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Amit M. Agarwal
arihant
Skills m
Mathematics for
JEE MAIN &
ADVANCED
ilif
Integral
Calculus
With Sessionwise Theory & Exercises
v a
Skills in
Mathematics for
JEE MAIN &
ADVANCED
Integral
Calculus
With Sessionwise Theory & Exercises
Amit M. Agarwal
sjcarihant
ARIHANT PRAKASHAN (Series), MEERUT
I 1
I
f
Skills in Mathematics for
Illi JEE MAIN & ADVANCED
arihant
ARIHANT PRAKASHAN (Series), MEERUT
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i
Skills in Mathematics for
10 j JEE MAIN & ADVANCED
PREFACE
“YOU CAN DO ANYTHING IF YOU SET YOUR MIND TO IT, I TEACH CALCULUS
TO JEE ASPIRANTS BUT BELIEVE THE MOST IMPORTANT FORMULA IS
COURAGE+ DREAMS = SUCCESS”
It is a matter of great pride and honour for me to have received such an overwhelming
response to the previous editions of this book from the readers. In a way, this has inspired
me to revise this book thoroughly as per the changed pattern of JEE Main & Advanced. I
have tried to make the contents more relevant as per the needs of students, many topics
have been re-written, a lot of new problems of new types have been added in etcetc. All
possible efforts are made to remove all the printing errors that had crept in previous
editions. The book is now in such a shape that the students would feel at ease while going
through the problems, which will in turn clear their concepts too.
A Summary of changes that have been made in Revised & Enlarged Edition
•
Theory has been completely updated so as to accommodate all the changes made in JEE
Syllabus & Pattern in recent years.
•
The most important point about this new edition is, now the whole text matter of each
chapter has been divided into small sessions with exercise in each session. In this way the
reader will be able to go through the whole chapter in a systematic way.
•
Just after completion of theory, Solved Examples of all JEE types have been given, providing
the students a complete understanding of all the formats of JEE questions &. the level of
difficulty of questions generally asked in JEE.
•
Along with exercises given with each session, a complete cumulative exercises have been given
at the end of each chapter so as to give the students complete practice for JEE along with the
assessment of knowledge that they have gained with the study of the chapter.
•
Last 10 Years questions asked in JEE Main &Adv, IIT-JEE &. AIEEE have been covered in all
the chapters.
I
However I have made the best efforts and put my all calculus teaching experience in
revising this book. Still I am looking forward to get the valuable suggestions and criticism
from my own fraternity i.e. the fraternity of JEE teachers.
I would also like to motivate the students to send their suggestions or the changes that they
want to be incorporated in this book.
All the suggestions given by you all will be kept in prime focus at the time of next revision of
the book.
Amit M. Agarwal
Skills in Mathematics for
Lil J JEE MAIN & ADVANCED
CONTENTS
1. INDEFINITE INTEGRAL
LEARNING PART
Session 1
• Fundamental of Indefinite Integral
Session 2
• Methods of Integration
/
Session 3
• Some Special Integrals
Session 4
• Integration by Parts •
1-82
Session 5
• Integration Using Partial Fractions
Session 6
• Indirect and Derived Substitutions
Session?
• Euler’s Substitution, Reduction Formula and
Integration Using Differentiation
PRACTICE PART
• JEE Type Examples
• Chapter Exercises
2. DEFINITE INTEGRAL
LEARNING PART
Session 1
• Integration Basics
• Geometrical Interpretation of Definite
Integral
• Evaluation of Definite Integrals by
Substitution
Session 2
• Properties of Definite Integral
Session 3
• Applications of Piecewise Function Property
Session 4
• Applications of Even-Odd Property and Half
the Integral Limit Property
83-164
Session 5
• Applications of Periodic Functions and
Newton-Leibnitz’s Formula
Session 6
• Integration as Limit of a Sum
• Applications of Inequality
• Gamma Function
• Beta Function
• Wallis Formula
PRACTICE PART
• JEE Type Examples
• Chapter Exercises
Skills in Mathematics for
[11 j JEE MAIN & ADVANCED
3. AREA OF BOUNDED REGIONS
LEARNING PART
Session 1
• Sketching of Some Common Curves
• Some More Curves Which Occur Frequently
in Mathematics in Standard Forms
• Asymptotes
• Areas of Curves
165-232
Session 2
• Area Bounded by Two or More Curves
PRACTICE PART
• JEE Type Examples
• Chapter Exercises
4. DIFFERENTIAL EQUATIONS
LEARNING PART
Session 1
• Solution of a Differential Equation
Session 2
• Solving of Variable Seperable Form
• Homogeneous Differential Equation
Session 3
• Solving of Linear Differential Equations
• Bernoulli’s Equation
• Orthogonal Trajectory
-4
j
-
1
233-308
Session 4
• Exact Differential Equations
Session 5
• Solving of First Order & Higher Degrees
• Application of Differential Equations
• Application of First Order Differential
Equations
PRACTICE PART
• JEE Type Examples
• Chapter Exercises
Skills in Mathematics for
Illi JEE MAIN & ADVANCED
.. -J
SYLLABUS FOR JEE MAIN
Integral Calculus
Integral as an anti - derivative. Fundamental integrals involving
. algebraic, trigonometric, exponential and logarithmic functions.
Integration by substitution, by parts and by partial fractions.
Integration using trigonometric identities.
Evaluation of simple integrals of the type
dx
x2 ± a2
dx
a - x2
dx — >
x2 ± a 2
2
dx
dx
ax2 + bx + c
(px + q) dx
ax2 + bx + c
ax2+bx + c
(px + q)dx
r‘
--------- — -
ax2 + bx + c
- 9
a2 ±x2 dx
dx ----->
J a2-x 2
and
>
1/7 - a2 dx
Integral as limit of a sum. Fundamental Theorem of Calculus.
Properties of definite integrals. Evaluation of definite integrals,
determining areas of the regions bounded by simple curves in
standard form.
Differential Equations
Ordinary differential equations, their order and degree. Formation of
differential equations. Solution of differential equations by the method
of separation of variables, solution of homogeneous and linear
differential equations of the type
= q(x)
i
■
■
Skills in Mathematics for
I III 1 JEE MAIN & ADVANCED
SYLLABUS FOR JEE ADVANCED
Integral Calculus
Integration as the inverse process of differentiation, indefinite
integrals of standard functions, definite integrals and their
properties, application of the Fundamental Theorem of
Integral Calculus.
Integration by parts, integration by the methods of substitution
and partial fractions, application of definite integrals to the
determination of areas involving simple curves.
Formation of ordinary differential equations, solution of
homogeneous differential equations, variables separable
method, linear first order differential equations.
I
I
I
i
,1
CHAPTER
£
I
I
Indefinite
Integral
Learning Part
Session 1
• Fundamental of Indefinite Integral
Session 2
• Methods of Integration
Session 3
• Some Special Integrals
Session 4
• Integration by Parts
Session 5
• Integration Using Partial Fractions
-
Session 6
• Indirect and Derived Substitutions
Session 7
• Euler's Substitution, Reduction Formula and Integration Using Differentiation
Practice Part
-
i
• J EE Type Examples
• Chapter Exercises
Arihant on Your Mobile!
Exercises with the @ symbol can be practised on your mobile. See inside cover page to activate for free.
2
Textbook of Integral Calculus
It was in this aspect that the process of integration was
treated by Leibnitz, the symbol of j being regarded as the
initial letter of the word sum, in the same way as the
symbol of differentiation d is the initial letter in the word
difference.
Definition
§ Example 1 *f^xn+1
+ c] = (n+1) xn, then find
j xn dx.
Sol. As, —[xn + 1 + C] = (n + l)x"
dx
(xn + * + C) is anti-derivative or integral of (n + 1) x".
=>
.n +1
If f and g are functions of x such that g' (x) = f (x), then
the function g is called a anti-derivative (or primitive
function or simply integral) of f w.r.t. x. It is written
symbolically, J f (x) dx = g (x), where, — g (x) = f (x)
f x" dx = -— + c
J
n +1
(j
I Example 2 If — (sin x+c) = cos x, then find
j cos x dx.
Remarks
1. In other words, j/(x) dx = g (x) iff g' (x) = f (x)
2. J /(x) dx =g (x) + c.wherecisconstant,
(g(x) + C)'=g'(x) =/(*)] and Cis called constant of integration.
Sol. As,
x + C) = cos x
=> sin x + C is anti-derivative or integral of cos x.
jcosxdx = (sinx) + C
Session 1
Fundamental of Indefinite Integral
Fundamental of Indefinite Integral
^-{g(x)} = f (x)
dx
| f(x)dx = g(x)+C
Since,
Therefore, based upon this definition and various standard
differentiation formulas, we obtain the following
integration formulae
rxn+1'
x"41
=xnn,n#-l=>Jx" dx (i) —
dx ^n + 1,
n+1
(ii)
dx
dx = log | x | + C, when x * 0
(log | x |) = -=> j[ —
x
x
(iii) -y-(ex) = ex
dx
R
ax
dx I log
-ax,a >Q,a *1
(iv) —.
=>
dx = ex +C
ax
Jax dx = -------+C
loge a
(v) — (- cos x) = sin x
dx
=> j sin x dx = - cos x + C
(vi) — (sin x) = cos x
dx
=>
(vii) — (tan x) =sec2 x
dx
=> [ sec2 x dx = tan x + C
J
cos x dx = sin x + C
(viii) — (-cot x) =cosec2x =»fcosec2x dx - -cot x + C
dx
J
d
(ix) — (sec x) =sec x tan x
dx
=> j sec x tan x dx = sec x + C
, . d
(x) —(-cosec x) = cosec x cot x
dx
=> J cosec x cot x dx = - cosec x 4- C
(xi)
(log |sin x |) =cot x
dx
=^> j cot x dx = log | sin x | + C
Chap 01 Indefinite Integral
x3/2 + 5x1/2
(xii)— (- log | cos x |) = tan x
dx
=> j tan x dx = - log | cos x | + C
,n + 1
X3/2 + !
5x1,2 + i
+--------~3/2 + l 1/2+1
=> jsecxdx = log|sec x + tanx| + C
=> j cosec x dx = log | cosec x - cot x
( ■
dx
I = j(x6 + 15x4 + 75x2 + 125) dx
___ 1
. x7
15xs
75x
15x
75x 3 + 125x+ C
/= — +------ +------7
5
3
7?
= sin
-7
I = — + 3x5 + 25x3 + 125x + C
7
-i
-x2
(xvi) — cos -i£
dx
a
/ T?7?
Sol. (i) I = j tan2 x dx => I = j(sec 2 x - 1) dx
2 .
a)
dx
1= jsec2 xdx- jldx [usingJ sec2 xdx = tan x + C]
+C
• =- tan"1
a2 + x2 a
/ ...\ d r 1
—i X
(xvm)— -cot ■
dx <a
a
=> / = tanx-x + C
i
(ii) 1= j • 2
-dx
sin x cos 2 X
a +x
■sin2 x + cos 2
-1
1
dx - - cot
a2+x2
a
(xix)
1
-sec -i £
dx a
a
dx
X
’-I sin2 x cos2 x£dx
1
1
= -sec
a
-a2
d (1
(xx) — - cosec -i £
dx
'=/
7 -a2
x<
4?
7 -a2
r(sin2x)3 + (cos2x)
2 ~J
• 2
2
sin x cos x
[using (a + i>).33 = a,33 + b3 + 3ab (a + &)]
p(sin x+cos x) -3sin xcos x(sin x + cos x)
I—J
~2
J
sin x cos x
(ii) j(x2 +5)3 dx
2
2
1-3 sin x cos x
x2 + 5x - 1 J
Sol. (i) 1= J ------ 7=----- dx =
x
cos2 X
7 X cos2 -dx
J sin
X
(iii) 1 = f*sin
— 6 x + cos6 x dx
~2
2
sin x cos x
I Example 3 Evaluate
VX
J-
1 = tan x - cot x + C
1
= = - cosec
aM
—a
(i) fxL+^x-1dx
-dx +
[Using sin2 x + cos2 x = 1]
I = jsec2 x dx + J cosec2 x dx
+c
-dx
X
sin2 x
sin2 x cos2 X
_-_l
X
• 2
smz xcos 2 x
sin6 x +cos6 x
cos x-cos 2x
(iii) /
dx (iv) J
dx
sin2 xcos2 x
1-cosx
1
X
dx
(i) j tan2 x dx
— dx =cos-1
2
-x
. ... dd fl
(xvn) — I - tan
dx \ a
=> J
I Example 4 Evaluate
-l
-J
-1/2 + 1
5
3
(ii) I = J(x2 +5)3 dx [using(a + h)3= ai3+3a2b + 3ab2+b3]
(xiv) — (log| cosec x - cot x |) = cosec x
dx
\ d
-1/2 +1
—------- + c
2 5x«
/ = ?x« + i.
-2x1/z + C
=>
(xv) — sin -i £
dx (
a
- X ~U2)dx
using f xn dx = —— + c
J
n+1
(xiii) -7-(log|secx + tanx|)=secx
dx
f
3
x2
5x’
1 1
---- +----------ax
xU2 x.1/2
U2 xU2J
sin2 x cos2 x
dx
4
Textbook of Integral Calculus
1
I Example 6 Evaluate
'■[ sin2 x cos2 x dx - j3 dx
(ii)j2l0^x dx
(i) J5108'x dx
(sin2 x + cos2 x)
2=J sin2 x cos2 x dx - 3x + C
X
Sol. (i) l = |5logfX
dx
[Using alogr = feloge 4]
Iog' 5 dx
Ylogr 5 + 1
I = jsec2 x dx + j cosec2 x dx - 3x + C
= —------- + c
(loge5+l)
7 = tan x - cot x - 3x + C
,
r cos x - cos 2x
(iv) 1= I—-------------- dx [using cos 2x = 2 cos2 x -1.]
J
1 - cos X
f I
x10ge5 + l
[5 0g'Xdx = —----------+ C
J
loge5 + l
(ii) I = J2log< x dx = J2\°^
1 X dx = |21/2,og2Xdx
cos x -(2 cos2 x - 1) ,
---------- ---------------- - dx
1 - cos x
1 .
using logfc„ X = - log/, X
n
2
•J
- 2 cos x + cos x + 1
dx
■
1 - COS X
= J21O8^ (dx = J 4x dx
- (2 cos x + 1) (cos x - 1) fa
„3/2
- (cos X - 1)
[as -2 cos2 x + cos x -1 = -(2 cos x +1) • (cos x -1)]
= —+ C
3/2
J
2 3/2
j 2log< x dx
+C
=-x
=* I = J(2 cos x +1) dx
3
I =2sinx + x + C
I Example 7 Evaluate J
Remark
In rational algebraic functions if the degree of numerator is
greater than or equal to degree of denominator, then always
divide the numerator by denominator and use the result of
integration.
V7(
'=[
x2
dx
(OjA
J x+2
1=
=> 1= J
1=
x3
x2 +5
x3+8-8
(x + Vx + 1)
= f(Vx + l)(Vx-l)(x + Vx +1) fa
J
(x + Vx + 1)
dx
,
[Using, a3 -b3 = (a- b)(a2 +ab + ft2)]
8
k x+2
x + 2,
x+2
r
x2
= I (x - 1) dx =------ x + C
J
2
dx
\x + 2)(x2-2x +4)
8
x +2
x+2
x2 -2x + 4-
8 >
x + 2,
I Example 8 Evaluate
dx
.
r dx
a ___
dx = x - 5 —-—
J x +.(V5)2
x2 + 5)
r
5
I = x —-j= tan
V5
+C
I = x-^5 tan-1
+C
X6 -1
... f 1 + 2x2
W Jx\i + X2) dx
dx
3x3
1 = — - x2 + 4x - 8 log | x + 2 [ + C
3.
x2
f x2 + 5-55-dx = \ ( x2+5
(ii) I=J
J x2 +5
2 +5
x2 +5
1=
(J7+i)[(J7)3-i3l dx
dx = j ----------- dx
x+2
r(x3+23)
(Vx + 1)(x2 - Vx)
dx.
x7x + x + Jx
Sol. Here, I = J (V X + L)-yJX{X —— dx
X+1
I Example 5 Evaluate
Sol. (i) 1= J
[using alog’b = b]
(X2 4-1)
■1 + x2 + x 2
1 + 2x2
Sol. (i) Here, l = f
x2(l + x2)
5
- f
\
X2+5y
dx
1 + *2
x2(l + x2)
dx
dx = j
dx + J
x2(l + x2)
dx
X2
x2(l + X2)
dx
1__
= J-ydx
+ J-J:1 + x2 dx = - —x + tan-1 x + C
J x2
(ii) Here, I = J
x6 + 1-2
dx
dx
=
J
x2 +1
x2 +1
x6-l-
f(*2)3 + i3
J X2 +1
—dx
x2 + 1
5
Chap 01 Indefinite Integral
I
=f
+1)((x
*42 4-1)--+1) _ 2Jf dx
J
[Using, a3 4- b3 = (fl 4- b)(a2 - ab 4- b2)]
X24-l
(i)” J^>/2
if—x -1/2
-x
-x^i/T 7
f 1-x-2
So/, (i) Here, l = j ^x*"-x"“
.1/2
„-l/2
l-2x
2
VX------ 7=
- X
-x
2
x3/2
3/2
So/.
dx
dx
1
rcosKx-b)-(x-fl)} fa
cos (a - b) * sin (x - fl) cos (x - b)
dx = J 1-x
x-1
2
dx
x3/2
Vx
3
y/x
1
4x2(2x+1)'
x2
x~*~ 64
44-2x-14-x-2 4-4x-14-x
l-64x6
x6
2
X
l-2x
dx
(l-2x)
2
cos (a - b)
7
4x2(2x +1)
x2
1
dx
>
____ 4x2(2x 4-1)'
l-(4x2)3
x6
vX6 •(4x2 +2x + l)
(4x2 -4x + l)
(l-2x) >
(1 — 4x2)(l 4- 4x2 4-16x4)
4x2(2x 4-1)
v(4x2 4-2x 4-l)(4x2 -4x4-1)
(l-2x)
(1 —4x2)-(4x24-2x4-l)(4x2—2x4-1)
(4x2 4-2x 4-l)(2x - l)2
dx
sin (x-a) cos(x-b)
j {cot (x - fl) 4- tan (x - b)} dx
{log | sin (x - a) | -log| cos(x - b) |) + C
log.
sin(x-a)
cos (x - b)
4-C
(ii) I = f------------- -------------- dx
J cos (x - fl) cos (x - b)
1
f_____ rin<g-fl)_____ dx
sin (a - b) J cos (x - fl) cos (x - b)
1
r sin {(x-b)-(x-a)} j*
sin (a - b) J cos (x - fl) cos (x - b)
-A-fJ sin (x - b) cos (x - a)
sin (fl - b)J
dx
cos (x - a) cos (x - b)
cos (x - b) sin (x - a)
4x2(2x4-l)
(l-2x)
dx
[using, 16x4 4- 4x2 4- 1 = 16x4 4- 8x2 4- 1 - 4x2
cos (x - a) cos (x - b)
1
SS ------------------------- J {tan (x - b) - tan (x - fl)} dx
sin (a - b)
(l-2x)2 (l-2x) ?
4x2(2x 4- if
dx
dx
sin (a - b) [’ log|cos(x - b)|4- log | cos(x - a) | ] + C
= (4x2 4- 1)2 - (2x)2 = (4x2 4-1 4- 2x) ]
(• [(l-2x)(14-2x)(4x2 -2x4-1)
\dx
cos (a - b)
cos (a - b)
X
/
cos (fl - b) J
+ sin(x - b)-sin (x - a)
+ C = -?x3/2 + -^= + C
4x2 4-2x4-1 4x2 - 4x4-1
/
q
>
(ii) Here
/
—--- 1 sin (x - a) cos (x - b)cos (x - b) • cos (x - )
1
x~V2
------- 23/2
-1/2
1=
(i) J=f------------- -------------- dx
J sin(x-a)cos(x-b)
j - cos (a ~
[ ________ dx________
cos (a - b) sin (x - a) cos (x - b)
7
\
k
vx
7
(- Vx - 2x-3/2 )dx
1______ dx
(ii) J --------cos (x-o)cos (x-b)
7
/
\
*.3/2
dx
X
2
X
x3/2 + *1/2
=/ (1 — x-2) 4-(x-2 — x)
x1/2 - x-V2
1-x
r
1
dx
4x2(2x4-lf
4-4x-1 + x
v4 + 2x‘
dx
I Example 10 Evaluate
CO j ---- 7—7—Trdx
Jsm(x-o)cos(x-b)
x
-X
___ xj_
X-6 -64
(l-2x)
= r(2x4-l)(l-2x)dbf
'
(l-2x)
= j(2x 4-1) dx = x22 4-X4-C
I Example 9 Evaluate
2
X'
x3/2+x1/2
(l-2x)
f (2x 4- l){4x2 - 2x 4-1 - 4x2} dx
J
l-2x
x5 x3
---------- + x -2 tan x + C
5
3
t
4x2(2x 4-1)
-J k
x2 + 1
= f(x4 - x 2 4-1) dx - 2j
f(14-2x) (4x2-2x4-1)
cos (x - a)
------------ 1log
cos (x - b)
sin (a - b)
4-C
6
Textbook of Integral Calculus
■ r-
I
—
rSin(x+o) .
.
Important Points Related
to Integration
I Example 11 Evaluate —-—— dx.
J sin x + b
Sol. Let I = [
dx. Put x + b = t => dx = dt
J sin (x + b)
1. J k f (x) dx = k [ f (x) dx, where k is constant. i.e. the
j _ pin (t - b + a)
dt
■'
sin t
integral of the product of a constant and a
function = the constant X integral of the function
cos t sin (a - b)
- dt
sin t
sin t cos (a - b)
sin t
2. J{/i(x)±y2(x)±...±/n(x)}dx
= j fi(x)dx± j f2(x)dx±...± j fn (x)dx.
= cos (a - b) j Idt + sin (a - b) j cot (t) dt
= t cos (a - b) + sin (a - b)log | sin 11 + C
= (x + b) cos (a - b) + sin (a - b) log) sin (x + b)|+C
I Example 12
x 2
5
(i) If f'(x) = - + - and /(I) = -, then find /(x).
2 x
dy
3
(ii) The gradient of the curve is given by ~ = 2x —
dx
x2
The curve passes through (1, 2) find its equation.
Sol.
Y*
i.e. the integral of the sum or difference of a finite
number of functions is equal to the sum or difference
of the integrals of the various functions.
3. Geometrical interpretation of constant of
integration By adding C means the graph of
function would shift in upward or downward
direction along y-axis as C is +ve or - ve respectively.
e.g.
2
x2
y = jx dx = — + C
2
(i) Given, f'(x) = - +
2 x
On integrating both sides w.r.t. x,
2L
we get J f '(x) dx = r(x
- + - dx
x)
X2+C3
x2+Cg
x2+Cj
1 x2
/(x) = --— + 21og|x| + c
M
Now, as /(I) = — (called as initial value problem
4
5
5
i.e. when x = l,y = - or f(l) = -)
4
4
Putting, x = 1 in Eq. (i),
- = - + C => C = 1
4 4
x2 + 21og|x| + l
y(x) = ^4
' = „2x--^3 or dy = f2x - 3 1 dx,
(ii) Given, —
dr
JI'
I
dx
X
X 1
■ 4
On integrating both sides w.r.t. x, we get
J‘'y=f(2x--^jdx
2x2 3 _
y —---- + — + C
2
x
Since, curve passes through (1,1).
=>
1 = 1+3 + C=>C = -3
/(x) = x2 + --3
X
0
Ct
/(1)= Y+21og|l| + C. but/(l) = —
4
4
I
(x.O)
Figure. 1.1
y = J/(x)dx = F(x)+C
=>
F/(x) = /(x);F'(xI) = f(x1)
Hence, y = j /(x) dx denotes a family of curves such
that the slope of the tangent at x = Xi on every
member is same i.e. F'(xi) = J(x) [when Xj lies in
the domain of /(x)]
Hence, anti-derivative of a function is not unique. If
gi(x) and ga(x) are two anti-derivatives of a function
/(x) on [a, b], then they differ only by a constant.
ie.
gi(x)-g2(x) = C
Anti-derivative of a continuous function is
differentiable.
4. If /(x) is continuous, then
f/(x)<fc = F(x) + C
=>
F'(x) = /(x)
=> always exists and is continuous.
=>
F'(x)
Chap 01
Indefinite Integral
5. If integral is discontinuous at x = xit then its anti-derivative at x = Xi need not be discontinuous.
e.g.
p
-1/3
l3dx=-x
<l.
213 + C is continuous at x = 0.
~',3dx. Here, x‘1/3 is discontinuous at x =0. But | x-'
J
2
6. Anti-derivative of a periodic function need not be a periodic function, e.g. f(x) =cos x +1 is periodic but
[ (cos x +1) dx = sin x + x + C is a periodic.
Daily Life Applications
The Integral
The Derivative
In symbols
It’s Anti­
derivative Function
Function
In symbols .
Function i
Its derivative function
Distance (s)
Velocity (v)
ds
v-—
___ dt_
Velocity
Distance
s = j v(r) dt
Velocity (v)
Acceleration (a)
dv
a =—
___ dt
Acceleration
Velocity
v = j a(t) dt
Mass (p.)
Liner Density (p)
^4
Linear Density
Mass
p = Jp(x) dx
dx
Population (P)
Instantaneous growth
dP
dt
Instantaneous Growth
Population
Cost (C)
Marginal cost (MC)
dC
MC =---dq
Marginal Cost
Cost
Revenue (P)
Marginal Revenue (MR)
MR = —
dq
Marginal Revenue
Revenue
dq
Here, q is quantity of products.
Exercise for Session 1
■ Evaluate the following integration
x2 + 3
1. f
dx
3-1
(1 + x)‘
dx
x(1 + x2)
4- J 1+ x2 dx
X4 4~ X2 4-1
6-1
x
,2
5- J 2(14-x2) dx
7.
x2
dx
(a-f-bx)2
9- J
3x j. ^5x
e"
+a
— dx
ex + a"
1 + cos 4x .
dx
ft. J ---------------cot x - tan x
(■ sin4x ,
13. 1------- dx
J sinx
15. j sin3xcos3xdx
2- I x6(x2 + 1) dx
x4
(x2 + sin2x)sec2x
(1+x2)
dx
8. J2xexdx
10. j(eal09X 4-exl09a)dx
/ tan x tan 2x tan 3x dx
”■1 cos3 x dx
dt
»>-!(?)
dq
7
I
Session 2
Methods of Integration
Methods of Integration
If the integral is not a derivative of a simple function, then
the corresponding integrals cannot be found directly. In
order to find the integral of complex problems.
sin x dx, f^^dx, f -^—dx
e-g.J
J x
J log x
x
we substitute g (x) = t and g' (x) dx = dt
The substitution reduces the integral to J f (t) dt. After
evaluating this integral we substitute back the value of t.
I Example 13 Prove that
f(oxWdx=(^,0+1
(n+1)o
J
Some Integrals
which Cannot be Found
Any function continuous on an interval (a, b) has an
anti-derivative in that interval. In other words, there exists
a function F (x) such that F' (x) = f (x).
