
16. Find the area of the figure bounded by the curves
y = (x + 1)2 , x = siny, y = 0 (0 y 1)
17. Find the area of the figure bounded by the curves
y = e–x |sinx|, y = 0 (x 0) (assume that the area of
this unbounded figure is the limit, as A , of
the areas of the curvilinear trapezoids
corresponding to the variation of x from 0 to A).
Thus, if g (y) 0 for y [c, d] then the area bounded
by the curve x = g(y), y–axis and the abscissa y = c and
d
y = d is
g(y)dy .
yc
Sometimes, a curve in the xy-plane can be defined
as the graph of a function of y and not as a function
of x. An example is the parabola defined by the
equation (y – 1)2 = x – 1 and illustrated in the figure.
3.22
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Although this equation cannot be solved uniquely for
y in terms of x, it is easy to do the opposite. We get
x = (y – 1)2 + 1 = y2 – 2y + 2, and so the curve is the graph
of the function f defined by f(y) = y2 – 2y + 2. The area
of the region bounded by the parabola, the two coordinate
axes, and the horizontal line y = 2 is given by
Area
=
2
2
f(y) dy = (y2 – 2y + 2) dy
0
0
2
y3
8
2
= y 2y .
3
0 3
Example 1. Find the area bounded between
y = sin–1x and y–axis between y = 0 and y = .
2
–1
Solution y = sin x x = sin y
2
Further, if we wish to find the area between the curves
x = F(y) and x = G(y) where F(y) G(y) for some values
of y but G(y) F(y) for other values of y, then we split
the given region S into several regions.
Approximation by horizontal strips of width
dy
The required area = sin y dy = cos y 02 = – (0 – 1) = 1.
0
Alternative :
The area in the above example can also evaluated by
integration with respect to x. The required area = (area
of rectangle formed by x = 0, y = 0 , x = 1, y = ) – (area
2
bounded by y = sin–1x, x–axis between x = 0 and x = 1)
1
=
=
=
1
× 1 – sin x dx
2
0
– (x sin–1x + 1 x 2 )1
2
– 0 0 1 = 1.
2
2
Note: If F and G are continuous functions and
if F(y) G(y) for all y in [c, d], then the area of the region
bounded on the left by x = G(y), on the right by
x = F(y), below by y = c, and above by y = d is
d
A = (F(y) G(y))dy
c
Figure Area by horizontal strips
Note that regardless of which curve is “ahead” or
“behind,” the horizontal strip has length |F(y) – G(y)|
and area dA = |F(y) – G(y)| dy. However, in practice, we
must make sure to find the points of intersection of the
curve and divide the intergrals so that in each region
one curve is the leading curve (“right curve”) and the
other is the trailing curve (“left curve”). Suppose the
curve intersect where y = b for b on the interval [c, d],
as shown in the figure, then
b
A=
d
F(y)]dy [F(y) G(y)
[G(y)
.
c
G ahead of F
b
F ahead of G
Example 2. Find the area enclosed by the line
y = x – 1 and the parabola y2 = 2x + 6.
Solution By solving the two equations we find
that the points of intersection are (–1, –2) and (5, 4).
3.23
AREA UNDER THE CURVE
1
y3
4
y
= 2
= .
3
3
0
Example 4. Find the area of the region between
the parabola x = 4y – y2 and the line x = 2y – 3.
Solution The figure shows the region between the
parabola and the line, together with a typical horizontal
strip.
We solve the equation of the parabola for x and notice
from the figure that the left and the right boundary
1
curves are xL = y2 – 3 and xR = y + 1.
2
We must integrate between the appropriate y-values,
y = –2 and y = 4. Thus
4
A = (xR – xL) dy
4
2
=
[(y + 1) – (1/2 y2 – 3)] dy
=
(–1/2 y + y + 4) dy
=
=
2
4
2
2
4
1 y3 y 2
4y
2 3 2
2
1
4
– (64) + 8 + 16 – 2 8 = 18.
6
3
Example 3. Compute the area of the figure
bounded by the parabolas x = – 2y2 , x = 1 – 3y2.
x
Solution We rewrite the equations as y2 = – 2 and
1 x
.
3y2 = 1 – x y2 =
3
The graph is shown below :
Solving the two equations :
x 1 x
= 2x – 2 = 3x
3 2
x = –2 y = 1, –1.
Since 1 – 3y2 – 2y2 for –1 y 1, the required area
=
1
[(1–3y – (–2y )] dy
1
2)
2
To find where the line and the parabola intersect, we
solve 4y – y2 = 2y – 3 to obtain y = –1 and y = 3
Throughout the interval [–1, 3], the parabola is to the
right of the line (test a typical point between –1 and 3,
such as y = 0). Thus, the horizontal strip has area
[(4y y 2 ) (2y 3)]dy
dA =
Right curve
Left curve
and the area between the parabola and the line is given
by
3
A = [(4y – y2) – (2y – 3) dy
1
3
=
(3 + 2y – y ) dy = (3y + y – y )|
=
(9 + 9 –9) – (–3 + 1 + 13 ) = 10 23 .
1
2
2
1
3
3 3
1
Alternative :
In this example, the area can also be found by using
vertical strips, but the procedure is more complicated.
Note in the figure that on the interval [–5, 3], a
representative vertical strip would extend from the
bottom half of the parabola y2 – 4y + x = 0 to the line
y = 21 (x + 3), whereas on the interval [3, 4], a typical
vertical strip would extend from the bottom half of the
parabola y2 – 4y + x = 0 to the top half. Thus, the area
is given by the sum of two integrals. It can be shown
that the computation of area by vertical strips gives
the same result as that found by horizontal strips.
3.24
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
3
Q = (2, – · 2) = (2, – 3). Consequently, a cross
2
section of the region is determined for each number
y in the interval [–3, 4]. The area of the region is
therefore
4
x dy . For 0 y 4, the cross section is
–3
determined by the line x = 2 and the parabola
y . Thus for 0 y 4, x1 = 2 – y ,the large
minus the smaller. For –3 y 0, the cross section
x=
Example 5. (i) Find the area of the region shown
in the figure. The region is bounded by the curve
y = x2, the line y = – 3/2 x, and the line x = 2.
is determined by the line x = 2 and the line
x2 = – 2/3y. Thus for – 3 y 0,
2y
2
x = 2 – – y = 2 +
3
3
The integral now breaks into two separate
integrals,
0
4
x dy x dy
1
–3
0
0
2
2y
4
2 3 dy (2 – y ) dy
=
–3
0
y2 0
2 3/2 4
+ 2y – y 0
2y 3
3
–3
8
17
= 0 – (– 3) + – 0 =
.
3
3
(i) Needed only one integral, but (ii) needed two.
Moreover, in (ii) the formula for the cross-sectional
length when 0 y 4 involved y , which is a
little harder to integrate than x2, which appeared
in the corresponding formula in (i). Although both
approaches to finding the area of the region in are
valid, the one with cross sections parallel to the y
axis is more convenient here.
=
(ii) Find the area of the same region, but this time use
cross sections parallel to the x axis.
Solution (i) Here, f(x) = x2 and g(x) = – 3/2x. For
x in [0, 2], the cross-sectional length is
x2 – (– 3/2 x). Thus the area of the region is
2
x 2 – – 3x dx 2 x 2 3x dx
0
0
2
2
x3
3x 2 2 23
3·22
03
3·0 2 17
= 3 4 0 =
– 4 3 .
4 3
3
(ii) Since the cross-sectional length is to be expressed
in terms of y, first express the equations of the
curves bounding the region in terms of y. The
curve y = x2 may be written as x = y , since we
are interested only in positive x. The curve
y = – 2/3 x can be expressed as x = – 2/3 y by
solving for x in terms of y. The line x = 2 also
bounds the region. The point P in the figure lies
on the parabola y = x2 and has the x coordinate 2.
Thus P = (2, 22) = (2, 4). The point Q lies on the line
3
y = –
x and has x coordinate 2. Thus,
2
Example 6. Compute the area enclosed between
the curves y = tan–1x , y = cot–1x and the y-axis.
Solution The required area
1
A=
(cot x tan x)dx
1
1
0
1
1
= 2 tan x dx = ln 2
2
0
AREA UNDER THE CURVE
3.25
Alternative :
4
4
A=2
x dy = 2 ·
tan y dy = ln 2.
0
0
E
1.
The area of the region that lies to the right of the
y-axis and to the left of the parabola x = 2y – y2 is
4.
2
2
given by integral (2y y )dy . (Turn your head
0
clockwise and think of the region as lying below
the curve x = 2y – y2 from y = 0 to y = 2.) Find the
area of the region.
5.
6.
7.
8.
2.
3.
The boundaries of the shaded region are the
y-axis, the line y = 1, and the curve y = 4 x . Find
the area of this region by writing x as a function of
y and integrating with respect to y.
Find the area of the shaded region
Find the area of the region in the 1st quadrant that
is bounded above by y = x and below by the
x-axis and the line y = x – 3.
Compute the area enclosed between the curves
y = sec–1x , y = cosec–1x and line x – 1 = 0.
Find the area of the region bounded by x = y2
and x = 3y – 2y2.
Find the area of the region given by
(i) y-axis, x = y3 – 3y2 – 4y + 12.
(ii) y
1
1 x
2
,y
2
, y axis .
x 1
(a) If f(y) = –y2 + y + 2, sketch the region bounded
by the curve x = f(y), the y-axis, and the lines
y = 0 and y = 1. Find its area.
(b) Find the area bounded by the curve x = –y2 +
y + 2 and the y-axis.
(c) The equation x + y2 = 4 can be solved for x as
a function of y, or for y as plus or minus a
function of x. Sketch the region in the first
quadrant bounded by the curve x + y2 = 4,
and find its area first by integrating a
function of y and then by integrating a
function of x.
10. Find area common to circle x2 + y2 = 2 and the
parabola y2 = x.
2
11. Find the area of the region bounded by y + 4x = 0
2
and (y + 4)x + 8 = 0.
9.
3.26
3.7
AREA OF A REGION
BETWEEN SEVERAL
GRAPHS
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
If it is required to find the area of a plane figure of more
complicated form, then we try to express the sought
for area in the form of an algebraic sum of the area of
certain curvilinear trapezoids. For instance, the area of
the figure represented in the figure below is computed
by the formula
S = SaABb –SaACc – ScCBb
Example 2. Find the area bounded by the
curves y = sin –1 |sin x| + cos–1 cos x and y = cos x +
and the lines x = 0 and x = 4.
Solution Let f(x) = sin–1 |sin x| + cos–1 cos x
f(x) = sin–1 (sin x) + cos–1 (cos x) (for 0 x )
2x,
0x
2
3
x
f(x) = ,
2
2
3
4 2x, 2 x 2
The area bounded is shown in the figure
/2
Let the curves AB, BC and AC be the respective graphs
of the following functions :
y = f(x), x [a, b] ; y = g(x), x [a, c]
and y = h(x), x [c, b]. Then
A = 4 ( + cos x – 2x) dx
0
3 / 2
+2
( – ( + cos x)) dx
/2
b
c
b
=
4[x + sin x – x2 ]0 /2 2[sin x]3/2/2
a
a
c
=
2
2
4 1 – 2 [–2]
4
2
2
2
2 + 4 – + 4 = (2 + 8).
S = f(x)dx g(x)dx h(x)dx
Example 1. Compute the area of the plane
figure bounded by the curves y = x , x [0, 1],
y = x2, x [1, 2] and y = –x2 + 2x + 4, x [0, 2].
Solution
=
Example 3. Find the area of the region enclosed
between the curves 4|y| = |4 – x2| and |y| (x2 + 4) = 12.
Y
Solution
(0,1)
O
The required area
2
1
2
S = ( x 2x 4)dx
0
=
The required area
2
xdx x dx
=
2 2 2
2 2 12
2 4 x2
x 4
dx
dx
dx
4
2
4
4
0 x 4
0
2
=
2
2 2
2 2
12
1
x3
1 x3
1 x
tan
4x
4x
4
2
20
4
3 0 4 4
2
2
0
3 2
1
1
X
3 2
2
x
2
x
19
2
x 2 4x 0 x3/2
.
0
3 0
3
3
3
0
1
AREA UNDER THE CURVE
4 42 2
4 6 tan 1 2
3
3
4
1
(18 tan
2 2 2 ).
=
3
Example 4. Find the area bounded by the region
inside the ellipse, x2 + 2 y2 = 2 and outside the graph of
the function , f (x) = 1 x2
=
1
. (8a)2 + [area O P M + area M P R]
2
4a 3
1
=
. (8a)2 + 2 [
x dy, for y2 = 12ax]
0
2
=
+2[
3a
4a 3
= 32 a2 + 2
The graph is shown below :
Solution
3.27
4a 3
0
x dy, for x2 + y2 = 64a2]
8a
y2
( 4a 2 y 2 ) dy
dy + 2
4a 3
12a
4a 3
= 32a2 +
1 y3
6a 3 0
8a
1
64a 2 1 y
2
2
sin
+ 2 y (64a y )
2
8a 4a 3
2
= 32a2 +
1
A
=2
2 2 y 1 y dy
2
1/2
1
=
2 2
1
1 y dy
1/2
1/2
=
1 y 2 dy 2
2 2
3 3
8 9 3
=
.
3
2 2
6 2
Example 5. Show that the larger of the two areas
into which the circle x3 + y2 = 64a2 is divided by the
parabola
16 2
a [8 – 3 ].
y2 = 12ax is
3
Solution x2 + y2 = 64a2 is a circle with centre (0, 0)
and radius 8a and y2 = 12ax is a parabola whose vertex
is at (0, 0) and latus rectum 12a. Both the curves are
symmetrical about x-axis. Solving the two equations,
1 64 3 3a 3
6a
3
+ 2[{0 – 8a2 3 } + 32a2 {sin–11 – sin–1 ( 3 / 2 )}]
32 2
32 3a 2
– 16a2 3
a
= 32a2 +
3
3
128 2
16 2
16 2
=
a –
a 3
a (8 – 3 ).
3
3
3
Example 6. Show that the area included between
8
the parabolas y2 = 4a(x + a), y2 = 4b (b – x) is (a + b)
3
ab .
Solution y2 = 4a (x + a) represents a parabola whose
vertex is (–a, 0) and latus rectum is 4a. Also y2 = 4b(b – x)
represents a parabola whose vertex is (b, 0) and latus rectum
4b. Both the curves have been shown in the figure.
the co-ordinates of the common point P are (4a, 4a 3 ).
S
Y
R
M
(0,8a)
(4 ,4
P
3)
O
A
X
T
Q
Now the area of the larger portion of the circle (i.e. the
shaded area) = the area PRSTQOP = the area of the
semi-circle RST + 2 area OPR
Equating the values of y2 from the two given equations
of parabolas, we get 4a(x +a) = 4b (b – x) or, x = b – a
i.e. the abscissa of the point of intersection P is b – a.
Now both the curves are symmetrical about x-axis.
The required area
= 2 [Area APM + Area PMB], by symmetry
3.28
=
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
b a
2 a ydx, for the parabola y2 = 4a(x + a)
+
b
b a
ydx, for the parabola y2 = 4b(b – x)
=
b
b a
2 a {4a(x a)} dx ba {4b(b x)}dx
=
4 a
b a
a
b a
=
=
=
b
(x + a)1/2 dx + 4
b
(b x) dx
1/2
b a
b
2
3/2
4 a (x a) – 4
3
a
1
1
8 a.b3/2 + 8 ba 3/2
3
3
8
(a + b) (ab) .
3
bx
b 1
1 x
(a 2 x 2 ) + . a2 sin
a
2
2a
a 2
bx
(a 2 x3 )
–
2a
ab
ab
x
1 x
sin
=
=
cos–1
2 2
a
2
a
=
2
b (b x)3/2
3
ba
ab
x
cos–1 .
2
a
x 2S
x
2S
2S
cos–1
or cos
or x = a cos .
a ab
a
ab
ab
Thus, S =
Also y =
Example 7. If P(x, y) be any point on the ellipse
x2/a2 + y2/b2 = 1 and S be the sectorial area bounded
by the curve, the x-axis and the line joining the origin
to P, show that x = a cos (2S/xb), y = b sin (2S/ab).
Solution The given ellipse is shown in the figure. We
have
b 2
b 2
2S
2S
a x2
a a 2 cos2
bsin .
a
a
ab
ab
Example 8. Find the area common to the
circle x2 + y2 = 4 and the ellipse x2 + 4y2 = 9.
Solution The equation of the circle
...(1)
x2 + y2 = 4,
and the equation of the ellipse is
x2 + 4y2 = 9.
...(2)
Both the curves (1) and (2) are symmetrical about both
the axes and have been shown in the figure.
Y
B
P(x,y)
b
O
a
M (a,0)
A X
S = sectorial area OAP (i.e. the shaded area)
= area of the OMP + area PMA
a
1
= OM . MP + ydx, for the ellipse
x
2
a b
1
(a 2 x 2 )dx ,
= xy +
x
a
2
1 b
(a 2 x 2 )
= x
2 a
a
1 2 1 x
b x
2
2
+ (a x ) a sin
2
a x
a 2
bx
(a 2 x 2 )
=
2a
a2 x
1
x
b
(a 2 x 2 ) a 2 sin 1
0
4
2
2
a
a
Solving (1) and (2) for x, we have
x2 + 4(4 – x2) = 9 or 3x2 = 7 or x2 = 7/3
The x-coordinate of the point of intersection P
lying in the first quadrant is (7 / 3) . Also putting
y = 0 in x2 + y2 = 4, we get x = 2 at C.
Now the required area is symmetrical about both the
axes.