However, not every anti-derivative F (x), even when it
exists, is expressible in closed form in terms of elementary
functions such as polynomials, trigonometric, logarithmic,
exponential functions etc. Then, we say that such
anti-derivatives or integrals “cannot be found”.
Some typical examples are
cos x ,
(ii) J -----dx
dx
x
(i)J^
—dx
(■>>) J ~-~
log X
(iv) jjl - k2 sin2 x dx
(v) j -Jsin x dx
(vi) Jsin(x2) dx
(vii) jcos(x2) dx
(viii) j x tan x dx
(ix) Je"x’ dx
(x) jex2 dx
(xi) J 1 +x2x5 dx
(xii) j 3-Jl + x2 dx
Sol. Putting, ax + b = t, we get
adx = dt or dx = -dt
a
,n+l
+i +c
I = j(ax + l,rdx^-^^—
l(ax + b)n + ldx
a(n + l)
Remarks
1. If J7(x) dx ~g(x) + Gthen J7 (ax + b) dx = - g (ax + b) + C
2. If J-l dx = log | x | + C, then j —1— dx = - log | ax + b | + C
ax + b
a
Thus, in any fundamental integral formulae given in article
fundamental integration formulae if in place of x we have
(ax + b), then same formula is applicable but we must divide by
coefficient of x or derivative of (ax + b) i.e. a
Here is the list of some of
frequently used formulae
b) — 4- C, n * -1
(i) J (ax + b)n dx= (ax
—+
+a (n +1)
(ii) J 1 dx= — log | ax + b | + C
ax + b*
a
(iii)= —
Je"
+‘
dx
eax+b
a
+C
(xiii) J71 + *3 dx etc-
.
la bx + c
(iv) J a bx + c dx-=
----+c
b log a
Integration by Substitution
[or by change of the independent variable)
(v) j sin (ax + b) dx = - - cos (ax + b) + C
If g (x) is a continuously differentiable function, then to
evaluate integrals of the form,
1 = ff (g(x)).g'(x) dx,
(vi) | cos (ax + b) dx = - sin (ax + b) + C
(vii) jsec2 (ax 4- b) dx--tan (ax + b) + C
Chap 01 Indefinite Integral
(viii) j cosec2 (ax 4- b) dx = - — cot (ax + b) + C
(iii) Here, I = J(7x - 2)^3x 4- 2 dx = 7
=7
"3
(ix) J sec (ax 4- b) tan (ax 4- b) dx = — sec (ax + b) + C
(x) f cosec (ax4- b) cot (ax 4- b) dx =-—cosec (ax 4- b) +C
J
a
(xi) | tan (ax 4- b) dx = - — log | cos (ax 4- b) 14- C
3x-‘
3x 4- 2 - 2 - y L/3x + 2
3
= - j(3x 4- 2)^3x 4- 2 dx - y/j3x + 2<ix
(xii) j cot (ax + b) dx = — log | sin (ax 4- b) 14- C
= j(3x 4-2)3/2 dx— yp3x+2<ix
,5/2
(xiv) j cosec(ax4-b)dx = - log|cosec(ax4-b)-cot(ax4-b)|4-C
7 (3x 4- 2):
20 (3x 4- 2)’/2
3
3
?X3
I 2
(i) [-=---- 1 -------dx (ii) [ 8x4-13 dx.
j73x + 4-73x+1
j yfix+7
X2 (14-X2)
/
(iv)Here,/ = f(2^3y2)^ - f2 + 2*Z + x2
dx
J x2(14-x2)
J X2(14-X2)
dx.
dx
fox 4- 4 - ^3x 4-1
<2(l + x2) |
x2
.X2(l + X2)
X2(14- X2)
f f 22
f (V3 X 4- 4 4- 7?x 4- 1) j*
J (3x 4-4) - (3x 4-1)
1
= J" \x'
S + 14-x2>
_ r_______ (^3x4-4 4-73x4-1)_______
~ J (^3x 4- 4 - 73x4-1) (73x 4- 4 4- 73x4-1) dx
'
dx
dx = 2$ x~2dx 4- J ^-dx
14-X
x~'
i
—2
•
= 2----- 4- tan- x 4- C =------- 4- tan- x 4- C
-1
x
I Example 15 Evaluate
(i) |sin(|o8
= ~ f & 4- 4 dx 4- J 73x + 1 dx
3/2x3
+C
24-3x2
(iii) j(7x - 2) 73x4-2 dx. (iv) j
3
5x3
k 2
= —(3x4-2),5/2
: -^(3x4-2)l:3/2+C
45
27
I Example 14 Evaluate
1 [(3x 4- 4)',3/2
Y
/
/
(xiii) J sec (ax4- b)dx =— log | sec (ax 4- b) 4- tan (ax4- b)|4-C
Sol. (i) Here, J = j
U3x 4-2 dx
3x + 2 dx
7
7
x-
9
,3/2
, l (3x + 1)
3S 3/2X3
„mtan
e
>+c
(ii)
^+c
3sinx-i-4cosx
dx
4sinx-3cosx
X
(iv) Jxsin(4x2 +7)dx
(iii) J 14-X2
Using, j (ax 4- b)n dx _ (ax 4- b)1
(n4-1) a
= — [(3x 4- 4)3/2 4- (3x 4-1)3/2] 4- C
27
r 8x 4-14 - 1
(ii) Here, Z = j 8x + 13 ,
dx
dx = J I
dx
— dx
sol. (i) 1= i^^dx
J
x
We know that, T-(log !r) = dx
x
Thus, let log x = t
=>
— dx = dt
(i)
1= Jsin(t)dt = -cos(t)4-C
=2J-giZ=
= 2 J(74x4-7)dx - j(4x 4-7)r>/2
-,/zdx
=2
f(4x 4-7)3/2>
3/2X4 J
(4x4-7)V2^
+C
^1/2X4
1
= -(4x4-7)3/2 —-(4x4-7)1/Z 4-C
2
3
= - cos (log x) 4- C
....
r 3 sin x 4- 4 cos x ,
(u) I = --------------------- dx
J 4 sin x - 3 cos x
[using Eq. (i)]
We know, — (4 sin x - 3 cos x) = (4 cos x 4- 3 sin x)
dx
Thus, let 4 sin x - 3 cos x = t
=> (4 cos x 4- 3 sin x) dx = dt
•(>)
x
10
Textbook of Integral Calculus
/= J^ = log| f| + C
= — [2 cos x cos 5x + 2 cos 3x cos 5x}
4
= log | 4 sin x - 3 cos x | + C [using Eq. (i)]
(iii) /= J e
m tan 1 x = t
dx Let
1 + X2
m
dx = dt =>
1 + x2
r
cosxcos2x cos5x = -{cos2x+cos4x+cos6x+cos8x|
4
I = J (cos x cos 2x cos 5x) dx
,
1
dx = — dt
m
= 4 J(C0S
I t dt
=> I = - fe' dt
m J
J
m
I = — e,+C =
m
m
(iv) I = | x sin (4x2 + 7) dx
m tan" x
1
4x +7 = t => 8x dx = dt => x dx - - dt
8
I = f sin (t) — = - - cos (t) + C
8
8
1
,
= --cos(4x +7) + C
I Example 17 Evaluate
(i) Jsinxcosx-cos2x-cos4xdx
2
Jl + tan2x
(iv) J
Remarks
While solving product of trignometric, it is expedient to use the
following trigonometric identities
1. sin mx cos nx = 1 {sin (m - n) x + sin (m + n) x}
+ cos ^x"+ cos 6x + cos 8x) dx
sin 2x sin 4x sin 6x sin 8x
=------- +-------- +-------- +--------+ C
8
16
24
32
+C
,
Let
= — {(cos 4x + cos 6x) + (cos 2x + cos 8x)}
4
m tan"1 x
J1+cos2x
cos 2x
------ 2
. 2
i
fsec2x-1
v ----------- dx
Jsec2x+1
dx
cos xsin x
Sol. (i) Here, I = j sinx cosx- cos 2x cos 4x dx
2. cos mx sinnx = 1 {sin(m + n) x -sin(m - n) x}
= - j2sinx- cosx cos2x cos 4x dx
[using, sin 2x = 2sin x • cos x]
3. sin mx sin nx = 1 {cos (m - n) x - cos (m + ri) x}
= —— [ 2sin 2x ■ cos 2x ■ cos 4x • dx
2x2J
4. cos mx cos nx = 1 {cos (m-n) x + cos (m + n) x }
= — [sin 4x cos 4xdx = ■ * - [2sin4x cos4x dx
4J
2x4J
= lfsin8xdx = ^^ + c
8J
64
I Example 16 Evaluate
(i) j cos 4x cos 7x dx
(ii) Jcos x cos 2x cos 5x dx
Sol. When calculating such integrals it is advisable to use the
trigonometric product formulae.
(i) j cos 4x cos 7x dx
Here, cos 4x cos 7x = - (cos 3x + cos llx),
[using cosmxcosnx = -{cos(m-n)x + cos(m + n)x]
2
1 = jcos 4x cos 7x dx - | j(cos 3x + cos llx) dx
= - [cos 3x dx + - [cos llx dx
2J
2J
sin3x sin llx ~
=------- +--------- + C
6
22
(ii) j cos x cos 2x cos 5x dx,
We have(cosxcos 2x)cos 5x=-(cosx + cos3x)cos 5x,
2
[using cosmx.cosnx = i{cos(m-n)x + cos(m + n)x]
2
(ii) Here, Z = j^^dx
J 1 + tan2 x
2
„
1 - tan x
Using, cos 2x =------- -—
l + tanzx
I = j cos 2x dx
sin 2x
=-------- F c
2
/•••\ u
t
f1 + cos2x ,
(ui) Here, I = ------------dx,
J1 + cos 2x
2
1 + cos2 x .
1 + COS X
f
J 1 + 2cos2x-1 dx = j --- 5—dx
2 cos x
[Using, cos 2x = 2 cos2 x-1]
1 f/
2
1
= - J (sec x +1) dx = “(tan x + x) + C
2x
,
(iv) Here, I = j ------ cos
2----------cos x-sin x
(cos2 x - sin2 x) dx
• 2
cos 2 x-sin
x
[Using, cos 2x = cos2 x - sin2 x]
Chap 01
r 1
<sin2 x
1
(
cos2 X
dx
So/.
Indefinite Integral
2
Here, I = J cotz2x-l - cos 8xcot 4x dx
k 2cot2x
,2
coseczx - sec2 x) dx — - cot x - tan x 4- C
I = J (cot 4x - cos 8x • cot 4x) dx
cos 2x ,
(v) Here,f= fsec-2—J.rfx = fl—- ------- dx,
J sec 2x 4-1.
' 1 4- cos 2x
using cot 2A =
2sinzx
We get, l = f^^-dx
[using, 1 - cos 2x = 2sin2 x and 1 4- cos 2x = 2 cos2 x]
= j tan2xdx
r cos 4x „ . 2
As, tan2x = sec2x -1
=---------2sinz4xdx
J sin 4x
1=
= j2sin 4x• cos 4xdx,
sec2 x -1) dx = tan x - x 4- C
I Example 18 Evaluate
using
sin 2A = 2sin A cos A
rcot2 2x -1
V
------------ cos8x-cot4x dx.
2cot2x
7
= J sin 8x dx = - cos 8x + C
8
Exercise for Session 2
■ Solve the following integration
dx
f cos x - sin x
o
,
;----- (2 4-2sm2x)dx
J cos x 4- sin x
1- J 1 + sinx
9
3. J (3 sin xcos2x -sin3x)dx
4. Jcosx°dx
5. J sin^/1x+ +sincos2xx ■dx, here (sin x 4- cos x)>0
a t cos 2x - cos 2a ,
o.------------------- dx
sin3 x 4- cos3 x ,
x^
4/
J sec2 x ■ cosec2x dx
rSin6X4-C0S6X
dx
J sin2 x cos2 x
9. j ^/1 - sin 2x dx
9rc
.8
J cos x - cos a
8.
• 2
sin*
x cos*2 x
11.
7n+ —x
sin2 — 4-- -sin.2! --
13.
sina sin(x -a)4-sin2
.8---- 4,
X
2
dx
cos 4x 4-1
dx
«• J ---------------cot x - tan x
dx
- sin 3x
14. J sincos2xx+4-1sin—5x2sin
dx
22x
4- sin4 x
15. J cos^14+xcos
dx, here cos 2x >0
4x
cot2 A -1
2cot A
= | cot 4x(l - cos 8x) dx
[using 1 - cos 2A = 2sinzA]
= J cot 4x -2sinz(4x) dx
J 2cos x
7- /
11
Session 3
Some Special Integrals
Some Special Integrals
... f dx
l-i
(0 J“—7=-tan
+ a2 a
Jx2+a
(ii) f 2
dx
„2
Type I
+c
2a
x +a
dx
1 ,
a+x
dx
= = sin -1
Ja2-x 2
dx
(V) J
(vi) j &
+c
= log | x + y]x2 + a2 | + C
2 b
c
ax2+bx + c = a x+-x+a
a
= log I X + 7 X 2 -a2 | + C
b
make the substitution x H---- = t.
2a
— a2
2 .
1
- x dx = - x
2
(vu)
• -1
-x2 + -1 a 2 sin
2
+C
2
1 2
+ -a2
2
1
-------- dx
J 2x2 + x—1
dx * (iv) |^2x2 -3x + 1 dx
J X2 - X + 1
Some Important Substitutions
a2 + x2
Substitution
x = a tan 0 or a cot 0
a2-x2
x = a sin 0 or a cos 0
x2-a2
x = asec0 or a cosec0
a-x
|a+ x
------ or
a- x
ya+ x
X = a cos 2 0
or 7(* -a)(x-p)
(II) f
-dx
Jx -x+1
So/, (i) 1=1-5-^-
2 log|x+7*2-a2 | + C
= -2 x 7? -a2 --a
2
Expression
4a2
Jx2-2x +3
(ix) jT*1 -a2 dx
VP7
(i) J
(iii) J
2
b2
=a<
I Example 19 Evaluate
(viii) j7fl2 +x2 d*
x-a
yax 2 +bx+c
easily done by the method of partial fractions (explained
later). If the denominator cannot be factorised, then
express it as the sum or difference of two squares by the
method of completing the square
2
_dx
ax2 +bx+c
If ax2 + bx + c can be factorised, then the integration is
+c
iog
(Hi) J a 2 -x 2 = —
a-x
2a
(iv) j
dx
(iii) j yjax2 +bx + c dx
= ^-i°g x-a +c
x“ - a
dx
completing x2 - x +1 into perfect square.
dx
= J1—
—
—
------f
x2 —x + 1/4 —1/4 + 1 J (x-1/2)2 +3/4
dx
(x-1/2)2 +(V3/2)2
c
= -2— tan"1
V3/2
.•.I = 4=tan"’ 2x-l + C
V3
using J
x=acos20+Psin20
(ii) I = J
Application of these Formulae
The above standard integrals are very important Given
below are integrals which are applications of these.
x-1/2
+C
^3/2 .
1
dx = — tan 11 — | + C
a
x2+a2
aJ
1
dx = - ( ____ 1____ dx
2 J x2 +X/2-1/2
2x2 + x - 1
1
a
= -1 f —z---------- - -------------------dx
2 J x2 + x/2 + 1 /16 — 1 /16 — 1 / 2
Chap 01 Indefinite Integral
_ 1 r______ dx______ _ 1 f______ dx______
~ 2 J (x + 1 / 4)2 - 9 /16 ” 2 J (x + 1/4)2 - (3/4)2
i2
2a
‘dl
J
x + 1/4-3/4
1
1
+C
log
x + 1/4 +3/4
2 2(3/4)
1 ,
dx
using J ~x^a
T = —log
x-a
+C
x+a
tl -1
= ii°g
x-1/2
1,
+C
= "log
Z = 310g
(iii) 1=J
dx
jx2 ~2x+3
- t dt
dx = -
=>
-JJ yfxr 2- 2xdx+ 1-1+3
{Jt2 12-l
i1 i I '-1
=—
log ----- + C
21
2-1
I t+1
1 - e2x -1
2x
(ii) Here, I = J
71-x2-x<
dx
Let, x2 = t, 2x dx = dt
dt
■■■
■71-t-t
-f ,
__
J7(x-l)!+(72)!
75/4-(t + 1/2)2
= sin 1
(iii) Here, I = J
(iv) I = J(72x2 -3x + l) dx = 42 j(7^2 -3x/2 + l/2)dx
-(4x - 3) -J*2 -3x/2 + l/2
=^2- 8
• +C
- — log | (x - 3/ 4)+ Jx2 -3X/2 + 1/2
. 32
/ = —— sin-1 (ax) + C
log a
(iv) Here, I =
Let,
(iii) /
ax
2x
—
dx
2x
Sol. (1) Here, I = J
Let,
Then,
x3/2 = t,-xu2 dx = dt, xuz dx = -dt
3
2
2 *
-dt
3
x3/2>
2 .-i t
= --sin | — + C = -sin-1 3/2 + C
3
3
a
J
= dx
-X4
(iv) J J X 3 dx
a -x
=
dx
2x
xv2dx
x
ydx=J
a3 - x
V(a3/2)2 -(x3/2)2
7(*3/2)2-(o2
I Example 20 Evaluate
e
dt
log a
7^
- (x - 3/4)7(x-3/4)2 - 1/16
2
• +C
1 log/(x - 3/4) + 7(x-3/4)2-1/16
. 16x2
1
a
= dx. Let, ax = t
yjl-a 2x
... i=J __ 1 2 logdt a _ log1 a •sin-1 (t) + C
= 42 j(7(x-3/4)2 -1/16) dx
r
f2x2 +1
t + 1/2
+ C = sin-1
+C
V5/2
\
ax log a dx = dt, ax dx =
= 72 j(7x2 — 3x/2 + 9 /16 — 9 /16 + 112) dx
4
,
dt
1 V(Vs/2)!-(l + l/2)!
d‘
1=1.
dx
7*2 + a,2‘
I = log I (x - 1) + 7^2 - 2x + 3 1 + C
■
I 1 +--------i i t-t22
4
= log | (x -1) + 7(x-l)!+(s6)2 |
using J
+C
1 - e2x + 1
Lt
2x-l
+C
2(x + l)
13
dx
7
1 - e2x = t 2
-2e2xdx = 2tdt
I Example 21 Evaluate
cosx
-j=
■= dx.
7sin2 x - 2 sin x - 3
2 sin 2x-cosx
(iii) J -------- ------------- dx.
6-cos x-4sinx
sin(x-a) ,
(ii) j —
;---- 7 ax.
sin(x+a)
14
Textbook of Integral Calculus
cos x dx
Sol. (i) Let I = j
(4t -1) dt
(t2-4t+5)
ysin 2 x - 2 sin x - 3
Now, let (4t - 1) = A (2t - 4) + p
Comparing coefficients of like powers of t, we get
2X = 4, - 4X + p = - 1
X =2,p =7
(ii)
2.<2t--4->*7*
[using Eqs. (i) and (ii)]
t2 - 4t + 5
cos x dx ~dt
Put sin x = t
dt
■=J
d>
=> / = J yjt2 -2t-3
7t2 -2t + 1-1-3
dt
_________
= log |(t —1) +- I)2 — (2)2 | + C
2t - 4 dt + 7 J
dt
t2 - 4t +5
t2 -4t + 5
/ = log ] (sin x -1) + ^sin2 x-2sinx-3| + C
(ii)
Let 1 =
1=
=J
Isin (x - a) ,
------------ dx
sin (x + a)
dt
t2- 4t + 4-4 + 5
dt
(t-2)2+(l)2
'sin (x - a) x sin (x - a)
sin (x + a) sin (x - a)
= 2 log 112 - 4t + 51 + 7 tan-1 (t - 2) + C
sin (x - a)
-Jsin2 x - sin2 a
= 2 log | sin2 x - 4 sin x + 5|+7tan-1(sinx-2) + C
sin x cos a - cos x sin a ,
------ ,
---- dx
■Jsin2 x - sin
sin 2 a
= cosaj
sin x dx
cos x dx
-sinaj
7sin2 x-sin2 a
7sin2x-sin2a
sin x dx
= cos a j
71 - cos2 x - sin2 a
cos x dx
- sin a j
tjsin
x -eir.
:in“2 x
- sin 2 a
= cosaj
sin x dx
cos x dx
-sinaj
7cos2a-cos2x
Vsin2x-sin2a
In the first part, put cos x = t, so that - sin x dx = dt
and in second part, put sin x = u, so that cos x dx = du
dt
j
du
I = - cos a j i
- sin a
7cos2 a-t2
-sin2 a
= - cos a sin-1
= - cos a -sin"1
t
- sin a • log
cos a
|u - -Jiw22 - sin2 a)| + C
COS X
...(i)
- sin a • log
cos a
|sin x - -^/sin2 x - sin2 a | + C
(iii)/=J 2 sin 2 x - cos x dx
6- cos2 x - 4 sin x
(4 sin x - 1) cos x
dx
6 - (1 - sin2 x) - 4 sin x
r (4 sin x - 1) cos x
J sin2 x - 4 sin x + 5
Put sin x = t, so that cos x dx = dt
Type II
(i)f(£phL
(px+q)
(ii)f
J ax2 +bx+c
dx
ax2 +bx +c
(iii) J(p* + <7)7ax 2 + bx + c dx
The linear factor (px + q) is expressed in terms of the
derivative of the quadratic factor ax2 +bx + c together
with a constant as px + q = — (ax2 + bx + c) + p
dx
px + q = X(2ax + b) + p
=>
Here, we have to find A. and p and replace (px + q) by
{X (2ax + b) +p} in (i), (ii) and (iii).
I Example 22 Evaluate
CO fj^dx
Jya + x
(ii) Jx
a2 — x 2
a2
TP dx
Sol. (i) Let Z = J a - x dx
a-x a-x .
r
------ X------- dx =
a+x a-x
J
a-x
T2
a
X
Ja2 -X2
- X
= a • sin 1
x
\a
dx
-x2
dx
rt dt
+h~
Put a,22 - X,22 = t2 => - 2x dx = 2t dt => x dx = - t dt
15
Chap 01 Indefinite Integral
=JH(1'2x)+;
= a - sin-11 — I +1 + C
la J
= - - f(1 -2x)71 + x-x2dx + - j7i+x- x2dx
2J
2
'a2-x2+C
= a • sin-1
11— —1)1)dx
= -lj7tdt + ljJ-(x2-x + ---------dx
2
(ii) Let
/ = fx a2 - x dx
Put
x2 = t=$2x dx = dt
-
1 + x - x2 dx
7T72
£
4
V
[where t = 1 + x - x 21
5 V
<^3/2
a2-t a2 -1
a2 -1
a2+t
■
■
4J
1
=-a
2
2 ”2J
- t_
7
1
2(3/2
dt
a2 +1 a2 - t
dt
_1 f
7(«2)2-(o2
= -a2sin 1
2
J?-t 2
du
(?)• 41 Jr Vu
t
1
+ -•
4
72
x
3
, 1 [(x-1/2)
2
<^U2 >
+C
2 \
a7
Integrals of the Form j
-X4 +C
2
*(X)
ax2+bx+c
dx,
where k(x) is a Polynomial of Degree
Greater than 2
To evaluate this type of integrals we divide the numerator
by denominator and express the integral as
QW +
RW
ax2 +bx +c
2
= --(l + x-x!)!,!
[where t = x2 and u = a4 - x4]
= - a2 ■ sin 1
2
, where R (x) is a linear function of x.
(ii) j(x+1)A/l-x-x2 dx
Put
Then, comparing the coefficients of like powers of x,
we get
1 = -2\ and X + Jl = 0
I
1 + x - x2 dx
X = - l/2,|l = l/ 2
7s
+C
Put, (x + l) = X j — (l-x-xz2 )l + p
\dx
J
Then, (x + 1) = X (- 1 - 2x) + p comparing the
coefficients of like powers of x, we get - 2X = 1 and
p-X = l=>X = -l/2 and p = l/Z
...
(x + l) = _l(_1_2x) + l
2
2
So, j (x +1) 1 - x - x2 dx =
2
2
- x - x72 dx
1 r
I------------= -- (-l-2x) Jl-x-x 2 dx + - f 1 - x - x2dx
2J
v
2J
= - - f (— 1 — 2x) Jl - x - x2 dx
2J
v
l-|x2 + x + — - - > dx
I
4 4
= -- 7t dt + - j7(75/2)2 — (x + 1/2)2 dx
2
2
[where t = 1 + x - x 2]
^3/2
d
2\l
x = X — (l + x-x2)l+p
dx
J
2x -1
(ii) Let I = j(x + 1) ^1 - x - x2 dx
44
I = jx 1 + x - x2 dx
Sol. (i) Let
+ - sin 1
8
2
I Example 23 Evaluate
(i) Jx^1 + x-x2 dx
I 2 /
(x-1/2)
2
f dt
where a* - t2 = u => -2t dt = du
I
2
1
x — I dx
2
_ _1 p/2 + -1
3
2
dt
-1
= - a2 -sin-1
2
4
1 1
1
2^3/2 + -2 ( 2
1
1 - X - X2
2
1 5
+----- sin
2 4
= -1(1-x-x2)372
3
8
! f x + 1/2
«+C
75/2
x - x2
5 . _1|2x + l
+ —sin 1
+C
16
5
16
Textbook of Integral Calculus
X2 + X+3 ,
I Example 24 Evaluate J —--------- dx.
I Example 25 Evaluate j
x2 + x + 3 ,
--------- dx
2-x-2
Sol. Let I = j 2x. 2 + 5x + 4dx,
yjx2 + x +1
x2 -x-2
Sol. Let
/
2x4-5
'=/ 1 + -
'
d
Put 2x2+ 5x + 4 = X (x2+ X + 1) + p. ~ (x2 + x + 1) • + r
dx '
or 2x2 + 5x + 4 = X (x2 + x + 1) + p (2x + 1) + y
dx
l x 2 - x-2,
+ 5,
/ = Jldx +J —2x
--------- dx
Comparing the coefficients of like terms, we get
2 = X, 5 = X+2jl, 4 = X+|l+y
.
X=2, |i=3/2,
y = l/2
Hence, the above integral reduces to
x2 - x-2
2x + 5 ,
—--------- dx
x-x-2
d
t
2
. ..
_1_
rr*l
___ r
Put, 2x + 5 = X ] — (x2 - x - 2) > +11. Then, obtaining
r
J
dx'
2x + 5 = X (2x - 1) + p, comparing the coefficients of like
terms. We get, 2 = 2X and 5 = p - X
X = 1 and |1 = 6
X (2x - 1) + p dx
x2-x-2
2x-l dx + 61
x2 - x - 2
x2 - x - 2
1
= x + j-dt+6j2
1 1 n
4 4
[where t = x2 - x - 2]
J
= x + log 111 + 6 J
1 3
x--------
1
2 2 +C
log
1 3
2(3/2)
X-- + 2 2
x-2
= x + log I x2 - x - 21 + 2 • log
x +1
+C
{px2 + qx+r)
(ax2 + bx +c)
yjpx2 + qx +r
(2x +1)
+1
1
dx
Jx2 + X + 1 2 ^x2 + x + 1,
dx
yjx2 +X + 1
dt
If
7= + = 2/7x+x + ~ldx + -32Jf| —Vt
2J
[where t = x2 + x + 1]
- --------------------------------------
Al 2
r.
= 2J7(x + 1/2)z +(73/2)2 dx + -----•
=2
1
1
x+—
2
2
1 r
dx
2*’A/(x + l/2)2+(x/3/2)2
4J
x2+x + l +-•— log|[x + - ] + Jx2+x+I|
2 4
I
2
2J
r2
+ 3y[x''2 ------+ X + 71 + -i,log IJ X + -1) + y]X
2
2
/=jx+V
2
r I X+7
=>/=
l
dx
dx
3. J(ax2 +bx + c)y]px
=jf2.
I---------- 3
x2+ X +1 + - log
4
2
1
x + - +7*2+x+i
+ 3 7x2 + x +1 + - log <
2
Integrals of the Type
ax2 + bx +c
r f 2x2 + 5x + 4 ,
I=
,
dx
■jx2 + x + 1
dx
dx
(x-1/2)2-(3/2)2
= x + log| x2-x-2|+6-
2x2+ 5X + 4 •
dx.
ylX2 + X+1
2
1
+ x2 + x + 1 > + C
2
x2+x + l+—log' I X + - I + Jx2+x + l ► + C
4
I
2)
2
Trigonometric Integrals
2
In above cases; substitute
d
(a) Integrals of the Form
1
fJ ----1
----dx, j
a cos2x+ b sin2x
a + bsin2x
1
a+b cos2 x
dx
Find X, p. and y. These integrations reduces to integration
of three independent functions.
dxj ---------
a + b sin2 x + c cos x
dx
Chap 01 Indefinite Integral
17
denominator by (I + tan2 x) and put tan x = t. So that
To evaluate this type of integrals we put
2 tan x / 2
,
1 - tan2 x 12
sin x = —-------—'"
- and cos x =
-2------=
----------------- and replace
„ ,„
1 , »__ 2 „ / o
I + tan2 x
/2
I + tan2 x / 2
sec2 x dx = dt.
tan x / 2 = t,by performing these steps the integral reduces
To evaluate this type of integrals, divide numerator and
denominator both by cos2 x, replace sec2 x, if any, in
to the form J
I Example 26 Evaluate
... r
sinx
1
2
dX
4sin x+9cos x
methods discussed earlier.
GO J sm3x
2
I Example 27 Evaluate
dx
Sol. (i) I = j
v dt which can be evaluated by
at2 +bt+c
4 sin2 x + 9 cos2 x
Here, dividing numerator and denominator by cos2 x.
We get
(i) JJ—
-—
2+sinx+cosx
Sol. For this type we use, sin x =
,2
sec
.
X
------- 2--------dx
4 tan x + 9
dt
t2 + (3 /2)2
4t2 +9
1
1
X
---------tan -1
4 3/2
I = - tan-1
3
6
l + tan2 x/2
sin 3x
2 x J
=j------—5----2 + 2 tan2 — + 2 tan — + 1 - tan
+C
2
I
dx
J
Dividing numerator and denominator by cos2 x, we get
r
sec2 x dx
J 3(1 + tan2 x)- 4 tan2 x
sec2 x dx
= 2 f____ dt
dt
= 2jJ t +2t + l + 2
r
J(t + 1)2 +(V2)2
I
dt
J(^)2-(t)2
1 .
2>/3 °g
'3 — t
/=^10g 73 + tan x + C
73 - tan x
(b) Integrals of the Form
t+i
= 2- —7= tan-1
+C
7z
V2
tan x/2 +1
I = 72 tan-1
+C
+C
V2
(a)U,, = ^sin* cosx
=J
___________ dx
_______
73 -2 tan x/2 | 1-tan2 x/2
1 + tan2 x/2__ 1 + tan2 x/2
- ------- dx,
jJ —
■—u----- *^’1J ------a + bsin x
a sin x + b cos x
J a+ b cos x
2
sec2 — dx
2 x
, x
2dt
1
2 x .
j
r
=> - sec — dx = dt =
2
2
J t2 +2t +3
Let tan x = t => sec2 x dx = dt
J— 1
2 X
Put tan — = t
2 -
3- tan2 x
r
2
tan2 — + 2 tan —1-3
2 2
3—4sin2 x
sec2 x dx
/=f
2
2
1
J 3 sec x - 4~
tan x
1 + tan2 x/2
sec — dx
sin x
,
------------------ 7—dx
J 3 sin x - 4 sin x
(ii) Let I = J
■
dx
= f—
1-tan2 x/2
2 tan x/2
+--------- ------2+
fe)+c
2 tan x
1 + tan2 x/2
(i) Let I = f--------- —--------J 2 + sin x + cos x
=> sec2 x dx = dt
dt
2 tan x/2
1 - tan2 x/2
cos x =--------- -------1 + tan2 x/2
tan x = t
Put
—
sinx +cosx
00 J
dxj -------------- - -------------- dx
a sin x + b cos x + c
2 x j
sec — dx
= J—-2—
275 tan — + 1 - tan 2 *
2
2
18
Textbook of Integral Calculus
1
x
x
Put tan — = t => - sec2 — dx = dt
2
2
2
dt________
2dt
2 + 2y/it -3 + 3 + 1
-t2
. . /=J
-Jdt
dt
= 2f
r
2
J
4-(t-V3)
(2)z — (t —-Ts)2
; log
Rule for (i) In this integral express numerator as,
X (denominator) + p (diffh. of denominator) + y.
Find X, p and y by comparing coefficients of sin x, cos x
and constant term and split the integral into sum of three "
integrals.
(denominator)
dx+y f------- --------Xjdx + p J d.c.of
J asinx+ bcosx+c
denominator
(c) Alternative Method to Evaluate
the Integrals of the Form
f--------- --------- dx
J a sin x + bcos x
Rule for (ii) Express numerator as X (denominator) +p
(differentiation of denominator) and find X and p as above.
To evaluate this type of integrals we substitute
a = r cos 6, b = r sin 0 and so
r = t/a2 + b2, 0 = tan -i
I Example 29 Evaluate J
a sin x + b cos x = r sin (x + 0)
So,
f
J a sin x + b cos x
dx =-1 f|
—r—dx
r J sin(x +0)
If
z
,
1,
(x
0
X
= - J cosec (x + 0) dx - - log tan I — + 2
f
1
,
J asinx + bcosx
2
+C
1
la2 +b2
(X
i
+C
r pcos x +q sin x + r dx, fPcosx-hgsmx^
J a cos x + b sin x + c
' a cos x + b sin x
2 - V3 + tan x/2
+C
2 + ^3 - tan x/2
1
11 i
x
i x
it
= - log tan— + —
12 12
2
+C
(d) Integrals of the Form
2 + t - V3~
+C
= 2 • —log
2— t + yfi
2(2)
f
it x n
= - log tan — 4- — — —
4 2 6
2
1
log tan — + - tan -112 2
a
+C
I Example 28 Evaluate f-=---- ---------dx.
J
sinx+cosx
Sol. Let-73 = r sin 0 and 1 = r cos 0.
(2+ 3 cos x)
sinx+2cosx+3
dx.
Sol. Write the numerator = X (denominator) + p (d.c. of
denominator) + y
=> 2 + 3cosx=X(sinx + 2cosx + 3)+p(cosx-2sin x) + y
Comparing the coefficients of sin x, cos x and constant
terms, we get
0 = X - 2p, 3 = 2X + p, 2 = 3X + y
X =6/5, p=3/5, y = -8/5
cos x - 2 sin x ,
Hence, l = -jldx + |j• -------------------dx
5 J sin x + 2 cos x + 3
dx
sin x + 2 cos x +
6
3
8
= - • x d- - log | sin x + 2 cos x + 31 — I3
5
5
5
Where, I3 = f--------- —--------J sin x + 2 cos x + 3
' Then, r = J(-\/3)2 + (I)2 = 2 and tan 0 = — => 0 = —
1
3
dx____________
2(l-tan2x/2)
2 tan x 12
+---------- -------- + 3
1 + tan2 x/2
1 + tan2 x/2
/=[-,----- 1-------dx
J V3 sin x + cos x
2 X j
= f------------ ------------- dx
J r sin 0 sin x + r cos 0 cos x
= - f-----—---- = - f sec (x - 0) dx
r J cos (x - 0) r J
1.
= - log tan
r
7t
4
x
2
0
2
— 4----------
+c
sec —dx
= f----------------- 2-----------J
X
x
2 tan - + 2 - 2 tan2 - + 3 + 3 tan 2 £
2
2
2
■> x
2 x J
sec — dx
2
-, let tan — = t
2
tan2 — + 2 tan — + 5
2
2
X
X
(
Chap 01 Indefinite Integral
=>
2dt
1
-sec 2-dx = dt= f
2
J t2 +2t +5
2
= 2f (t 4-1)dt2 4-22
(
1 tan-> Ltl
i3=2.l.
= tan 1
2
I 2
x
tan -4-1
2
2
= cot A■J -
A
(ii)
cos x-cos (x4A)
( tan—
x 4-1 }
2
2
\
+c
7
I Example 30 The value of j {1 + tan x • tan (x + A)} dx
is equal to
sec x
+C
sec (x + A)
. (b) tan A-log | sec (x + A) | + C
sec (x + A)
+C
(c) cot A-log
sec (x)
= cot A • {log | sec (x 4- A) | - log | sec x |) 4- C
I Example 31 The value of f C—S— dx, is equal to
J sin x
(a) log | cot x 4- -Jcot2 x -114- 72 log | cos x
4- COS2 X -1 /2 | 4- C
(b) — fog | cot x + ^/cot2 X -1 I 4-72 log I COS X
-l-^COS2 X —1/2 I 4- C
(c) log I cot X 4-^/cot^x^ 1 I 4-2 log I COS X
-i-^COS2 X — 1 /2 I 4- C
(d) - log [ cot X 4- 7cot2 X-l I 4-2 log I COS X
So/. Let l =
(d) None of the above
J
sin x
dx=J
r 1 - 2 sin2 x
Sol. Let I = J{1 + tan x-tan (x 4- A)}dx
J sin x 7cos 2x
sin x-sin (x 4- A)
cos x-cos(x + A)
dx
=J
4- ’cos2 X —1/2|4-C
cos 2x
sin x yjcos 2x
[■ cos (x 4- A - x) dx
J cos x-cos (x + A)
dx
cos x-cos (x 4- A)
dx
dx
=—
- dx-2 f
2 x - sin2 x
|> cos x-cos (x 4- A)4-sin x-sin(x 4- A)
J
cos x-cos (x 4-A)
= cos A•j
dx
Hence, (c) is the correct answer.
From Eqs. (i) and (ii),
(a) cot A -log
cos (x + A)-sin x
cosxcos(x-l-A)
= cot A - {j tan (x 4- A)dx - J tan xdx]
\
6
3
8
1= -x + - log|sinx + 2cosx + 3| — tan
5
5
5
sin(x4-A)-cos x
19
sin x
Tzcos2 x-l
sin x
dx
dx
cos2 x - 1/2
- ds
2
2-l
Js2 -(1/72)2
[where t = cot x and s = cos x]
=-iog|t+7^ -114- 72 log I s 4- 7? -1/2 14- C
Multiplying and dividing by sin A, we get
= cot A•J
= cot A•j
sin A dx
= - log I cot X + 7cot2 X - 11+V2 log I COS X
cos x cos (x 4- A)
sin (x 4- A - x) dx
cos x-cos (x + A)
+ -^cos2 x-l/2| + C
Hence, (b) is the correct answer.
20
Textbook of Integral Calculus
Exercise for Session 3
■ Evaluate the following integrals
x2
z J 9 + 16x6 dx
J 9-16x4
x3dx
4.
3- J 16 8-25
x
x4
5.
dx
1___
a6 + x6
%
8x -11
8- I ■^5 4- 2x - x= dx
7- J ^25= dx
2
dx
9- I —x25-----------+2x + 2
x-3
dx
10. J 3-2x
-x2
3x-1
5------------- dx
11. f —
4x2-4x + 17
Vx dx
12. J yl2x+3
X 4-2
,
a -x .
------- dx
x-b
13.
14.
dx
15. fx/ + 2V-3 dx
16. J 1 + sin xdx+ cos x
17. f----- -------
cos2 x sin x ,
18. J ---------------dx
sin x - cos x
19. Jf-■/ - *XJ I
X - COS X
20. J COS(1-cos
dx
3 x)
J yjx2 4- X 4- 1
J sin x 4- V3 cos x
I
X
a3 -x 3
dx
IJ 4e -9e*
6------- 6 dX
2*
I
3
= dx
J -^5 -4e* -a 2x
21. Evaluate f3slnx + 2cosxdx
J 3cos x 4- 2sin x
23
f
22. Evalute | (2x - 4)^4+ 3x -x22 dx.
*5* + 9)dx
(X 4- 1) yjx2 + X 4- 1
24. The value of f-------- —-------- , is equal to
J sec x -t- cosec x
(a) ■ (sin x + cos x) + -X log tanx/2- 1-V2
V2
tan x / 2 - 14- 42
+C
(b) 2 • (sin x + cos x) + —L log tanx/2- 1-V2
tanx/2- 1+ V2
V2
+c
(c) - • (sin x - cos x) 4- —1L log tanx/2- 1-J2
tan x - 1 + V2
2
2
V2
+c
(d) None of these
Session 4
Integration by Parts
Integration by Parts
Theorem If u and v are two functions of x, then
juvdx = ujvdx-p^-jvdx>dx
i.e. The integral of product of two functions = (first
function) x (integral of second function) - integral of
(differential of first function x integral of second function).
Proof For any two functions /(x) and g(x), we have
4- LfW •«(*)}=/(x) ■ ± {&)} + g(x) • 7- {/(x))
dx
dx
dx
M ■ y-{g(x)}+g(x)
•j- {/(x)}ldx=J
d
/(x)g(x)dx
dx
=>
d {«(X)}1 dx + J|g(x)■ ± {f (X))k
fW-4-
dx
J
dx
J\
)
=jf(x)-g(x)dx
Usually we use the following preference order for
selecting the first function. (Inverse, Logarithmic,
Algebraic, Trigonometric, Exponent).
In above stated order, the function on the left is always
chosen as the first function. This rule is called as ELATE.
I Example 52 Evaluate
(i) Jsin-1 xdx
(ii) J loge | x|dx
So/, (i) I = |sin-1 x dx = jsin 1x• 1 dx
i
n
Here, we know by definition of integration by parts that
order of preference is taken according to ILATE. So,
‘sin-1 x’ should be taken as first and ‘1’ as the second
function to apply by parts.
Applying integration by parts, we get
I = sin-1 x (x)- j
fW~{g(x)}]dx
dx
J
= J f • g (x) dx " j
(*) •
= • x dx
y/l - X 2
. -1 , 1 f dt
= x • sin
x+- —
2 Jt172
{/(*)} j dx
Let
f(x) = u and y-{g(x)}=v
dx
So that, g (x) = j v dx
1
1 - x2 = t
Let
-2xdx = dt => xdx = --dt
2
• -i
juvdx = u- jvdx-j-
■ j v dx > •dx
While applying the above rule, care has to be taken in the
selection of first function (u) and selection of second function (v).
Normally we use the following methods:
1. If in the product of the two functions, one of the functions is
not directly integrable (e.g.
log | x |, sin"1 x, cos-1 x, tan-1 x.....etc.) Then, we take it as the
first function and the remaining function is taken as the
second function, i.e. In the integration of Jx tan-’ x dx,
tan-1 x is taken as the first function and x as the second
function.
2. If there is no other function, then unity is taken as the second
function, e.g. In the integration of J tan-1 x dx, tan-1 x is taken
as first function and 1 as the second function.
3. If both of the function are directly integrable, then the first
function is chosen in such a way that the derivative of the
function thus obtained under integral sign is easily integrable.
n
2 1/2
I = x sin"1 x +
Remarks
1 tUZ
- x2 + C
.". jsin-1 x dx = x sin-1 x +-Jl- x2 +C
(ii) I = j log, | x | dx = J log, | x | • 1 dx
I
n
Applying integration by parts, we get
= log | x | • x — f — • x dx
J X
= X log I X I - J 1 dx
I = xlog|x|-x + C
I Example 33 Evaluate
(I) jxcosxdx
Sol. (i) J x cos x dx,
'• J
(ii) jx2 cosxdx
x cos x dx
1 II
22
Textbook of Integral Calculus
Applying integration by parts,
I = x (J cos x dx) - — (x)l {[ (cos x) dx} dx
dx
J J
J = x sin x - j 1 • sin x dx = x sin x + cos x + C
= —- sin-1 4x (1 - 2x) + - • 4x jl- x
2
2
From Eqs. (i) and (ii), we get
2 = 1 —- (sin-1 Vx) (1 - 2x) + - Vx Jl - x -x+C
n
j x22 cos x dx
(ii) I =
I n
Applying integration by parts,
2
Integral of Form j’ex{f(x)+f(x)}dx
= x2 sin x - j2x -’(sin x) dx
Theorem Prove that
Je’{y(x)+y'(x))dx = e*y(x) + C
= x,2£ sin x - 2 J x (sin x) dx
We again have to integrate J x sin x dx using integration
Proof We have, jex {/(x) + f'(x)} dx
by parts,
= Jex -f(x)dx + jex - f'(x)dx
.
n
i
= f(x) • ex - J /'(x) • exdx + j ex- f\x) dx+C
= x • sin x - 2 J x ♦ sin x dx
I
n
= x2 sin x -2 - x(Jsinxdx)-
J^)(Jsin xdx}dx-
= /(x)-ex+C
= xz sin x - 2 {- x cos x - J1 • (- cos x) dx}
Thus, to evaluate the integrals of the type
je-(/(x) + f'(x))dx,
I = x2 sin x + 2x cos x - 2 sin x + C
we first express the integral as the sum of two integrals
J ex f (x) dx and jexf' (x) dx and then integrate the
sin-1 Vx -cos-1 x ,
I Example 34 Evaluate [—
— dx.
J sin -1 Vx + cos-1 x
integral involving ex f (x) as integral.by parts taking ex
Sol. Let Z = j sin-1 x - cos-1> ~ dx
sin-1 lx + cos T' -Vx
_ j sin-1
=$
2
= -(V
x~x2 - (1 - 2x) sin 1 Vx} -x + C
7t
1 = x2 (j cos x dx) -
2
.(ii)
as second function.
x- -(rt/2-sin-1 Vx)
tc/2
[•/sin-1 0 + cos 10 = k/2]
I = — [(2 sin-1
n J
The above theorem is also true, if we have eta in place of e*
i.e.
{f (kx) + f'(kx)} dx = eF f (kx) + C
x -n/2)dx
General Concept
Je ‘M {f(x)g'(x)+f'(x)]dx
= 1 [sin-1V7dx- f Idx
71 J
Remark
J
I = — [sin-1 7x dx - x + C
7t J
...(i)
Let x = sin2 0, then dx = 2 sin 0 cos 0 d0 = sin 20
jsin-1 Vx dx = |0 sin 20 d0
I
n
Applying integration by parts
/.
r -i r~.
_ cos 20
[sin 1Vxdx = -0-------- + [-cos20d0
J2
J
2
—0
1
=----- cos 20 + — sin 20
2
4
= ——- • (1 - 2 sin2 0) + - • sin 0 • Jl - sin2 0
2
2
Proof I = [ eg^x\ f(_x)/(x)dx + pw f'(x) dx ■
I
II
n
as it is
Using, J eg«(*)
^ .■ g'(x) dx = egg(x)
^, we get
= f(x) • e«w- J f '(x) ■ eMdx + JesM-f'(x) dx
=/(x)-e*w + C
e.g.
_ J g(x sinx + cos x) | x2cos2x -(xsinx + cos x)
I
J g(xsinx + cos x)
x sinx +cos x
cos2 x-------------------- dx
x
dx
7
Chap 01
Je(xsinx + cos x)
g(xsinx + cosx)
/
xcosx
COS X
x
cosx
cos x Y
X
X
2
dx
1
+C
= Je (sinx - sec x) dx
= Jetanx sin x dx - J e tanx sec x dx
=> -etanx • COS X + Jetanx sec2xcosxdx
dx
(1 + x2)2
2x
2x
1
.
d
■dx as —
(1 + x2)2
dx<l + x.2
1 + x2
tanx
e.g.
2x
(1 + x2)2
23
Indefinite Integral
(1+x2)2.
ex 2+C
1+X
ex
1=
sec xdx
1 + x2
+C
=> -etanx •cosx
Integrals of the Form
I Example 35 Evaluate
+ sinx cos xA
ex
dx
cos2 x
2x
J+cos2x>
f 1 + sin x cos x X
Sol. (i) / = Je
dx
dx
cos2 x
/ = Jex-
r 1 + sin 2x
f eax sin bx dx, e“ cos bx dx
Let
I - J e“ (sin bx) dx
Then,
1 = J sin bx • e “dx
i
sin x cos
1
+------- x • dx
cos2 X
cos‘ X
fe“
= sin bx • ----
Ia
I = J ex {tan x + sec2 x} dx
I = tan x e - J sec'.2 xex dx + Je •sec2 x dx + C
= - sin bx -e“ a
a2
I = ex tan x + C
(ii) 1 = fe
1 + sin 2x
- [ b cos bx------dx
J
a
buL
e
r
e
= —sinbx-e“ — scos bx------ I (- b sin bx)----- dx •
a
a
a J
a
I = Jex • tan x dx + Jex (sec2 x) dx
n i
2x
1 + cos 2x
b2
= Je2x
1 + 2 sin x cos x
= Je2x
2 sin x cos x
.2 +--------- 2------ ^dr
2 cos x
2 cos x
I
= Je2x -sec',2 x + tan x • dx
or
= Je2x • tan x dx + -2 Jfe2x ■sec2 x dx
n I
e2x J
1
e 2x
= tan x----- - f sec 2 x----- dx +- Je2x •sec2 x dx
Thus, Je
T
1
2x
2
Sol. 1= fe X
' 1-x
ax
dx
4-
1=
ax
e“
a2+b2
e
(a sin bx - b cos bx) + C
ax
sin bx dx= 2„
(a sin bx - b cos bx) + C
a2+b2
Similarly, J e“ cos bx dx= —
e
ax
7 +b2 (a cos bx + b sin bx) + C
a
Aliter Use Euler’s equation
Let
(1-x \2
1 + x2
V ’ "
I
2
•tan x + C
I Example 36 Evaluate J e x
1-e
J sin bx • e
(a2+b2} e™
=
(a sin bx - b cos bx)
a
2
\
7
2
I=-e
2
'a2
,+7,= a2 • (a sin bx - b cos bx)
• dx
1
2
cos bx ■
1
ax
I = -sinbx-e - -- cos bx-e"
a2
a
■ dx
2 cos2 x
n
^ax
dx.
7
Hence,
P
cos bx dx and Q = J e
ax
sin bx dx
ax ■eibx
2
(l-2x + x2)
dx
I
dx=
Je
x
J + X2
(1 + x2)2
1
(a + ib)x
P + iQ = --------e
a + ib
a2+b2
e
ax
(cos bx + i sin bx)
24
Textbook of Integral Calculus
(ae^cos bx+ be™ sin bx) - i (ae^sin bx-be^cos bx)
77b2
ax
r
x2
Sol. Let I = j------------------- - dx
J (x sin x + cos x)
(a cos bx + b sin bx)
77b2
QX
Q=—
X2dx
r
! Example 38 Evaluate ------------------- 2
J (xsin x + cos x)
Multiplying and dividing it by (x cos x), we get
\ ----------------(* cos x)—- dx
IT = fz
I (x sec x)
J
i
(x sin x + cos xf
(a sin bx - b cos bx)
a2 + b2
n
I Example 37 Evaluate
(i) jex cos2 xdx
I = x sec x ■ J
(xsin x + cos x)2
■/f£(xsecx)IV
(ii) Jsin(logx)dx
Sol.
I = J ex • cos2 x dx - j ex •
(i)
X cos X
1 + cos 2x
dx
2
X cos X
(x sin x + cos x)2
dx >dx
-1
= x sec x--------------------(x sin x + cos x)
I = lje dx + - f cos 2x ex dx
[(x sec x■ tan x + sec x) ■ -------- 4------ dx
(x sin x + cos x)
1 = 1 ex + -1 Ar
- x sec x
(x sin x + cos x)
2J
2J
2 1
2
where,
= j cos 2x ex dx
- x sec x
n
i
+ Jsec2 xdx
,
- x sec x
_
I =------------------- + tan x + C
(x sin x + cos x)
= ex • cos 2x + 2 jsin 2x- ex dx
I
4 cos2(xx sin• (x xsin+ xcos+ x)cos x) &
(x sin x + cos x)
Ji = j cos 2x • ex dx = cos 2x ■ ex - j- 2 sin 2x-ex dx
n
- e~ -cos 2x + 2 {sin 2xex - j2cos 2xex dx}
I Example 39 The value of
= ex-cos 2x + 2sin 2x ex - 4
1
I------- sin -1
J\3+x
,V6 ’
L=- {ex cos 2x + 2sin 2x ex}
5
From Eqs. (i) and (ii), we get
...(ii)
l = hx + 4 - - {ex cos 2x + 2 sin 2x • ex)
2
4
I
11
(
4
I
(b) -< - 3 cos
2 5
l = -ex + — ex (cos 2x + 2 sin 2x} + C '
2
(ii) I = jsin (log x) dx
(
11
(c) — <-3 sin -1
4
Let log x = t
1 = [(sin t)• e‘ dt = sin t-ef - [ cos t • e' dt
n
2
0)
0
•COS
+C
+ 2y)9-x2 sin
}+c
+ 2-J9-X2 sin
+ 2x •+c
2
2
(d) None of the above
x = ef or dx = e‘ dt
J i
dx, is equal to
(a) — J _ 3 [ cos 1
4
=>
dx
J
I
n
Sol. Here, I
Put
=>
I = e‘ sin t - ef -cos t - I
:. 1 =
I = — {sin (log x) - cos (log x)J + C
2
W6
x = 3 cos 20
dx = - 6 sin 20 dB
I = sin t-ef - {cos t-e' - j-sin t -e* dt}
I = - ef (sint - cos t) + C
• sin"1 f -4= d3- x dx
3 - 3 cos 20 .
------------ sin
3 + 3 cos 20
-4= 73-3cos20 I(-6sin 29
K *6
)
_ f sin 0
J cos 0 ■ sin-1 (sin 0) • (- 6 sin 20) d0
= - 610 (2sin2 0)d0 = — 6 Je (1 - cos 20) de
■
25
Chap 01 Indefinite Integral
02
Sol. Here, we have only one function. This can be solved
easily by applying integration by parts taking unity as
second function.
If we take u = log (71 - x + -J1 + x) as the first function
= - 6----- 10 cos 20 d0
2
J
-6 X
sin 20
y------- —
2
2
and v = 1 as the second function.