The area common to the circle and the ellipse
= 4 × (common area lying in the first quadrant)
= 4 × area OCPB
= 4 [area OBPM + area CPM]
(7/3)
=4 0
y dx, for the ellipse
2
+ (7/3) y dx, for the circle
3.29
AREA UNDER THE CURVE
(7/3) 1
= 4
(9 x 2 )dx +
2
0
2
(4 x 2 )dx
(7/3)
x (9 x 2 ) 9 1 x
sin
=2
2
2
3
0
(7/3)
2
x (4 x 2 )
x
2 sin 1
+
2
2
(7/3)
1 7 20 9 1 1 7
= 2 2 . 3 3 2 sin 3 3
1 7 5
1 7
sin 1
+ 4 2sin 1 (1)
2 3 3
2 3
7
7
2
2 35
+ 9sin–1
+ 4–
– 8sin–1
3 35
3
3 3
2 3
1 7
1 7
= 4 + 9 sin–1 3 . 3 – 8 sin–1 2 . 3 .
=
x
2x
y = sin cos and x-axis (where [.] is the greatest
4
4
integer function).
x
2x
Solution y = sin 4 cos 4
x
x
1 < sin2 + cos < 2, for x (–2, 2]
4
4
sin 2 x cos x
y=
=1
4
4
Now we have to find out the area enclosed by the
2
3
1
x ,
4
2
line y = 1 and x-axis. The required area is shaded in the
figure.
circle x2 + y2 = 4, parabola y
Y
(0,2)
Find the area enclosed between the
1
curves y = ln (x + e), x = ln and x-axis.
y
Solution The given curves are y = ln (x + e) and
Example 9.
1
1
x = ln = ex y = e–x
y
y
Using transformation of graph we can sketch the
curves.
(–2,0)– 3 –1 –1/2 O
3 (2,0) X
(0,–2)
Hence the required area
=
0
3 × 1 + ( 3 – 1) × 1 + (x x 1) dx
1
+2
2
2
( 4 x )dx
2
3
0
x3 x 2
= (2 3 – 1) + x
2
2
1
2
Hence, the required area
=
0
1 e
0
ln (x + e) dx + e dx
e
= ln t dt +
1
x
e dx (putting x + e = t)
x
0
= (t ln t – t) 1e – (e–x) 0 = 1 + 1 = 2.
Example 10. Find the area enclosed by circle
x2 + y2 = 4, parabola y = x2 + x + 1, the curve
x 4 x 2 2sin 1 x
+2
2
2 3
1 1
= (2 3 – 1) + 0 1
3 2
2
+ 2 (0 ) 2 3
5 2
2
1
3
3 .
= (2 3 – 1) +
6 3
6
3
3.30
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
F
1.
2.
3.
4.
5.
6.
Find the area included between the ellipses
x2 + 2y2 = a2 and 2x2 + y2 = a2.
The circle x2 + y2 = a2 is divided into three parts
a2
. Determine the
by the hyperbola x2 – 2y2 =
4
areas of these parts.
Find the area of the shaded region in the figure. The
curves are parabolas. The inscribed square has area
4, and the circumscribed square has area 16.
2
A figure is bounded by y = x + 1, y = 0, x = 0, x = 1.
2
At what point of the curve y = x + 1, must a
tangent be drawn for it to cut off a trapezoid of the
greatest area from the figure ?
2
A figure is bounded by the curves y = (x + 3) ,
y = 0, x = 0. At what angles to the x-axis must straight
lines be drawn through the point (0, 9) for them to
partition the figure into three parts of the same size?
The figure here shows triangle AOC inscribed in
the region cut from the parabola y = x2 by the line
y = a2. Find the limit of the ratio of the area of the
triangle to the area of the parabolic region as a
approaches zero.
Y
2
C y=a
2
(a,a )
A
(–a,a )
2
–a
7.
2
y=x
0
aX
Let the sequence a1, a2, a3,... be in G.P. If the area
bounded by the parabolas y2 = 4a nx and
3.8.
DETERMINATION OF
PARAMETERS
Example 1. Find the area of the figure bounded
by the parabola y = ax2 + 12x – 14 and the straight line
y = 9x – 32 if the tangent drawn to the parabola at the
point x = 3 is known to make an angle – tan–16 with
the x-axis.
y2 + 4an(x – an) = 0 be n, prove that the sequence
1, 2, 3,... is also in GP.
8. What part of the area of a square is cut off by the
parabola passing through two adjacent vertices of the
square and touching the midpoint of one of its sides?
9. For what value of the parameter a > 0 is the area of
the figure bounded by the curves y = a x ,
y = 2 x and the y-axis equal to the number b?
For what values of b does the problem have a
solution?
10. Show that the area bounded by the semi-cubical
parabola y2 = ax3 and a double ordinate is 2/5 of
the area of the rectangle formed by this ordinate
and the abscissa.
11. Prove that the area common to the two ellipses
x2 y2
x2 y2
2 1, 2 2 1 is 4ab tan–1 b/a.
2
a
b
b
a
12. The area between the parabola 2cy = x2 + a2 and
the two tangents drawn to it from the origin is
1 2
a /c.
3
13. Let A and B be the points of intersection of the
parabola y = x2 and the line y = x + 2, and let C be
the point on the parabola where the tangent line
is parallel to the graph of y = x + 2. Show that the
area of the parabolic segment cut from the parabola
by the line four-thirds the area of the triangle ABC.
14. Compute the areas of the curvilinear figures formed
x2
by intersection of the ellipse
+ y2 = 1 and the
4
2
x
hyperbola
– y2 = 1.
2
15. Find the value of c for which the area of the figure
4
bounded by the curves y = 2 , x = 1 and y = c is
x
9
equal to .
4
Solution
–2
y = ax2 + 12x – 14
Y
O
2 3
X
AREA UNDER THE CURVE
dy
dy
6a 12 .
= 2ax + 12
dx
dx x 3
Hence, tan( – tan–16) = 5a + 12 – 6 = 6a + 12
a=–3
Hence , y = – 3x2 + 12x – 14
(note that D < 0, so y < 0 x R)
The point of intersection of the line with parabola are
x = – 2 or 3.
3
125
Hence A = [ 3x 2 12x 14] (9x 32)]dx =
.
2
2
Find the value of 'a' (a > 2) for which
the reciprocal of the area enclosed between y = 12 ,
x
1
, x = 2 and x = a is 'a' itself and for what
y=
4(x 1)
values of b (1, 2), the area of the figure bounded by
1
the lines x = b and x = 2 is 1 – .
b
Example 2.
Solution
3.31
We get x = 1.
a
1
4
1
A = x 2x 1 dx = ln
5
2
a
4
1
ln x ln (2x 1) = ln
2
2
5
a
x2
2
2 ln 2x 1 = ln 16
a
2
5
4
16
ln
ln 3 ln 5
2a 1
a2
64
a2
64
ln
2a 1
15
2a 1 15
15a2 – 128a + 64 = 0
8
a=8 ; a= .
15
Thus, a = 2 since a is greater than 2.
ln
x2 = 4 (x – 1) (x – 2)2 = 0
The curves touch other.
a
1
1
1
4(x 1) x 2 dx =
a
2
a = e2 + 1.
2
1
1
1
Also
1 – = 4(x 1) x 2 dx
b
b
–2
b=1+e .
For what value of 'a' (a > 2) is
1
,
the area of the figure bounded by the curves, y =
x
4
1
y=
, x = 2 and x = a equal to ln
?
5
2x 1
Example 3.
Solution
Solving the equations y =
1
1
and y =
x
2x 1
Example 4. If the line y = mx divides the area
enclosed by the lines x = 0, y = 0, x = 3/2 and the curve
y= 1 + 4x – x2 into two equal parts, then find the value of m.
Solution The given curve is y – 5 = – (x – 2)2.
Thus given curve is a parabola with vertex at (2, 5) and
axis x = 2
Given that area CBFC = Area CDEBC
So area CDEBFC = 2 Area CBFC
3/2
Area CDEBFC
(1 4x x )dx
2
0
3/2
x3
3
9 9 39
x 2x
2 .
3 0
2
4 8 8
2
3/2
Area CBFC
9m
mxdx 8
0
3.32
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
So we must have
39 18m
13
or m
8
8
6
Example 5. Consider the two curves C1 : y = 1 + cosx
& C2 : y = 1 + cos (x ) for 0, ; x [0, ]. Find
2
the value of , for which the area of the figure bounded by
the curves C1, C2 and x = 0 is same as that of the figure
bounded by C2 , y = 1 and x = . For this value of , find
the ratio in which the line y = 1 divides the area of the figure
by the curves C1, C2 and x = .
Solution
Example 6. The possible values of b > 0, so that
the area of the bounded region enclosed between the
parabola
x2
is maximum.
y = x – bx2 and y =
b
Solution The given parabolas are
...(1)
y = x – bx2
x2
...(2)
and
y=
b
The abscissae of their points of intersection are given
by
x2
= x – bx2
b
b
x2 (1 + b2) – bx = 0 x = 0, x =
1 b2
Thus, the two parabolas intersect at O (0, 0) and
b
b
A 1 b2 ,
2
2
1
b
The region enclosed by the two parabolas is shaded
in the figure.
1 + cos x = 1 + cos(x – ) x = – x
x=
2
2
Now
cos x cos(x ) dx
0
= –
cos(x ) dx
2
or, sin x sin(x )0 2 = sin(x ) 2
sin sin [0 sin( )]
2
2
= sin sin
2
2sin – sin = 1 – sin ,
2
.
Hence, 2sin = 1 =
3
2
To find the area of this region, let us slice it into vertical
strips. The approximating rectangle shown in the figure
has length = y2 – y1, width = x and it can move between
b
x = 0 and x =
. So, the area A enclosed by the two
1 b2
parabolas is given by
b
1 b2
A=
b
1 b2
y y dx
2
1
0
b
x 2 bx3 x3 1 b2
x2
2
=
x bx dx = 2 3 3b
b
0
0
2
3
1
b
1
b
2
=
1 b
2 1 b2 3b
1 b2
3.33
AREA UNDER THE CURVE
2
=
2
1 b 1 b
1
b2
2 1 b2 3 1 b2
6 1 b2 2
Now, A =
a3
dA
da = 0
1 a4
gives a = 31/4. The greatest area is bounded when a = 31/4.
=
1
b2
6 1 b 2 2
Example 8. Compute the area of the figure
bounded by the straight lines, y = 1 , y = 2 and the
1
curve y = a x2 and y = a x2 . Find also the values of
2
the parameter a (a 1) for which the area is greatest.
2 2
2
2
dA 1 1 b ·2b b ·2 1 b ·2b
2 4
db 6
1
b
dA
b
1 b2 2b 2
db 3 1 b2 3
Y
Solution
2
dA b 1 b 1 b
3
db
3 1 b2
1
For maximum value of A, we must have
dA
= 0 b (1– b) (1 + b) = 0
db
1–b= 0
b=1
b 0
dA
Since b > 0. Therefore, 1 + b > 0. Thus,
changes its
db
sign from positive to negative in the neighbourhood
of b = 1.
Example 7. Find the value (s) of the parameter
''a' (a > 0) for each of which the area of the figure
a2 a x
bounded by the straight line, y =
and the
1 a4
x 2 2 a x 3a 2
parabola y =
is the greatest .
1 a4
Solution
Y
x1
x2
A=
x2
X
2
2y
a
1
A=
y
a dy
A=2
4 53 2
3 a
Since a 1, the greatest area is obtained when a = 1,
and the greatest area is
4 53 2
.
3
Example 9. If the area bounded by y = x2 + 2x – 3
and the line y = kx + 1 is the least, find k and also the
least area.
Solution
X
1 a4
Let x1 and x2 be the roots of the equation
x2 + 2x – 3 = kx + 1
x2 + (2 – k) x – 4 = 0
x1 + x2 = k – 2, x1 x2 = – 4.
where x1 and x2 are the roots of ,
x2 + 2 a x + 3 a 2 = a 2 a x
x = 2a or x = a
a
y=1
(a 2 a x) (x2 2 a x 3a 2 ) dx
x1
A=
y=2
2a
(a 2 a x) (x 2 2 a x 3a 2 )
1 a4
dx
x2
Now, A =
[(kx 1) (x 2x 3)]dx
2
x1
3.34
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
dA
=0
dc
f(c) [2c – (a + b)] = 0 (as f(c) 0)
x2
x 2 x3
(k
2)
4x
=
2
3
x1
For maxima or minima
x 22 x12 1 3
x 2 x13 4(x 2 x1 )
= (k 2)
2
3
(k 2)2 1
(x 2 x1 )2 x1x 2 4
= (x2 – x1)
3
2
(k 2)2 1
(k 2)2 4 4
2
3
= (x 2 x1 )2 4x1x 2
=
(k 2)2 16 1
16
(k 2)2
6
3
6
2
=
.
A is minimum if k = 2 and Amin =
32
.
3
Least value of a variable area
If y = f(x) is a strictly monotonic function in (a, b), with
f '(x) 0, then the area bounded by the ordinates x = a,
x = b, y = f(x) and y = f(c) (where c (a, b)) is minimum
when c =
Proof
Hence, A is minimum when c = a b .
2
We have a similar proof when f is strictly decreasing.
Example 10. If the area bounded by
x3
– x 2 a and the straight lines x = 0, x = 2 and
3
the x-axis is minimum, then find the value of a.
Solution f ' (x) = x2 – 2x = x (x – 2). (note that f (x)
is monotonic in (0, 2)). Hence, for minimum area, f (x)
must cross the x-axis at
02
=1.
x=
2
f(x)=
32
[(k 2) 16]
6
ab
.
2
Since f is strictly increasing f(c) > 0.
a b dA
dA
Hence, for c <
,
< 0 and for c > a b ,
>0
2
dc
dc
2
Hence, c =
ab
.
2
Assume that f is strictly increasing in (a, b)
c
b
A = (f(c) – f(x)) dx + (f(x) – f(c)) dx
a
c
c
=
b
f(c) (c – a) – (f(x)) dx + (f(x)) dx – f(c) (b – c)
a
c
b
c
c
a
A = [2c – (a + b)] f(c) + (f(x)) dx – (f(x)) dx
Hence, f (1) = 1/3 – 1 + a = 0 a = 2/3.
Example 11. Find the value of the parameter 'a' for
which the area of the figure bounded by the abscissa
axis, the graph of the function y = x3 + 3x2 + x + a, and the
straight lines, which are parallel to the axis of ordinates
and cut the abscissa axis at the point of extremum of the
function, is the least.
Solution f (x) = x3 + 3x2 + x + a
6
3
Hence for minimum area, f (x) must cross the x-axis at
f ' (x) = 3x2 + 6x + 1 = 0
Differentiating w.r.t c,
dA
= [2c – (a + b)] f(c) + 2f(c) + 0 – f(c) – (f(c) – 0)
dc
x=
x=–1±
1
6
6
1
= – 1
1
2
3
3
Thus, f (–1) = – 1 + 3 – 1 + a = 0
a = – 1.
AREA UNDER THE CURVE
3.35
G
1.
2.
3.
4.
5.
6.
7.
8.
Find the value of a for which the area of the figure
bounded by the curve y = sin 2x, the straight lines x
= /6, x = a, and the x-axis is equal to 1/2.
(i) Find the area of the region enclosed by the
parabola y = 2x – x2 and the x-axis.
(ii) Find the value of m so that the line y = mx
divides the region in part (i) into two regions
of equal area.
Find the values of c for which the area of the figure
bounded by the curve y = 8x2 – x5, the straight line
x = 1 and x = c and the abscissa axis is equal to
16/3.
Find the values of c for which the area of the figure
bounded by the curves y = 4/x2, x = 1 and y = c is
1
equal to 2 .
4
Find the value of k for which the area of the figure
bounded by the curves x = /18, x = k, y = sin6x and
the abscissa axis is equal to 1/6.
At what value of d is the area of the figure bounded
by the curves y = cos5x, y = 0, x = /30 and x = d
equal to 0.2 ?
For what value of a does the straight line y = a
bisects the area of the figure bounded by the lines
y = 0, y = 2 + x – x2?
Find all the values of the parameter b (b > 0) for each
of which the area of the figure bounded by the
3.9
SHIFTING OF ORIGIN
Since area remains invariant even if the coordinates
axes are shifted, hence shifting of origin in many cases
proves to be convenient in computing the areas.
Example 1. Find the area enclosed between the
parabolas y2 – 2y + 4x + 5 = 0 and x2 + 2x – y + 2 = 0.
Solution (y – 1)2 = – 4(x + 1) ; (x + 1)2 = y – 1
Y2 = – 4X
X2 = Y
curves y = 1 – x2 and y = bx2 is equal to a. For what
values of a does the problem have a solution?
9. For what value of a is the area bounded by the
curve y = a2x2 + ax + 1 and the straight lines y = 0, x
= 0, and x = 1 the least?
10. For what value of k is the area of the figure bounded
by the curves y = x2 – 3 and y = kx + 2 is the least.
Determine the least area.
11. Find the least value of the area bounded by the line
y = mx + 1 and the parabola y = x2 + 2x – 3, where m
is a parameter.
12. For what positive a does the area S of a curvilinear
x 1
trapezoid bounded by the lines y = 2 , y = 0,
6 x
x = a, x = 2a assumes the least value?
13. For what values of a (a [0, 1]) does the area of the
figure bounded by the graph of the function f(x)
and the straight lines x = 0, x = 1, y = f(a), is at a
minimum, and for what values is it at a maximum, if
f(x) = 1 x 2 ?
14. For what value of a does the area of the figure,
bounded by the straight lines x = x1, x = x2, the
graph of the function y = |sin x + cos x – a|, and the
abscissa axis, where x1 and x2 are two successive
extrema of the function f(x) = 2 sin(x + /4), have
the least value?
Solving the two equations we get the points of
intersection as (0, 0) and ( – 41/3, 42/3).
0
( 4X X )dX
2
The required area =
1/3
4
0
1
4
3/2
3
.