Then,
6 sin 20 + cos 26
------+c
= - 302 + 6
2
4
= | - 3[C0S~^f + 2 ^9-x
2 • cos
J = j 1 • log (71^ x +
X
3
+ 2x
+c
+ x)dx
1
sec x (2+sec x)
dx, is
(1+2sec x)2
sinx
(a) - ------- + C
2 + cos x
-sinx
(c) -------- + C
2 +sinx
Sol. Let I = j
(b)
(d)
sec x (2 + sec x)
(1 + 2 sec x)2
(2 + cos x)z
1 f (1 — x) + (1 + x) — 2 71 ~ x 2
-dx
(2 + cos x)2
II
= -2x dx
2 -1
= x log (71 - x + 71 + x) - - J 71 - x
dx
2
7i - x7
1
= x log (71 - x + 71 + x) - - J1 dx + - j
= dx
2
2 -Ji-x2:
dx
1
1
sinz x
1
dx
dx + j
(2 + cos x)2
(2 + cos x)
I
Applying integration by parts to first by taking cos x as
second function, keeping J
i
7i-x.2:
(l-x)-(l + x) .
2
2
= J COS X •
- • r dx
= x log (71 - x + 71 + x)
c
2 cos x +1 .
------------- -dx
(cos x + 2)
sin2 x
1
if i-x + 71+X 71-x2
2 + sin x
r cos x (cos x + 2) + sin 2
_ f cos x
J 2 + cos x
1 - x - Jl + x
,
-—.
•
2 + COS X
*=J
*— ■xdx
2y/l + x 7
=X
COS X
COS X
+
-x
2
equal to
. .
+ 7T77
V1 “
Hence, (a) is the correct answer.
I Example 40 The value of j
1______
= {log (71-X + 71 + x )}■ X - j
X - dx as it is.
2
2
Hence, (c) is the correct answer.
f
X4+2
I Example 42 The value of Je X
dx, is
7
sin x
I =----- ------ • (sin x) - j sin x •
dx
(2 + cos x)2
2 + cos x
equal to
ex (x +1)
(a)
ex (1-x + x2)
sin2 x
+J (2 + cos x)2 dx (c) ex (1 + x)
2 + cos x
Hence, (a) is the correct answer.
(d) None of these
/
Sol. Let I = | e X
x'4 +2
dx
((1 + x2)5'2,
I Example 41 The value of jlog (7l-x + 71+x)dx,is
equal to
(a) x log (7l - x + 71+ x) + x - sin"1 (x) + C
(b) x log (7l — x + 71+ x) + -- x + sin-1 (x) + C
(c) x log (71-x + 71+ x) - x + sin"1 (x) + C
1
(1 + x2)1/2
l-2x2
+ (l + x2)s'2 dx
1
X
Jl + x2)V2
(1 + x2)3'2
x
l-2x 2
+(l+x2)’'2 +
xex
ex
+ (l + x2)3'2 + C
(1 + x2)1'2
Hence, (d) is the correct answer.
(d) None of the above
\
(1 + x2)5'2J
ex {1 + x 2 + xJ
crTT2)372 + C
dx
26
Textbook of Integral Calculus
I Example 43 If J(sin 30+sin0)esln0 cos 9 de
./Ax2+ 8x4-13
0 • cos 0 d0
2x + 2
dx
<y]4x2 +8x 4-13/
f
\
2x4-2
= j sin-1
dx
J(2x 4-2)2 4-32,
...(i)
As, I =( Asin3 0 4- B cos2 0 4- Csin0 + D cos0 4- E)ean 0 ■+ F
= (A sin30 - Bsin204- Csin04- Dcos0 4- B4- E)e,sin6+p
When, sin 0 = t
I = (At3 - Bt2 4- Ct 4- B 4- E)e‘ 4- F
as by Eq. (i) D = 0 ...(ii)
From Eqs. (i) and (ii),
J(3t - 4t3)e' dt = (At3 -Bt2 + Ct+ B+E)e' + F,
77)
'
differentiating both sides
(3t - 4t3) e'=(At3- Bt2+ Ct + B + E)e‘ + (3At2 - 2Bt + C)e‘
=>
A = — 4 and 3A = B
Hence, (b) is the correct answer.
B = — 12
= - {0 tan 0 - j tan 0 70}
3
= - {0 tan 0 - log | sec01} + C
2
/
. 3 2x + 2
=> /=-< ------ tan
2
3
=>I = (x +1) tan
3
x2 cos2 x
dx, is equal to
7
3
X COS X
7.
3
2
3
x - x sin x + cos x
2
2
X COS X
= f(x-e(x”x + co‘x)-xcos x )
X
dx
J
jg(x sin x + co»x)
(x cos x)2
Applying integration by parts
_{x.e(x«nx + c°«x)_ j e(x«mx + c«x)
11
X COS X
_ f g(x an x + e°» x)
■*
(
=> I = —
x sin x - cos x
x + c0’ x) •
+1
3
2
13
- (x + 1)1 - -log (4xz + 8x + 13) 4- C
x sec x + tan x
(x tan x + l)2
(d) None of the above
COS
TA
x2 (x sec2 x+tan x)dx
Sol. Here, I = j x2 • x sec x + tan x dx
(x tan x + 1)\:2
f
1
1
= x2
J2xdx ...«
(x tan x + 1)
(xtanx + 1)
+c
4
2x + 2
(x tan x+1)2
X
X
1+
-log
2
1
(c) e(xsinx+cosx) • X------------+C
— .
2x + 2
I Example 46 Evaluate j
1 }
(a) e(xsinx+cos x) X +---+C
X COS X J
Sol. LetI = Je(xrinx + C0,x)-
3 tan 0^ 3
2 n m 3 fa
--------- | - sec 0 dQ = - 10 sec 2 0 <70
3sec0 j 2
2J
= Jsin
3 2
-1^(x +1)'] - log ^4x2 +8x + 13 - -H
= -<-(x + l)tan
2 3
x cos x — xsmx+cos x
k
Put 2x 4-2 = 3 tan 0=>2dx=3 sec2 0 d0
X
I Example 44 The value of
X COS X 4- -
>dx.
[IITJEE200
f
Put sin 0 = t => cos 0 d0 = dt
= J(3t — 4t3) ef dt
(bj e(xsinx + cosx) #
2x + 2
I Example 45 Evaluate jsin -1
Sol. Here, I = J sin"1
Je(xsinx+cosx)
7
Hence, (c) is the correct answer.
= J(3sin 0 - 4 sin3 0)-e $m0-COS0d0;
/ 4
+C
X COS X
= (A sin3 0 + Bcos2 0+C sin 0 +Deos 0+E)e “n’ + F,
then
(a) A = -4,B = 12
(b) A = -4,B = -12
(c)A = 4,B = 12
(d)A = 4,B = -12
Sol. Let / = J (sin 30 4- sin 0)
1
x-----
= e(xdnx + cWx)
dx
2
2
*=j4=-i= (x tan1 x 4-1
\
x
kx tan x +1,
4
J t2
t
2x (cos x)
x sin x + cos x
dx
[put, x sin x + cos x =
=> (x cos x + sin x - cos x) dx - dz
_
X2
„ rdu
I---------- + 2
J u
(x tan x 4-1)
x. 2- + 2 log | u | + C
x tan x 4-1
x2
= —-----x tan x 4-1
=—
Chap 01
Exercise for Session 4
1. [x2exdx
2. jx2smxdx
3. jlogx-dx
4. J(logx)2dx
5. j(tan"1x)dx
6. J (sec 1x)dx
7. j x tan xdx
»• I ^5-dx
xz
o fx-sinx .
y. ---------- dx
10. Jlog(1 + x2)dx
11. je*(tanx + logsecx)dx
x cos x
72. Je*{-1 + sin
dx
coszx
log x
J 1 - COS X
13.
log (logx) +
1 '
dx
(logx)2,
1 + sin 2x
dx
1 + cos 2x
-------5~^dx
15. Jex --(14X2)2
16. J ey(2-x2) dx
17. J a" •cos(bx + c)dx
18.
19. j sinTx dx
20. j(sin1x)2dx
21. jcot‘1(1-x + x2)dx
22. fsin’’07 dx
22
(1-xyJi-x2
r a/x2 + 1{log (x2 +1) - 2 log x}
dx
J
x4
r
22
\
COS X
7
25. Jeslnx xcos^x5- sin x dx
I
14. Je*
sec3xdx
cos2 x 4- sin 2x
24. J -----------------^dx
(2cos x -sinxf
Indefinite Integral
27
Session 5
Integration Using Partial Fraction
This section deals with the integration of general algebraic
f(x)
rational functions, of the form----- .where f(x) and g(x)
g(x)
are both polynomials. We already have seen some
examples of this form. For example, we know how to
_
_ . r
1
L(x)
P(x)
mtegrate functions of the form------or------ or-----Q(x) Q(x) Q(x)
where L(x) is a linear factor, Q(x) is a quadratic factor and
P(x) is a polynomial of degree n > 2. We intend to
generalise that previous discussion in this section.
We are assuming the scanario where g(x) (the
denominator) is decomposible into linear or quadratic
factors. These are the only cases relevant to us right now.
Any linear or quadratic factor in g(x) might also occur
repeatedly.
Thus, g(x) could be of the following general forms.
• g(x) = L1(x)L2(x)... Ln(x)
(n linear factors)
n linear factors; the zth
. ^x) = L,(x)...l5(x)...L„(x)
factor is repeated k times
linear factors, the zth
• g(x) = L‘'(x)L‘!(x)...4(x)
^factor is repeated kt times
• g(x) = L1(x)L2(x)...Ln(x)Q(x)Q2(x)...Qm(x)
f n linear factors and
quadratic factors
a particular quadratic factor
repeats more than once
• g(x) = ...Qf(x)...
• A combination of any of the above
Suppose that the degree of g(x) is n and that of f(x) is m.
If m > n,we can always divide f(x) by g(x) to obtain a
quotient g(x) and a remainder r(x) whose degree would
be less than n.
s(x)
g(x)
(i)
Ifm<^
is termed a proper rational function.
g(x)
The partial fraction expansion technique says that a proper
rational function can be expressed as a sum of simpler
rational functions each possessing one of the factors of
g(x). The simpler rational functions are called partial
fractions.
From now on, we consider only proper rational functions.
If/W is not proper, we make it proper r(x) by the
g(x)
g(x)
procedure described in (1) above. Let us consider a few
examples.
Let g(x) be a product of non-repeated, linear factors :
g(x) = Lj(x)L2(x)...Ln<x)
f(x)
Then, we can expand ----- in terms of partial fractions as
g(x)
f(x)= Aj [ A2 t
An ,
g(x) Lj(x) L2(x) ”* Ln(x)
where the A/s are all constants that need to be determined
Suppose /(x) = x +1 and g(x) = (x -1) (x - 2) (x - 3). Let
f(x)
us write down the partial fraction expansion of -—-:
g(x)
/(x)^
x +1
A
B
C
g(x) (x — 1) (x -2)(x -3) x-1. x-2 x-3
We need to determine A, B and C. Cross multiplying in the
expression above, we obtain :
.
(x +1) = A(x-2)(x-3) + B(x -l)(x-3)
+ C(x-l)(x-2i
A, B, C can now be determined by comparing coefficients
on both sides. More simply since this relation that we
have obtained should held true for all x,we substitute
those values of x that would straight way give us the
required values of A, B and C. These values are obviously
the roots of g(x).
x=l
2 = A(-1) (-2) + B(0) + C(0)
A=1
x=2
3 = A(0) + B(1)(-1) +C(0)
B = -3
x=3
4 = A(0) + B(0) +C(2) (1)
C=2
Thus, A = 1, B = - 3 and C = 2.
f(x)
We can therefore write ----- as a sum of partial fractions.
g(x)
2
3
+-----g(x) x-1 x-2 x-3
29
Chap 01 Indefinite Integral
T
•
/X)
• is now a simple matter of integrating the
Integrating
Again, let x = - 1
2(— 1) 4-1 = A(— 1 - 2) 4- B(0)
partial fractions. This was our sole motive in writing such
an expansion, so that integration could be carried out
easily. In the example above :
A=l.
3
2x + l
(x 4-1) (x-2)
[/(*)
J g(*) dx = In (x -1) - 3 In (x - 2) + 2 In (x - 3) 4- C
Now, suppose that g(x) contains all linear factors, but a
particular factor, say Lj (x), is repeated k times.
Thus,
g(x) = L^(x)L2(x)...Ln(x)
f(x)
----- - can now be expanded into partial fractions as follows
£(x)
/(*) _ -^i + A2 + A,
A3
4----- —
g(x) Lx(x) Lf(x) L
3(x)
f-i(x)
Af.
L*(x)
B2
L2(x)'
Bn
Ln(x)
1
I Example 48 Resolve (x-1)(x + 2)(2x4-3)
(x-1)2(x-2)
A
B
C
can be expanded as; ------ +-------- 4-------X-1 (x-1)2 x-2
°
partial fractions.
B
C
Sol. Let------------- -------------- = -^— 4--------+------- ,
(x — l)(x 4-2)(2x 4-3) x-1 x 4-2 2x4-3
where A, B, C are constants.
1 = A(x 4-2)(2x 4-3) 4-B(x -1) (2x 4-3) 4-C(x - l)(x 4-2) ...(i)
For finding A, let x - 1 = 0 or x = 1 in Eq. (i), we get
k partial fractions corresponding to Li (x)
This means that we will have k terms corresponding to Lx (x).
The rest of the linear factors will have single corresponding
terms in the expansion. Here are some examples.
1
1/3 [ 5/3
x-2
1 = A (1 4- 2) (2 4- 3) 4- B (0) 4- C (0)
A=±
15
Similarly, for getting B, let x 4- 2 = 0 or x = - 2 in Eq. (i), we
get
1 = A (0) 4-B (—2 — 1)(—4 4-3) 4-C (0)
=>
B=1
3
For getting C, let 2x 4- 3 = 0 or x = - 3 / 2 in Eq. (i), we get
1
( 3
1 = A (0) 4-B(0) 4-C [---I -3- + 2
2
I 2
(x-l)3(x-2) (x-3)
can be expanded as
C
D
C
A
B
=>
------ 4----------- H-----------7 + (x-2)+(x-3)
X-1 (x-1)2 (x-1)3
c = -l
5
„
114
1
Hence,---------------------- =---------- 4---------------------(x-lXx4-2)(2x4-3) 15(x —1) 3(x4-2) 5(2x4-3)
____ 1
(x-1)2(x+5)3
I Example 49 Resolve
can be expanded as
A
B
x-1
C
+ ~----
(x-1)2
I Example 47
E
D
>-------- + (x+5)3
(x+5) (x4-5)2
2X4-1
(x+1)(x-2)
into partial fractions
2x4-1
has Q(x) = (x + 1) (x - 2) i.e linear
(X 4-1) (X-2)
and non-repeated roots.
2x 4-1
A
B
---------------- —------ 4-------(x 4-1)(x-2) x + 1 x-2
Sol. Here,
=>
(2x 4-1) = A(x -2) 4- B(x 4-1)
On putting, x = 2 we get
5 = A(0)4-B(3) => B = ^
3x5 + 2x2 + x+1
(x+D(x+2)
into partial
fractions.
Sol. This is not a proper fraction. Hence, by division process it
is to be expressed as the sum of an integral polynomial
and a fraction.
Now, 3x3 4- 2x2 4- x 4-1 = 3x (x2 4- 3x 4- 2)
- 7 (x2 4- 3x 4- 2) 4- (16x 4-15)
So, the given polynomial
3x3 4-2x2 4-x 4-1 /o
(16x4-15)
(3x—-7)4
7) 4-------------------- ==(3x
(x41)(x+2)--------------- (xfl)(x + 2)
Now, the second term is proper fraction hence it can be
expressed as a sum of partial fractions.
16x 4-15
A
B
---------------- —------ +-----(x-Fl)(x4-2)
X 4-1
X4-2
...(i)
30
Textbook of Integral Calculus
To find A put x 4-1 = 0, ie, x = - 1 in the fraction except in
the factor (x + 1).
16 (-1)4-15
•■•(ii)
A => A = — 1
(-1+2)
To find B put x 4- 2 = 0, ie, x = - 2 in the fraction except in
the factor (x + 2).
16 (- 2) + 15
= B => B = 17
(iii)
(-2 + 1)
1
17
=> The given expression = (3x — 7) — " — +-----x + 1 x+2
o= A + C + D
5
4(x + l)
--------------- 4-
Ax + B
assume that
I Example 52 Resolve
Q (*)
(* ~a)
A2
t
(x - a)2
+ ...+
-, where A and B are
ax~ + bx +c
constants to be determined by comparing coefficients of
similar powers of x in numerator of both the sides.
Br
b2 + ... +-----+---- 1— +
—
(x-ar)
(x-ai) (x-a2)
x+5
(x-2)2
has repeated
(twice) linear factors in denominator, so find partial
fractions.
Sol. Let
A
B
(x —2) + (x —2)2
x 4- 5
(x-2)2
(x 4-5) = A (x - 2) 4-B
Comparing the like terms, A = 1,-2A4-B = 5 or B = 7
X4-5
1
7
(x-2)2 “(x-2) + (x-2)2
I Example 51 Resolve
_____ 3x-2
(x-1)2 (x + 1)(x + 2)
, r
3x - 2
Sol. Let-------------------------(x — I)2 (x + l)(x+ 2)
c
D
=-A_+_1_ d---------+--------(x-1)2
2x + 7
(x + 1)(x2 +4)
into partial
2x 4- 7------ -------A + Bx-----4- C
----------(x4-l)(xz 4-4) x + 1 xz 4-4
(x + 1)
2x 4- 7 = A (x2 4- 4) 4- (Bx 4- C) (x 4-1)
x = -l
Put
5 = 5A or A = 1
Comparing the terms, 0 = A + B
=> B = - 1
7 = 4A 4-C => C = 3
2x4-7
1 | (-X4-3)
(x 4-l)(x2 4- 4) x + 1
X2 4-4
Aliter To obtain values of A, B and C from
2x 4-7 = A (x2 4-4) 4-(Bx 4-C)(x 4-1)
2x 4-7 = (A 4-B) x2 4-(B 4-C) x 4-4A 4-C
i.e.,
Equating the coefficients of identical powers of x, we get
A 4- B = 0, B 4- C = 2 and 4A 4- C = 7.
Solving, we get A = 1, B = - 1, C = 3
into
partial fractions.
(x-1)
.2
fractions.
(x - a)k
Sol. Let
I Example 50 Expression
8
9 (x 4-2)
Case III When some of the factors in denominator are
quadratic but non-repeating. Corresponding to each
quadratic factor ax2 +bx +c, we assume the partial
fraction of the type
Aj
13
8
9
4
36
3x-2______
13
|
1
” (x — I)2 (x 4-l)(x 4-2) ~ 36(x - 1) 6(x —1)z
[using Eqs. (i), (ii) and (iii)]
Case II When the denominator g (x) is expressible as the
product of the linear factors such that some of them are
repeating. (Linear and Repeated)
LetQ(x)=(x -a)k (x-a1)(x-a2)...(x-ar).Then,we
P(x)=
=> A=’
(x + 2)
3x - 2 = A (x - 1) (x 4-1) (x 4- 2) 4- B (x 4-1) (x 4- 2)
+ C (x - l)2(x + 2)4- D (x - l)2(x +1)
Putting(x - 1) = 0, we get B = 1 /6,
Putting (x 4-1) = 0, wegetC = -5/4
Putting(x + 2) = 0, we get D = 8/9
Now, equating the coefficient of x3 on both the sides, we get
I Example 53 Find the partial fraction
2x+1
(3x + 2) (4x2 + 5x4- 6)
2x 4-1
A
Bx 4- C
=-------------- 1-----------------------t
(3x4-2)(4x2 4-5x4-6) (3x4-2) (4xZ4-5x4-6)
then 2x 4-1 = A(4x2 4-5x 4-6)4-(Bx 4- C) (3x 4- 2)
Sol. Let
where A, B, C are constants.
For A, let
3x 4- 2 = 0,
ie.,
x = -2/3
2
2
3
+1=x{4H+6HbHH(0)
10
3
1 . ( 40 A
.
3
= A — => A =
= ----V 9 )
40
3
31
Chap 01 Indefinite Integral
. 3 log 11 - 2x I
1
= -x + log|x|
—----- 1 + C
log |x| + --^
-
Comparing coefficients of x2 and constant term on both the
sides for B and C, we get
4 A + 3B = 0,
=> B = — and 6A +2C = 1,
10
z. 29
=> C = —
40
29
4
-3
B=--A
3
1-6-A
C=
2
______ 2x +1
+
” (3x+ 2)(4x2+5x+ 6) - 40(3x+ 2) ' 10(4xz+5x + 6)
Case IV When some of the factors of the denominator
are quadratic and repeating. For every quadratic repeating
factor of the type (ax2 + bx + c)k, we assume :
Axx+A2
A3 x + A4
ax2+bx+c
(ax2+bx+c)2
I Example 54 Resolve
+...+
Lt
(ii) Let,
1
3
=- x + log | x|---- log |1 - 2x| + C.
2
4
3x-1
A
B
"(x-2) + (x-2)2
(x-2)2
3x — 1 = A(x - 2) + B,
=>
(i)
On putting x = 2 in Eq. (i), we get B = 5.
On equating coefficients of x on both sides of (i), we
get A = 3,
/
3__ __ 5
(x -2)2
J <(x-2) + (x-2)2> dx
f3x~‘ <fr=f
•••
A2k-i x + A#
= 3 log |x - 2| -
(ax2+bx+c)k
2x4+2x2 + x + 1
x(x2 + 1)2
into partial
Lt
Lt
-?+c
x-2
x2 + x + 1 A
B
C
(iii) Let, —;--------- = — +—7 +
x2 (x +2)
x X2 x + 2
x2 + x + 1 = Ax(x + 2) + B(x +2) + C(x2) ...(i)
fractions.
On putting, x = - 2 and x = 0 in Eq. (i), we get
2x4 + 2x2 + x + 1 _A_
Dx + E
A Bx + C
=—
Sol. Let
x(x2 + l)2
x +
+ x2+ 1 +(x2+ 1)2
x
or 2x4 + 2x2 + x + 1 = A (x2 + l)2 + (Bx + C) x (x2 + 1)
+ (Dx + E)x
Comparing coefficients of x4, x3, x2, x and constant term
C =3/4 and B =1/2
On equation Coefficient of x2 on both sides of (i),
we get 1 = A + C => A = 114.
t x2 + x + 1 ,
----- dx~
J x2 (x + 2)
A + B = 2, C = 0,2 A + D + B = 2, E = 1, A = 1
A we get A = 1, B = 1, C = 0, D = - 1, E = 1
(iv) Let,
Hence, the partial fraction,
2x4 + 2x2 + x +1
x(x2 +1)2
1
=—+
X
x
x2
3/4
x+2
8 = A(x2 + 4) + (Bx + C) (x + 2)
1 + x2 + d+x7
-.(i)
On putting x = -2 in Eq. (i), we get A = 1.
3x~l
------- ydx
(x-2)2
(IJ J x(1-2x)
1/2
dx
x+2
1
13
= - log |x| - — + - log |x + 2| + C
4
2x 4
_ A
Bx + C
(x+2)(xz + 4)
1-x
X
I Example 55 Evaluate the following integrals:
... r(1-x2)dx
8
1/4
On equating coefficient of x2 on both sides we get,
0 = A + B=>B = -1
On equating constant term on both sides, we get,
8 = 4A+2C=>C = 2
(m)
fx2+x+1
Sol. (i) Let,
=>
dx
8dx
(iv) f (x+2) (x2 +4)
1 , (-x + 2)
dx
... Jf ( +2)(*,2+4) *4 x+2
x2 + 4
x
x
1-x2
1 A
B
= —I---- H----------x(l-2x) 2 x (l-2x)
(1 - x2) =- x (1 - 2x) + A(1 - 2x) + B(x)
2
Jx + 2
2Jx2 + 4
+2J
dx
x 2+ 4
= log|x + 2| - - log|xz + 4| + 2. - tan
2
2
X
2
On putting x = 0 and x =^, we get
l = Aandl-l = B.i=>A = l,B = 4
22
1-x2
x(l - 2x)
dx =
l2 +lx +3-'l dx
2(l-2x)J
I Example 56 Evaluate f----- —---- dx.
r
Jsinx-sm2x
1
dx = f------------ ------------- dx
Sol. Let 1= fJ sin x - 2 sin x cos x
•J di
sin x - sin 2x
= [----- 1—
J sin x (1 - 2 cos x)
j
f
sin x
dx = I —-------------------- dx
sin2 x (1-2 cos x)
32
Textbook of Integral Calculus
I Example 58 Evaluate Jsin 4x e tan2 x dx.
= f---------- ----------------dx
1 (1 - cos2 x) (1 - 2 cos x)
(put cos x — t => — sin x dx = dt)
-dt
= [-------- zl-------- df.„(i)
J (1-0(1 + 0(1-20
(1 - f2)(l - 2f)
Here, in Eq. (i) we have linear and non-repeated factors thus
we use partial fractions for,
-1_______
A
B
C
(i -1) (i + 0(1 - 2t)" (i - 0 + (i + 0 + (1 ~ 20
or - 1 = A(1 + 0 (1 - 20 + B(1 - 0 (1 - 20 + C (1 - t) (1 +1)
-(ii)
Putting
(t +1) = 0 or t = -1, we get
1
- 1 = B (2) (1 + 2) => B = 6
Putting
(1 - 0 = 0 or t = 1, we get
-1 = A(2)(-1)
=>
=> A = |
2
Putting
(1 - 20 = 0 or t = 1 / 2, we get
Sol. The given integral could be written as,
I = J 4 sin x • cos x • cos 2x ■ eUn Xdx
= 4 J tan x-cos2 x(cos2 x - sin2 x) e “"'*dx
= 4 |tan x-cos4 x(l - tan2 x)-eUn x dx
= 4f-ta?X- (l-tan2x)e“"!l dx
J (sec2 x)2
Put tan2 x = t
=> 2 tan x-sec2 x dx = dt
=> / = 4f<L^.*=2j ^-dt
. .
2(1
So, Eq. (i) reduces to
1 r 1 ,
lr 1 ,
I = - ------dt —- ----- dt
2Jl-t
6J 1 + t
+
t)
P
= -2
4 f
[ —— dt
3 J l-2t
Sol. Here, I =
Put
x(1-x3e3c0SX)
Jx(l-(xecotx)3)
——<I t(l-t)(l + t + t2)
t(l-t’)
c(a
b
— +------ +
Ct + D '
2
i + t + rJ
dt
Comparing coefficients, we get
1
2 _ = -i
A. = .l,B„ = i,C
= -^,D
3-3
3J1-t
1
e‘ +C
(i + 02
[using Je'(/(x)) + f'(x))dx = e *./(x) + c]
2etan’x
=------------------ + C
(1 + tan2x)2
J = -2cos4 x-e*"1’* + C
J Example 59 Solve j
Sol. Let I = j
1+xcosx
-dx.
x(1-x2e2sinx
1 + X cos X
dx
x(l - x2e2 sin x)
Differentiating both the sides, we get
(xeMX'cos x + eanx)dx = dt
=>
e™x (x cos x +1) dx = dt
i=f—
= 1--- ---t
J t
dt>
Put(xerinx) = t
xe CMX =t
(xecosx • (- sin x) +k ecosx)dx = dt
dt
=
(1 + o3
J (i + t)3
2
3(1-20
(1 - x sin x) dx
2
= 2 P (i + 03 (1 +1 t)2 dt
11.1.
e*. I ■ ✓s
1. .