= 4( X) X
1/3
3
3
4
Example 2. Find the area enclosed by the
parabola (y – 2)2 = x – 1 and the tangent to it at (2, 3) and
x-axis.
Y
Solution
(1,1)
X
(–5,–2)
–2
3.36
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Put x – 1 = X and y– 2 = Y
Hence the parabola becomes Y2 = X
Also
x= 2
X=1
and
y= 2
Y= 1
Also
x-axis means y = 0 Y = – 2
1
Equation of tangent : YY1 = 2· (X + X1)
4
2YY1 = X + X1
2Y = X + 1
2
0
=
16 3 3 4
.
3
Find the area of region enclosed by
Example 4.
2
(x y)
(x y)2
= 2 (a > b), the line y = x
2
a
b2
and the positive x-axis.
(x y)2 (x y)2
= 2 is an
Solution The given
a2
b2
ellipse whose major and minor axes are x – y = 0 and
x + y = 0 respectively.
The required area is shaded in the figure.
the curve
1
Hence, A =
A = 2 4 2 3 X 2 16 X 2 dX
[Y (2Y 1)]dY
2
2
1
Y3
Y2 Y 9 .
=
3
2
Example 3. Find the area enclosed between the
smaller arc of the circle x2 + y2 – 2x + 4y – 11 = 0 and the
Y
y=–x
y=x
parabola y = – x2 + 2x + 1 – 2 3 .
Solution
Circle : (x – 1)2 + (y + 2)2 = 16 ...(1)
X
O
Parabola : y = – [x2 – 2x – 1 + 2 3 ]
= – [(x – 1)2 – 2 + 2 3 ]
y + 2 = (4 – 2 3 ) – (x – 1)2
...(2)
Instead of directly solving the problem we can solve
an equivalent problem with an ellipse whose axes are
along x-axis and y-axis.
The equivalent region is shown as (OABO) where
x2 y2
equation of the ellipse is 2 2 = 1.
...(1)
a
b
The required area = area (OAA+ A AB)
ab
ab
,
where A = 2
2
2
2 . Note that A' is
a b
a b
obtained by solving the equation (1) with the line y = x.
ab
ab
1
Area OAA = ×
2
2
2
2
a b
a b2
=
Let x – 1 = X and y + 2 = Y
Hence, Circle : X2 + Y2 = 16;
Parabola : Y = 4 – 2 3 – X2
Solving the circle and parabola
X = 2 or – 2
and Y = – 2 3 ; Y = 1 + 2 3 (rejected)
1 a 2 b2
2 a 2 b2
...(2)
Y
A
O
A
B X
AREA UNDER THE CURVE
Area ABA =
=
a
ab
a
ab
a2 b
x2
b 1 2 dx
a
2
a 2 x2
a 2 b2
dx
3.37
inverse. Find the area bounded by g (x), the x-axis and
the ordinates x = – 2 and x = 6.
Solution The required area will be equal to area
enclosed by y = f (x) and the y-axis, between the
abscissa at y = – 2 and y = 6.
a
b x 2
a 2 1 x
2
= a x sin
a 2
2
a ab/ a 2 b2
2
ab
a 2 b2
b 0 a .
. a2 2
2 2 2 a 2 b2
a b2
a
=
a 2 1
a
sin
2
2
2
a b
b a 2 a 2 1
b
a3b
sin
=
2
2
a 4
2
a 2 b2 2(a b )
2 2
b
a b
ab ab
=
sin–1 2
2
2 ...(3)
2 –
4
2
a b 2(a b )
Hence the required area = sum of areas (2) and (3)
b
ab ab
sin–1 2
.
4
2
a b2
=
f (0) = 2 ; f (–1) = – 2 and f (1) = 6.
Note that f(x) is monotonic in [ – 1, 1].
1
Let f (x) = x3 + 3x + 2 and g (x) be its
3.10 AREA BOUNDED BY A
CLOSED CURVE
Let us now find the area of a closed curve, such as
that represented in the figure,
Let PM = y2, P'M = y1, the elemental area is represented
by (y2 – y1) dx, and the entire area by
OB'
OB
(y 2 – y1 )dx ,
1
0
Area bounded by inverse function
Example 5.
0
Hence, A = 6 f(x) dx + f(x) ( 2) dx
1
=
0
5
(4 x x)dx + (x 3x 4) dx = 4 .
3
3
1
0
The graph is symmetric to both axes.
Solution
Since y = ± 4 x 2 , it is clear that the total graph lies
between x = –2 and x = 2, The point ( 2, 2) is a
maximum point on the graph . We can plot a few points
and sketch the rest of the curve by symmetry, as
indicated in the figure.
Let us find the area of the shaded region R bounded
by the loop between x = 0 and x = 2. The equation of
in which OB, OB' are the limiting values of x.
the top half of this loop is y = x
Example 1. Find the area of the region bounded
by one loop of the graph of the equation y2 = 4x2 – x4.
of the lower half is y = – x
Thus, A =
2
4 x 2 , whereas that
4 x2 .
[x 4 x ) (x 4 x )]dx
0
2
2
3.38
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
2
Note: that in the equation (1), for the portion
2x 4 x dx
2
0
In order to evaluate this integral, we let
u = 4 – x2,
du = –2x dx.
Then u = 4 when x = 0, and u = 0 when x = 2. Hence,
0
0
A = – u du = 2 u 3/2 16 .
4
3
3
4
(a x)
CPA, y = – x
]
(a x)
x(a x)
0
= 2 a
1/2
Example 2. Find the area of a loop of the curve
xy2 + (x + a)2 (x + 2a) = 0.
Solution The curve is symmetrical about x-axis.
Putting y = 0, we get x = – a and x = – 2a
(a 2 x 2 )
[multiplying the numerator and the
denominator by (a x) ]
( a sin )(a a sin )
. (–a cos ) d,
a cos
[putting x = –a sin and dx = –a cos d]
=2
0
/2
Y
0
(sin – sin ) d
= – 2a2
P
(–2a,0)
= 2a2
(–a,0)
A O
X
y2 =
x 2 (a x)
ax
...(1)
Note: that the shifting of the origin only
changes the equation of the curve and has no effect
on its shape. now the origin being at the point A, the
new limits for the loop are x = – a to x = 0.
The required area of the loop = 2 × area CPA
0
=2
y dx , [the value of y to be put from (1)
=2
x a x dx ,
a
0
a
a x
2
/2
/2
(sin – sin ) d
2
0
1 1
= 2a2 1 . = 2a2 1 .
4
2 2
The loop is formed between x = –2a and x = –a.
To find the area of the loop, we first shift the origin to
the point (–a, 0). The equation of the curve then
becomes
(x – a) y2 + {(x – a) + a}2 (x – a + 2a) = 0
or y2(x – a) + x2 (x + a) = 0
or
dx,
Example 3. Trace the curve y2 (2a – x) = x3 and
find the entire area between the curve and its asymptotes.
Solution Tracing of the curve y2(2a – x) = x3.
(i) Since in the equation of the curve the powers of y
that occur are all even, therefore the curve is
symmetrical about the x-axis.
(ii) The curve passes through the origin. Equating to
zero the lowest degree terms in the equation of
the curve, we get the tangents at the origin as
2ay2 = 0 i.e., y = 0.
(iii) The curve cuts the coordinate axes only at the
origin.
(iv) Solving the equation of the curve for y, we get
x3
.
2a x
When x = 0, y = 0.
When x 2a, y2 . Therefore x = 2a is an
asymptote of the curve.
When 0 x 2a, y2 0 i.e., y is real. Therefore,
the curve exists only in this region.
Combining all these facts, we see that the shape
of the curve is as shown in the figure.
y2 =
3.39
AREA UNDER THE CURVE
Y
=2
A
0
2 a y3 / 2
( 2a y )
dy, from (1).
a
Putting y = 2a sin2 , we get,
x = 2a
O
2a
xdy
=2
X
=
0
2 /2
(2a)3/2sin3. (2a).cos .4a sin cos d
a 0
= 32a2
Now the required area
= 2 × area in the first quadrant
0
=2
0
3/2
3
x
x
2
dx, y
x
2x
(2a x)
Now put x = 2a sin2 , so that dx = 4a sin cos d.
The required area
/2
(2a sin 2 )3/2
0
(2a 2a sin 2 )
. 4a sin cos d
=
2
=
16a2
=
3.1
16a .
. , by Wallis formula
4.2 2
3a2.
=
/2 sin 3
0
cos
0
sin4 cos2 d
3 .1 . 1
. by Wallis formula
6 . 4 .2 2
Thus, the required area = a2.
2a
2a
/2
= 32a2 .
y dx
=2
sincos d = 16a2
/2
0
sin4 d
Example 5. Find the area between the curve
x2y2 = a2(y2 – x2) and its asymptotes.
Solution The given curve is symmetrical about
both the axes and passes through the origin. The tangent
at (0, 0) are given by y2 – x2 = 0 i.e., y = ± x are the
tangents at the origin. From the equation of the given curve,
y2 = a2x2/(a2 – x2). Equating the denominator to zero, the
asymptotes parallel to y-axis are given by x2 – a2 = 0 i.e.,
x = ± a.
Y
Example 4. Find the area bounded by the curve
a3x2 = y3(2a – y).
Solution The given curve is a2x2 = y3(2a – y) ...(1)
It is symmetrical about y-axis and it cuts the y-axis at
the points (0, 0) and (0, 2a). The curve does not exist
for y > 2a and y < 0.
A
(–a,0)O
MNA
X
(a,0)
The required area
= 4 × area lying in the first quadrant
=4
Y
A (0,2a)
Q x=a
P
x=–a
2
a a 2x2
y dx = 4 0 2
2 dx,
0
a x
a
a
axdx
a
2 xdx
(a x ) = – 2a (a x )
=4 0
B
O
2
2
0
a
X
The required area = 2 × area OBA
(a 2 x 2 )1 / 2
= 2a
1/ 2
0
3
= –4a [0 – a] = 4a .
2
2
3.40
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 6. The curves y = 4x2 and y2 = 2x, meet
at the origin O and the point A, forming a loop. Show
that the straight line OA divides the loop into two
parts of equal area.
Solution Solving the equations of the two given
1
curves, we have 16x2 = 2x or 16x4 – 2x = 0 i.e., x = 0 and x = .
2
1 ,1
.
Thus, the points of intersection are (0, 0) and
2
dy
3
= 1 + x1/2 > 0.
dx
2
y is a strictly increasing function.
For the branch y = x + x3/2
For the branch y = x – x3/2 ,
dy
3
= 1 – x1/2
dx
2
1
dy
4 d2y
4
3
= 0 x = , 2 x 2 < 0 at x =
dx
9 dx
9
4
At x =
4
, y = x – x3/2 has a maxima.
9
The graphs of y = x + x x and y = x – x x are shown
in the figure.
x
Y
y=
x+
x
y=x – x x
1 0
(x – 0) i.e.
1
0
2
y = 2x. Now the area between the parabola y2 = 2x
and the line y = 2x
O
The equation of the line OA is y – 0 =
=
1/2
1/2
(y y )dx = ( (2x) 3x)dx
0
1
2
0
1/ 2
2 2
3/2
2
= 3 x x
0
2 2 1
1 1 1 1
.
=
...(1)
3 2 2 4 3 4 12
Again the area between the parabola y = 4x2 and the
line y = 2x
=
1/2
4
1
4/9
2
X
x=1
Hence, the required area
1
{(x x x ) (x x x )}dx
4
= (2 x x )dx = 2 x dx = .
5
=
0
1
1
0
0
3/2
Example 8. Find area contained by ellipse
2x2 + 6xy + 5y2 = 1.
Solution The ellipse 5y2 + 6xy + 2x2 – 1 = 0 is
centred at origin, with slanted principal axes.
(2x 4x )dx = [x ] 3 [x ]
2
0
2 1/2
0
3 1/2
0
1 1 1
...(2)
4 6 12
From (1) and (2) we observe that the straight line OA
divides the loop into two parts of equal area.
=
Example 7. Determine the area of the figure
bounded by two branches of the curve (y – x)2 = x3 and
the straight line x = 1.
Solution Two curves are given by (y – x)2 = x3
y = x ± x 3/2
i.e.
y – x = ± x x , x > 0.
On solving the equation for y, we have
y=
6x 36x 2 20(2x 2 1)
3x 5 x 2
=
10
5
AREA UNDER THE CURVE
y is real, 5 – x2 0
If x = – 5 , y = 3 5
–
5 <x<
5
If x = 5 , y = –3 5
The required area
3x 5 x 2 3x 5 x 2
dx
5
5
5
5
=
5
2
4
5 x 2 dx =
=
5 5
5
5
5 x dx
2
0
Solution The given equation denotes an ellipse,
because 0 and h2 < ab. The equation of curve can
be re-written as 2y2 + 6(1 + x) y + 5x2 + 7x + 6 = 0
Solving for y,
3(1 x) (3 x)(x 1)
y1
,
2
3(1 x) (3 x)(x 1)
y2
2
Also, the curves y1 and y2 are defined for values of x
for which (3 – x) (x – 1) 0
i.e., 1 x 3 .
Put x = 5 sin , dx = 5 cos d
When x = 0 = 0
When x = 5 , =
2
Hence, the required area
4
=
5
2
5 5sin
5 cos d
0
2
=4
2
0
The required area is given by
3
1
cos d = 4
=
2 2
3.41
3
A (y1 y 2 )dx A
1
2
(3 x)(x 1) dx
1
Put x = 3 cos + sin i.e., dx = – 2sin2 d
2
/2
2
A 2 sin 2 2 d
Example 9. Find the area of the figure enclosed
by the curve 5x2 + 6xy + 2y2 + 7x + 6y + 6 = 0
0
.
2
H
1.
2.
3.
4.
5.
6.
Find the area bounded by the curve g (x), the
x-axis and the ordinate at x = – 1 and x = 4 where
x3
x2
+
g (x) is the inverse of the function f (x) =
24
8
13x
+
+ 1.
12
Find the smaller of the two areas enclosed between
the ellipse 9x2 + 4y2 – 36x + 8y + 4 = 0 and the line
3x + 2y – 10 = 0.
Prove that the area included between the curve
x = y2 (1 – x) and the line x = 1 is .
Find the area of loop y2 = x (x – 1)2.
Find the ratio in which the curve x2/3 + y2/3 = a2/3
divides the area of the circle x2 + y2 = a2.
Find the area bounded by the curve
x4 + y4 = x2 + y2.
7.
Compute the area of the figure bounded by two
branches of the curve (y – x)2 = x5 and and the
straight line x = 4.
8. Compute the area of the figure bounded by the
curve (y – x – 2)2 = 9x and the coordinate axes.
9. Find the area of the figure enclosed by the curve
y2 = (1 – x2)3.
10. Find the area of the figure enclosed by the curve
x4 – ax3 + a2y2 = 0.
11. Find the area of the finite portion of the figure
bounded by the curve x2y2 = 4(x – 1) and the
straight line passing through its points of
inflection.
12. Find the area of the figure enclosed by the curve
(y – sin–1x)2 = x – x2.
3.42
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
13. Find the area of the figure contained between the
curve xy2 = 8 – 4x and its asymptote.
14. Sketch the curve | y | + (|x| – 1)2 = 4, and also find
the area enclosed by this curve.
15. Find the area enclosed by curve y2 = x2 – x4.
16. Show that the area of a loop of the curve
y2 = x2(4 – x2) is 16/3.
3.11 AREAS OF CURVES
GIVEN BY PARAMETRIC
EQUATIONS
Now let us compute the area of the curvilinear trapezoid
bounded by a curve represented by parametric
equations :
x = (t), y = (t), t being parameter ...(1)
where t and () = a, () = b. Let equation
(1) define some non-negative function y = f(x) on the
interval [a, b] and, consequently, the area of the
curvilinear trapezoid may be computed from the formula
b
A = f ( x) dx
a
We change the variable in this integral :
x = (t), dx = (t) dt
From (1) we have y = (t)
A = (t)(t)dt
Consequently,
This is the formula for computing the area of a
curvilinear trapezoid bounded by a curve represented
parametrically.
Example 1. Compute the area of a region
x2
y2
bounded by the ellipse 2 + 2 = 1 whose parametric
a
b
equations are x = a cos t, y = b sin t.
Solution We compute the area of the upper half
of the ellipse and double it. Here, x varies from – a to a,
and so t varies between and 0.
0
A = 2 (t)(t)dt = 2 (bsin t)(–a sin tdt)
17. Find the area enclosed by the curve xy2 = a2(a – x)
and y-axis.
18. Find the area between the curve y2(a + x) = (a – x)3
and its asymptote.
19. Show that the total area included between the two
branches of the curve y2 = x2/[(4 – x) (x – 2)] and
the two asymptotes is 6.
0
= – 2ab sin 2 t dt 2ab sin 2 t dt
0
1 cos 2t
t sin 2t ab
= 2ab
dt = 2ab
.
4 0
2
2
0
Now consider a closed curve represented by the
parametric equations
x = (t), y = (t)
We suppose that the curve does not intersect itself.
Suppose that as the parameter ‘t’ increases from a value
t1 to the vaue t2, the point P(x, y) describes the curve
completely in the anti-clockwise sense. The curve
being closed, the point on it corresponding to the value
t 2 of the parameter is the same as the point
corresponding to the value t1 of the parameter. The
area of the region bounded by such a curve is given
by the formula
t2
1 x dy y dx
A=
dt dt
2 t dt
t2
=
1
1 (t) '(t) (t) '(t) dt
2 t1
Y
O
X
The above formula gives the area enclosed by any
closed curve whatsoever, provided only, that it does
not intersect itself ; there being no restriction as to the
manner in which the curve is situated relative to the
coordinate axes. For example, we again find the area
enclosed be ellipse x = acos t, y = bsin t. The ellipse is
a closed curve and is completely described while t
varies from 0 to 2.
3.43
AREA UNDER THE CURVE
dy
dx
–y
= ab (cos2 t + sin2 t) = ab
dt
dt
Therefore, the required area
2
1 2 dy
1
dx
ab dt = ab.
dt
=
=
x
y
2 dt
2 0
dt
We have x
0
Example 2. Compute the area bounded by the xaxis and an arch of the cycloid x = a (t – sin t) y = a
(1 – cos t)
Solution The variation of x from 0 to 2 a
corresponds to the variation of t from 0 to 2. We have
A = (t)(t)dt
2
=
0
2
2
2
dt
–
2
cos
t
dt
cos2 t dt
= a2
0
0
0
2
2
cos t dt 0 ,
Since, dt 2 ,
0
0
2
0
=4
=4
0
a
0
1 cos 2t
dt = , we get
2
0
A = a2 (2 + ) = 3a2.
Example 3. Find the area bounded by the curve
given by the equations x = a cos3t, y = b sin3t.
Solution Eliminating t from the equations the
cartesian equation of the curve is obtained as (x/a)2/3 +
(y/b)2/3 = 1
Since the powers of x and y are all even, the curve is
symmetrical about both the axes. It does not pass
through the origin. It cuts the axis of x at the points
(±a, 0) and the axis of y at the points (0, ±b). The tangent
at the point (a, 0) is x-axis.
Y
B(0,b)
(putting for y and dx/dt)
/2
= 12ab
= 12ab.
3 . 1. 1 3
. ab.
6 .4 .2 2 8
0
Example 4.
sin4t cos2t dt
For anyreal t, x =
bounded by the hyperbola and the lines joining the
centre to the points corresponding to t1 and – t1.
Solution
where x =
We sketch the hyperbola x2 – y2 = 1,
et et
et et
,y=
.
2
2
Y
P(t)1
A
A
C
B(0,–b)
N
X
Q(–t)1
We have to find the area of the region bounded by
the hyperbola and the lines joining the centre x = 0,
y = 0 to the point (t1) and (– t1).
The required area
= 2 [area of PCN – area of PANP]
= 2 [area of PCN –
A
(a,0) X
O
et et
et et
, y=
2
2
is a point on the hyperbola x2 – y2 = 1. Find the area
xN
1
A
(–a,0)
dx
. dt
dt
b sin3 t . (–3a cos2 t sin t) dt,
/2
2
cos 2 t dt =
y dx = 4 t / 2 y.
x0
2
a(1 cos t)a(1 cos t)dt a 2 (1 cos t)2 dt
0
1
.
2
At the point A, x = a and t = 0.
The required area = 4 × area OAB
At the point B, x = 0 and t =
ydx ]
1 e t1 e t1 e t1 e t1 t1 dx
=2
1 y dt
2
2
dt
2
2
e 2t1 e 2t1
t1 e t e t
dt
0
=2
8
2
3.44
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
e 2t1 e 2t1 1 t1 2t
(e e 2t 2)dt
=
4
2 0
x = asint
y = bsin2t
t = /4
Y
t1
=
=
e 2t1 e 2t1 1 e 2t e 2t
2t1
4
2 2
2
0
e
2t1
e
4
2t1
2t1
1 e
e
2t1 = t1.
2 2
2
When constructing the curve one should
bear in mind that it is symmetrical about the axes of
coordinates. Indeed, if we substitute – t for t, the
variable x remains unchanged, while y only changes
its sign; consequently, the curve is symmetrical about
the x-axis. When substituting + t for t the variable y
remains unchanged, and x only changes its sign which
means that the curve is symmetrical about the y-axis.
Furthermore, since the functions x = a sin t; y = b sin 2t
have a common period 2, it is sufficient to confine
ourselves to the following interval of variation of the
parameter : 0 t 2. From the equations of the curve
it readily follows that the variables of the parameter :
0 t 2. From the equations of the curve it readily
follows that the variables x and y simultaneously retain
non-negative values only when the parameter t varies
Solution
on the interval 0, , therefore at 0 t we obtain
2
2
the portion of the curve situated in the first quadrant.
The curve is shown below.
3.12 AREAS OF CURVES
GIVEN BY POLAR
EQUATIONS
If r = f() be the equation of a curve in polar coordinates
where f() is a single valued continuous function of
, then the area of the sector enclosed by the curve
and the two radii vectors = 1 and = 2 (1 < 2), is
1 2 2
r d .
2 1
X
2t1
Example 5. Compute the area of the region
enclosed by the curve x = a sin t, y = b sin 2t.
equal to
t =0
t=
0
t = (3/4)
As in seen from the figure, it is sufficient to evaluate
the area enclosed by one loop of the cur ve
corresponding to the variation of the parameter t from
0 to and then to double the result
S = 2 yx' dt 2 bsin 2t × acost dt
0
0
2
= 4ab cos t sin tdt
0
cos3 t
8
= 4ab
3 ab .
3
0
Find the area enclosed by the curve
2t
1 t2
y = f(x) defined parametrically as x =
.
2 , y=
1
t2
1 t
Solution Clearly t can be any real number.
Let t = tan
1 tan 2
2 tan
sin 2
|x| =
= cos 2 and y
1 tan 2
1 tan 2
x2 + y2 = 1.
Note that the variables x and y take both positive and
negative values, so that the given equation represents
the complete circle. Thus, the required area = .12 = .
Example 6.
Example 1. The equation of an ellipse in polar
coordinates, the centre being pole, is
1 cos 2 sin 2
. Find its area.
r2
a2
b2
1 2 2
Solution The area is equal to 2 0 r d
1 2
a 2 b2
=
d
2
2
2 0 a sin b2 cos2
1
d
d
2a 2 b 2 2 2 2
0 a sin b 2 cos 2
= ab
AREA UNDER THE CURVE
Hence the required area is ab.
Compute the area bounded by the
Example 2.
lemniscate r = a cos 2 .
Now, the required area = Area OQAPBO
= 2(area OAPBO), (by symmetry)
= 2[Area OAP + Area OPBO]
1 /4 2
=2 0
r d, for the circle r = a 2
2
Solution
+
=
The radius vector will describe one fourth of the
required area if varies between 0 and /4 :
1
1
A
4
2
/4
r d
2
0
/4
2
2
1 2 cos 2 d a sin 2 /4 a
a
2 2 0
4
2
0
Find the area common to the circles
r = a 2 and r = 2a cos .
Solution The given equations of circles are r =
a 2 and r = 2acos . The first equation represents a
circle with centre at pole and radius a 2 .
The second equation represents a circle passing
through the pole and the diameter through the pole as
the initial line. Both these circles are symmetrical about
the initial line. Eliminating r between the two equations,
we have at the points of intersection
a 2 = 2a cos , i.e., cos = 1/ 2 , i.e., = ± /4.
Y
p
2 p
r=a 2
4
B P
O
/4
A
X
Q r = 2acos
At P, = /4. For the circle r = 2a cos , at O, r = 0 and
1
so cos = 0 i.e., = .
2
/4
0
1 /2 2
r d, for the circle r 2a cos
2 /4
(a 2 )2 d +
/4
2
= 2a2 []0 2a
/2
/4
/2
/4
(2a cos )2 d
(1 + cos 2) d
/2
sin 2
= 2a2 + 2a2
4
2 / 4
=
1
a 2
+ 2a2
2 4 2
2
=
1
1
a2 + a2 – a2 = a2 – a2 = a2( – 1).
2
2
Hence, A = a2.
Example 3.
3.45
Example 4. Find the area common to the circle r
= a and the cardioid r = a (1 + cos ).
Solution Eliminating r between the given
equations, we get a(1 + cos ) = a or cos = 0 or = ± /2.
Thus the two curves cut each other at the points where
= ± /2.
Y
= /2
B
C
=
A O
A
= 0
D X
B
Both the curves are symmetrical about the initial line.
Hence, the required area = 2 × area ABCOA
= 2 × (Area OABO + Area OBCO)
...(1)
Now area OABO
=
1 /2 2
r d, for r = a
2 0
=
1 2 /2
1
a
d a 2 []0 /2
0
2
2
3.46
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1 2 a 2
a
.
2 2
4
And area OBCO
1 2
r d, for r = a (1 + cos )
=
2 /2
1 2
a (1 + cos )2 d
=
2 /2
1
1
1 2 cos (1 cos 2 ) d
= a2
/2
2
2
1 2 3
1
= a /2 2 cos cos2 d
2
2
2
=
Y
Solution
=
a 2 3
sin 2
sin
2 2
4 /2
a 2 3 3
a2
2
(3 – 8)
2 2
4
8
Hence, from (1), the required area
5
a 2 a 2
2 .
(3 8) = a2
=2
8
4
4
=
=/4
=3/4
O
X
The original graph is closer to the origin than its
3 5 7
reflection for ,
,
, and the
4 4 4 4
region is symmetric about the origin. Therefore, the
area we wish to find is the polar integral
3 /4 1
4
(2 cos 2 )2 d
/4 2
2
=2
3 /4
(4 4 cos 2 cos2 2)d
/4
3 /4
4 4 cos2 1 (1 cos 4) d
/4
2
3 /4
1
9 4sin 2 sin 4
4
/4
Example 5. The graph of r = 2 + cos 2 and its
reflection over the line y = x bound five regions in the
plane. Find the area of the region containing the origin.
27
9
9
4
4
8.
4
4
2
I
1.
2.
Find the area of the loop formed by the curve
given by x = a(1 – t2), y = at(1 – t2), –1 t 1.
If the curve given by parametric equation x = t – t3,
y = 1 – t4 forms a loop for all values of t [–1, 1],
then find the area of the loop.
3.13 AREAS OF REGIONS
GIVEN BY INEQUALITIES
3.
4.
Find the area of the region bounded by the curve
r = a cos 4.
Find the area common to the cardiod r = a(1 + cos)
3
and the circle r = a, and also the area of the
2
remainder of the cardiod.
The bounded region is shown below :
Example 1. Find the area of the region bounded
by | x2 + y2 – 2x + 4y – 6 | < 8 .
Solution
| (x – 1)2 + (y + 2)2 – 11 | < 8
– 8 < (x – 1)2 + (y + 2)2 – 11 < 8
3 < (x – 1)2 + (y + 2)2 < 19
It is an annular region having area 19 – 3 = 16
3.47
AREA UNDER THE CURVE
1
(MR + NS) . MN
0
2
– [Area of square ONSGB – Area of quarter OSGO]
2
2
1
8xdx (4 + 3).1 – 32 .3 9 1 .
=
0
4 4 6
2
Example 2. Find the area of the region defined
by 1 | x | + | y | and x2 – 2x + 1 1 – y2 .
=
1 | x | + | y | represents the region outside
the square and x2 – 2x + 1 1 – y2 represents the region
inside the circle, as shown in the figure.
Solution
Y
8xdx
Example 4. Find the area of the region represented
1
Solution log2(logyx) > 0
Case-I : y (0, 1), x < y
Case-II : y (1, ), x > y
3
=
.
4
4
Example 3. Find the area given by
x + y 6, x2 + y2 6y and y2 8x.
Solution Let us consider the curves :
...(1)
P y2 – 8x = 0
C x2 + y2 = 6y i.e., x2 + (y – 3)2 – 9 = 0 ...(2)
and L x + y – 6 = 0
...(3)
The intersection points of the curves (2) and (3) are given
by
(6 – y)2 + y2 – 6y = 0
y = 3, 6
Hence, the points are (0, 6) and (3, 3). The intersection
points of the curve (1) and (3) are given by
y2 = 8(6 – y),
i.e. y = 4, –12
Hence, the point of intersection in 1st quadrant is (2, 4)
Y
P(0,6)
R(2,4)
The required area
=
1 1 1 1
1 1 5
(1)(1) .
2 2 2 2
8 2 8
Example 5. A point P moves in the xy plane in
such a way that [|x|] + [|y|] = 1, where [.] denotes the
greatest integer function. Find the area of the region
representing all possible positions of the point P.
If [|x|] = 1 and [|y|] = 0
then 1 | x | 2, 0 | y | 1
x (2, 1] [1, 2), y (1,1)
If [|x|] = 0, [|y|] = 1 , then
x ( 1, 1), y ( 2, 1] [1,2] .
Solution
Y
G
O
by the inequality log2(logyx) > 0 where x 2 , 2 .
X
The required area = –
2
S(3,3)
MN
Q
2
1
X
–2
1
–1
O
Now,
C 0 denotes the region, inside the circle C = 0.
P 0 denotes the region, inside the parabola P = 0.
L 0 denotes the half-plane containing origin.
The required area = Area of OMRO
+ Area of MNSR – Area of ONSO.
2
X
–1
–2
Area of the required region = 4(2 – 1) (1 – (–1)) = 8.
3.48
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Example 6. Find the area of the region
represented by [x + y] + [x – y] = 5, for x y, x 0, y 0.
[x + y] + [x – y] = 5
For [x – y] = 0, [x + y] = 5
0 x – y < 1, 5 x + y < 6
Similarly for 1 x – y < 2, 4 x + y < 5 and so on.
Solution
3
=
2
2
3 cos · 3 cos d
6
3
3
= 2 (cos 2 1) d = 2 1 sin 2
6
36
3 2
2 1 3 1 3 2
= 3 2· 2 3 2· 2 6 = · =
9
3 6
1 3 3 a 3 b
=
=
3 9
9
9
Hence, a = 3, b = 1
and a + b = 4.
A=
Example 8. Find the area of the region given by
|x – 2y| + |x + 2y| 8 and xy 2.
The required area
= area of rectangle (ABCD + DEGF + GHJI)
Solution We have |x – 2y| + |x + 2y| 8
...(1)
i.e. |y – x/2| + |y + x/2| 4
1
3
= 3 .1.1 square units.
2
2
Example 7. Consider the following regions in the
plane : R1 = {(x, y) : 0 x 1 and 0 y 1}
R2 = {(x, y) : x2 + y2 4/3}
The area of the region R1 R2 can be expressed as
Let us divide the reference frame into four regions (see
figure) using the lines y = x/2 and y = – x/2.
a 3 b
, where a and b are integers. Then find the
9
value of (a + b).
x
x
Region I : y 0 and y 0
2
2
The inequality (1) reduces to
x
x
– y + y 4 x 4.
2
2
Solution
x
x
Region II : y 0 and y 0
2
2
The inequality (1) reduces to
The required area A =
1
A1 =
4 3 x dx , put x = 3 sin
2
1
1
1 A1 where
3
3
2
x
– y x + y 4 y 2.
2
2
Similarly, for regions III and IV, the inequality
(1) respectively reduces to x – 4 and y – 2.
Thus, the region covered by the inequality (1) is a
rectangle. The following figure shows the region
satisfied by the inequality (1) and xy 2.
AREA UNDER THE CURVE
| 3y x |
2
Case I : If y
y
The required area, is
x 3y
,
2
A = 2 2 2 dx = 4[x – ln x] 14 = 4(3 – ln 4).
1
x
1
be
Example 9. Let O (0, 0), A(2, 0) and B 1,
3
the vertices of a triangle. Let R be the region consisting
of all those points P inside OAB which satisfy
d(P, OA) min {d(P, QB), d(P, AB)}, where 'd' denotes
the distance from the point to the corresponding line.
Sketch the region R and find its area.
(2 + 3 ) y x
y (2 – 3 )x
y x. tan 15º
...(3)
( y = x. tan 15º is an acute angle bisector of AOB)
Y
Let the coordinates of P be (x, y).
Equation of line OA y = 0
Solution
O
Equation of line OB 3 y = x
Case II : If y
Equation of line AB : 3 y = 2 – x
d(P, OA) = distance of P from line OA = y
d(P, OB) = distance of P from line OB
| 3y x 2 |
2
| 3y x 2 |
2
1
B1,
3
A(2,0) X
Given : d (P, OA) min. {d(P, OB), d(P, AB)}
=
| 3y x | | 3y x 2 |
,
y min.
2
2
1
(base) × (height)
2
=
1
(2) (1 tan 15º) = tan 15º
2
y
| 3y x |
2
| 3y x 2 |
and y
2
( 3 y + x – 2 < 0)
From (3) and (4) we conclude that P moves inside the
QOA, (Q is the incentre of ABC), as shown in the
figure.
As QOA = OAQ = 15º, QOA is isoceles.
OC = AC = 1 unit
Area of shaded region
= area of QOA
P
O(0,0)
C(1,0) A(2,0) X
(2 + 3 )y 2 – x
y – (2 – 3 ) (x – 2)
y –(tan 15º) (x – 2)
...(4)
(y = (x – 2) tan 15º is an acute angle bisector of CA)
| 3y x |
2
d(P, AB) = distance of P from line AB
Y
1
B1,
3
Q
2y 2 – x – 3 y
=
=
( 3 y – x < 0)
i.e., x > 3 y
4
3.49
...(1)
...(2)
= (2 –
3 ) sq. units.
Example 10. Consider a square with vertices at
(1, 1), (–1, 1), (–1, –1) and (1, –1). Let S be the region
consisting of all points inside the square which are
3.50
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
y2 = 1 – 2x, y2 = 1 + 2x, x2 = 1 – 2y, x2 = 1 + 2y
nearer to the origin than to any edge. Sketch the region
S and find its area.
Solution Let the square be ABCD where the
equations of the sides of the square are as follows :
AB : y = 1,
BC : x = –1,
CD : y = –1,
DA : x = 1
Let the region be S and (x, y) is any point inside it.