4
= — “ log | 1 — t | — y log | 1 + t | — — X — log | 1 — 2t | + C
6
'32
3
“'
'
1x
1
2
= — log 11 - COS X | — log I 1 + COS X | + - log
2
6
3
11 - 2 cos x | + C
I Example 57 Evaluate J
J (1 + 0
.
_ [f2et-(l + Qet
-i = cfi-- , 1 + - => C = -l
3
2) V 2
_______ -1_______
1
J _4
(1-0(1 + 0(1-20
i
----- 3/dt
J 1 + t + t2
1+ 1
2(1-t)
1
\dt
2(l + t)
= log 111 - - log 11 - 11 - - log 11 + 11 + C
2
= log (11 - - log 11 - 11 - - log I 1 +1 + t I
3
[where, t = xe CO»X]
[using partial fraction]
2
= log | x e*" x |--log 11 - x2e2dn x
2
Chap 01
r
I Example 60 Evaluate;
j- {log e6* . log e®2* . log ee’x} dx
AC
Sol. We have,
1
1__
1
dx
x{l + log ex} {2 + logex} {3 + logex}
[using partial fraction]
dx
=| log |1 + t| - log |2 + t| + log |3 + t| + C.
= f_____________
= | log |1 + logex| - log |2 + logex| + log |3 + logex| + C.
J x{logee + logex} {loge*2 + logex’} {logee + logx2}
Exercise for Session 5
■ Evaluate the following Integrals :
Y2
1. J(x-1)(x-2)(x
f------------------dx
-3)
3- J1-^x(x + 1)
c fI-------------------------cosx
■ 5.
dx
■I (1 + sinx)(2 + sinx)
7
33
Put logex = t =* —dx = dt
x
1 1
1
1
T f
dt
I = -------------------- =
----------------- +------- dr
J (l+t)(2 + t)(3 + t) J ^2(l + t) (2 + t) (3 + t)
I = J—{loge“ . logee X.log€e3x} dx
Jx
<7 log 6“ log/’' loge€’x
Indefinite Integral
, secx dx
J 1 +cosec x
9. f_____ _______
J x|6(logx)2 + 7logx + 2|
2- J 1+dxx3
2x
4- J (x2 + 1)(x
2+3)
O.
----------------------- dx
J sin x (3+ 2cos x)
tan x + tan3 x
8- J 1 + tan3 x dx
tan
10. J V
.dx
Session 6
Indirect and Derived Substitutions
Indirect and Derived Substitutions
(i) Indirect Substitution
If the integral is of the form f (x) • g (x), where g (x) is a
function of the integral of f (x), then put integral of
/(x) = t.
(
I = a log
where,
+C
.(i)
+C
(given) ...(ii)
x5/2 + l
0
X
k
'
< 1 + X* t
From Eqs. (i) and (ii), we get
X
4^+2
a log
(b) ^x2 + 2 + C
U+
J
x5/2
2
+c
+ C = - log
Vl + x5/2 ,
5
=>
a = 2/5 and k = 5/2
Hence, (a) is the correct answer.
(d) None of these
_ . TT
r f d(x2 +1)
So/. Here, I =
- J ylx2+2
5x4+4x 5
I Example 63 Evaluate j
dx.
(x5 + x+1)2
We know, d(x2 + 1) = 2x dx
So/. Here, I = j 5x4 + 4x5
2x dx
5 ,
•Jx2 +2
, ,\2
dx = J — x4 (5 + 4x)
x.10
]
<
1
x2 + 2 = t2
=*
2x dx = 2t dt => I = f
J
J = 2 -Jx2 +2 + C
1
1 '
x
X .
+ “ + —
\
Put,
5
t
Put
/
(^)s
(Vx)7 + x
6
5/x6 + 4/x
— dx
'
1
1 '
1 + —+ —
<
x
X .
1
1
= 2t + C
Hence, (a) is the correct answer.
I Example 62 If J
x5/2
2,
= -log
dx = a log
xk
1+xk y
1+7+7=1
=>
x5
X6 )
/=j-4=-+
J t2
t c=
then oandk are
_i
1
(a) 2/5,5/2
(b) 1/5,2/5
1 + n- + 4
(c) 5/2,1/2
(d) 2/5,1/2
.5
So/. Here,
(V?)5
(V?)’ + 7^
=1-
,7/2
X
+c
1+y
5
5
d(x2 + 1) .
I Example 61 The value of j -1
, is
(cjx^/x 2
1 •
2
2
= - - log 11 + y I + C = - log
x
~‘
+c
X
x
—+c
x3 + X + 1
dx
(V?)2+(V?)’
I Example 64 For any natural number m, evaluate
dx
1 + x5/2
J(x3m + x2m + xm) (2x2m + 3xm + 6)1/m dx, x > 0
[HTJEE2002]
1
-^=y
5
dx = dy
So/. Here, I = j (x 3m + x:2m + xm)(2x2m + 3xm + 6)1/m dx
x3m
5J 1 + y
2m + xm)
(2x3w+3x2'”+6xffl)1/w
dx
X
+ xm-l)(2x3m+ 3x2m+ 6xm)Vmdx
:
x■3n~1+x,2m-1
(i)
Chap 01 Indefinite Integral
2x3m + 3x2m + 6xm = t
Put
=>
6m(x3r""1 + x2m-1
I Example 67 The evaluation of
rpxp+2q-1 -qx9"1
dx is
x2p+2q+2xp+q+1
m~1)dx = dt
Eq. (i) becomes,
(l/m) -r 1
I
=J?'« A
— ----------+ C
6m 6m (1 / m) + 1
1 =----- ----- (2x3m + 3x2m + 6xm} m
6(m +1)
*q
(a)- ---- 1- c
(b) ------------ + C
*q
xp
(d) ----------+ C
xp + <? + 1
xP^ + 1
m +1
+c
xP^ + 1
(c)- ----- F C
xP^ + 1
I Example 65 j — ■■ xdx
■P + 2q-l _
(b) 2 -Ji + -^1 + x 2 + C
2
(c) 2(1 + 71 + x2) + C
Sol. j
Put
1 + x2 ) Jl+Jl + x 2
= dx = 2t dt
zjl + x 2
x dx
■Jl + x'.2
dx is equal to
- J-ln(ln2x-x2) + C
(a)^ln
t
Hence, (b) is the correct answer.
(2X+1)
(b) -1 In In x-xy —1 ttan
4
Jnx + x, 2
dx
(b)
+C
(c) J- In Inx + x'l + -1 ttan
4 Jnx-xj 2
... if. ^nx-x' + tan
(d)- In
(c)
+C
+C
4
—
dx = J[ —
,3/2
. 4
1 1
x3 l + - + ~y
x x2)
.
+
4
“
X
+
lnx4x-x4
dx
x2 (1 - Inx)
i y/2
Inx
Put----- = t =>
X )
>
1-lnx
X
14
■>
Now, put — + — +1 = t =>
x
x
-t dt
t3
(_2__ 4
I x3
x2
=-+C
t
x
+C
■Jx2 + 4x + l
dx = 2t dt
*=J-
-1
dx
—
+c
x2(l - In x)
X4
2x~2 + x"3
1
3/2
(t)J
Jnx + x,
Sol. Here, 1 = J
Sol.
dx
Hence, (c) is the correct answer.
2
+c
px',-~1t-qx~<?~1
(xp + x"’)2
- 2t dt
J
I Example 66 J
r
1
Taking xq as x2q common from denominator and take it in
numerator.
=> (pxp ~1 - qx ‘ ~l)dx = dt
Put
/
r dt
i _
xq
I =J-=--+C=+C
J t2
t
\xp + q + 1
x dx
2x
-1
, 1\2
(d) None of these
1 + /1 + x2 = t2
V
Sol. Here, I = j px^24?-1-^-1
x2p + 2q+2xp + q+l
is equal to
x2 + A/(1 + x2)3
(a) — In (1 + -Ji + x2) + C
dt
(t4-l)
x2
X2
= dt
= Jf--------(t2 + l)(t2-l)
= 1 f(t2 +l)~(t2 ~ 1)
2 J (t2+l)(t2-l) dt
1 (1 ,
Hence, (b) is the correct answer.
35
t -1
tan t
-I— -2 1^2- In-------t+1
2
t2 - 1 Jt2 + 1
36
Textbook of Integral Calculus
Inx
1
Inx - x
— tan
= — In
2
Inx + x
4
x
1
fx"-2g(x)dx = J (l +x""
dx
nxn)’/n
+c
= ^~ fn2-xn‘1-(l + nx•n
n)-)-,/n dx
n2 J
Hence, (b) is the correct answer.
X2-1
I Example 69 J-3
J X
-- ------ + c
(b)
x.3:
Hence, (a) is the correct answer.
dx
Derived Substitutions
2~7.2’+7'
Some times it is useful to write the integral as a sum of
two related integrals which can be evaluated by making
suitable substitutions.
_4_
Examples of such integrals are
2
dx
xz -1
2
i
2~
L
4
x 3 ~ x5
In
.1-1
_(l + HXn) "
n(n -1)
-2x2 + 1
X5
1---n
.
-2x2 +1
——+c
2x2
(d)
X2 -1
Sol. Let 1= f-
•
X
,
1 A L
2
1
dx = ~2 .2---- r + —
4
1
2
x2
V
x4
..rHx) dx = 2 ylf(x) + C
72^4 -2x2 +1
2?
/ J77w
Type I
(a) Algebraic Twins
X4 +1
X4 + 1
_2
X2 +1
+1
2x2
«
— dx,’Jf
[IITJEE 2007]
1-1
!
1-1
1
(1 + nxn)I n +C (b) —— (1 + nxn) "+C
n-1
n(n-1)
1
(1 + nxn) + " +C (d) — (1 + nxn) +" +C
n(n+1)
n+ 1
X
X
y = (l + xn)1/7‘ (14-2x")1//l
X
(1 + 3xn)1/n
and (fofofo...of)(x) = g(x) =
n times
J
(x4+l —
+ fcxr
2) dx
6
i
6
sm x+cos x
p ± sin x ± cos x
d
J a + b sin x cos x
y
f(f(x)) = (1 + x")
17"'
■
+1
jJ —
7—-—7— dx, JJ ------ i------ dx,
(sin4 x +cos4 x)
f(*) = (l + xn),/n
*- -
r.::
dx
| -^/tan x dx, | ^/cot x dx,
•
Similarly,/(f(f(x))) =
x2-l
(b) Trigonometric Twins
X
Where,
dx
1+1
1+1
Sol. •/
dx + j
<fc-J X4
J x4 +1
x4 + l + kx 2
n times
(c)
x’ +1
— dx = J
x4 ■
X
for n > 2 and
(1+x")1/n
g(x) = fofo...of(x), then xn“2 g(x)dx equals to
(a)
x2-l
dx
I Example 70 Letf(x)=
J
x2 +1.
+c
Hence, (d) is the correct answer.
V-
i-i
"
-------r—+ C
n
1
[IITJEE 2006]
Jlx1* -2x22 +1
7^
1 (l + nxn)
t
'2x4-2xz + l
J2x4 -2x2 +1
---- + C
(a)
x
-2x2 +1
---------- + C
(0
x
=f
dx is equal to
4
l
Method of evaluating these integral are illustrated by
mean of the following examples :
Integral of the Form
1.
X_____
_ n \l/n
Putx + - = t =>
X
iV
,
1----- - dx=dt
x2 J
Chap 01 Indefinite Integral
.2
1
1 + — dx.
=> I = - f x4+5x2+1 dx + -2Jf
2J
x22 J
1
Put x----- = t =>
x
3J
x4 +5x2 + 1
dx
1 f 1 + 1/x2
-dx-lf 1 - 1 / x2 dx
2 J x2 + 5 + 1/x 2
2J x2 + 5 + 1/x2
(1 + —
1L
.
\dx = dt
I x2 J
[(dividing Numerator and Denominator by x2)]
x2 +1
x4 +kx2 +1
dx.
1 f (1 + 1/jr) dx-lf
dx
2 J(x + l/x)2 +3
2 (x- 1/x)2 + 7
Divide numerator and denominator by x2.
du
4bu2 +(Ji)2
1 [
x2 -1
x4 +kx2 +1
2h2 +(J1)2
dx
i
1
where t = x---- and u = x + —
x
x
t
-•4-ftan‘1-^>| + C
1 1tanL —-i
It - -----f= f
2 Jl
2 Ji
13 J
7
Divide numerator and denominator by x .
I Example 71 Evaluate J
iol. Let Z = J
1-x2
37
5
?
.4 dx = -25JJ
1+x
5
1+X4
2
1 + x4
dx.
x - 1/x
1 1
—j= tan
2 V7
dx
X + l/x
—U tan
Ji
Ji
V3
+c
i Example 73 Evaluate j"^tan x dx.
, 1 - X2
_5 p 1 + x'2
~2 J 1 + x4 dx + J —
1 + x4
Sol. Here I = j J tan x dx
r x2 - 1
_ 5 r x,2 + 1 ,
----- dx
~2 J x4 +1 dx - JI —
x4 +1
Put
tan x = t2
=> sec2x dx = 2t dt
2t dt
=> dx-
1 + t4
Remark
Here, dividing Numerator and Denominator by x2 and
converting Denominator into perfect square so as to get
differential in Numerator
i.e.
z = p. 2t ,
1 + 1/x2
j
f
&
' " 2 [J t2 + (Ji)2
j
a
1
2
,
tan
1
-7= tan
•2 V2
ds
u2 - (Ji)2 _ ’
-blos
-i
1
and u = x + —]
x
x
1
1
x
Ji
2ji
x + —+ 5/2
x
„
1
■>
dx-
dt + j
f
(t + l/t)2 -2
1
dr
1_
Ji
tan 1
t-i/t
Ji
1.
+ -i°g
dt
,
• 1
G+1-V2
’ 7+17^
+C
t
k
+C
[(where t = Jtan x)]
I Example 74 Evaluate j—
- 2
x4 + 5x2+1
2
pl. Let I = - f
dx
2 J x4 +5x2 +1
1-1/t2
= -J= tan 1 s | + 1- . r - V2 + c
Ji ) ^
2Ji 10g 77T2
Ji
u-Ji
+c
u + Ji
x-1/x
dt
'■J s2+ (Ji)2+h2-(Ji)2,. [s=t—t and r=t + -t
1
I Example 72 Evaluate j
1 + 1/t2
■’(t-l/t)2 +2
[where t = x
55
2
= f
du
-/
J t4 + 1
dt
_ f 1 + 1/t2
1-1/t2
dt
Jt2+l/f2 dt + J
t2 + l/t2
— ■ dx
~2 J (x-1/x)2 +2 dx - J -----------(x + 1/x) -2
_5 r
t4 + 1
= Jf—
t4+1
. 5 r 1 + 1/x2
I = - —--------- - ■fc-J \ —2 dx
2 J x2 +l/x
J x2 + l/x2
_5 r
r 2t2
So/. Let
I=4j-
4
4
x+cos
dx
sin4 x + cos4 x
• Cl'
4
-dx.
X
Textbook of Integral Calculus
38
i+4
1
1+~
,2
y
v+4
y
1
y-_ I
.
y.
Dividing numerator and denominator by cos4 x, we get
y2 +1
r sec4 x
/ = 4 I---- ----------dx
J tan4 x + 1
Z=4J
Put
y4 +1
sec2 x (1 + tan2 x)
tan x = t
1 + tan4 x
dx
r
Put, y - — = t
y
sec2 x dx = dt
=>
___ y_
i
,2
J t4 + 1
■Ji
t2 +2
= 4(1 + 1/'2
dt = 4 j
1 + 1/t2
(1-1/t)2 +2
Z=4J
Ji
+C
/
xx
+ C=-U tan
■Ji
__ y
= -7= tan
■Ji
5/2
dz
=
7
(x2-l)dx
j
tan x -1/ tan x
+C
■Ji
I = 2V2 tan"1
'JL +
I Example 77 Evaluate
tan 1
V2
z2 +2
1 )
*2
7
z=t--
Again, put
t
tan
( y-_0
dt
+2
11
1 + — dy = dt
y
dt
dt
4
2
1
(x4 + 3x2 + 1) tan-1 X + X
I Example 75 The value of J
equal to
(a) - sin
c
(c) sin
1 Hh
(ax2 -b)dx
, is
x^c2x2 - (ox2 +b)2
(b)csin
(x2-l)
Sol. Here I = j
(x4 + x3 +1) tan
The given integral can be written as
j
(1 - l/x2)dx
1
(x2 + 3 +1/x2) tan-11
bA L
ax + - + k
xj
x 7
ox + b / x'
+ k (d) None of these
c
xyjc2x2. - (ax2 + b)2
(dividing numerator and denominator by
(l-l/x2)dx
’
'-I {(x + l/x)2 +1} tan-1
b V
a---- -\dx
x )
(ax2 - b) dx
Sol. Here, 1 = j
1_ j
b
ax + —
x
Jc2
2
X
dx = dt
. Put, X + — = t =>
X
Put ax + — = t
dt
(t2 +l)tan-1(t)
X
(
b
a---- 2 dx = dt
\
x .
'=1
=>
sin
Now, make one more substitution
dt
= = sin
■Jc2 -1 2
ax + b/ x\
+k
c
)
t_
c
-i ♦
+k
dt
t = u. Then, —— = du
t‘+l
tan
Eq. (i) becomes, I = f — = log | u | + C
J u
=> I = log | tan"111 + C = log | tan-1 (x +1 / x) | + C
Hence (c) is the correct answer.
I Example 76 f-- (X
X(x2x + l)(lnx-n) dx
Sol.
x4x + l
Put xx = y
=> xx (In x + 1) dx = dy
I Example 78
(x
:
x /3 (x2 + x+1)/2 -x/2 (x2 + x+l)
Sol. I = j
x7/6(x-7/6 -x5,6)dx
x7/6 • x1/3 (x2 + x + l)1/2 _ X-J/2 •„7/6/..2
- \A
f x3/2
(1- x2)dx
1),/2 _ x5/3
.1/3
Chap 01 Indefinite Integral
-H
■
1 iY/2
x + —+ 1
X
'-I
J
dx
1/3
X+-+1
Putting x + — = t
x
i
=> 1---- 2- dx = dt
The following examples illustrate the procedure :
I Example 80 Evaluate J-—
Sol. Let I = j
1
dx
(x +1) Jx-2
Here, P and Q both are linear, so we put Q = t 2
Substitute, (t + 1) = u6
u3
6u5 du
-f. - u
3
,
= -6j ------ du,
2
u-1
Le.
x - 2 = t2
So that dx = 2t dt
■2t dt = 2 j dt
t2 +3
(t2+2 + l) Vt2
1
u -1 = z
Put
I3 .
dz = — 6 j (z + 1)— dz
=-6
2
1
z + 3z + 3 + z
= 2 x —j= tan
V3
.1/6
(b) Integrals of the Form j
dx, where P is a
To evaluate this type of integrals we put Q = t2 Le. to
[.] denotes fractional part of x and greatest integer
function, is equal to
(b) 1
fe)+c
quadratic expression and Q is a linear expression
I “I
I Example 79 The value of j{{[x]}} dx, where {.} and
(a) 0
Sol. Let
-i
Z = -^tan-1 n/x-2 + C
V3
V3
dz
3
o_2
3z + 3z + iog|z| L + C
Z
= -6 . £_
+ f±_
2
3
z=
1
«
z
„ f z3 + 3z2 + 3z + 1 ,
= - 6 I--------------------- dz
J
z
where,
dx.
I x2)
X
‘
39
evaluate the integrals of the form
1—
---- L-7== dx
(ax + bx + c) -Jpx + q
put px + q = t2.
(d)-l
(c) 2
I = J{{W}}dx
x+2
I Example 81 Evaluate J
dx.
where, [x] = Integer and we know {nJ = 0; n e Integer.
■ H 0dx=0
•••«
Sol. Let
Hence, (a) is the correct answer.
______ x + 2______
(x2 + 3x + 3) y/x + 1
Put x + 1 = t2 => dx = 2t dt
I = J----------- (-f2 ~1) —------- j= ■ (2t) dt
Type II
{(r2 -i)2 +3(t2 ~i) + 3) Vt2
[2
| |
=4 t4 +t2
Integration of Some Special
Irrational Algebraic Functions
In this case we shall discuss four integrals of the form
J
dx
+1
.
dt = 2 j
1 _i_ 1 / *2
1 + 1/t
dt
t2 + 1 + 1/t2
= 2[__________ dt = 21 2 du
J (r - i/r)2 +(73)2
u2 +(73)2
dx, where P and Q are polynomial functions of x
and<}>(x) is polynomial in x.
(a) Integrals of the Form J
-y=l
dx, where P and Q
To evaluate this type of integrals we put Q = t2, i.e. to
evaluate integrals of the form f------------
&
cx + d = t2.
= 4. tan’1
V3
Z = ^=tan
V3
are both linear of x
- ■= dx, put
J(ax + fc) Jcx+d
,
1
where u = t —
t
2
= —f= tan
V3
ft2-f + C
/
X
J3(x + 1),
+c
40
Textbook of Integral Calculus
(c) Integral of the Form J
1
dx
(ax + b) JpX2 + qx + r’
r dx
where in | —f= P is linear and Q is a quadratic put,
/ NQ
ax + b = -'
t
2z dz
/—1J
2 J (z2 +1 + 1)7?
Z = -itan-1
V2
J+C
V2
= —7= tan 1
V2
Sol. Let I = f----------
j(x-d7 x2 + x + 1
Put x - 1 = - => dx = - -vdt
t
t2
- 1 /12 dt
1+1
J
*'4
- + 11 +1
t
J
1 [•
dt
73
17
“
J t+- +—
dt ’
73? + 3t +1
K 2> 12
= -log | (t + 1 /2) + 7(t + 1/2)2 +1/12 | + C
3
2
1
1
12 — + - I +1
Vx-1 2
+C
12
/
1
1
=-^10g
+— +
Kx-1 2
(d) Integrals of the Form J
, where P and Q both
are pure quadratic expression in x, i.e.P = ax2 + b
2 J . f
dx
and Q = cx + d, ue. I-------------- =
(ax2 +b) yjcx 2+d
To evaluate this type of integrals of the form we put x = j
dx
I Example 83 Evaluate j
Put
/
\
7^
+C
\
/
Aliter Put x = cos 0, dx = - sin 0 d0
de
sin0d0
___ r
(1 + cos2 0) sin 0
2 1 + cos2 0
fix-
V2
... mJ—
I=-J
0d9
_ f sec? ddB __ r
■* sec2 0 +1
2 tan2 0 + 2
Put tan 0 = t. => sec2 0 d0 = dt
1
dt
h----x+2= --7=rtan
1
-i
2 J. o------- ^2
= —L tan 1 tan 0 + C
72
’
V2
= - ’ tan'1
V2
f^~x2 A
/2x
where, cos 0 = x
. 7i - x2
I Example 84 Evaluate
3
2
r(x-1)7x4 + 2x"-x
2+2x + 1
dx.
•I
x2(x + l)
(x2 - 1) + 2x3
7
- x2 + 2x + 1
dx
(x + lF ’
2
,2
n
1 I
x2 x2+2x-1 + X +—
X2)
- (1 + x2) 71-X2
1
J
dx
1
,
x = -, so that dx = —- dt
t
t2
-l/t2dt
t dt
+ 2|x + — | -1
7 = 4J (t2 +1) Jr2 -1
I
1 f
du
= which reduces to the form j dx
2 J (u + 1) /a -1
PjQ
where both P and Q are linear so that we put u - 1 = z 2 so
that du = 2z dz
xJ
—r--------------- dx
Again, t2 = u => 2t dt = du.
_
sin 0 = 71“ x2
+c
tan 0 = —-------x
Sol. Here, I = J
(1+X2)7iT< 2
mJ-----------
\
+C
-1
= —L tan 1
1
Sol. Let
O
I = —U tan 1
I Example 82 Evaluate f------- , ---j (x-1)7x2+ X + 1
[• dz
2 (z2 + 2)
1 1 dx - dt
Put x + — = t, i.e. 11 x '
I x 2J
-
"J
-2)+2t-l
(t + 2).
41
Chap 01 Indefinite Integral
-J 71t2+dt12 =J dt
t2+2t-3
.
■ - at
(t+2)712 +2t-3
-------------- .
1(1 + 2)
(t + 2)712+ 2t-3
dt - 3 J
dt
(t + 2)7t2+2t-3
=log 11+7i+t21 - - 7i+f2 - - log 11+-ji+t21+c
2
^2i
=1 log 11+7i+121 - - 7i+12+c
1 = 1,-312
dt '
Where, A = f , *
~ and 12 = J
J 7t2 +2£-3
(t + 2) 7t2 +2t-3 f
2
2
7x2 -6x + 10
= | log 1 + ^x2 -6x + 10
I x-3|
I
•*
r(z — l)dz
+C
|x-3|2
Lt
- f
f dt
J 7(1 + l)2 - 4
Put, t +1 = z =
jji + t2 dt
ax2 +bx+c
(f) Integrals of the Form j
(dx + e) 7/x2 +gx + h
dx
Here, we write
zdz
dz
H
ax2 + bx + c = Aj (dx + e) (2fx + g) + Bj (dx + e) + Cj
= yjz2 - 2Z - log I z + 7z2 - 4 I
= 712 + 2t - 3 - log I (t + 1) + 712 +21 +3 I -(ii)
dy
Also, /, = f--------
I Example 86 Evaluate J
\2
/■i 1-2 + 2(1-2 -3
ly J
y i y .
dy _
dy
Put t + 2 = 1 = f
y J ^-2y-3y2
( n
= -L sin 1
2
V3
Where , B, and C, are constants which can be obtained
by comparing the coefficients of like terms on both the
sides.
=>
...(Hi)
A = 1,B = 2,C = 6
r
2x2 + 5x + 9
’■* (x + l)Jx2+x + l dx
(X + 1)(2X + 1) ^ + 2I
(x + l)7x2+x + l
dx
(x-k)r Jax 2 + bx + c
where r > 2 and r g I
,
, 1
Here, we substitute, x - k = t
J+2
1FT dx
X.+
7x2 + x + l
+4 (x + 1)7dxx2+x + 1
! p+5
,
1
where, t = x + —
x
(e) Integrals of the Formj
dx.
or 2x2 + 5x + 9 = x2 (2A) + x (3A + B) + (A + B + C)
2
L3 '
?
+
2t
3 - log (t + 1 + 7t2 + 2t —3)
1=
sin"
(x+1) 7x2 + x+1
Sol. Let 2x2 + 5x + 9 = A (x + l)(2x + 1) + B (x + 1) + C
1 • -i 5 + t
= -7= sin |-----2+t
-V3
-
2x2 + 5x+9
2y + 1
7*2+x+i
= f-^ + 2 f~7______
V“
y](x + l/2)2 + (3/4)
- f
dx
(x + 1)7x2 + x + 1
-dt
+l
[where u = x2 + x + 1 and - = x + 1]
= 2-Jx2 + x + 1 +2-1 log|(x +1 /2) + 7x2 +x + l |
dt
—6 f
I Example 85 Evaluate [--------- /X
j(x-3)37*2-6x + 1O
Sol. Substitute (x - 3) =
J y/(t-l/2)2 +3/4
1
= 2<Jx2 + x + 1 +21og
x + — + 7x2 +x + l
2.