Then according to given conditions,
y2 + 2y + 1
2
2
2
y < 1 – 2x, y < 1 + 2x, x < 1 – 2y and x2 < 2y + 1
Y
A(1, 1)
(0, 1)
2
y = 2x – 1
G
I
Ex22y1
O
H F
A(1, 1)
(0,1/2)
(–1/2,0)
(1/2,0)
X
0
(0, –1/2)
x 2 y 2 < |1 – x|, |1 + x|, |1 – y|, |1 + y|
x2 + y2 < x2 – 2x + 1, x2 + 2x + 1, y2 – 2y + 1,
Y
B(–1, 1)
C(–1, –1)
D(1, –1)
Now, S is symmetrical in all four quadrants
S = 4 × area lying in first quadrant
Now y2 =1 – 2x and x2 =1 – 2y intersect on y = x
The point of intersection is E ( 2 – 1, 2 – 1)
Area of region OEFO
= area of OEH + area HEFH.
1/2
1
1 2x dx
= ( 2 – 1)2 +
2
2
1
2
= (2 + 1 – 2 2 ) + [(1 – 2x)3/2 ]1 /221
2
3
1
1
= (3 – 2 2 ) + (3 – 2 2 )3/2
2
3
1
1
= (3 – 2 2 ) + ( 2 – 1)3
2
3
1
1
= (3 – 2 2 ) + (5 2 – 7)
2
3
1
= [4 2 – 5]
6
1
Similarly, area OEGO = (4 2 – 5)
6
2
so, area of S lying in first quadrant = (4 2 – 5)
6
4
Hence, S = (4 2 – 5).
3
2y – 1
(1, 0) X
Now in y2 = 1 – 2x and y2 = 2x + 1, the first equation
represents a parabola with vertex (1/2, 0) and second
equation represents a parabola with vertex (–1/2, 0)
and in x2 = 1 – 2y and x2 = 1 + 2y, the first equation
represents parabola with vertex at (0, 1/2) and second
equation represents a parabola with vertex at (0, –1/2).
So, the region S is the region lying inside the four
parabolas
J
1.
2.
3.
Find the area of the region represented by
x y 2,
x y 1,
x 0,
y 0
Find the area enclosed by |x| + |y| < 3 and xy > 2.
Find are bounded by x2 + y2 < 2ax and y2 > ax, x > 0.
4.
5.
Consider the closed figure C made by the line
|x| + |y| = 2. Let S be the region inside the figure
such that any point in it is nearer to the side
x + y = 2 than the origin. Find the area of S.
Give a rough sketch of the region R consisting of
points (x, y) satisfying |x ± y| 2 and
x2 + y2 2 and find the area of the region.
AREA UNDER THE CURVE
6.
Calculate the area of a plane figure bounded by
parts of the lines max (x, y) = 1 and x2 + y2 = 1 lying
in the first quadrant :
x, if x y,
max (x,y) =
y, if x y.
7.
Find the area of the bounded region represented
by |x + y| = |y| – x and y x2 – 1.
Find the area between the curve y = x2 + x – 2 and
8.
Find the area of the region bounded
Problem 1.
by the curve C : y =
x 1
x2 1
and the line L : y = 1.
Y
2
(0,1)
Solution
–1 O (0,0) (1,0)
On solving C : y =
y = 2x for which |x2 + x – 2| + |2x| = |x2 + 3x – 2| is
satisfied.
9. Let P (x, y) be a point satisfying these three
inequalities, |x + y| 1, |y – x| 1 and 3x2 + 3y2 1,
sketch the region in which ‘P’ can move, also find
the area of the region.
10. Find the area of the region which consists of all
the points satisfying the conditions |x – y| + |x + y|
8 and xy 2.
Q
Solution A
X
x 1
and the line L : y = 1, we have
x2 1
x 1
= 1 x + 1 = x2 + 1 x(x – 1) = 0 x = 0, 1
x2 1
Thus, the points of intersection are at x = 0 and x = 1.
1
x 1
1 dx
The required area = 2
x
1
0
1
1
1
1
2x
1
dx
dx
= 2 x2 1
x 2 1 dx
0
0
0
1
1
1
1
2
= ln(x 1) (tan x) x = ln 2 1 .
2
0 2
4
Find the area of the shaded region if APB and DQC are
semicircles.
F
P
C
The required area is double of the area QEFQ.
r
1
.
In EPG, cos =
2r 2
3
Since area QEFQ = area of segment PEQFP area of
1 2 2 r r
tan 30°
triangle PEF = r
2
3 2 2
1
3 2
r
Area QEFQ = r 2
3
4
3 r2
The shaded area = 2r 2
[4 3 3] .
3 4 6
Problem 3. Find the area of the curve enclosed
by the curves y|y| + x|x| = 1, y|y| – x|x| = 1 and y = |x|.
Solution In the first and second quadrants, we
have y2 + x2 = 1 and y2 – x2 = 1. In the third and fourth
quadrants, we have x2 – y2 =1.
2
2
x–y = 1
2
2
x–y = 1
Y
ABCD is a rectangle with 2AD = AB = 2r.
Problem 2.
B
G
E
D
–2
3.51
B(0,1)
A
Q
B
C
2
2
x+y = 1
(–1,0)N
D
P
C
x–y = 1
2
2
O
A
2
2
x+y
=1
M(1,0) X
2
2
x–y = 1
3.52
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
1
1
Area OABO = (area of OMABO) = 2 4 8
2
7 8 11
7
44 7 37
= 1 12 3 =
–
=
=
.
3
12
12
12
1
1
(area of OBCNO) = 2 4 8
2
The required area OABCO = area OABO + area
OBCO
= .
8 8 4
Problem 5. Find the area of the region bounded
by the parabola y2 = 4x and a normal drawn to it with
gradient – 1.
Area OBCO =
Solution
Sketch the graph of the cubic
y = x – x – x + 1 and find the area of the region bounded
by the curve and the line y = x + 1.
Problem 4.
3
2
y = (x – 1)2(x + 1) i.e. two concident
roots are x = 1, 1 and other root is x = – 1.
dy
= 3x2 – 2x – 1 = (x – 1)(3x + 1) = 0
dx
x = 1, x = – 1/3
Solving y = x + 1 and x3 – x2 – x + 1 = x + 1
x(x2 – x – 2) = 0
x = 0, x = 2, – 1
Solution
Y
Equation of normal to the parabola is y = mx – 2am – am3
y = – x + 2 + 1 (since a = 1 and m = – 1)
y = 3 – x . Solving it with y2 = 4x,
(3 – x)2 = 4x
9 + x2 – 6x = 4x
x2 – 10x + 9 = 0
x = 1 or x = 9
The values of y are 2 and –6.
2
2
64
y2
.
Now, A = x dy = (3 y) dy =
3
4
6
6
Prove that the area of a sector of the
ellipse of semi-axes a and b between the major axis and
1
ab( – e sin ),
a radius vector from the focus is
2
where is the eccentric angle of the point to which the
radius vector is drawn.
Solution Let the equation of the ellipse be
2/a2 + y2/b2 = 1. Then, O is its centre and S(ae, 0) is the
focus. Let be the eccentric angle of any point
P(x, y) on the ellipse. Then x = a cos , y = b sin .
Now SP is the radius vector of P drawn through the
focus S and SA is the radius vector along the major
axis. At the point A, x = a and = 0.
Problem 6.
0
X
0
A=
x
[(x3 x 2 x 1) (x 1)]dx
1
2
+ [(x 1) (x3 x 2 x 1)]dx
0
0
2
= (x 3 x 2 2x) dx + (2x x 2 x3 ) dx
1
Y
B
0
0
2
x
x
x
x
x2 + x2
=
3
3
4 0
4
1
4
3
3
4
8
1 1
= (0) 4 3 1 4 3 4 0
b
A
O
P(a cos,
b sin )
A
M (a,0) X
AREA UNDER THE CURVE
Draw PM perpendicular to the x-axis. The required area
of the sector SAP
= area of the SMP + area PMA
1
= SM . MP +
2
=
=
=
a
a cos
...(2)
Line (2) cuts the x-axis at (–4, 0) and y-axis at (0, 2).
y d d
0
1
(a cos – ae) b sin + b sin . (–a sin ) d
0
2
[x = a cos and y = b sin ]
1
ab(cos – e) sin +
2
0
ab.
The equation of tangent at P(2, 3) is
y–3 =
dx
1
(1 – cos 2)d
2
3
The required area, RQPAR = (x1 x 2 ) dy
0
1
1
1
ab (cos – e) sin + ab ( – . 2 sin cos )
2
2
2
1
= ab[cos sin – e sin + – sin cos ]
2
=
1
ab ( – e sin ).
2
Find the area enclosed by th e
parabola (y – 2) = x – 1, the tangent to the parabola at
(2, 3) and the x-axis.
Solution The given parabola is (y – 2)2 = x – 1 ...(1)
Its axis is y = 2 and vertex is (1, 2). Let P be the point
(2, 3).
Problem 7.
2
3
3
0
0
[(y 2)2 1 (2y 4)]dy (y2 6y 9)dy
sin 2
1
1
= ab (cos – e) sin + ab
2 0
2
2
=
dy 1
dx 2
1
(x 2)
2
or, x 2y 4 0
ydx, for the ellipse
1
(OM – OS) . MP +
2
At P (2, 3),
3.53
3
y3
3y 2 9y = (9 – 27 + 27) – 0 = 9.
3
0
Each line passing through the origin O
Problem 8.
(0, 0) and with slope > 0 cuts the curve y = xn ,
(x 0, n –1) at the point A(x1, y1) which projects the
point B(x1, 0) on the x-axis. OAB is divided into two
regions by the curve y = xn. Demonstrate that ratio of the
areas of these two regions is independent of . For which
value of n are the areas of these two regions equal ?
Solution
Y
y = x
A(x,y)
1 1
y = xn
B
O
X
Since x1n = y1 = x1, the area of the triangle OAB
=
y1x1
x12
=
and the area under the curve
2
2
x1
n
= x dx =
0
From (1), 2(y – 2)
dy
1
dx
dy
1
dx 2(y 2)
x1n 1 x12
, and therefore the other
n 1 n 1
portion of the triangle has area
x12 x12 x12 (n 1)
.
2
n 1
2n 2
The ratio of both portions is therefore
3.54
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
x12
n 1 2 , independent of .
2
x1 (n 1) n 1
2n 2
2
= 1 then n = 3 and so the portions will have
If
n 1
equal area for n = 3.
Problem 9. Let C1 and C2 be the graphs of the
functions y = x2 and y = 2x, 0 x 1 respectively. Let
C3 be the graph of a function y = f(x) ; 0 x 1,
f(0) = 0. For a point P on C1, let the lines through P
parallel to the axes, meet C2 and C3 at Q and R
respectively (see figure). If for every position of P on
(C1), the areas of the shaded regions OPQ and ORP are
equal, determine the function f(x).
Solution On the curve C1, i.e. y = x2 , let P be
(, 2). Hence, ordinate of point Q on C2 is also 2.
Now on C2 , i.e. y = 2x, the abscissae of Q is given by
x
y 2
.
2
2
2
= 0 (x1 x 2 )dy 0
y
y 2 dy
2 3 4
3
4
Area ORP =
2 3 4
(x2 f(x))dx
0
3
4
Differentiating both sides with respect to we get
22 – 3 = 2 – f()
f() = 3 – 2
f(x) = x3 – x2.
Problem 10. Prove, without using a calculator, that
k 2
k
9.5
10
10
k 1
9
Solution The inverse of the function
f : [0, [0, f(x) = x2 is :
f–1 : [0, [0,
f–1(x) =
x
In the diagram each rectangle Vk has its lower left corner
k
k
1
and height
.
0, 10 , base
10
10
k
Each rectangle Hk has lower left corner at 10 , 0 ,
2
2
Q is 2 ,
And R on C3 is (, f()). Now, Area OPQ
2
Thus,
1
k
and height .
base
10
10
The collective area of these rectangles is
2
2
2
2
1 1
1
2
3
9
2
3
9
...
10 10
10 10
10 10
10
10
10
Since these grey rectangles do not intersect with the
black squares on the corners, their collective area is
less than the area of the unit square minus these squares
i.e.
2
0 (y1 y2 )dx 0 (x f(x))dx
2
2
1
4
95
1 2
1– = 1
.
100 100 100
10 10
We thus conclude that
3.55
AREA UNDER THE CURVE
2
2
2
2
1 1
1
2
3
9
2
3
9
...
10 10
10 10
10 10
10
10
10
95
< 100
k 2
k
9.5 .
10
10
k 1
9
For each positive integers n > 1, An
represents the area of the region restricted to the
Problem 11.
following two inequalities :
x2
y
2
n2
x2
n
2
y 2 1 and
lim A
1 . Find n n .
Solution
Y
–n
B
O
I
2 ab h 2
(x )( x) dx.
b
To find this, assume x – = ( – ) sin2 then
– x = ( – ) cos2 and we get
n X
–n
The intersecting points of the ellipses are :
n
n
(x. y)
.
2
n2 1
,
n
1
n
n
n n
,
,
n2 1 n2 1 , n2 1 n2 1
n
n
,
n2 1 n2 1
Notice that the region is symmetric with respect to :
x-axis, y-axis, y = x and y = –x.
Hence, An = 8 . Bn,
where Bn = area between the y-axis, y = x, and
n2 x2
.
n
As n , lim
y2 – y1 = 2
(h 2 – ab)x 2 2(hf – bg)x f 2 – bc
b
Also, the limiting values of x are the roots of the
quadratic expression under the radical sign.
Accordingly, denoting these roots by and , and
observing that h2 – ab is negative for an ellipse, the
entire area is represented by
n
A
Find the area of an ellipse given by
the general equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
Solution Here, solving for y, we easily find
Problem 12.
2 /2
(x )( x) dx 2( – ) 0 sin cos d
=
n
2
2
(ab h 2 )2
Hence, the area of the ellipse
=
(af 2 bg 2 ch 2 2fgh abc)
(ab h 2 )3/2
.
Compute the area enclosed by the
curve y = (1 – x ) .
Problem 13.
2
2 3
The curve is symmetric about x–axis as
well as y–axis. When x = –1,1 y = 0, and
when x = 0 y = ±1
Solution
Y
1
1 and Bn approaches the
area of AOB where O = (0, 0), A = (1, 0) and B = (1, 1).
(hf bg)2 (f 2 – bc)(ab h 2 )
(ab h 2 )2
4b(af bg ch 2fgh abc)
2
y
n n2 1
2
( )2 .
8
Again, ( – )2 = 4.
=
2
1
lim A n 8 . . 1 . 1 4 .
2
n
–1
O
–1
1
X
3.56
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
But the indefinite integral is equal to
1
1 = 0 (1 – x2)3/2 dx
e x
e . sin x dx = – 2 2 (sin x + cos x) + c
–x
1
= 41 = 4 0 (1 – x ) dx
2 3/2
Consequently
(n 1) /
Put x = sin
/2
/2
=4 0
=4 0
/2
0
e x
( sin x cos x)
2
n
+
1
Sn = (–1)
2
n /
cos3 . cos d
=
cos 4 d
cosn x dx =
n 1 (n 3)
3 1
.
........ .
n
n2
4 2 2
=
3 1 3
= 0.75 .
4 2 2 4
Prove that the areas S0, S1, S2 ... S3
.... bounded by the x-axis and half–waves of the curve
y = e– x sin x x 0 form a geometric progression with
common ratio q = e–.
Solution The curve intersects the positive x-axis
Problem 14.
at the points where sin bx = 0. Hence xn =
n
, n W..
2 2
[e– (n + 1) (–1)n + 1 – en (–1)n]
e–dn (1 + e)
2 2
Hence, the ratio of area of two consecutive half waves
=
if n is even
=4×
( 1)n 1
Sn 1 e (n 1) / –
=
=e
,
e n /
Sn
which is independent of n. Hence, the areas form a
geometric progression.
Problem 15.
Find out the ratio of areas in which
x3
x
divides the circle
the function f(x) =
100 35
x2 + y2 – 4x + 2y + 1 = 0 ([.] denotes the greatest integer
function).
Solution
Circle : x2 + y2 – 4x + 2y + 1 = 0
or, (x – 2)2 + (y + 1)2 = 4 = 22
Now for 0 x 4 ,
The function y = e–x sin x is positive in the intervals
(x2k x2k + 1) the sign of the function in the interval
(xn + xn + 1) coincides with that of the number (–1)n.
Sn =
(n 1)
(n 1)
n
| y | dx = (–1) n e–x. sin x dx
n
0
...(1)
x3 x
x3 x
1
0 .
100 35
100 35
So, we have to find out the ratio in which x axis divides
the circle (1). Now, at x-axis, y = 0. So, (x–2)2 = 3,
Hence, the circle cuts the x axis at the points (2 – 3 , 0)
and (2 + 3 , 0)
3.57
AREA UNDER THE CURVE
2 3
Let A
4 (x 2)2 1 dx 4 3 3
3
2 3
The required ratio is
For what value of 'a' does the area of
the figure bounded by the straight lines, x = x1 , x = x2,
the graph of the function y = sin x + cos x a and
the abscissa axis where x1 and x2 are two successive
extrema of the function, f (x) = 2 sin x 4 , have
the least value . Also find the least value .
2 cos x = 0
4
f (x) =
3
x + = or
4
2
2
x1 =
5 / 4
5 /4
2 , then A =
a 2 sin 4 x dx
/4
A = a Amin =
2 where a = 2 .
Let a 2 ,
5 / 4
2 sin 4 x a dx = a
/4
Hence Amin =
Let 2 < a <
5 /4
A=
/4
Let x +
/2
3 /2
2 when a =
2
= 2 cos y
sin 1 a
2
2
+ a y
2
This on simplification reduces to
2 a 2 + 2 a sin 1
A = 2
1 ( 2 a)
dA
= 2.
+ 2a
da
2 2 a2
=
sin 1
2a
2a
a
2
2
+
a
2
1
2a
2a
2
1
2
2
1 a
2
a
2
+ 2 sin 1
.2
a
2
=0
= 0 a = 0 and Amin = 2 2 .
Finally, the least value of area is 2 2 and the value of
a is 0.
Problem 17. Consider the collection of all curves
of the form y = a – bx2 that pass through the point
(2, 1) where a and b are positive real numbers. If the
minimum area of the region bounded by y = a – bx2 and
the x-axis is
m , find the value of M.
Solution
2 sin y a dy
a
2
2
3 /2
/2
1
sin 1 a
2 sin x a dx
4
= y =
4
a 2 sin
3
2 cos y 2
2
a 2 sin y dy
sin 1 a
+ sin 1
2 sin x a dx
4
/4
A=
2 sin y a dy
5
and x2 =
.
4
4
Hence, A =
Let a
=
2
A
4 3 3
.
4 A 8 3 3
Problem 16.
Solution
sin 1 a
We have y = a – bx2
3.58
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Since the point (2, 1) lies on the curve,
1 = a – 4b
a = 1 + 4b
a b
A= 2
bx3
(a bx 2 ) dx = 2 ax
3
0
0
...(1)
a b
where x1 and x2 are roots of x2 – mx + (m–c) = 0.
x1 + x2 = m, x1 x2 = m – c
m(x x )
= (x2 – x1)
4 (1 4b) 1 4b
= ·
,
b
3
...(2)
1 1
12 3
b · 1 4b ·4 (1 4b)3 2 ·
dA
4
2
2 b
=
db
3
b·2 b
32
4 12b 1 4b (1 4b)
=
=0
2b3 2
3
2
2
x 2 x 22 x1x 2
(c m) 1
3
Putting these values in (2), we get
4· 3 · 2
4
·(3 2)· 3 2 ·2 2 =
=
2
3
=
m 2 4(c m)
m 2 4(c m)
6
=
[m 2 4(c m)]3 2 [m 2 4m 4c]3 2
=
6
6
[(m 2)2 4(c 1)]3 2
6
A is least if m = 2
=
Aleast =
12b = 1 + 4b
8b = 1 and hence
b = 1/8 and hence a = 3/2 using (1).
m =
1
m2
m 2 (m c)
2
m
4(c
m)
(c
m)
=
3
2
2 3a a a a 2 2a a
=
= 3
b 3 b
b
Hence,
A = (mx (c m) x 2 )dx ,
x1
a a b a a
·
= 2
b 3 b b
Amin =
x2
48 .
48 which m 48.
Problem 18. Let 'c' be the constant greater then
1. If the least area of the figure given by the line passing
through the point (1, c) with gradient 'm' and the
parabola y = x2 is 36 sq. units find the value of (c2 + m2).
Solution Equation of the line through (1, c) is
y – c = m(x – 1)
y = mx + (c – m)
...(1)
4
[4(c 1)]3 2
23 (c 1)3 2
=
= · (c – 1)3/2
3
6
6
4
· (c – 1)3/2 = 36
3
(c – 1)3/2 = 27
(c – 1) = (33)2/3 = 9
c = 10
Hence, c = 10, m = 2
c2 + m2 = 100 + 4 = 104.
Problem 19. Let 0 a 4. Prove that the area of
the bounded region enclosed by the curves with
equations y = 1 – |x – 1| and y = |2x – a| cannot exceed
1
.
3
Solution In the situation that 0 a 1, the two
curves intersect in the points (a/3, a/3) and (a, a), and
the bounded region is the triangle with these two
vertices and the vertex (a/2, 0). This triangle is
contained in the triangle with vertices (0, 0), (1/2, 0)
and (1, 1) with area 1/4. Hence, when 0 a 1, the area
of the bounded region cannot exceed 1/4.
Let 1 a 3. In this case, the bounded region is a
AREA UNDER THE CURVE
quadrilateral with the four vertices (a/3, a/3), (a/2, 0),
((a + 2)/3, (4 – a)/3) and (1, 1). Nothing that this
quadrilateral is the result of removing two smaller
triangle from a larger one, we find that its area is
1 a a 1 (4 a)
a
1 . . .
. 2
2 3 2 2
3
2
1
2
a
1
(4 a)2
12 12
a 2 4a 2 1 (a 2)2
6
3
6
whence we find the area does not exceed 1/3 and is
equal to 1/3 exactly when a = 2.
The case 3 a 4 is the symmetric image of the case
0 a 1 and we find that the area of the bounded
region cannot exceed 1/4.
Problem 20.
Case II : If x 0, y 0, then locus of the point P is
– xy + 2x – 2y + 2 = 0
y (–2 – x) = –2(x + 1)
y=
y = 4 x and the locus of the point P which moves
such that the sum of its distances from the co-ordinate
axes is equal to its distance from the curve y = 4 x 2 .
Solution
2(x 1) 2(x 1)
.
(2x x)
x2
2(x 1)
is passing through the points
x2
2(x 1)
(1, 0) and (0, 1) and the curve y =
is passing
x2
through (–1, 0) and (0,1).
Hence the required area = Area of circle – Area
under the locus of the point P.
The curve y =
Area =
0
2(x 1)
x 2 dx
1
= 2 – 2 (1 + n 1 – n 2) – 2(1 – n 2 + n 1)
= 2 – 2 + 2 n 2 2 + 2 n 2 = 2 – 4 + 4 n 2
Problem 21. Let T be the triangle with vertices
(0, 0), (0, c2) and (c, c2) and let R be the region between
y = cx and y = x2 where c > 0 then show that
Area (R) =
Y
c3
and lim Area (T) = 3.
6
c 0 Area (R)
Solution
M
P
x
1
4
2(x 1)
dx
2
x2
0
Find the area between the curve
2
3.59
2
y= 4 – x
y
O
X
Let the point P be (x, y)
|x| + |y| = PM = OM – OP
|x| + |y| = 2 – x 2 y 2
x2 + y2 + 4 – 4 |x| – 4 |y| + 2 |xy| = x2 + y2
2|xy| – 4|x| – 4|y| + 4 = 0
|xy| – 2 |x| – 2|y| + 2 = 0
Case I : If x 0, y 0, then locus of the point P is
xy – 2x – 2y + 2 = 0
y (x – 2) = 2(x – 1) y =
2(x 1)
x2
Area (T) =
c·c 2
c3
=
2
2
c3
Area (R) =
–
2
c
0
x2 dx =
c3
c3
c3
–
=
2
3
6
Area (T)
c3 6
= lim
= 3.
·
3
c 0 2 c
c 0 Area (R)
lim
3.60
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Problem 22.
1 x
Let f(x) = 0
2
(2 x)
0 x 1
if
if 1 x 2
2x3
if
x
and F(x) = f(t) dt then find the area enclosed by the
0
curve y = F(x) and x-axis as x varies from 0 to 3.
x
Solution
F(x) = f(t)dt F(x) = f(x)
0
x
For 1 < x 2, F(x) = (1 t)dt 0dt
0
1
1
0
2
F(x) is constant
1
2
x
1
2
(x 2)3
3
For 2 < x 3, F(x) = f(t)dt f(t)dt f(t)dt
0
1
2
1
x
2
0 (t 2)2 dt
0
2
1
3
x2
1 1 (x 2)3
x
dx
dx
=
2
2 2
3
0
2
1
3
x 2 x3
1 x (x 2)4
= 2 6 2 2 12
0
2
1 1 1 3 1
(1)
2 2 6 2 12
Let f (x) be a differentiable function
and satisfy f (0) = 2, f ' (0) = 3 and f '' (x) = f (x). Find the
area enclosed by y = f (x) in the second quadrant.
Solution Given f '' (x) = f (x) or 2 f ' (x) f '' (x) = 2 f (x)
f ' (x) [multiplying both sides by 2 f ' (x)]
Problem 23.
x
1
F(x)dx
2
= 1 1 7 6 4 7 17 .
2 3 12
12
12
1
x2
and F(1) =
2
2
1
3
F(x)dx
0
x
2
For 0 x 1, F(x) = (1 t)dt t t
2
0
0
F(x) = x –
1
Area =
2
d
d 2
dx f '(x) = f (x)
dx
Integrating, (f(x))2= f 2(x) + C
...(1)
Put x = 0, (f(0))2 = f 2(0) + C
9=4 +C
C = 5.
Hence equation (1) becomes
2
x2
x
2
1
F(x) =
2
1 (x 2)3
3
2
if
dy
dy
and f (x) = y)
dx = y2 + 5 (where f ' (x) =
dx
0 x 1
if
1 x 2
if
2x3
dy
y 2 5 (cannot be –
dx
Integrating,
2
ln y y 5 = x + C1
F(x)
Put x = 0, y (0) = 2
ln(2 + 3) = 0 + C1
C1 = ln 5
5/6
1/2
1/2
1/6
0
dy
y2 5 dx ,
1
5/12
2
3
x
y 2 5 as f '(0) = 3)
3.61
AREA UNDER THE CURVE
x
2
ln y y 5 = x
5
y 2 5 y = 5ex
Rationalizing ,
...(2)
5
y2 5 y
= 5ex
y 2 5 y = e–x
Now
1
f y f(x) f y = f(x) – f(y)
...(3)
(2) – (3), gives 2y = 5e – e
x
–x
f(x h) f(x)
h
h0
Now, f(x) = lim
h
f 1 f(1)
1 xh
x
f
lim
= hlim
h
0 h x h 0
x
x
f (1) 2
f(x) = 2lnx + c c = 0
x
x
(since f(1) = 0)
f(x) = 2lnx
=
5e x e x
5e x e x
, f (x) =
> 0 x R
2
2
f is increasing.
The graph of y = f (x) is :
f(x) =
Y
(0,2)
1
ln5
2
0
X
We have f(0) = 2.
Now f(x) = 0, When 5ex – e–x = 0
5e2x = 1 e2x =
The required area
1
1
x=
ln 5.
5
2
Area in the second quadrant =
0
5 x
1 x
dx
2 e 2 e
1
ln 5
2
5 x 1 x 0
= e e 1
= 3 5 1 5 = 3 5 .
2
2
2 5
ln 5
2
2
Problem 24.
Let f(x) be a function which satisfy
the equation f(xy) = f(x) + f(y) for all x > 0, y > 0 such
that f(1) = 2. Find the area of the region bounded by
the curves y = f(x), y = |x3 – 6x2 + 11x – 6| and x = 0.
Solution Take x = y = 1 f(1) = 0
Now, put y =
1
1
1
0 = f x x f(x) f x
x
1
f x = – f(x)
1
= (x3 6x 2 11x 6)dx
0
e
0
y/2
7
dy .
4
(Using horizontal strip for the area below the x-axis)
Let f(x + y) = f(x) + f(y) – xy x,
Problem 25.
f(h)
= 3. Find the area bounded by
h 0 h
the curves y = f(x) and y = x2.
y R and lim
In f(x + y) = f(x) + f(y) – xy
Put x = y = 0, so that f(0) = 0.
Solution
Also, f(x) = lim
h0
lim
h0
f(x h) f(h)
h
f(x) f(h) hx f(x)
f(h)
–x
lim
h
h0 h
f(x) = 3 – x
f(x) = 3x –
x2
+ c and f(0) = 0 c = 0
2
3.62
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
f(x) = 3x –
x2
.
2
h0
x2
x2 0 x 2
2
The area is bounded before x = 0 and x = 2.
f(x) x2 3x –
2
x2 2
3x
x dx
Area =
2
0
2
2
x 2 x3
x2
3 x dx 3
= 2.
2
6 0
2
0
x
f(x)
Problem 26. If f y f(y) x, y R and f(1)
exists, and area under the curve f(x) bounded by x-axis
n
r
1
er / nf
x = 0 and x = 1 is , then find lim
n .
n
3
r 1
Solution
f(1 h) f(1)
h
f(1 h) 1
f(1) = lim
...(1) ( f(1) = 1)
h
h0
f(x h) f(x)
Now f(x) = lim
h
h 0
xh
h
f
f 1 1
1
x
x
= lim
f(x) = f(x) lim
h0
h
h
h 0
x
x
f(x)
=
f(1)
[ from equation (1)]
x
f(x)
f(x) =
. k where f(1) = k
x
f(x) = xk + c
For x = 1, f(x) = 1 c = 0 f(x) = xk.
f(1) = lim
x f(x)
If f
x, y R
y f(y)
1
1
0
0
1 x k 1
1
Now, area = f(x)dx
k = 2.
3 k 1
3
So, f(x) = x2.
n
n
r
r/n r 1
lim
er / nf
e . n . n
n
n n r 1
r 1
Now, lim
1
= e x .x.dx 1 . (using integration by parts)
0
1.
Curve Sketching
(a) Extent
(b) Intercepts
(c) Sign Scheme of f(x)
(d) Symmetry
(i) The graph of F(x, y) = 0 is symmetric
about the y-axis if on replacing x by – x,
the equation of the curve does not
change. i.e. F(x, y) = 0 implies F(– x, y) = 0.
(ii) The graph of F(x, y) = 0 is symmetric
about the x-axis if on replacing y by – y,
the equation of the curve does not
change. i.e. F(x, y) = 0 implies F(x,–y) =
0.
(iii) The graph of F(x, y) = 0 is symmetric
about the origin if on replacing x by – x
and y by – y, the equation of the curve
does not change.
i.e. F(x, y) = 0 implies F(– x, –y) = 0.
(iv) The graph of F(x, y) = 0 is symmetric about
the line y = x if on interchanging x and y,
the equation of the curve does not
change. i.e. F(x, y) = 0 implies F(y, x) = 0.
(v) The graph of F(x, y) = 0 is symmetric
about the line y = –x if on replacing x
by – y and y by – x, the equation of the
curve does not change.
i.e. F(x, y) = 0 implies F(– y, – x) = 0.
(e) Periodicity
(f) Monotonicity
(g) Local Maximum and Minimum Values
(h) Concavity and Points of Inflection
(i) Asymptotes
(i) If either lim f(x) = L or lim f(x) = L,
x
x
then the line y = L is a horizontal
AREA UNDER THE CURVE
asymptote of the curve y = f(x).
(ii) The line x = a is a vertical asymptote if at
least one of the following statements is
true :
lim f(x) =
x a
lim f(x) = –
x a
below by y = g(x), on the left by the line x = a, and
on the right by the line x = b is
b
A = (f(x) g(x))dx
lim f(x) =
xa
lim f(x) = –
a
5.
xa
| f(x) g(x) | dx .
a
6.
lim [f(x) m1x] c1 ,
x
then the straight line y = m1x + c1 will be
an asymptote (a right inclined asymptote
or, when m 1 = 0, a right horizontal
asymptote).
If there are limits
lim
x
lim [f(x) m 2 x] c2 ,
2.
then the straight line y = m2x + c2 is an
asymptote (a left inclined asymptote or,
when m2 = 0, a left horizontal asymptote).
If f(x) 0 for x [a, b], then the area bounded
by curve y = f(x), x-axis, x = a and x = b is
b
A=
f(x) dx .
a
3.
Let f(x), x [a, b], be a continuous function on
[a, b] whose graph intersects the interval [a, b]
of the x-axis at a finite number of points. Then, the
area of the plane figure bounded by the graph of
the function f(x), the interval [a,b] of the x-axis,
and line segments x = a and x = b is computed by
the formula
b
A = f(x) dx
a
4.
If f and g are continuous functions on the interval
[a, b], and if f(x) g(x) for all x in [a, b], then the
area of the region bounded above by y = f(x),
If g (y) 0 for y [c, d] then the area bounded by
the curve x = g(y), y–axis and the abscissa y = c
d
and y = d is
g(y)dy .
y c
7.
f(x)
m 2 and
x
x
The area bounded by curves y = f(x) and y = g (x)
between ordinates x = a and x = b is
b
(iii) If there are limits
f(x)
m1 and
lim
x x
3.63
If y = f(x) is a strictly monotonic function in (a, b),
with f '(x) 0, then the area bounded by the
ordinates x = a, x = b, y = f(x) and y = f(c) (where
ab
.
2
Since area remains invariant even if the
coordinates axes are shifted, hence shifting of
origin in many cases proves to be convenient in
computing the areas.
The area of the curvilinear trapezoid bounded by
a curve represented by x = (t), y = (t), where
t and () = a, () = b. is
c (a, b)) is minimum when c =
8.
9.
A = (t)(t)dt .
The area of the region is also given by the formula
t
1 2
(t) '(t) (t) '(t) dt
A=
2 t
1
10. If r = f() be the equation of a curve in polar
coordinates where f( ) is a single valued
continuous function of , then the area of the
sector enclosed by the curve and the two radii
vectors = 1 and = 2 (1 < 2), is equal to
1 2 2
r d .
2 1
3.64
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
SINGLE CORRECT ANSWER TYPE
1.
2.
3.
4.
5.
6.
7.
8.
{x}, x I
If f(x) =
and g(x) = {x}2, (where {.}
xI
1,
denotes fractional part of x), then the area bounded
by f(x) and g(x) for x [0, 10] is
(A) 5/3
(B) 5
(C) 10/3
(D) none of these
The area enclosed by the curve | y | = sin 2x, when
x [0, 2] is
(A) 1
(B) 2
(C) 3
(D) 4
2
Let f(x) = x , g(x) = cos x and , () be the
roots of the equation 18x2 – 9x + 2 = 0. Then the
area bounded by the curves y = fog(x), the
ordinates x = , x = and the x-axis is
1
(B)
(A) ( – 3)
3
2
(C)
(D)
4
12
The area common to the region determined by
y x and x2 + y2 < 2 has the value
(A)
(B) (2 – 1)
1
(C)
(D) none of these
4 6
The graph of y2 + 2xy + 40 | x| = 400 divides the plane
into regions. The area of the bounded region is
(A) 400 sq. units
(B) 800 sq. units
(C) 600 sq. units
(D) none of these
The area of the region defined by || x | – | y || 1 and
x2 + y2 1 in the X-Y plane is
(A)
(B) 2
(C) 3
(D) 1
The area defined by 1 |x – 2| + |y + 1| 2 is
(A) 2
(B) 4
(C) 6
(D) none of these
The area enclosed by the curve | y | = 5 – (1 – | x |)2 is
8
(A) ( 7 5 5 ) sq. units
3
2
(7 5 5 ) sq. units
(B)
3
2
(5 5 7) sq. units
3
(D) none of these
A square ABCD is inscribed in a circle of radius 4.