—61og |t-i| + 7t2-t + l + C
= 2a/x2 + x + 1 +21og
x+—
2
=4> dx = - -1 dt
dx
We get, J
- 6x +10
=J
-l/t2dt_________
1/t3 7(1/1 +3)2-6(l/t + 3) + 10
1
-6 log
X2 + X + 1
2(x + l)
+C
42
Textbook of integral Calculus
Type III
I Example 89 Evaluate j
Integration of Type J (sinmx -cos" x) dx
Sol. Let i = J
(i) Where m, n belongs to natural number.
(ii) If one of them is odd, then substitute for term of even
power.
(iii) If both are odd, substitute either of them.
(iv) If both are even, use trigonometric identities only.
m +n -2
(v) If m and n are rational numbers and
2
dx
2 sin x + sec x
negative integer, then substitute cot x = p or tan x = p
which so ever is found suitable.
dv
,
=’f—*—+’ jr —
, where, v = sm x + cos x
-7= sin x + —r= cos x
V2
V2
dx
--------- ---------- + C
sin f x + — 2 (sin x + cos x)
k
=>
- sin x dx = dt
l-t2)tsdt
I=-
.8
1=
8
X
8
6
cos
cos 6 X
6
4
1 , .
f itK)
(
It
Ttl
= —7=log|cosec x+— -cot x+— | --------- ---------- + C
2V2
\ 44J
V
44J 2 (sin x + cos x)
.6
I = f t7 dt - f t5 dt = — - — + c
J
J
8
6
COS
■ i
2
1 .
- 2^2
’ Jj------*---------+c
2v ’ ~
1 ■
1
Sol. I = jsin3 x-cos5 x dx
cos x = t
1 r 2 cos x dx
2 J 1 + sin 2x
1 r cos x + sin x
■fr + ‘ f
5dx
2 J (sin x + cos x)2
2 J (sin x + cos x)z
I Example 87 Evaluate j sin 3 x-cos5 x dx.
Let
<■ cos x dx
J sin 2x + 1
1 r (cos x + sin x) + (cos x - sin x)
dx
2 J (sin2 + cos2 x + 2 sin x cos x)
2 1 sin x + cos x
is a
dx
2 sin x + sec x
+c
Type IV
Integrals of the FormJ xm(a+bxn)pdx
7 = J/?3 (1 - R2)2 dR,
Aliter
Case I If PeN. We expand using binomial and integrate.
if
sin x = R, cos x dx = dR
I = ^R3 dR- J2/?5 dR + J R7dR
CaseII IfPeI
(ie,negative integer), write x = t ,
where k is the LCM of m and n.
1=
sin4 x
4
sin8 x
2 sin6 x
----------+
+C
6
8
Remark
This problem can also be handled by successive reduction or by
trigonometrical identities. Answers will be in different form but
identical with modified constant of integration.
I Example 88 Evaluate Jsin"11/3 x-cos'1/3 xdx.
i „
1
------ 2
3
3
= -3
2
( i_i
11/3 x-cos”'3 x dx i.e.
.. h
Case III
is an integer and P <-> fraction, put
n
(a + b xn) = tk, where k is denominator of the fraction P.
Case IV If
m + l + p | is an integer and P e fraction.
n
We put (a + b xn) = tkx n, where k is denominator of the
fraction P.
I Example 90 Evaluate Jx1/3 (2+ x1/2)2 dx.
1/3
cos“*" X
sin’1/3 x - sin4 x
dx = j(cot-1/3 x) (cosec2 x)2dx
I = j(cot“ 1/3 x)(l + cot2 x) cosec2 x dx
= -Jr’'3 (l + t2)dt = -f(t”'3 +ts/3)dt
[Put cot x = t, => - cosec2 x dx = dt]
_ _ I 3 ^2/3
[2
+c
8
Since, P is natural number.
I = J x1'3 (4 + x + 4x1/2)dx
= J(4x1/3 + x4/3+4x5/6)dx
x
x7/3
7/3 4x”/6
• + -— + —— -{• c
11/6
7/3
4/3
4x4/3
3
= -J-(cot2'5 x) + - (cot
2
Sol. 1 = jx1'3 (2 + x1'2)2 dx
8
x)-+C
7/3 + 24?„1
= 3x4'3 + — X '
7
11
+c
’ •»
Chap 01 Indefinite Integral
I Example 91 Evaluate jx-2/3 (1 + x2^3) dx.
2 3 *16/3 _ 3 f10/3
3 16
10
Sol. If we substitute x = t3 (as we know Pe negative integer)
3t2
or
t2(l + t2)
=>
dt = 3 j
I Example 95 Evaluate jx-11 (1+x4)"1/2 dx.
dt
2
■
+c
= -(1 + x33))8/3
u -l(l + x’)s'3 + c
8^
:. Let x = tk, where k is the LCM of m and n.
x = t3 =»dx = 3t2 dt
43
= 3 tan *(t) + C
t* +1
)
771 + 1
Sol. Here,
----P I=
n-----)
I = 3 tan-1 (x1/3) +C
-11 + 1
1
4
2
= -3
If we substitute (1 + x4) = t2 x4,
I Example 92 Evaluate j x 2/3 (1+ x1/3)]/2 dx.
Sol. If we substitute 1 + x,/3 = t2, then
i
dx = 2t dt
3x2/3
Z=J
...
. / = |L^ = 6Jt2dt=2t3+C
2
t8 + 4t10 + 6t12 +4t14 +tl6)dt
t9 4tn 6t13 4t15 t17
= 6( — +----- +----- +------- f- — > + C
9 11
13
15
17
5
3
1___
Sol. Let
3,
1
,3/6+ — x■5/2
; +—x.17/6
^+C
15
13
17
x + Vx
1
dx.
Vx + Vx
dx
Put x1/12 = t, => x = t12 and dx = 12t" dt
’-I
r 7®
1
—------ • 12 tn dt = 12 f------dt
t4+t3
Jt + 1
T— X
I Example 94 Evaluate Jx5 (1+x3)2/3 dx.
2
Jx5 (l + x3)2/3 dx have m = 5, n = 3 and p = m +1
6 „
[an integer]
n
3
So, we substitute 1 + x3 = t2 and 3x2 dx = 2t dt
Again put(t + l) = y
:. dt = dy = 12 ^y-
J
V
= 12
= J(t2-l)(r2)2/3^tdt
3J
y
8y7
28y6
56y5 + 70y4>
7
6
5
4
56y3 28y2 o
, , ,
-_2- + -£-_8y + log|y|
8
\
= ^f(t2-l)t7/3dt = -f(t13/3'-t
- 7,3)dt
I8 .
-dy
, o r (/- 8y7+28/-56/ + 7°/~ 56y3+28?2- 8y+*)
= 12 -------------------------------------------------------- - —-dy
y
[using binomial]
= 12 J (y7- 8y6+ 28y5- 56y4+ 70y3- 56y2+ 28y -8+ l/y)dy
Jx’tl + x1)2'3 dx = fx3 (l + x3)2/3x2dx
3J
—- —+ t + C
I Example 96 Evaluate J
.11/6, 6
Sol. Here,
1 f2tdt
4 J 77
Wheret = 1 + —V
x4
I = 6 ft8 (1 + 4t2 + 6t4 + 4t6 + t8) dt
1=6 x2/3‘+-x:
11
=fJ ----x11 -x2 (l-+----l/x4)V2
1 j(t2 -l)2 dt = -| j(t4 -2t2 + l)dt
2
1
I = Jt3(l + t2)4 6t5 dt
=6
dx
x"(l + x4),/2
— •“
I Example 93 Evaluate jVx (1 + x1/3)4 dx.
Sol. Here, m = -1 and, n = -i
2
3
Put x = t6 =>dx = 6ts dt
— 4
_ f
dx
J x13(l + l/x44)')1/2“
I = 2(1 + x1/3)3/2 + C
or
■>
1
then 1 + — = r and —— dx = 2t dt
x
x
3
2
Where y = x1/12 + 1
+ C,
/
44
Textbook of Integral Calculus
Exercise for Session 6
3.
■ Evaluate the following integrals
1- J x2(x4 x+4x-12 + 1)V2 dx
dx_______
3-1 (x + 1)l1/3 + (x + 1)1/2
5. J
(x + 2) dx
2-1 (x2 + 3x + 3) 7* + 1
______ dx_______
4-1 (x +a)0/7(x -b)6'7
r
sec x . dx
^/sin (x + 2A) + sin A
6. The value of J [{x}]dx; (where [.] and {.} denotes greatest integer and fractional part of x) is equal to
(a)0
(c)2
(b)1
(d)-1
1 2 (x)+ C, then f(x)can be
• 7. If Jf(x)cos x dx = ^f
(a)x
(b)1
(d) sin x
(c) cosx
8. The value of, J sin x + cos x dx is
9+ 16 sin 2x
(a) -L log 5+4 (sin x - cos x) + C
40
5 - 4 (sin x - cos x)
(b) log
(c)^'°g 5+4 (sin x + cos x) + C
(d) None of these
5-4 (sinx + cos x)
9. The value of J cos 7x - cos 8x dx, is
1 + 2 cos 5x
sin 2x + cos 3x
(a)
+C
2
:3
sin 2x cos 3x
+0
(c)
3~
2
5+4 (sin x - cos x)
+C
5-4(sinx - cos x)
.iT.fTV
(b) sin x - cos x + C
(d) None of these
10. The value of J cos 5x + cos 4x dx, is
1 - 2 cos 3x
(a) sin x + sin 2x + C
(c) - sin x -
sin 2x
+C
2
(b) sin x -
sin 2x
+c
~ 2
(d) None of these
Session 7
Euler's Substitution, Reduction Formula
and Integration Using Diffrentiation
Euler's Substitution, Reduction and
Integration Using Diffrentiation
<5 + 2t2>
Hence, I = j
Integration Using
Euler's Substitutions
( 3t >
3
G + t2,
calculated with the aid of one of the three Euler’s
substitutions
if a >0.
(ii) Jax2 +bx + c =tx+4c, if c > 0.
=-J
ax2 + bx + c = a (x - a) (x - 0), i.e. If a is real root of
(ax2 + bx + c).
Remark
The Euler's substitutions often lead to rather cumbersome
calculations, therefore they should be applied only when it is
difficult to find another method for calculating a given integral.
x dx
t .
f
I? ’
-2 f-5
— + 2t + C,
• • f (•^7x -x10dx- x2)3 —
9
t
J
,
Jlx - 10- x 2
where, t = —--------------x-2
(iii) yjax2 +bx +c =(x - a) t, if
I Example 97 Evaluate /I - J[ , ■ --=—
(7?x-10- x2 )3
Sol. In this case a < 0 and c < 0. Therefore, neither (I) nor (II)
Euler’s Substitution is applicable. But the quadratic
7x -10 - x2 has real roots a = 2, p = 5.
.•. We use the substitution (HI)
i.e.
=J
5 + 2t 2
-dt
27 J t2
-2 -5
_-2 r 5
9 J — + 2 | dt------9 —t + 2t + C
Integrals of the form j f (x), ^jax2 +bx + c dx are
(i) yjax2 +bx +c =
x dx
(^7x - 10 - x2 )3
— 6t
kl + t2 J (l + t22\2); dt
I Example 98 Evaluate j
dx
x+Jx2 — x+1
So/. Since, here c = 1, we can apply the second Euler’s Substi­
tution.
yjx2 - X + 1 = tX - 1
2t — 1
Therefore, (2t -1) x = (t2 -1) x2 => x = -----t2 -1
2(t2 - t + 1) dt
t
dx = and x +
t-1
(t2 -1)2
r — 2t2 + 2t — 2 .
dx
= I------------------ -dt
J t(t-l)(t + l)2
X + yjx2 - X +
7
Using partial fractions, we have
— 2t2 + 2t — 2 _ A
B +
yjlx - 10 - x2 = 7(x-2)(5-x) = (x - 2) f
t(t —l)(t + l)2
t
t-1
C
D
+
(t + 1)
(t + 1)2
(- 2t2 + 2t — 2) = A (t — l)(t +1)2 +'Bt(t +l)2
Where
(5-x) = (x-2)t2
or
or
5 + 2t2 = x(l + t2)
+ C (t - 1) (t + 1) t + Dt
we get A =2, B = -l/2, C = -3/2, D = -3
dt
tt
T n [dt 1 f dt
3 f■ dt
J t
2J t-1 22J
J (t + 1)
(t + 1)2
5 + 2t2
x =------ 1 + t2
5 + 2t
-3f
2
>
3t
-2 t=
1 + t2
J + t2 >
— 6t
dx = /. . *2\2 dt
(x-2)f =
3.
= 21oge| t l-|log<| t-11. -^loge|
t + 11+
—+ C
Z&
2
(t + 1)
r 2
where t =
X -X+1+1
X
Textbook of Integral Calculus
46
Introduction of Reduction
Formulae (a recursive relation]
Over Indefinite Integrals
where, tan x = t => sec2 x dx=dt
I J— “ In-2
n-l
4=
Reduction formulae makes it possible to reduce an integral
depending on the index n > 0, called the order of the
integral, to an integral of the same type with smaller
index, (i.e. To reduce the integrals into similar integrals of
order less than or greater than given integral).
Application of reduction formula is given with the help of
some examples.
Reduction Formula for J sin" x dx
Let In = J sin" x dx = j sin" 1 x sin x dx
I
n
= -sin" 'xcosx +
n-l) sin"-2 x cos2 x dx
= -sin"-1 xcosx+(n-l
= -sin"-1 xcos x+(n-l) I„_2 “(n-l) In
:.nl„ = -sinn~ 1 x cos x +(n -1) In_2
sin"- 1 x cos x n-l
------------+------- In-2
n
n
n-l
Thus, J sin" xdx=——
x cos x +------n-l Jsin"-2
--------x dx
n
n
or
cos"-1 xsin x n-l
Jcos" x dx = ------------------- +
Jcos n-2 x dx
n-------------- n
Reduction Formula for J tan” x dx
Let In - Jtan" x dx
In
n
I
= cosec"-2 x (-cot x) - J(n - 2) cosec"-2 x (cosec2 x -li
=-cosec"-2 x cot x - (n - 2) J (cosec" x - cosec n~2 x);
= -cosec"-2 xcot x -(n-2) In +(n-2) In_2
(n -1) I„ =-cosec"-2 x cot x +(n -2) I„_2
or
In=~
cosec"-2 x cot x
n-l
___ __ n-2
j cosec nxdx = - cosec
x cot x
n-l
n-2 T
+----- T In-2
n-l
n-2 f
+------ I cosec n~2xc
n-l
Reduction Formula for fsec” x dx
Let /„ =J sec n x dx = j sec -2 x sec2 x dx
I
n
—-2 x tan x — J(n — 2) sec n-3
= sec
xsec x tan x-tan xi
x
tan
x
-(n-2)
Jsec"
-2 x (sec2 x-l)(
= sec"-2
= sec"-2 x tan x -(n-2) In +(n-2) Int —2
=> (n -1) In = sec "-2 x tan x +(n -2) In_2
or
In =
n
= cos'*
-1 x sin x+(n -1) I„_2 -(n-i)r„
nln = cos"-1 xsinx+(n-l) I„_2
J tan " 2 x dx
Let I„ = Jcosec"n x dx = j cosec" 2 x cosec2 x dx
Let /„ =j cos" x dx = J cos" 1 x cos x dx
—-1 x sin x +(n -1) Jcos"-2
=
cos
x (1-cos2 x) dx
~I„-2
n-l
Reduction Formula for f cosec" x dx
Reduction Formula for J cos” x dx
i
n
-1 x sin x + J (n -1) cos"-2
x sin2 x dx
= COS
tan"-1 x
J tan n x dx = tan"-1 x
n-l
sin"-2 x(l - sin2 x) dx
= - sin" -1 x cos x + (n -1) J (sin n~2 x - sin" x) dx
9
n
sec"-2 xtanx (n-2)
(n-l)
+(^1) "-2
x dx =
sec.n-2 x tan x (n - 2)
(n-l)
+ (n -1) J sec n~2 x dx
Reduction Formula for f cot" x dx
Let I„ = Jcot" x dx = Jcot n -2 x cot2 x dx
= J cot"-2 x (cosec2 x -1) dx
= Jtan
tan ”-2 x tan2 x dx = J tan"'2 x (sec2 x -1) dx
= Jcot"-2 x (cosec2 “ x -1) dx = J cot n-2
= Jtan"-2x sec2 x-In-2 = J tn~2 dt -1„-2
= Jt""2
x dx
where t =co!
Chap 01 Indefinite Integral
/„=j cot" x dx = -
cot" 1 x
n-1
Reduction Formula forj cos'" x sin nx dx
~In-2
cot"-1 x
=J
jcot n -2 x dx
n-1
cos'" x sin nx dx
n
I
___ m
cos
x cos nx
m
n
n
cos m x cos nx
m
n
n J
m
Reduction Formula for jsin xcos" xdx
let A = sinm-1 xcos"+1 X
dA
„ Q
— =(m-l)sinm"2 xcos”*2
dx
cos'" x cos nx
xcos" X
dA .
_o
=> — =(m-l)sin 2 x cos" x -(m + n) sin"1 xcos" X
dx
Im.n
Integrating with respect to x on both the sides, we get
A = (m -1) jsin”1 ~2 x cos" x dx - (m + n)
cos"" x dx
Jsin"x---sin”1 x cos" x dx =(m -1)
fsin"1-2 x cos" x dx -P
(m “ 1)
Jsin" x cos x dx =---------- sin"’2 x cos" x dx
(m + n)
sin"1-1 xcos"+1 X
m
cos'" x sin nx dx
n
m
m -1
+ — j cos
x sin (n -1) x dx
n
_m
cos x cos nx
m
m
f m, n
-l,n-l
n
n
n
cos x cos nx m
n
sin”1 xcos" x
m+n
f tn, n
or
m,n
(m -1)
(m + n) •* m -2, n
sinm -1 xcos"*1 x
(m + n)
n
m
cos x cos nx
_
...
m
r
m+n
Remarks
• Similarly, we can show
1. loos'” x cos nx dx = cos'” x sin nx + m
m+n
m+n
jcos1
x cos (n- 1) x dx
msin'”"’ x cos /cos nx
2. Jsin01 x sinnx dx = nsin"1 x cos nx
nr2 - m2
m2 - n2
m(m-1) jsirf”2
m 2 - n2
•
f
—+
m+ 1
jsin'i”1 *2 x cos” x dx
•
f
i”1*1 x cos”*1 x
3. jsin”1 x cos” x dx = —
n+ 1
m + n+ 2
n+ 1
jsin”1 x cos”*2 x dx
i”1-1 x cos”*1 x + m -1
4. jsin”1 x cos” x dx = - —
n+1
n+1
i
”
1
jsin' "2 x cos” *2 x dx
sin”1*1 x cosn "1 x
m+1
n-1
m+ 1
jsin”1*2 x cos”■2 x dx
m -n
2
n?(m-1)
Jsin'r "2 x cos nx dx
m2 - n2
+1
sin”1*1 xcos
x cos”
” *1 x . m+n+2
m +1
2
rm - m
i”1*1 xcos”*1 x
n-1
1. jsin”1 xcos” x dx= —-------------- + ------ jsin^xcos” 2xdx
m+n
m+n
x sin nx dx
msin'”-1 x cos x cos nx
3. jsin"1 x cos nx dx =■ nsin"21 x sin2 nx
Similarly, we can show
5. jsin”1 x cos” x dx =
n
-------- + ——7 m -1, n -1
m+n
2 m, n ~
Remarks
2. jsin'” x cos” x dx =
f/n - l,n -1
*
n
m+n
or
x
=> sin x cos nx = sin nx cos x - sin (n -1) x]
m
= (m-l)sin"‘2 xcos" x-(m-l + n + l)
,
-1
cos
[using sin (n -1) x = sin nx cos x - cos nx sin x
-(n +1) sin
n
1 x sin x cos nx dx
cos'"
{sin nx cos x - sin (n -1) x} dx
xcos" x
= (m-l) sin"1-2 x cos" x (1-sin2 x)
=> (m + n
47
dx
g Example 99 Evaluate ln = J
(x2 + o2)n
dx
Sol. Here, In = j
i_____
(x2 + a2)"
(x2 +a2)"
• 1 dx
Applying Integration by parts, we get
(2x)
(-n)-(x)dx
(x2 + a2)"
(x2 +a2)"*’
1
H
x
+ 2n j
(x2+n2)"
x
(x2 + q2)n
+ 2n j
x2
(x2 + a2)"*‘
x2 + a2 - a,2
(x2 +a2)n*1
dx
dx
Textbook of Integral Calculus
48
in =
In =
1
dx
x
dx-2a2nf
+ 2nJ
(x2+a2)"
(x2 +a2)"+1
(x2+a2)"
x
(x2 + fl2F
+ 2nl„ — 2n a2 In
x
2na2In + x =
or
Inf
+ (2n - 1) In
(x2 +a2)"
1
, (2n - 1) 1 T
2n
a2 n
2n a2 (x2 +a2)"
I.
X
Remark
Above obtained formula reduces the calculations of the integral
+1 to the calculations of the integral ln and consequently, allows
us to calculate completely an integral with natural index, as
2 =21 tan-1
J x2 + aa'2 a
2a2 a
1
x
2a2 x2 + a2
1 .
2a3
= —n * "j------ o + 71 ‘ tan
TT .2
_i
a + b sin x
(b + asinx)2
■0
dx,... we follow the following method
, T
sin x
t
cos x
1. Let A —---------------- or A =----------------- according to!
a + b cos x
a + b sin x
integral to evaluated is of the form
(a + b cos x)2.
+ C
dx
0-c
_ 1
x
4a2 (x2 + a2)2
1_
x
+ 3
4a2 (x2 + a2)2 8a4
I Example 101 Evaluate J
__ x
3 . -1
+
tan
x2 + ai:2 8a3
(F
dx
I Example 100 Derive reduction formula for
sinn x
\n.m) ~ J cos'” x dx.
cosm ~1 x
Zn. m ~■ 7
7’
- j (n - 1) sin" " 2
. (m-1) cos
1 x
is required reduction formula.
dx
4 (5 + 4 cos x)
4 J
(cos x)
1 ~2 x
(” - 1) r sinn
dx
(m-1) 1
~ 2
COS
x
TZTi
m
5
1_______ 9
1
4 (5 + 4 cos x)2‘
£
-m t1
sin^"1 x
sin"
1
dA
(5 + 4 cos x)2
r
dx
A=-J1 5+4dx
2
cos x 4 Jz (5 + 4
dx
—*--------- A
-=-f
-JI
2
4
J
5 + 4 cos x
4 J (5 + 4 cos ‘
dx
sin x
=-J
4 J — (1 - tan2 x
(5 4 cos
5+4
(m-1)
m-1
(5 + 4 cos x)2
4
X• COS X■
sin” "1 x
dx
Integrating both the sides w.r.t. ‘ x’, we get
(m-1)
1
dA
(5+4 cos x)2
25
5
— (z cos x + 5) + 4 - —
5 cos x + 4 _ 4
4
we get
sin x
f . n -1
x-----dx
•f(n. m) = J sin
COS
x
I
n
(cos x)
(5+4 cos x)2
dA _ (5 + 4 cos x) (cos x) - sin x (- 4 sin x)
...and so on.
= sin" "1 X ■
dx
sin x
Sol. Here, A — —, then
5+4 cos x
3 /
J?'
Sol. Using Integration by parts for
(a + bsinx)2
3. Integrate both the sides of the expression obtained
step 2 to obtain the value of the required integral
Let n = 2
Z3 = J (x2 + a2)3
*J
dx
dA
1
....
2. Find — and express it in terms of ---------------- or
dx
a + b cos x
1
as the case may be.
a + b sin x
771—is2 + --2ao2 ' A1
2a2 x2 + a2‘
dx
dx
(a + bcos x)7 J (a + b sin x)7-J
.
X
1 4 x
= JL._2L_
+A—
tan
[■
dx
+ C
.-. From above formula
Let n = 1
/,=
f_ft =JL.
2 J (x2 + a2)2 2a2
Integration Using
Differentiation
(n -1)
7
r
77 ‘ l(n - 2. m - 2)
(m-1)
cos x)
x)
/2)
+
x)
(1 + tan2 x/2)
1 + tan2 x/2
sin. J
dx-f----:
=
-f
|2
9
J
9 5 +4 c?.
J (5 + 4 cos x)
9 + tan2 x/2
f
dx
(5 + 4 cos x)(2
2dt
9 J 9 + t2
4
sin x
9 5 + 4 cos x
(where tan x/2
49
Chap 01 Indefinite Integral
=4 (5 + 4dxcos x)2 —109 . -3i tan _xfr| -3
4
sin x
9 5 + 4 cos x
t
tan x/2
dx
10 .tan
=—
(5 + 4 cos x)2 27
1 Example 102 Evaluate f
Sol. Let
=>
= 2j
...(i)
dA _ (16 + 9 sin x) (- sin x) - cos x (9 cos x)
dx
(16+ 9 sin x)2
— _
(16+9 sin x)2
dA
- — (9 sin x +16) + — - 9
9_______________ 9
(16 + 9 sin x)2
dx
dx
9 J
=>
175 r
=>—
9 J
=>
(1 + tan2 x/2)dx
dx
. 16 f
-= A+—
I
2
9 J 16+16 tan2 x/2+18 tanx/2
(16+9sinx):
175 r
dx
A
2dt
16 f
- Tf (16 + 9 sin x)2 = A+ — 16t2 + 18t + 16
[where tan x / 2 = t]
dt
175 f
dx
= A + 9*JJ
=> ---- 9 J (16 + 9 sin x)2
t*+’t+l
8
dt
=>
=>
(2a + sin 2x)2
dA
- 4a sin 2x - 2
dx
(2a + sin 2x)2
dA
- 4a (sin 2x + 2a) - 2 + 8a2
dx
(2a + sin 2x)2
dA
4a
"fa
(2a + sin 2x)
(8a2 - 2)
t
(2a + sin 2x)2
A = - 4a f------ --------- + (8a2 -2) 7,
(2a + sin 2x)
241
(+’ I ++ -----16
I 16 J
= 71+2 j
t2+-t + l,
a
dt
(+l
2a
cos x
9
=* I (16 + 9 sin x)2 175 (16+ 9 sin x)
=A+2
2
16 tan x/2 + 9
+C
+ (175)3'2 tan"1
7175
t
Sol. Here, I = [--------- —--------J (sin x + a sec x)2
'-a
(2a)
tan 1
yjia2 -1
cos 2x
=>(8a2 -2)1,= ------------------- +
2a + sin 2x
,■
(
(2at +1)
\
v
4a
--
.-1
dx
I Example 103 Evaluate ---------------J (sin x+osec x)
when |o|>1/2.
or I = j
cos2 x dx
(sin x cos x + a)2
cos2 x dx
=La + 2a sin x cos x + sin x cos x
2
dA _ (2a + sin 2x) (- 2 sin 2x) - cos 2x (2cos2x)
= A + ^[- dt___
2
2
cos 2x
2a + sin 2x
2
, r
sec2 x dx
=>(8a2 - 2) 7, = 71 + 4a -------------------------- —
J 2a + 2 tan x + 2a tan x
16 ♦
-i 16t + 9
2
= 71 + -x 7= tan
7175
9
'175
dx
_
-.(i)
Integrating both the sides w.r.t. ‘x\ we get
9 J
=A +J
9 J
(2a + sin 2x)2
A = J (2a + dxsin 2x)2
dx
...(h)
(16+ 9 sin x)2
9
cos 2x dx
Jr
Integrating both the sides of Eq. (ii) w.r.t. ‘ x’, we get
dx
dx
! 175 r
A = --f 16 + 9 sin x
(2a + sin 2x)2
Put
_16
1
t
175
9 (16 + 9 sin x) 9(16 + 9 sin x)2
dA
+ 2j
=> I = 27, + f — [where (2a + sin 2x)=t, (2 cos 2x) dx=dt]
where
•
dx
dx
=> 1=2^- ------- --------+ C
(2a + sin 2x)
- 16 sin x - 9
dA
■■
•J =>
dx
(16+ 9 sin x)2
cos X
A =-------------16 + 9 sin x
~2
1
a +asin2x + —sin 2x
4
f
4 cos2 x dx
_2 r (1 + cos 2x)dx
(4a2 + 4asin2x+sin22x)
(2a + sin 2x)2
4
sinx
+C
9(5 + 4 cosx
3
cos
cos"2 x dx
[■
" ■* ~2
*22
tan-1
From Eqs. (i) and (ii)
1
cos 2x
1=
(2a tan x + 1)
.