A point P moves inside the circle such that d(P,
AB) min (d(P, BC), d(P, CD), d(P, DA)) where
d(P, AB) is the distance of a point P from line AB.
The area of region covered by tracing point P is
(B) 8
(A) 4
(C) 8 – 16
(D) None of these
(C)
9.
10. The parabolas y2 = 4x and x2 = 4y divide the square
region bounded by the lines x = 4, y = 4 and the
coordinate axes. If S1, S2, S3 are respectively the
areas of these parts numbered from top to bottom,
then S1 : S2 : S3 is
(A) 2 : 1 : 2
(B) 1 : 1 : 1
(C) 1 : 2 : 1
(D) 1 : 2 : 3
11. Consider the graph of continuous function
y = f(x) for x [a – b, a + b] a, b R+ and b > a. If
the origin is shifted to (a + b, 0) such that new
axes are parallel to the old axes, then the area
bounded by the given curve, the X–axis and the
new ordinates X = –a, X = –b can be written as
(A)
ab
2
f(x) dx
f(x) dx
(B)
f(a b x) dx
1
(D) 2
ba
2
b
(C)
b
a
a
ab
f(x) dx
ab
12. The area of the region bounded by two branches
of the curve (y – x)2 = x3 and the straight line x = 1 is
3
2
(B)
(A)
5
5
4
(C)
(D) 1
5
13. If the tangent to the curve y = 1 – x2 at x = , where
0 < < 1, meets the axes at P and Q. Also varies,
the minimum value of the area of the triangle OPQ
is k times the area bounded by the axes and the part
of the curve for which 0 < x < 1, then k is equal to
AREA UNDER THE CURVE
(A)
2
3
(B)
75
16
25
2
(C)
(D)
18
3
14. Let f is a differentiable function such that
f(x + y) = f(x2 + y2), where x, y R. If 4 points
A, B, C, D, are selected on curve y = f(x), then
area of (ABC + BCD) can be equal to
(A) 2
(B) 1
(C) > 2
(D) None
15. Suppose y = f (x) and y = g(x) are two functions
whose graphs intersect at the three points (0, 4),
(2, 2) and (4, 0) with f (x) > g (x) for 0 < x < 2 and
4
f (x) < g (x) for 2 < x < 4. If [f(x) g(x)]dx =10
4
0
and [g(x) f(x)]dx =5, the area between two
2
curves for 0 < x < 2, is
(A) 5
(C) 15
(B) 10
(D) 20
16. 3 points O(0, 0), P(a, a2), Q(–b, b2) (a > 0, b > 0) are
on the parabola y = x2. Let S1 be the area bounded
by the line PQ and the parabola and let S2 be the
area of the triangle OPQ, the minimum value of
S1/S2 is
(A) 4/3
(B) 5/3
(C) 2
(D) 7/3
17. The area bounded by the curve f(x) = |tan x + cot x|
– |tan x – cot x| between the lines x = 0, x = and
2
the x-axis, is
(A) ln 4
(B) ln 2
(C) 2 ln 2
(D)
2 ln 2
18. The area bounded by the curves
|x||y| |x||y|
1
and
2 is
2
2
2
(A) (7 + ln 4) sq. units
(B) (7 – ln 2) sq. units
(C) (14 + 2 ln 2) sq. units
(D) (14 – 2 ln 2) sq. units
|y| = e–|x| –
3.65
19. A function y = f (x) satisfies the condition f '(x) sin x
+ f (x) cos x = 1, f (x) being bounded when x 0.
2
If I =
f (x)dx then
0
(A)
2
<I<
2
4
(B)
2
<I<
4
2
(D) 0 < I < 1
2
20. Area enclosed by the graph of the function
y = ln2x – 1 lying in the 4th quadrant is
(C) 1 < I <
(A)
2
e
(B)
4
e
1
1
(C) 2 e
(D) 4 e e
e
21. The area bounded by the curve y = f(x), the
co-ordinate axes & the line x = x1 is given by
x1 . e x1 . Therefore f (x) equals :
(A) ex
(B) x ex
(C) xex ex
(D) x ex + ex
22. The slope of the tangent to a curve y = f(x) at (x , f(x))
is 2x + 1. If the curve passes through the point (1 , 2)
then the area of the region bounded by the curve,
the x-axis and the line x = 1 is
5
6
(B)
6
5
1
(C)
(D) 1
6
23. Area of the region enclosed between the curves
(A)
x = y2 – 1 and x = |y| 1 y 2 is
(A) 1
(C) 2/3
(B) 4/3
(D) 2
24. The area bounded by the curve y = x e–x, xy = 0
and x = c where c is the x-coordinate of the curve's
inflection point, is
(A) 1 – 3e–2
(B) 1 – 2e–2
–2
(C) 1 – e
(D) 1
25. Area enclosed by the curves y = lnx , y = ln | x | ,
y = | ln x | and y = | ln | x | | is equal to
(A) 2
(B) 4
(C) 8
(D) cannot be determined
3.66
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
26. If (a, 0); a > 0 is the point where the curve y = sin2x
– 3 sinx cuts the x-axis first, A is the area
bounded by this part of the curve , the origin and
the positive x-axis, then
(A) 4A + 8 cos a = 7
(B) 4A + 8 sin a = 7
(C) 4A – 8 sin a = 7
(D) 4A – 8 cos a = 7
27. A function y = f (x) satisfies the differential
dy
equation
– y = cos x – sin x, with initial condition
dx
that y is bounded when x . The area enclosed
by y = f (x), y = cos x and the y-axis in the 1st quadrant
2 1
(A)
(B)
2
1
(D)
2
(C) 1
28. If the area bounded between x-axis and the graph
of y = 6x – 3x2 between the ordinates x = 1 and
x = a is 19 square units then 'a' can take the value
(A) 4 or – 2
(B) two values are in (2, 3) and one in (–1, 0)
(C) two values one in (3,4) and one in (–2,–1)
(D) none of these
29. If area bounded by the line y = x, curve y = f(x) and
lines x = 1, x = t, is
then f(x) =
t2 1 t t
/2
t 1
1/2
4 x2 1
x2 1 x x
x
4 x 1
x2 1 x 2 x2 1
4 x2 1
x2 1 x x
(B)
(C) x
3
2
3
1/2
x2 1 x + 2 x2 1
(D) None of these
×
30. The area bounded by the curve y =
y + |2 – x| = 2 is
(B) 2 – ln 3
(D) none of these
31. The area bounded by y = x2 + 2 and y = 2|x|
– cos x is equal to
(A) 2/3
(B) 8/3
(C) 4/3
(D) 1/3
32. The graphs of f(x) = x2 and g(x) = cx3 (c > 0) intersect
1 1
at the points (a, 0) & , 2 . If the region which
c c
lies between these & over the interval [0, 1/c] has
the area equal to 2/3 then the value of c is
(A) 1
(B) 1/3
(C) 1/2
(D) 2
33. The area between curve y = 2x4 – x2, x-axis and the
ordinates of the two minima of the curve is
(A)
7
sq. units
120
3
and
|x|
(B)
11
sq. units
120
13
sq. units
(D) None of these
120
34. If A(n) represents the area bounded by the curve
y = nlnx, where n N and n > 1, the x-axis and
the lines x = 1 and x = e, then the value of A(n)
+ nA(n–1) is equal to:
(C)
n2
e 1
(C) n2
n2
e 1
(D) e.n2
(A)
(A) x
3
4 ln 27
3
(C) 2 + ln 3
(A)
(B)
35. If y = mx, equally divides the area bounded by
y = cos–1 (cos x) (x [0, ]) and y = 0, then m is equal
to
(A) 1/2
(C) 1/3
(B) 1– 2
(D) None of these
36. Area enclosed by the curve y = (x2 + 2x)e–x and the
positive x-axis is
(A) 1
(B) 2
(C) 4
(D) 6
37. Let S(t) be the area of the OAB with O (0, 0, 0),
A (2, 2, 1) and B (t, 1, t + 1). The value of the
e
2
definite integral (S(t)) ln t dt, is equal to
1
3.67
AREA UNDER THE CURVE
(A)
2e3 5
2
(B)
e3 5
2
2e3 15
e3 15
(D)
2
2
38. The length of sub-normal at any point P(x, y) on
the curve, which is passing through M(0, 1) is
unity. The area bounded by the curves satisfying
this condition is equal to
(C)
1
2
(B)
3
3
4
8
(C)
(D)
3
3
39. A triangle has one vertex at (0, 0) and the other
two on the graph of y = –2x2 + 54 at (x, y) and
(A)
(–x, y) where 0 < x < 27 . The value of x so that
the corresponding triangle has maximum area is
(A)
27
2
(C) 2 3
(B) 3
(D) None of these
40. The area of the region of the xy plane defined by
the inequality | x | + | y | + | x + y | 1 is
1
3
(B)
2
4
(C) 1
(D) None of these
41. The value of ‘a’ (a > 0) for which the area bounded
x 1
by the curve y = 2 , y = 0, x = a and x = 2a has
6 x
the least value, is
(A) 2
(B) 2
(C) 21/3
(D) 1
42. The area included between the curve xy2 = a2 (a x)
and its asymptote is
(A)
(1)
a2
2
(C) a 2
(B) 2 a2
(D) none
43. The area bounded by the curves y = x (1 ln x),
x = e 1 and a positive x-axis between x = e 1 and
x = e is
e 2 4 e 2
(A)
5
e 2 5e 2
(B)
4
4 e 2 e 2
(C)
5
44. I f
th e
5e 2 e 2
(D)
4
ar ea
e n c l os e d
1
1
be t we e n
f(x) = min. cos (cos x ), cot (cot x ) and
x-axis in x (, 2 ) is
2
where k N, then
k
k is equal to
(A) 4
(B) 6
(C) 8
(D) 12
45. Area enclosed by the curve , |x + y –1| + |2x + y + 1|
= 1 is
(A) 2 sq. units
(B) 1 sq. units
(C) 4 sq. units
(D) none of these
46. The area of the region consisting of all points
(x, y) so that x2 + y2 1 | x | + | y |, is
(A)
(B) – 1
(C) – 2
(D) – 3
47. The area enclosed between the curves
1
y = loge (x + e), x = loge and the x–axis is
y
(A) 2
(B) 1
(C) 4
(D) None of these
48. The area of the region bounded in first quadrant
by y = x1/3, y = –x2 + 2x + 3, y = 2x – 1 and the axis
of ordinates is
(A) 12/55
(B) 55/12
(C) 32/55
(D) None of these
49. Let C be a curve passing through M (2, 2) such
that the slope of the tangent at any point to the
curve is reciprocal of the ordinate of the point. If
the area bounded by curve C and line x = 2 is
p
expressed as a rational
(where p and q are in
q
their lowest form), then (p + q) is equal to
(A) 19
(B) 18
(C) 9
(D) 6
50. Let C be the curve passing through the point
(1, 1) has the property that the perpendicular
distance of the origin from the normal at any point
P of the curve is equal to the distance of P from
the xaxis. If the area bounded by the curve C and
k
square units,
x-axis in the first quadrant is
2
then find the value of k.
(A) 2
(B) 2 2
(C) 4
(D) 1
3.68
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
MULTIPLE CORRECT ANSWER TYPE FOR JEE ADVANCED
51. If area bounded by the curve y = cos[x], x-axis and
the lines given by x2 – 3x + 2 = 0 is A, then (where
[x] represents integer function)
/2
(A) A =
/2
cos[x]dx
/2
1
/2
(B) 12a
(C) A =
1
(C) 2
cos[x]dx
/2
(D) A < sin1
52. Let f(x) = x2 – 5x + 6 be a function x R. C1 and
C2 are two curves given by |y| = f(|x|) and x2 + y2
25
respectively. Then
4
(A) graph of C1 and C2 intersects each other at 8
distinct points
(B) area enclosed by C1, C2 and lying on left of
the line x – 2 = 0 in 1st quadrant is less than
5 sq. units
(C) graph of C1 and C2 intersects each other at 12
distinct points
(D) both (B) and (C)
=
53. The area bounded by a curve, the axis of coordinates and the ordinate of some point of the
curve is equal to the length of the corresponding
arc of the curve. If the curve passes through
the point P (0, 1) then the equation of this curve
can be
(A) y =
cos2t sin4t dt
a
2
cos[x]dx
0
(B) A = 2cos1
/2
cos4t sin2t dt
0
2
cos[x]dx
(A) 12a
1
(ex e – x + 2)
2
1
(ex + ex)
2
(C) y = 1
(B) y =
2
(D) y = x x
e e
54. The area enclosed by the currves x = a sin 3t
and y = a cos3t is equal to
(D) 4
(a – x ) dx
2/3
2/3 3/2
a
a
(a – x ) dx
2/3
2/3 3/2
0
55. If Ai is the area bounded by |x – ai| + |y| = bi, i N,
b
3
bi and bi+1 = i , a 1 = 0,
2
2
where a i+1 = a i +
b1 = 32, then
(A) A3 = 128
(B) A3 = 256
n
(C)
8
A i 3 (32)2
n
lim
i 1
n
(D) lim
n
4
Ai 3 (16)2
i 1
56. Sup pose f i s defi n ed fr om R [–1 , 1]
as f (x) =
x 2 1
where R is the set of real number..
x2 1
Then the statement which does not hold is
(A) f is many one onto
(B) f increases for x > 0 and decrease for x < 0
(C) minimum value is not attained even though f
is bounded
(D) the area included by the curve y = f (x) and
the line y = 1 is sq. units.
57. The curve x6 – x2 + y2 = 0
(A) is symmetric about both axes
(B) meets the x-axis at three points.
(C) is symmetric about y = –x
(D) bounds an area
2
AREA UNDER THE CURVE
58. Consider the functions f (x) and g (x), both defined
from R R and are defined as f (x) = 2x – x2 and
g (x) = xn where n N. If the area between f (x) and
g (x) is 1/2 then n is a divisor of
(A) 12
(B) 15
(C) 20
(D) 30
59. Which of the following statement(s) is/are true
for the function f (x) = (x – 1)2(x – 2) + 1 defined on
[0, 2]?
23
(A) Range of f is 27 , 1 .
(B) The coordinates of the turning point of the
5 23
graph of y = f (x) occur at (1, 1) and , .
3 27
(C) The value of p for which the equation f (x) = p
has 3 distinct solutions lies in interval
23
27 , 1 .
(D) The area enclosed by y = f (x), the lines x =0
7
and y = 1 as x varies from 0 to 1 is
.
12
cos x
0x
2
60. Consider f (x) =
such that
2
x
x
2
2
f is periodic with period , then
2
(A) The range of f is 0, 4
(B) f is continuous for all real x, but not
differentiable for some real x
(C) f is continuous for all real x
(D) The area bounded by y = f (x) and the x-axis
3
from x = – nto x = n is 2n 1 for a
24
given nN
61. Two circles have centres that are d units apart
be the
and each has diameter d . Let A(D)
area of the smallest circle that contains both these
circles, then
d2 d
(A) A(D) =
2
d d
(B) A(D) =
2
(C)
lim
d
(D) lim
A(d)
2
d
A(d)
d
62. Let
d
2
4
2
3.69
2
2
2
e x dx = and f (t) =
e
t x2
dx , (t > 0)
x
(B) f (x) is monotonic decreasing
(C) if area bounded by y = f (x), x = 2, x = 16 and
(A) f (x) is equal to
3
(D) equation of normal to curve y = f (x) at x = is
y – 1 = 2(x – )
y = c is minimum then c =
63. Let f(x) = x x2 & g(x) = ax . If the region above the
graph of g and below the graph of f has an area
equal to 9/2 then 'a' is equal to
(A) 2
(B) 4
(C) 2
(D) 3
64. The equation (y x 2)2 = 9x represents a curve
(A) which is an ellipse
(B) which cuts the x - axis at two distinct points
(C) which touches the y - axis
(D) which encloses an area equal to unity with
the co-ordinate axes .
65. The area of the figure bounded by the curve
y = 3x3 + 2x and the straight lines x = a and y = 0 is
unity . Then value of 'a' is :
(A) 4
(C)
2
3
(B)
3
(D)
2
3
66. Let f (x) be differentiable function on the interval
(– ) such that f (1) = 5
3.70
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
a f(x) x f(a)
2 , x R. Then which
ax
ax
of the following alternative(s) is/are correct?
(A) f (x) has an inflection point.
and Lim
(B) f '(x) = 3 x R .
2
(C)
f(x)dx = 10.
0
(D) Area bounded by f (x) with co-ordinate axes
2
is .
3
67. The parabola x = y2 + ay + b intersect the parabola
x2 = y at (1, 1) at right angle.
Which of the following is/are correct?
(A) a = 4, b = – 4
(B) a = 2, b = – 2
(C) Equation of the director circle for the parabola
x = y2 + ay + b is 4x + 1 = 0.
(D) Area enclosed by the parabola x = y2 + ay + b
1
.
6
68. If [ ] denotes the integral part of , and
f(x) = sec–1 [– sin2 (x)] then
(A) domain is x R – {n} where n I and
range is
(B) thefunction has removable discontinuities at
x = n n I
and its latus rectum is
(C) area bounded by y = x2 and y = f(x) is
(D) none of these
4
3
69. Which of the statement(s) are true ?
(A) The area bounded by the curve y = x | x |,
x – axis and the ordinates x = 1, x = – 1, is 2/3
(B) The area bounded by y = x2, y = [x + 1], x 1
and the y-axis where [.] denotes the greatest
integer function is 2/3
(C) The slope of the tangent to curve y = f(x) at
(x, f(x) ) is 2x + 1. If the curve passes through
the point (1, 2), then the area of the region
bounded by the curve, the x-axis and the line
x = 1, is 5/6.
(D) None of these
70. Which of the statement(s) are true ?
(A) The area bounded by the
curve
y = max, {2 = 2, 2, 1+ x} and the ordinates
x = –1 and x = 1, is 9/2
(B) The area bounded by the curve y = max, {x + | x |,
x – [x]}, whee [.] denotes greatest integer
function, and ordinates x = –2 and x = 2, is 1
(C) The area bounded by the cureve y = |x3 – 3x2
+ 2x| and ordinates x = 0 and x = 3, is 11/4
(D) None of these
71. Which of the statement(s) are true ?
(A) The area bounded by [x] = [y], 0 x, y 10 is 10
(B) The area bounded by max (|x|, |y|) 1 is 4
(C) The area bounded by |y + x| 1, |y – x| 1,
2x2 + 2y2 = 1is 2 –
2
(D) None of these
Comprehension - 1
Consider two curves C1 [f(y)]2/3 + [f(x)]1/3 = 0 and C2
[f(y)] 2/3 + [f(x)] 2/3 = 12, satisfying the relation
f(x – y) f(x + y) – (x + y) f(x – y) = 4xy (x2 – y2).
72. The area bounded by C1 and C2 is
(A) 2 – 3 sq. units
(B) 2 + 3 sq. units
(C) + 6 sq. units
(D) 2 3 – sq. units
73. The area bounded by the curve C2 and |x| + |y| = 12 is
(A) 12 – 2 12 sq. units
(B) 6 – 12 sq. units
(C) 2 12 – 6 sq. units
(D) None of these
74. The area bounded by C1 and x + y + 2 = 0 is
(A) 5/2 sq. units
(B) 7/2 sq. units
(C) 9/2 sq. units
(D) None of these
Comprehension - 2
Consider the function defined implicitly by the
–1
–1
equation y2 – 2 y esin x + x2 – 1 + [x] + e 2 sin x = 0
(where [x] denotes the greatest integer function).
75. The area of the region bounded by the curve and
the lines x = – 1 is
(A + 1 sq. units
(B) – 1 sq. units
(C)
+ 1 sq. units
(D)
– 1 sq. units
2
2
AREA UNDER THE CURVE
76. Line x = 0 divides the region mentioned above in
two parts. The ratio of area of left hand side of line
to that of right hand side of line is
(A) 2 + :
(B) 2 –
(C) 1 : 1
(D) + 2 :
77. The area of the region bounded by the curve and
lines x = 0 and x = 1/2 is
(A)
3
3
sq. units
sq. units (B)
4 6
2 6
3–
3–
sq. units (D)
sq. units
4 6
2
6
83. Range of the function sin–1
3.71
(fog(x)) , is
(A) 0, 2
(B) 0, 2
(C) 2 , 0, 2
(D) 2
Comprehension - 5
Consider the polynomial f(x) = x6 – 2a2x4 + a4x2 + ax. The
graph of the function is tangent to a straight line at
three point A1, A2, A3 where A2 lies between A1 and A3.
A variable line y = l(x) intersects the parabola
y = x 2 at points P and Q whose x-coordinates
are and respectively with < . The area of the
figure enclosed by the segment PQ and the parabola is
4
always equal to . The variable segment PQ has its
3
middle point as M.
78. The equation of the straight line is
(A) y = x + a
(B) 2y = a x
(C) y = ax
(D) None of these
84. The value of ( – ) is equal to
(A) 1
(B) 2
(C) 4
(D) 8
79. The ratio of length of segments A1A2 and A1A3 is
(A) 1 : 2
(B) 1 : 9
(C) 2 : 3
(D) None of these
80. The ratio of areas of the figures bounded by line
segments A 1A2, A 2A3 and the graph of the
polynomial is
(A) 1 : 4
(B) 1 : 9
(C) 1 : 1
(D) None of these
85. Equation of the locus of the mid-point of PQ, is
(A) y = 1 + x2
(B) y = 1 + 4x2
(C) 4y = 1 + x2
(D) 2y = 1 + x2
(C)
Comprehension - 3
Comprehension - 4
Consider f, g and h be three real valued functions
1, x 0
defined on R. Let f(x) = 0, x 0 ,
1, x 0
g (x) = x (1 – x2) and h(x) is such that h''(x) = 6x – 4.
Also h(x) has local minimum value 5 at x = 1.
81. The equation of tangent at M(2, 7) to the curve
y = h(x), is
(A) 5x + y = 17
(B) x + 5y = 37
(C) x – 5y + 33 = 0
(D) 5x – y = 3
82. The area bounded by y = h(x), y = g f(x)
between x = 0 and x = 2 equals
23
20
(B)
(A)
2
3
40
32
(C)
(D)
3
3
86. Area of the region enclosed between the locus of
M and the pair of tangents on it from the origin, is
(A)
8
3
(B) 2
(C)
4
3
(D)
2
3
Assertion (A) and Reason (R)
(A) Both A and R are true and R is the correct
explanation of A.
(B) Both A and R are true but R is not the correct
explanation of A.
(C) A is true, R is false.
(D) A is false, R is true.
87. f(x) is a polynomial of degree 3 passing through
origin having local extrema at x = ± 2.
Assertion (A) : Ratio of areas in which f(x) cuts
the circle x2 + y2 = 36 is 1 : 1.
Reason (R) : Both y= f(x) and the circle are
symmetric about origin.
88. Assertion (A) : The area enclosed between
the parabolasy2 –2y+ 4 x + 5 = 0andx2 + 2x – y+ 2 = 0is
same as that of bounded by curves y2 = – 4x and x2 = y.
3.72
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
Reason (R) : Shifting of origin to point (h, k) does
not change the bounded area.
89. Assertion (A) : The area of the region bounded
by the curve 2y = logex, y = e2x and the pair of
lines (x + y – 1) × (x + y – 3) = 0 is 2k sq. units.
Reason (R) : The area of the region bounded by
the curves y = e2x, y = x and the pair of lines x2 + y2
+ 2xy – 4x – 4y + 3 = 0 is k units.
90. Consider two regions
R1 : Point P is nearer to (1, 0) than to x = – 1
R2 : Point P is nearer to (0,0) than to (8,0).
Assertion (A) : Area of the region common to R1
128
and R2 is
sq. units.
3
Reason (R) : Area bounded by x = 4 y and y = 4
32
sq. units.
is
3
91. Assertion (A) : (A) : Let An be the area outside a
regular n-gon of side length 1 but inside its
circumcircle and Bn be the area inside the n-gon
A
but outside its inscribed circle. Then lim n = 2
n Bn
2
n
1
Reason (R) : An = cos ec cot and
n
4
n
2
2
An
n
1
cot
sec
Bn =
so lim B = 2
n n
4
n
n
2
92. Assertion (A) : A convex quadrilateral is such
that each of its vertices satisfies the equation
x2 + y2 = 73 and xy = 24. The area of this
quadrilateral is 110.
Reason (R) : The quadrilateral is a rectangle whose
side lengths are 5 2 and 11 2 .
93. Assertion (A) : Let f(x) = x + 2x + 2x + 1 and g(x)
be its inverse. The area bounded by g(x),
x-axis, x = –3 and x = 6 is given by
3
2
1
0
0 (5 – x – 2x – 2x) dx + 2 (x + 2x + 2x + 4) dx
b
Reason (R) : f(x) dx = bf(b) – af(a)
a
f (b)
f (a) f (y) dy
3
2
3
2
–1
94. Assertion (A) : The area of the region represented
by the expression 2 x y x y 2 2 is
equal to 6.
Reason (R) : This inequation gives a squared strip
having inner side 2 unit and outer side 2 2
unit.
95. Assertion (A) : Area enclosed by curve
(y–sin–1x)2 = x – x2 is equal to
sin x x x dx
1
1
2
0
Reason (R) : If y = f(x) and y = g(x) are two curves
such that f(x) g(x) for x [a, b], where
f(x) = g(x) x = a, b then area enclosed by these
b
two curves is given by
{f(x) g(x)}dx
a
96. Assertion (A) : Area enclosed by the curve
1
2
1 x 2
2
1 x 2 dx
y e
(1 x ) is equal to 2
1
Reason (R) : If 1, 2, ....., n domain of
continuous functions f(x) and
g( x) , are the
roots of equation g(x) = 0 and 1 < 2.... < n, then
area enclosed by curve (y – f(x))2 = g(x) equals to
3
n
2
2
g(x)dx
g(x)dx .......
g(x)dx
2
n 1
1
MATCH THE COLUMNS FOR JEE ADVANCED
97.
Column-I
(A) Area enclosed by y = [x] and y = {x},
where [·] and {.} represent greatest integer
and fractional part functions.
(B) The area bounded by the curves
y2 = x3 and |y| = 2x.
Column-II
(P) 16/5 sq. units
(Q) 1 sq. units
AREA UNDER THE CURVE
(C)
The smaller area included between the curves
3.73
(R) 4 sq. units
x | y | = 1 and |x| + |y| = 1.
x2
(D) Area bounded by the curves y = 64 2
(where [·] denotes the greatest integer function),
y = x – 1 and x = 0 above the x-axis.
98.
Column I
(A) A differentiable function satisfies f'(x) = f(x) + 2ex
with the condition f(x) = 0, then the area bounded
by y = f(x) and the x-axis is
(B) C1 : y = ex, C2 : y = ea – x, where a > 0. If A is the area
(C)
8A
bounded by y-axis, C1 and C2, then lim
a 0 a 2
is equal to
Let y = f(x) and y = g(x) are two continuous function
intersecting at (0,4), (2,2) and (4,0) with f(x) > g(x)
4
(S) 2/3 sq. units
Column – II
(P) 3
(Q) 2
(R) 15
4
x (2,4). If (f(x) g(x))dx 10 , (g(x) f(x))dx 5
0
2
then the area between the two curves for 0 < x < 2 is
(D) If the area bounded by the curves y = x(1 – lnx),
(S) 5
e a be a
x = e–1 and x = e is
, then roots of
4
2
x – (a + b)x + 2b = 0 is /are
99.
Column I
(A) Area bounded by the curves y = [cosA + cosB + cosC],
y = 7 sin
Column – II
(P) 0
A sin B sin C
, (where [.] denotes the greatest
2
2
2
integer function and A, B, C are the angles of a triangle)
and curve |x – 4| + |y| = 2 is
(B)
The area bounded by |x| + |y| = 1 and |x – 1| + |y| = 1 is
(Q)
1
3
(C)
The area bounded by y = f(x), y-axis and line 2y = (x + 1) where
(R)
1
2
sin 1 x tan 1 x cos1 x tan 1 1
is A, then [A2]
6
x
is (where [.] is greatest integer function and –1 x 1)
f(x) =
(D) Area bounded by the curve y = x and y = |x| is
(S) 3
100. Let consider the region max {|x|, |y|} 2 represented by f(x, y). Now, if the region f(x, y) undergoes the
following transformations successively, then match the region obtained in column –II by transformations
3.74
INTEGRAL CALCULUS FOR JEE MAIN AND ADVANCED
done in column–I after each step.
Column I
(A) The region f(x, y) is a
Column – II
(P) square of area 16 sq. unit
2
2
The region given by f1(x,y) = f x , y , is a
(C) The region given by f2(x,y) = f1(y, x), is a
(D) The region given by f3(x, y) = f2 (x + 3y, y), is a
(B)
(Q) rectangle of area 12 sq. unit
(R) parallelogram
(S) rhombus of area 12 sq. unit
(T) square of area 4 sq. unit
101. Column - I
Column - II
(A) The positive value of a such that the parabola y = x2 + 1
bisects the area of the rectangle with vertices (0, 0), (a, 0),
(a, a2 + 1) and (a, a2 + 1) is
(P) 2 +
(B)
(Q) 3 6
The graph of x2 – (y – 1)2 = 1 has one tangent at (a,b)
with positive slope which pases through origin. Then a + b is
(C)
The area of the largest rectangle that can be drawn inside
a 3 – 4 – 5 right triangle with one of the rectangles sides
along one of the legs of the triangle, is
(D) A ballon in the shape of a cube is blown up at a rate such that ,
at time t its surface area is 6t. The rate of pumping of air when
the surface area is 144, equals
1.
2.
Find the area of the closed figure bounded by the
following curves.
2(y 1)
3(x 1)
(i) y =
,x=
, x = 3, x = 5
y 1
x–2
Find the area bounded by the curve y = 1 + 8/x2
and x–axis from x = 2 to x = 4. If the ordinate x = a
divide the area into two equal parts, find ‘a’.
4.
Compute the area of the figure bounded by
straight lines x = 0, x = 2 and the curves y = 2x and
y = 2x – x2.
5.
The line 3x + 2y = 13 divides the area enclosed by
the curve, 9x2 + 4y2 – 18x – 16y – 11 = 0 into two parts.
3
(S) 3
Find the ratio of the longer area to the smaller
area.
6.
Find the area of the closed figure bounded by the
following curves.
1
(i) y = x2 – 2x + 2, tangents to the parabola
2
1
at 1, and (4, 2)
2
(ii) y = 25x + 16 , y = b.5x + 4 whose tangent at
x = 1 has slope 40 n 5.
(iii) y = x4 – 2x2 + 5, y = 1, x = 0, x = 1
7.
Find the area of the region bounded by the curves,
y = loge x, y = sin4x and x = 0.
8.
Find the area bounded by the curves y = 1 x 2
and y = x3 – x. Also find the ratio in which the
y–axis divided this area.
9.
If the area enclosed by the parabolas y = a – x2 and
(ii) y = 8 sin4x + 4 cos2x , x [0, ] and y = 0
| 4 x2 |
, y = 7 – |x|
(iii) y =
4
6
, y = |2 – x|
(iv) y = 4 –
|1 x |
x 1
and
Find the area bounded by the curve y =
x 1
its asymptote from x = 1 to x = 2.
3.
(R)
2
y = x2 is 18 2 sq. units. Find the value of ‘a’.
AREA UNDER THE CURVE
10. Through the point (x0, y0) of the graph of the
function y = 1 cos 2 x draw a normal to the
graph, if it is known that the straight line x = x0
divides the area bounded by the given curve, the
x-axis, and the straight lines x = 0 and x = 3/4 into
equal parts.
11. Find all the values of the parameter a (a > 0) for
each of which the area of the figure bounded by
2
the straight line y =
(a ax)
1 a4
and the parabola
(x 2 2ax 3a 2 )
is the greatest.
1 a4
12. Find the area of the smaller portion enclosed by
the curves x2 + y2 = 9, y2 = 8x.
y=
13. Compute the area of the figure bounded by the
curve y = ln x and y = ln2 x.
14. Find the area of the figure bounded by the
pareabolas, x = –2y2, x = 1 – 3 y2 and y–axis.
15. A normal to the curve, x2 + x – y + 2 = 0 at the
point whose abscissa is 1, is parallel to the line
y = x. Find the area in the first quadrant bounded
by the curve, this normal and the axis of ’x’.
16. Indicate the region bounded by the curves y = x ln
x and y = 2x – 2x2 and obtain the area enclosed by
them.
17. Let us designate as S(k) the area contained
between the parabola y1 = x2 + 2x – 3 and the straight
line y2 = kx + 1. Find S(–1) and calculate the least
value of S(k).
18. Find the area enclosed by the curve y2 (x + 1) = x2
(1 – x).
1
19. A rectangle of length and height 4 is bisected
4
by the x-axis and is in the first and fourth quadrants.
The graph of y = sin x + C divides the area of the
square in half. What is C ?
20. Calculate the area bounded by the curves
y = cos–1 (cos x) and y = 2 + x2 – 2x.
21. The area from 0 to x under a certain graph is given
to be A = 1 3x 1 , x 0 ;
3.75
(i) Find the average rate of change of A w.r.t. x as
x increases from 1 to 8.
(ii) Find the instantaneous rate of change of A
w.r.t. x at x = 5.
(iii) Find the ordinate (height) y of the graph as a
function of x.
(iv) Find the average value of the ordinate
(height) y, w.r.t. x as x increases from 1 to 8
22. Find the area included between the curve x2 + y2 = a2
and | x | + | y | = a (a > 0).
23. A figure is bounded by y = x2 + 1, y = 0, x = 0,
x = 1. At what point of the curve y = x2 + 1, must a
tangent be drawn for it to cut off a trapezoid of the
greatest area from the figure ?
24. Find the area of the region
{(x, y): 0 y x2 + 1, 0 y x + 1, 0 x 2}.
25. Find the value of c for which the area of the figure
bounded by the curves y = sin 2x, the straight lines
x = /6, x = c and the abscissa axis is equal to 1/2.
x
,
4
y = 0, x = 2 and x = 4. At what angles to the positive
x–axis straight lines must be drawn through (4, 0)
so that these lines partition the figure into three
parts of the same size.
26. A figure is bounded by the curves y =
2 sin
27. Find the value of p (p < 0) for which the area of a
figure bounded by the parabola y = (1 + p2)2x2 + p
and the line y = 0 attains its greatest value.
28. A figure is bounded by the curves y = (x + 3)2, y = 0,
x = 0. At what angles to the x-axis must straight
lines be drawn through the point (0, 9) for them to
partition the figure into three parts of the same size?
29. Let C be the curve passing through the point
(1, 1) has the property that the perpendicular
distance of the origin from the normal at any point
P of the curve is equal to the distance of P from
the xaxis.
0
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