-1
...(ii)
7
4a
+
(4a2 - 1)3/2
:
\
2a tan x + 1
tan-1
- -------- -------- + C
(2a + sin 2x)
(4a2 - 1) (2a + sin 2x)
/
50
Textbook of Integral Calculus
JEE Type Solved Examples:
Single Option Correct Type Questions
• Ex. 1
dx
The value of^
(a) tan-1 (2 cot 2x) + C
• Ex. 3
, is equal to
cos6w x + sin6
x
+ e2x
=J
(1 + t2)(i-r2 + r4)
1 + t2
“J
=J 1 -12 + t7*
‘
„
1
Put t — = z
t
(
1^
So/. J = j
e 3x
1 + — | dt = dz = j
•■•Hr
2 -
fl
du
l1 + ?
dt
r
J (t-l/t)2 + l’
-J
. -J
1
1 —-\du
k___ iZ
:z J
dt
f
t2 -1
u+ —
tan x
+c
+c
e2x +
+C
i
U,
= |l°g e2ytan2 x -1
t =u+ -
-1
u2 — u + 1
„
11
t-1
1 .
+ C = - log
= -log
u2 + u + 1
t+1
2
= tan-1 (z) + C
+ C = tan 1
(“ = '
du
1 + u2 + u4
\___ tr
. . 1
1 + -z + U2
u,2z
‘
(u2-l)
dx= |
+ e4y
z‘ + 1
I '>
— ^X
j-i=! 1 +
dt
(f2~1>
= tan-1
b7
1 + e2* + e',4x
(e3y-ey)
1
(1 + 1 / t2) dt
(1 / t2 - 1 + t2)
dz
+ex+r
U2'
Put tan x = t => sec2 x dx = dt = J
(1 + t2)2
/ 2x
+?
— dx
1
—6
+ e-2x + 1
f 4x - e2x
1
e
. . 1 ,
e
+i
+ C (b) — log
(a) - log
e4x + e2x + 1>
-ex +\
2
\e
'eAx + e2x + ?
(e2x- e" +1
1
, . 1 ,
+ C (d) -log
(c) - log
- e2x +1,
2
+ ex +1
dx
_ r sec6 x dx
Sol. Let I = J
cos6 x + sin6 x J 1 + tan6 x
1 + tan6 x
e~x
Then, for an arbitrary constant c, the value of J -1 equals t
[IITJEE20i
(b) tan-1 (cot 2x) + C
(c)tan 1 I-cot 2x1 +C (d) tan 1(-2cot2x) + C
k2
)
(1 + tan2 x)2 sec2
ex ------- dx, J = [
Let I = f ——
J e x +e
+1
J e~Ax
Hence, (c) is the correct answer.
= tan-1 (- 2 cot 2x) + C
• Ex. 4 Integral of^1 + 2 cot x (cot x + cosec x) w.r.t. x, i
Hence, (d) is the correct answer.
-tan
y
x
(sec-1 Ji + x2 )2 +cos
• Ex. 2j —
(x > 0) is equal to
(a) e,an ' * • tan'
<1+j.
dx,
(a) 2 In cos — + C
2
1
x
(c) - In cos — + C
2
2
(b) 2 In sin — + C
2
(d) In sinx - In (cosec x - cotx) + <
Sol. I = j -Jl + 2 cosec x cot x + 2 cot2 x dx
x+C
= j -Jcosec2 x + 2 cosec x cot x + cot2 x dx
etan * -(tan-1x)'
-+C
(b) —
= j (cosec x + cot x) dx
2
x
-(sec
-,(7l+x2))2 + C
(c)e,m'
=
(d) e,an x •(cosec-1 (-^1 + x2 ))2 + C
f1 + COS X
f
(x'i
------------ dx - cot | — | dx = 21og sin — + C
sinx
J
\2j
2
J
Hence, (b) is the correct answer.
Sol. Note that sec-1 -^1 + x,22 = tan 1 x, cos'
1 -x2
1 + x2
For x > 0
tan
I
Put tan
x=t
J 1 + X.2
• Ex. 5 lfln = | cot" x dx, then l0+f+ 2
(/2+/3+....+ /8) + /9+/io equals to (where u =cotx)
;
=1—
=*
= 2 tan' x,
{(tan-1 x)2 + 2 tan-1 x} dx,
= Je'(t2 + 2t) dt = e‘-t2 =etan
Hence, (c) is the correct answer.
y(tan-1 x)2 + C
2
9
(a) u + — + ....+
u
2
9
, .
U
(c) — u + — +
I 2!
/
/. x
u
U
2
U
9A
b)- u + — + ...+ —
U
9\
1
.u
2
2
2u2
9J
9u9
d) - + — +....+ —
Tl (d) ~ + — +
2
3
10
Chap 01
Sol. I„ = J cot" x dx = | cot" "2 x-(cosec2x - 1) dx
In ~~
iZ1"1
~ ~ ln-2
In + ln-2 ~ ~
n-1
u"'1
n-1
10]
51
sin 0 cos 0 dQ
f(b sin2 0 - a sin2 0) (b cos2 0 - a cos2 0)
= 2 (b - a
[put n = 2,3, 4,
Indefinite Integral
. r
• n r
sin0COS0d0
._
= 2 (b-a) -------------------- = 2 I ld0
I (b - a) sin 0 cos 0
I2+ A - _ ~
= 20 + C = 2sin-1
u2
h+h=-~
b-a
Hence, (a) is the correct answer.
(X“1)
.
j
• Ex. 8 The value of^ ---------- ,
= dx, is
(x + 1) fx3 +x22 +x
■
■
Ao + /9 = Adding, Io +
uf_
(a) 2 tan
Sol. Let
x
I=j
J (x + I)2 fx2 + X,22 + X
(n 1 V
function off. The value of g' —I- —j= has the value equal
x2 (1 -1 / X2)
=J (x2 + 2x + 1) 7x3 + x2 + x
2
(c)2Sol.
where
.2
(
(b)
•* x (x + 2 + 1 / x) • x fx+l + l/x
f(x) = y = x + sinx
dy
— = 1 + cosx
dx
1
1
+
cosx
dy
It 1 1
it
n
y=—+
= x + sinx => x = —
4
fi
4
v2
'
1
, Tt
1
f2j 1 +(I/5/2)
= y/2 (f2 - 1) =2 - y/2
=
y2 + 1
Hence, (c) is the correct answer.
• Ex. 7 The value o/J
dx
f(x-a)(b-x)
(a) 2 sin
(c) sin
b-a
-+c
(b) 2 sin
x2 (1 - 1 / x2)
<■
2
(d) V2 +1
2
(x-1)
.--------------dx
(x + 1) fx*.3 +■ X2 + X
-f----- -----------dx
• Ex. 6 Let f(x) = x + sin x. Suppose g denotes the inverse
(a) 41 -1
(d) None of these
1———-f-
Hence, (b) is the correct answer.
to
X
X2 + X + 1
(c) 2 tan
9J
■
x2 + x + 1
------------ + C
X
9
+ 2 (I2 + I3 + ....+ /8) + A + /10
(
2
9\
I
u
u
= -« + — + + —
I 2
(b)tan
, is
x-b
------ + C
b-a
(d) None of these
b-a
Sol. Let x = a cos2 0 + b sin2 0 in the given integral.
So that, dx = a (2 cos 0) (- sin 0) + b (2 sin 0) (cos 0) d0
dx = 2 (b - a) sin 0 cos 0 df)
<•
2 (b - a) sin 0 cos 0 d0
/'- cos.22 Q + b sin2 0 - a) (b - a cos2 0 - h sin2 0)
f(a
n
Put
dx
1
X + - = t,
X
=>(1 -1 / x2) dx = dt
dt
, which reduces to J
■
Let t + 1 = z2
dt = 2zdz = j
2z dz
(z2 + 1) Tz2
= 2 [ ■ -dz
aZ ■ = 2 tan 1 (z) + C
J z2 + 1
= 2 tan"1 (ft + 1) + C = 2 tan"1
Hence, (c) is the correct answer.
x2 + x + 1
+C
x
(1 + x2)dx
• Ex. 9 The value of J
=, is
(1-x2)a/1 + x22 + x 4
Vx4 + x2 + 1-Tx
(a)7'°s7x4 + x2
+C
x
2
(b) -7og 7x
. 4 +x
2^3
!x
+C
5/x4 + x +1 - -fix
Vx4-x2+1-V3x
(c)70g7777
(d) None of the above
x
+c
52
Textbook of Integral Calculus
So/. Let ,
cos 0 d0
(a2 4- b2 sin2 0) • cos 0
(1 4- x2) dx
/=J
J ---(l-xFf
2)Vl 4- x2 4- x4
= ^f------,
/
1
dx
__________ x2
x2 fl
-J-(x"X)Xl/? + 1 + l2
a
J a2 4-b2 sin2 0
dividing numerator and denominator by cos2 0, we get
X
r- r
sec2 0 dQ
= da —---- -------- ------ —, put tan 0 = t
J a2 sec2 0 4- b2 tan2 0
(14-1/ x2) dx
=-J (1-1/x) 7(x - 1 / x)2 4- 3
dt
a2 4- b2 J
Again, put t2 + 3 = s 2
2t dt = 2s ds = - j
s ds
r
ds
s (s2 -3)“_J s2-G/3)2
= -^10g
f
tan-1
t
ft + b2'
s-fj
^Tb2
-jt2+ 3- f3
xfa2 + b
1
•tan 1
4-C
yja(a2 4- b2)
jifb-ax2 j
-Jl2 4- 3 4- fi
a
7
a
\
7
v l=tan0 =
X
^b-ax2
Hence (a) is the correct answer.
-J(x - 1 / x)2 4- 3 - d3
■J(x — 1 / X)2 4- 3 4- fi
•i'.v
+c
dx
• Ex. 11 The value of I = j
2x t/1 - x ij(2 -x)+yfi-x
1 , (
3 r~2---------- \
1
1
rv~'—
= -—< logl z+—+fz +3z+3 I • +— logs- — + fs -s+1 4-I
3
=-^10g
2
Jx2 4-^41 +
V
x
=’^10g
a2
a2 4- b2
?+
y/a
=-^J,og s 4- fi + c
=-^510g
dt
(a2 4- b2) t2 4- a2
r >f a2{l + tdt2) + b2t2
1
dt
Put x----1 = 1 = 11 4—- | dx = dt => - J
x2
tyjt2 + 3
X
\
;;
2
2
•
k.....................
■Jx4 4- x2 4- 1 - fix
-Jx4 4- X2 4- 1 4- f3x
.Y.J- •
ands - z =—, then value ofk, is
x
4-C
Hence (a) is the correct answer.
(b)2
(a)1
(d)4
(c)3
dx
Sol. Here, I = j
2x fl - x -J(2 - x) + fl- x
i ' ;/A
0 ■
dx
• Ex. 10 The value of I = J
= , is
(a + bx2y)fb -ax 2
1
tan
(a) ■Ja(n2 4-b2)
(b)
x
\aylb-ax2‘
+c
put (1 - x) = t2 - dx = 21 dt
=-J __2 (121-dtI2_______
) • 1 4-12 4-1
=-[—*
J (1 - I2) ^t2 + t+l
X
1
tan
7(»!+b!)
(xJTTb2 +c
afb-ax2
V
(c) .
1
tan
yja(a2 +b2)
( xfa2 4- b2
I
a
dt
1
-J _________
(t-iMt + vJi2 4-14-1 2
if 1
i .
+c
2
(t-D(r + i)
Let
'
cos 0 dQ
dx =
jl^
sade
I=!r
\
a 4- — sin2 0
a
b - b sin2 0
dt
^jt2 + t+.
_______ 1
4*7
Sol. Substituting ax2 = b sin2 0
1
14-1
1
14-1
1____
-1) y{t2 + t +1
(d) None of the above
1
1-1
2 J (t + 1) -yjt2 + 1 + 1
1
2
dt
where,
'■=J (1-1) ylt2 + t+ 1
and
;2“J
dt
(14-1) -Jr2 + * +1
For Iit put (t -1) = i => dt = dz
z
z2
dt
Chap 01
r ______ dz
-1 /z2 dz
'■=J £
1+i |+1
z>
1+z
z
• Ex. 13 iff
.(ii)
dt = -~ds
s
y + C. Then^i|m - n| is equal to
(d)1
(c)2
-(0
-.(ii)
1
~2(2x)
x dx
(x2 + a2)2
(x2 + a2)3
I
n
x.24 + a2 - a 2
x
+ 4j
- dx
(x2 + a2)2
(x2 + a2)3
x
dx - 4a2 J dx
+ 4j 1
(x2 + a2)2
(x2 + a2)2
(x2 + a2)3
x
=> A =
+ 4Zi - 4a2 • I
[using Eqs. (i) and (ii)]
(x2 + a2)2
x
+ 3/j
=>4a.2zI =
= f—!— • 1 dx =
1
s—
+4
. 2 >I
j (x2 + a2)2
(
1
s—
•(iii)
2
I
4a2(x2 +a2)
and Ii = ( 2 1 2 , dx
J (x2 + a2)2
ds
= -log
(x2 + a2)3
(b)3
dx
(x2 + a2)3
So/. Let
0
=>
X
1
+ —r tan
2a2(x2+a2)I 2a3
(a) 4
For I2, put (t + 1) = |
m
dx
x
= -log | z + - ] + Jz2 + 3z +3
V
2J
53
Indefinite Integral
z + | + 7zZ+ 3z + 3J j
1 ,
+ -log | s - -| + Js2-s + 1 +C
k
2J y
_1____
1
and s =
where, z =
71 - x + 1
71 - x -1
1___
i
= -=>k = 2
s-z =
rl-"x-l X
’ (x2 + a2)2
1=
Hence (b) is the correct answer.
x
3
4a2 (x2 + a22))'2 + 4a2 1
(Hi)
[using previous example,
• Ex. 12 Iff
dx
x
1
+ — tan
A-J-A
J (x2 + a2)t
2 - 2a2 (x2 + a2) 2a3
(x2+a2)2
lx
1
_! x
r {- tan — + C. Then the value ofk, is
o.
a
ka2’ lx2 +a
i
1
= -tan-1 —+ C '
J x2 + a 2 a
a
1
__ 1
J
■1 dx=
x2 + a2
x2 + a2
x
= x2 + a2
J
x2 + a2 +2J
X
-2x
xdx
(x2 + a2)2
• Ex. 14
x2 + a2 - a.2
Ify (x-y)2 = x, then
r
dx
’(x-3y) = — ln[(x -y)2 -1].Then(m + 2n) is equal to
n
+ 21 (x2 + a2)2— dx
U
dx
-2a2 J dx
(x2 + a2)2
x2 + a2
■■•(ii)
(b)3
(a)1
Sol. Let
x
7T7
=3
or
dx
f (x2 + a2)2
dx
+ 2- tan-1 --2a2 f
a
a
J (x2 + a2)2
X
1
J (x-3y)
dP
dx
1
x-3y
2
(x-y) >4
______ [
dx
{(x-y)2-i]
-1 X
- + -tan *a
X2 + a z a
1
-i x
1
x
+ - tan — ■ + c
a
2a2 x2 + a2 a
1
_] xl .
x
1
+ -tan —>+C
a Ika2 [xv2 +. a„2 a
k =2
Hence, (b) is the correct answer.
(d) 7
(c)5
2-l]
P=f-^-== lto[(x-y)
1
From Eqs. (i) and (ii), we get
- tan'r’* =
a
a
2a2 j dx
(x2 + a2)2
01-
m = 3 and n = 4
|m-n|=|3-4|=|-l| = l
Hence, (d) is the correct answer.
I
dx
x2 + a2
x
1
+ —- tan
2a2(x2+a2) 2a3
3
+c
...(iv)
f—
Also,
x
4a2(x2+ a2)2
(d)4
(c)3
(a)1
(b)2
So/. Here, we know
=> 1 =
X
Given, y (x - y )2 = x, on differentiating both the sides, we get
dy = l-2y(x-y)
dx (x-y)(x-3y)
dP
dx
(x-y)
t
l-2y(x-y)
(x-y)(x-3y)
{(x-y)2-!}
••■(ii)
Textbook of Integral Calculus
54
(x-y)(x-3y)-14-2y(x-y)_
(x-y)-i
(x-3y) {(x — y)2 — 1}
(x -3y) {(x-y)2- 1}
r-.
•
” dx
1
-(iii)
(x-3y)
• Ex. 16 If [ ■‘C—-■ dx, where f(x) is a polynomial of
J x -1
degree 2 in x such thatf(Q) = /(1) = 3/(2) = - 3 and
j /(x) dx = - log | x -114- log |
x3 -1
which is true as given
= ;log«x-y)’-l).
J-*J (x-3y)
2x4-1 j
m
4- —j= tan
75 J 4- C. Then (2m 4- ri) is
y/n
:.
m = l,n=2
=>
m 4- 2n = 5
Hence, (c) is the correct answer.
(a) 3
(b)5
Sol. Let
(c)7
(d)9
/(x) = ax2 4- bx 4- c
Given,
• Ex. 15 If j(x4--Jl 4-x2 )n dx.
1
=----- ------ {x 4- 714-X2}n+1 4a(n 4-1)1
1
-6(n-1)
+C
Then (a 4- 6) is equal to
(a) 2
(c)4
(b) 3
Sol. Let I = j (x 4-
(d)5
X 4- ^/1 4- x.22 = t
/
(i)
X
1 4- ----r1 .
• 2x
2^ + x2
J
dx = dt
dx = dt
-(ii)
Idx
________dx
(x+l/2)z + (f3/2)I2‘
•o'
2
2x4-1
= - log I x-l I + log I x2+ x4-1 |+-=tan'
+c •
V3
/. On comparing m-2,n =3 => 2m + n=7.
Hence, (c) is the correct answer.
/
— = x - -J 1 4- X2
t
’
Subtracting, we get
1__
2t
71 4- x.2
t2 4- 1
...(iii)
From Eqs. (i), (ii) and (iii), we get
,
t2 4- 1
dx =---- — dt
2t2
t2 + l
z = J f • (2t2) dt = ^ J (t" 4- t"~2)dt
1
2
n4-l
J x2 + x 4 1
= - log| X-l [4-log )x2
-1___
t =---x - Jl 4- x
2 Jl 4- x2 = t 4- - or
t
Bx + C
(x2 + x + 1)
(x - 1)
J X2 + X + 1
x-
1
|
we get, A = -1, B =2, C =2
1
(2x 4- 2)
dx 4- j
dx
(x2 4- x + 1)
x-l
i------- 7 x 1 4- X2 X-----
+ x2
We know, t = x+
Using partial fractions, we get
(x2-x-3)
A
(x - 1) (x2 4- X 4- 1)
f yj}+ X2 + X A
< 71+*2 /
x2 - x - 3
dx
(x -1) (x2 + x + 1)
/(x)
x3-l
4- x2)” dx
Put
I
/(0) = /(l)=3/(2) = -3
/(0) = c = -3
J(l) = a 4- b 4- c = -3
3/(2) =3(40 4- 2b 4- c) = -3
On solving, we get a = 1, b = -1, c = - 3
/(x) = x2 - x - 3
f"1
+----n-1
I =-------- I [x + 7(1 + x.2^’
2 (n 4-1)
1
+ 2(n-l) (x+7(l + x2))""1 + C...(iv)
Then comparing the values of a and b by Eq. (iv) a = 2, b = 2
(a+ b) =(2 4- 2) = 4
Hence, (c) is the correct answer.
(1 + x)
• Ex. 17 The value ofj
x(!4-xe*)2
x
(a) log
14- xe*
xe*
(b)log
14- xe*
xe*
(c) log
14-e*
dx, is equal to
4------ ------ 4-C
(14-xe*)
1
4---------—
4-C
X
1 4- xe
1
4----------- + C
X
14- xe
(d) None of the above
Sol. Let I = j
(14-x)
dx = | (1 4- x) e*
dx,
x(l 4- xex)2
(xe*) (1 4- xex)2
put 1 4- xex = t
(1 4- x) ex dx = dt = J
we get
dt
applying partial fraction,
(t - 1) • t2 ’
1
ABC
------------------- ----------- *■ ~2
(t ~ 1) t2 t-1
t
t2
Chap 01 Indefinite Integral
=>
1 = A (t2) + Bt (t -1) + C (t -1)
For
For
t = l =>A = 1
t = 0 => C = -1 and B = -1
<[^-i--U<* = log|t-l|-log|f| + 7+C
:. I
t
J [t-i
1 . = - sin
2
1
+C
1 + xe
1
+------- + C
1 + xe
xe
1 + xe
1 . _
= -sin
2
t
rj
= log | xex | - log 11 + xex | +
= log
11 1
X
+ -log - +
a
2z
,2
|log|x+7a2-x2 l + G
Hence, (b) is the correct answer.
(x2 +1) 7x44 +1
, is equal to
X2+l1
+C
Vix 7
1
(a) —f= sec
+
log | x +
(b) — sin-1
2
-
log I x + -]a2 -X2
■42
z \ 1
(c) —/= cosec
x2+?
V2
2
(c) - sin 1
2
Sol. Let I = J
X1 1
(d) - cos 1 —
+2
a) 2
'a2
dx, is equal to
1 t cos 0 + sin 0 + cos 0 - sin 0
(■ cos 0 d0
- = - I--------------------------------- do
J sin 0 + cos 0I 2 J
sin 0 + cos 0
1-
| dx = dt
X )
dt__
1
= -7=sec
't2-2 ■42
= 1 fld9 +1 rcosfl-sine
=
= - • 0 + - log (sin 0 + cos 0) + C
2
2
+ C (d)V2 cosec
(1 -1 / X2) dx
Put x + — = t
x
2 J sin 0 + cos 0
1 •
sec
■42
x2 + f
Vix ,
+C
+C
Hence, (a) is the correct answer.
JEE Type Solved Examples:
More than One Correct Option Type Questions
• Ex. 20 j
^4 + x 2 J X(4 + x2)3'2 (fix2 -6)
+ C,
-dx................. :s
X6
X
then
4_
t= 1 +
=> t2 = 1 + .2
V
x
x‘
g
2t dt =---- r dx
Put
X
(a) A = —
120
(c) A = -
(b) B = 1
J_
I=— f(t2-t4)dx = —■ p
16 J
16 3
(d)B = -1
t+4
Sol. Here, I = j
X
X6
X5
4+c
5
1 (4+x2)372
(x2-6) + C
~ 120
x5
120
1+----- ^-dx
x2x3
x2 + ?
+C
7
RHP
----- •, Put x = a sin 0
a cos 0 d0
dx = a cos 0 d0 = J
a sin 0 + ^a2 - a2 sin20
(b) Vi sec
x2*? + C
Vix }
(x2-l)
x2 (1 -1 / x2) dx
dx = j —
(x2 + 1) ^x4 + 1
' 2
1
.2 (x + x2 + X
\
x
r
7? -x2
2J
X2-1
• Ex. 19 The value of j
(a) - sin 1
2
Sol. Let I = J
where Q = C - - log a
2
-X2
x+
dx
- x2 |}log a + C
it
Hence, (a) is the correct answer.
dx
♦ Ex. 18 The value ofj-----
+C
2'
■(a
55
A = — ,B = 1
120
Hence, (a) and (b) are the correct answers.
56
Textbook of Integral Calculus
• Ex. 22 If I = j (Vtanx +7cotx) dx = /(x) + c, then
• Ex. 21 The value of the integral
J e51”2 * (cos x + cos3 x) sin x dx is
/(x) is equal to
(a) 2e-'"’ x (3 - sin2 x) + C
(a) fl sin”’(sin x - cos x) (b) ~^= — fl cos-1(sin x - cos x.
fl
(b) esin2* 1 + -1 cos ■> x 1 + C
2
(in2 x
(c) esl
tanx - 1
n
)
(3 cos2 x + 2sin2 x) + C
(d) None of these
fl ftanx
Sol. I = f (-7tanx + Vcotx) dx = f f2 • s^ny + cosy
J
J .
f2sinx cosx
(d) esi"! x (2cos2 x + 3 sin2 x) + C
If sin x - cos x = p, then (cos x + sin x) dx = dp
I = f2 f -7-^—— = >/2sin-1 p + c = fl sin-1 (sinx - cosx) +
Sol. Put t = sin2 x
1
3
The integral reduces to I = - J e‘(2 -t)dt = -ce‘ - — + C
2
2
2
= ie“2x(3-sin2x) + C
[option (a)]
2
1
■>
1
= e rin2x 1 + - cos x + C
2
)
[option (b)]
n
sinx-cosx
^2 cos-1 (sinx - cosx) = f2 tan'
fi
fl-(sinx - cosx
^nJC ~ cosx =
f2 sinx cosx
= V2 tan
tanx - 1A
V2 tanx )
tan'
Hence, (a), (b) and (c) are the correct answers.
Hence, (a) and (b) are the correct answers.
JEE Type Solved Examples:
Passage Based Questions
,
2
2
dx = dt
Put x + — + 1 = t => 2
x2
x .
_1 rdt_ 1 t'12
— — I —7= = — ------ + C = ft + C
ft 2 1/2
~ 2 J Vt
Passage
(Q. Nos. 23 to 25)
I
a
aa 1
x----- • 1 + — dx, put x---- -t
*) \
X2 J
X
For integral j f^x + —a A
a
For integral
,--]dx,putx+--t
For integral J /(x2
For integral
a A (
x2
X2
a 1 ,
2
a
x---- - ax, put x + — = t
x3 J
x2
many integrands can be brought into above forms by
suitable reductions or transformations.
. Ex. 24 f----- <^L=
X
x4 -2
4
(a)tan
X + —+ 1I + C
(b)tan
(c) 2 tan
Jx + - + 1 + C
(d) None of these
xq +xz +2
(a) x2 + 1 + -~ + C
V
x
^+C
(c) x2
(b). x2+1 + A+C
V
x
(d)
)
X
x
x2 -1
,
f
Sol. J -----------.
dx =
.2
(x +1)2 ^x3 + x2 + x
J
dx
2
,
dx
J (x+1)Vx3 +x2 •
V
• Ex. 23 J-2
2
XZ
Hence, (b) is the correct answer.
a V
2
a
x + — ax, put x---- - = t
X3 J
X
r.
V
= x2+—+1+c
x2
+4
+c
x
Put X + - + 1 = t2
X
1
x+—+1+c
x
H)
dx
1 n
x+-+2
x+-+l
X
X
I dx = 2t dt
x J
_ f 2tdt = 2 f—— dt
J (t2 + l)t
J (t2 + 1)
2
Sol. Here, I = j
x'7
dx
] .22
2
x + 1 + -7
X
= 2 ■ tan-1 (t) + C = 2 tan
Hence, (c) is the correct answer.
x + - + l Uc
x J
Chap 01 Indefinite Integral
Divide numerator and denominator by x10, we get
5x4 +4x5
Ex. 25 J (x5+x + l)2 dx
I
(a) x5 + x + 1 + C
(b)
"5 +C
(d)
(c)x-4
Sol. Here, I =
57
x5
X5 + X + 1
+C
Put 1 + x
_ (• 5x~* + 4x-5
J (1 + x-4 + x-5)2 dx
= t =>(-4x-5 -5X-6) dx = dt
+X
=l + C =___
1
I = _f^
J t2 t
1 + xM
X5
-- + C
x5 +x + 1
+ x~5
+C
___ /___
+C
x5 + X + 1
1 5x4 4- 4x5
(x5 + x + I)2
Hence, (d) is the correct answer.
JEE Type Solved Examples:
Matching Type Questions
. • Ex. 26 Ifxe (0,1) then match the entries of Column I
with Column II considering c’ as an arbitrary constant of
integration.
Column II
Column I
/
(P) |^4 + C
(A) J tan 2 tan
=-------—X
x - Vx
(B) I = j cot 2 tan'
dx
X + >/ x
7
(
Gc - Vx
= cot 2 tan'
X + Vx
k
/
cot 2 tan
\
’
7
Isec 0 - tan0
sec 0 + tan0 ,
= cot (2 tan-1 -J(sec 0 - tan 0)2)
7
(B) | cot 2 tan
(q)
r^ + C
\
"1 J'
1 - tan -sin
1-Vx
1 + y/x
' "1^
(C) J i
1 sin•
1 + tan f—
(r) -x374 + C
3
dx
= cot | 2 tan
1 - sin 0^1
COS0 J
If 0 g(0, n/4), then sec 0 - tan0 > 0
it _ 0A
f It n 1
= cot | 2 tan' tan
= cot---- 0 = tan0
\2
J
4 2)
I = j tan0- 4tan3 0 sec2 0 (30
'1+Vx-1
(D) jVxtan 2 tan
71 + a/x-i
Sol. Let
= - tan5 0 + C = — (x5/4) + C
5
5
dr (s) -xy4 + C
5 •
x = tan20
-i .
x = tan4 0 => tan 0 = x1/4
Vx€(0,l)
dx = 4 tan3 0 sec20 <30
-Jl + Vx = sec 0
/
r
(A) I = j tan 2 tan'
x
x -1
=> tan 2 tan'
0 e (0, n / 4)
=—\
x -1
0
tan 2.
= tan0
I = j tan0• 4 tan30 sec2 0 (30
(it
= - sin-1 (cos 20) = - sin' sin | — - 20 | = — - 0
2
2
U
4
/
1 - tan - sin-1
1 + Jxf
dx
(
1 . -1 1 -VxY
1 + tan - sin
I2
I2
X
1 - tan I — -0
7
1 + Vx
= tan | 2 tan'
dx
/
=tan 2 tan'
= - tan5 0 + C = - (x5/4) + C
5
5
1 . . 1 - tan20
(C) - sin
-sin
2
2
1 + tan20,
4.
Isec 0 -1
sec 0 + 1 y
=J 1 + tann---- 0_ 4 tan30 sec2 0 <30
4
= J tan n
4
*-0
.4
= 14 tan4 0 sec2 0 dQ
4 tan3 0 sec2 0 (30
58
Textbook of Integral Calculus
f
= tan20 -tan 2 tan
= -tan50 + C = -(x5/4) + C
5
5
(D) Let Vx = tan20 => x = tan4 0
0
. 0Y|
cos — sm 2
2
= tan20- tan 2 tan
0
. 0
cos - + sin 2
2JJ
n 0Y)
= tan20- tan 2tan tan
4 2)j
dx
v xe(0,l)
/. 0 e (0, n /4)
/
2 tan
^7? + Vx -1
Vx tan 2 tan
J = J tan9-4tan30secz0 d0= -
+ Vx -1 >
x +1 +
(
= tan20 • tan I — -0 | = tan2 0• cot 0 = tan 0
k2
J
7
/
k7sec 0 + 1 + fsec 0-1 f
/
dx = 4 tan3 6 sec2 9 dQ
x = sec 0
(
' fsec 0 + 1 - fsec 0 - 1
- + C = -(x5/4) + C
5
(A) -»(q); (B) -> (q); (C) -»(q): (D) -> (q)
JEE Type Solved Examples:
Single Integer Answer Type Questions
• Ex. 27 If the primitive of the function
/(x) =
.
x2009
(1 + x2)1006
Integrating,
..if x2
.
w.r.t. x is equal to n ! + X2 y
+ C,
then — is equal to
m
x = 0,^=C => C = 1
Put
r(o)
Z(x) = f (1 + x2)1006 dx
Put
,2
1 + xi = t =>
-
2x dx = dt
Put
1
j
1
2
1
1 -t,
t-1
|
t 1
I 7a
„
+ c=——
2010
/(X)
/"(x)
/'(x)
x2
1 + x2
1005
I +c
11
then^- is equal to
U
Sol. (1) I = j {sin (100 x + x)-(sinx)99} dx
= j {sin (100 x) cosx + cos lOOxsinx) (sinx)"dx
i
0
[/(x)]2-/(x)r(x)^
[/(x)]2
n
sin (lOOx) (sinx)100
100
-
equal to
f(x)
/(X)
Again, integrating, In [_f(x)] = 2x + k
Put x = 0 to get, k = 0
/(x) = e21 => 1 + k=2 + 0 = 2
= 0 where f(x) is
continuous differentiable function with f'(x) ^0 and satis­
fies /(0) = 1 and/'(°) = 2, then f(x) = e^ + k, then “k + k is
d
“
= |sin (lOOx) cosx-(sinx)99 dx + | cos(100x)-(sinx)IO°dx
/'(x)
dxLr(x)
■
’
L X.
n 2010
— =----- = 2
m 1005
Sol. f(x)-f(x) - f(x)-f"(x) = 0 or
’
. Ex. 29 /{sindOlxJ.sin” y|= sin(10to)(sinx)\,
1005
m = 1005, n =2010
• Ex. 28 Suppose
1004
□
1 - - = y => — dt = dy
t2
t
1 y’005
i
dy
2 1005
1
2010
•(ii)
-From Eq. (i), 2j(x) = /(x)
.. . :.. f(x)
1 f (t - I)1004 dt
^1006
2J
*1
2
/(x) = 1
f(x) 2
Hence,
x2009
Sol.
•(i)
rw
j cos (lOOx) (sinx)100 dx+ j cos (lOOx) (sinx)’
sin(100x) (sinx)100
+C
100
=>1 = 100,^ = 100 =>
X _100 =I
g ~ 100 "
Chap 01
Indefinite Integral
59
Subjective Type Questions
aVz
• Ex. 30 lfln denotes j z"n e
Vz dz, then show that
x"-1 (a2 - x2)3/2 , (n-1) a 2
-In-2
3
3
n+2
3
(n+T)lln =/0 +eVz (1!z2 +2!z3 +... +n! zn + 1).
i.=-
! as
Sol. In = j zn eUz dz, applying integration by parts taking e,1/z
^-‘(^-x2)372 + (n - 1) a 2
-In-2
(n + 2)
(n + 2)
first function and z" as second function. We get,
Alz ■ Z
_nn +1
+1
/
.
\
(n + 1)
ei/x-zn + 1
4=
(n + 1)
• Ex. 32 lflm = j (sin x +cos x)m dx, then show that
------ dz
n+1
(n + 1)
eVz-zn + 1
Hence Proved.
z" + ‘ J
__ 1
+ (n + l)
mlm = (sin x +cos x)m-1 (sin x -cos x) + 2 (m -1) lm_2
•Ux -zn~'dz
Sol. v Im = j (sin x + cos x)m dx
= j (sin x + cos x)m “1 • (sin x + cos x) dx,
(n+1)
Applying integration by parts
= (sin x + cos x)m " ‘(cos x + sin x) - \m -1) (sin x + cos x)m " 2
eUz.zn + 1
e1/x-z"
1
+ (n + l)
n
(n + 1)
_et/z(z)n + 1 , eUt-(z).nn ,
1__
In-2
(n + 1) n
(n + 1) n
(n+1)
_ e,,z-(z)n+t + e--(z)"
1
^•(z)\f <ellz-(z)n~l +
(n + 1) + (n + l)n (n + l)n-(n-l) (n+l)n(n-l) 4-3
• (cos x - sin x) • (sin x - cos x) dx
= (sinx+ cosx)'"-’(sinx-cosx) + (m-l) j(sinx + cos x)m-2
(sin x - cos x)2 dx
As we know, (sin x + cos x)2 + (sin x - cos x)2 = 2,
,'.Im = (sin x + cos x)m ~1 (sin x - cos x) + (m - 1)
J (sin x + cos x)m ~ 2 • {2 - (sin x + cos x)2} dx
e‘,x •(z)n + 1
e1/z-(z)n
+---- — + ... +
(n + 1) n
e',z-(z)1
(n+l)n...3-2
________ 1________
I0
(n + 1) n(n -1)...3-2
Multiplying both the sides by (n + 1)!. We get,
(n + 1)! IB =(e,/x-z" + 1 • n! + e1/x ■ z" (n -1)! +...
= (sin x + cos x)m ~1 (sin x - cos x) + (m - 1)
| 2 (sin x + cos x)m “2 dx - (m -1) j (sin x + cos x)m dx
Im = (sin x + cos x)m ~1 (sin x - cos x) + 2 (m - 1)
Im-2-(m-l)^m
or(m - l)/m+ /m = (sinx + cosx)m"‘(sinx-cosx)+2(m-l)/m_2
or m Im = (sin x + cos x)m "1 (sin x - cos x) + 2 (m -1) Im _ 2
+ ... + e,/x • z3-(2)!+e1/x-z2 • 1!) + Zo
=>
4(n + l)! = Z0 + e1/z (l!z2 + 2!z3+... + n!zn + 1)
Hence Proved.
2
•Ex. 31
Hence Proved.
• Ex. 33 If I,
{m + n) /m> n = cosm x• sin nx + m
Sol. We have,
-x2 dx, prove that
• cos n x dx
x"-'(a2-x2)3'2
(n-D
/n=(n + 2)
+
n
(n + 2)
So/.I,=j x" ^o2-*2 dx = J
= (cosm x)
- {xy/ai2 - x2} dx
I
n
Applying integration by parts, we get
(a2-x2)3/2l (az-x2)3/2
■+j (n-l)x"-2-<3
3
3
^-1(a2 -x2)3/2
(n-1) a2 ,
r
(n-1)
In
-4=
—4--------+ —
7-4-23
(n-1)
3
x"-1(a2-x•:2)3/2 , (n-1)a2
In-2
3
3
sin nx
sin nx ,
- j m cos'” 1 x (- sin x) • ------dx
n
n
= — cos'" x• sin nx + — f cos'"-1 x {sin x • sin nxj dx
n
n J
As we have, cos (n -1) x = cos nx cos x + sin nx sin x
dx
x"~1(q2-x.2)3/2
i
+ !1Z1) jx’-!-(a!-x•i2) ^a2 - x2 dx
A In +
m
cos " x ■ cos nx dx, show that
• • Im, n = — cos'" x ■ sin x + - f cos"-’ x {cos (n -1) x
n
n J
Im.
m T
- cos nx • cos x} dx
m
= - cos'" x-sin x + — f cos'” "1 x • cos (n - 1) xdx - —
n
n J
n
fees'" x • cos n x dx
’
m T
= -cos x sinnx+-/m_i>n_1--Imn
n
n
n
60
Textbook of Integral Calculus
(x sec x)
| sec 2 x dx
(x sin x + n cos x)
(x sec x)
+ tan x + C
(x sin x + n cos x)
=-[cOSm X • Sin nx + m Im _ 1>n _ j]
/m.n +
n
n“
m + o')
/«.«=-[cos'” x-sinnx+ m Im -l.n-l]
n
„ n )
(m + n)
= cos'" x • sin nx + m
Hence Proved
• Ex. 36 Ifcos 0 > sin 0 > 0, then evaluate
tan
n
-----x
_4____
• Ex. 34 Evaluate J —
7
dx.
2 x J tan3 x + tan 2
cos
x + tan x
tan
dx
---- *
4
Sol. I = f —
log
1 + sin 2 0
/
1 + sin 2 0 '
Sol. Here, I = j < log
cos 2 0
+ sin 2 0
+ log
J-sin 2 0,
co»2e
+ log
1 - sin 2 0 >
2
,2
\cos20
cos 2 0
1 + sin 2 0 ,
de
cos' x J tan3 x + tan x + tan x
(1 - tan2 x) dx____________
-JJ -------7
(1 + tan x)2 cos2 x f tan3 x + tan22 x + tan x
—V
tan x
let
cos 0 - sin 0
V
sec2 x —
~2ydy
(y2 + l)-y
1
•sec2x
tan2 x
-2J
dy = —2tan“’y+ C
1 + y2
tan x + 1 +---tan x
• Ex. 35 Evaluate j
x~2 +n(n-1) .
(x sin x + ncos x)2
Sol. Here,
x2 + n (n -1)
(x sin x + n cos x)2
dQ
n
dx
i
= -2 tan"1
}d9
= J cos 2 0 • log cos 0 + sin 0 ] dQ, applying integration by parts
cos 0 - sin 0 )
I
cos 0 + sin 0
sin 2 0
r
2
sin 2 0
de
= log
2
J
cos 0 - sin 0
cos 2 0
2
tan x
y = tan x + 1 4---- -—
tan x
2y dy =
cos 0 + sin 0
= J (2 cos';2 0 — 1) log
sec2 x dx
'=J tan x + 2 + —-— I tan x + 1 + —-—
tan x J
. = J ■ 2 cos2 0 log cos 0 + sin 0 -log cos 0 + sin 0
cos 0 - sin 0
cos 0 - sin 0
sin 2 0
log
2
cos 0 + sin 0
cos 0 - sin 0
• Ex. 37 Evaluate j
+C
tan 1 x
x4
+ - log | cos 2 0 | + C
2
dx.
1 ,
So/. I = j tan-1 x dx = j tan 1 x—
dx
x4
dx.
x
n
I
dx
\
tan-1 x
Multiplying and dividing by x2" " 2, we get
{x2 + n(n -1)} x2""2
'-J (x" sin x + n x"-1 cos x)7^
3x )
i
J i + x2
1
dx
1 f
dx
+ 3 J x3 (1 + x2) ’
Put 1 + X2 = t,
tan"1 x 1 r
dt
2x dx = dt = - —+ 6 J (f - I)2-1
We know x* sin x + n x" '1 cos x = t
=> {(nx""1 sin x) + (x* cos x) + n (n -1) x"”2 cos x
I =-
- (n x" "1 sin x) dx = dt}
tan-1 x
3x3
•CO
=> xn" 2 cos x • {x2 + n (n -1)} dx = dt
Keeping this in mind, we put
{x2 + n(n - I)} - x"~ 2 • COS X
• x" • secxdx
(x" sinx + nx'1"1 cosx)'T~
n
I
Applying integration by parts, we get
= x7* sec x-1----- --------------- T_]
. |
(x* sinx+nx
cosx))
(• (x" sec x tan x + n x71"1-sec x)
dx
J
(x" sin x +n x*"1 cosx)
Where, A = J
1
A
t-1
B
(t-i)2
■*
Comparing coefficients, we get
A = -1, B = 1, C = 1
1
__ 1
1
A=J- (t-1)
+ (t-1)2 + - dt
t
= - log | t - 1 | -
1
+ log 111
(t-1)
.(iii
61
Chap 01 Indefinite Integral
From Eqs. (i) and (ii), we get
I=-
tan-1 x 1 1
+ | -log|x2|-4 + I°g|l + x2|
—' 6 t
6 L
tan-1 x
x
x2 +1
1 .
“7 log
o
3x3
+C
J
1
1
“A + C
6xz
1
2,
/=
3-9
3
8
2-7
2
X
3
X
,
.
2.
2 x4
x6
x8
log(l-x) = -- x + —+ — + — + ...<* •
2
3
4
=>
=>
x2 log (1 — x2)=—
x
r
rX8 Xr10
X6
+ — + — + — + ...<»
2
3
4
a
J—
„
— + — + — + ...«■ )+C
1-5 2-7 3-9
=>
dA
- b sin x - a
=> — =
b + a sin x
dx (b + a sin x)2
dA
b
dx
a
dA
b
dx
a
a2
a sin x + b +------ b
____________ b
(b + a sin x)2
n
1
X
,
„2
1
a — bk2
-------------- +
b + a sin x b (b + a sin x)2
i -
(a2-b2)
f—*__
J (b + a sin x)2
a
dx
a-,2 — pA 2 r
J
(b
+
a
sin x) 2
a
--------- A
- = --f
a J b + a sin
x
From Eqs. (i) and (ii), we get
o=o+ c
. c=o
j x2 log (1 - x2) dx = -
—(i)
(b + a sin x)
COS X
a J b + a sin x
Now, to find constant of integration, put x = 0
=>
=>
dx
b f
dx
5
7
9
XXX
(b + a sin x)
Integrating both the sides w.r.t *x’, we get
Integrating both the sides, we get
j x2 log (1 - x2) dx = -
----------------------</x
.2
J (b + a sin x)
4
Xd----- +---- + •---- + ...°° •
2
3
4
Put x2 instead of x in the above identity,
ti
a
9
X
Sol. We know, log (1 - x) = -
a2 — k
b2
Now, let A =
that— +---- +----- +...=-log 2--.
1-5
a2
------ b + (b + a sin x)
Sol. Here, I - J a + b sin x dx=~ r
(b + a sin x)2
a J
• Ex. 38 Evaluate j x22 log (1 - x2) dx, and hence prove
.
Evaluate j a + b sin x dx
(b + a sin x)2
• Ex. 39
5
X
7
X
dx
.
b r
dx
-------------------- A + a J (b + a sin x)
a J (b + a sin x)
9
------ F---- +------- F ... 00
1-5--- 2-7----3-9
=>
I=-A+C =>
I=-
cos x
b + a sin x
+C
Applying integration by parts, and taking limits 0 to 1 for LHS
r3
y log (1-x2)
=>
(
=>
3
k 3
I'-fJ"1 3 1-x
1
log (1-x*) I + -2
L 3
• Ex. 40 Evaluate J ------
(x + 2) 5/4
x3
1 .
1+x
3
2
1-x
----- X + - log
/Q
So/. Let I = f---------------------- —
J(x-1)3/4 (x + 2)5/4
=1—
(x + 2)2
Taking log (1 - x2) = log (1 + x) + log (1 - x)
and
1+ x
= log (1 + x) - log (1 - x)
1-x
1 , '
1, n 2 2 ..
log (1 - x)
- log 2 + - log 2--------- + lim
3
3
3 9 x->i <3
3>
log
=>
>
=>
v lim (x3 -1) log (1 - x) = 0
2
dx
(x-1)
(x + 2)
Let
So that,
x+2
3
dx = dt
(x + 2)2
J3,M
3j
3/4 dt
R
- log 2 - - = RHS
3
9
8
2
1
1
1
— + —+ —+ -- = 7log(2)-Z
9
3
1-5 2-7 39
3/4
1
„ 4 x —1
—•■
+ C=—
3 1/4
3 x+2
1/4
+C
-(ii)
62
Textbook of Integral Calculus
@ Indefinite Integral Exercise 1:
Single Option Correct Type Questions
1. Let/(x) = J
x2
dx and /(0) = 0. Then
(l + x2)(l + 71 + x2)
/(I) is equal to
X
dx is equal to
— 2x2 +1
72x4 -2x2 + 1
.5 _L 1
(b) log. (1 + 72)-4
(c) loge (1 + yfi) +
(d) None of these
2. If J /(x) dx = /(x), then | {f(x)}2 dx is equal to
(b)V(x))’
' '
(C)^Z x
(c)
3. If J /(x) dx = F(x), then J x3 f(x2) dx is equal to
5(e8 + l)
fd)
(d)
2e4
5(e8—1)
(b) - [x2F(x2) - f F(x2) d (x2)]
(a)
(c) 1[x2F(x)-1J (F(x)}2 dx]
(a)
n+1
2x2
+C
(b)
e
x-1
(c)
equal to
+C
^2x4 - 2x2 + 1
(d)
(x2 +41nx) -x3
(a) 1 [x2(F(x))2 - J (F(x)}2 dx]
4. If n is an odd positive integer, then J | x" | dx is
3
f"M = /(x). Then /(4) is equal to
(d){/(x)}2
(d) None of the above
X
9. Let /(x) be a polynomial satisfying f(fi) = 2, f' (0) = 3 and
(a)
{/(x)}3
(c)
3
3
(a)^Z - 2x’ + 1
(a) logr (1 + J2)
(a)|lf(x))2
x2 -1
2e4
5 (e8 + 1)
x2
— dx is equal to
e31n* -elnx e*2 +C
(b)
(*2-l) -e*2 +C
(d) None of these
I
2x
2x
(x -1) xe
—+ c
2
11. jtan4xdx = A tan3 x + B tan x + /(x), then
(a) A = |,B=-1,/(x) = x+C
+C
x”*1 1 c
(b)
n+ 1
+C
(d) None of these
(c) A = |,B = l,/(x) = x + C
3x + 2
2
(d) A=-,B = l,/(x) = -x+C
5. Let F(x) be the primitive of
y/x-9
w.r.t. x. If F(10) = 60,
then the value of F( 13) is
(a) 66
(b) 132
(c) 248
(d) 264
6, J (xx )x (2x logf x + x) dx is equal to
(a)x(xi) + C
(b)(xx)x + C
(c) x2 ■ logf x + C
(d) None of these
7. The value of J x log x(log x -1) dx is equal to
(a) 2(x logx-x)2 + C
(b) i(xlogx-x)2 + C
(c) (x logx)2 + C
(d) ^(x logx)3 + C
2
(b) A=-,B = -1,f(x) = x+C
12. If the anti-derivative of J
sin4 x
dx is /(x), then
x
fsin4 {(p + <?)x}
dx in terms of /(x) is
J
x
f{(p + q) *1
(a) f {(p + g)x}
p+ q
(d) None of these
(c) f{(P + <?) *} (p + q)
sinO
sin30
sin 90
13. J f-----4------------ 1- -
cos 30
cos 90
<cos 270y
dQ is equal to
. . 1 sec 270
(a) - log
+C
sec 0
(b)llog
(c)^log 2^sec 270
+C
sec 0
(d) None of these
sec 0
+C
sec 270
Chap 01 Indefinite Integral
nil -1, n € N. Then, the value of
14. Let x2
2sin(x2 + l)-sin2(x2 + 1) , .
18. The primitive of the function /(x) = xj cos x |, when
,
equal to
--------- ---------- --------- - dx is
b -----2sin(x2 +l) + sin2(x2
+1)
sec (x2 + 1) + C
(a) log
(c)
log | sec (x2 + 1)| + C
(b) log sec
63
X2 + 1
2
+c
(d) None of these
— < x < tc is given by
2
(a) cosx + xsinx + C
(b) - cosx-x sinx + C
(c) xsinx-cosx + C
(d) None of the above
19. The primitive of the function /(x) = (2x + 1)| sin x |,
15. [--------- —--------- is equal to
J cos (2x) cos (4x)
<a)^10g 1 + 72 sin 2x -1 (log |sec 2x - tan 2x |) + C
1 - 1/2 sin 2x
(b) ^log 1 + 5/2 sin 2x - |(log | sec 2x - tan 2x |) + C
I + V2 sinx
(c) ^10g 1 + 5/2 sin2x -|(log |sec 2x-tan2x|)+C '
1 - V2 sin2x
when 7t < x < 2n is
(a) -(2x+ 1) cosx + 2sinx + C
(b) (2x + 1) cosx - 2 sinx + C
(c) (x2 + x) cosx + C
(d) None of the above
0
20. Given,/(x) = sinx-x2
2 - cos x
(d) None of the above
dx _ fM + C, then f(x) is equal to
16. J 1~7cos27xx dx
sin7 x cos2 x
(sinx)7
cos x
(b) cosx
(a) sin x
(d) cot x
(c) tan x
3 x
<7. J (cos4 x+3 cos2 x sin
dx is equal to
+1) tan -1 (sec x + cos x)
(a) tan-1 (sec x + cosx) + C (b) log, |tan-1 (sec x + cosx) |+C
(c)-------- i------- - + C
(sec x + cosx)
(d) None of these
x2 -sinx
cos x - 2
0
l-2x
2x —1
0
, then
J f(x)dx is equal to
x3
(a) ----- x2 sinx + sin2x + C
3
x3
(b) ----- x2 sinx - cos2x + C
3
x3
(c) ----- x2 cosx - cos2x + C
3
(d) None of the above
g Indefinite Integral Exercise 2:
More than One Option Correct Type Questions
2H
dx
= A log (x +1) + B log (x - 2) + C, where
(x + l)(x-2)
9eXx -4e-x
(b) AB = 0
(d) None of these
(a) A + B = 0
(c)A/B = -l ’
22. If J
Jx
= k tan
(x2 +l)(x2 +4)
2
(a)fc = l
(b)/=-
x +1 tan
(a)A = |
— + C, then
2
11
(c)fc
(c)fc==-;
-
1 + X2
(c) V(x) = -
2
1 + x2
2
1 + x2
2
1 + x2
(d)<Xx) = 2
(b) y (x) =
(c) C is indefinite
(b)B = 36
19
(d)A + B=-^
36
25. If j tan5 xdx = A tan4 x + Btan2 x + g(x) + C,then
(d)Z
(d)/ == -~
-
3
3
6
3
23. If jx log(l + x 2 )dx = <J>(x)log(l + x2) + x(y) + C, then
(a)(Xx) =
24. If J 4e + 6e~x dx = Ax + B loge (9e2x - 4) + C, then
(a) A = -,B=-4
2
(b) g(x) = In | sec x|
(c) g(x) = ln|cosxl
(d) A = --,B=4
3
64
Textbook of Integral Calculus
g Indefinite Integral Exercise 3:
Statement I and II Type Questions
■ Directions (Q. Nos. 26 to 30) For the following
questions, choose the correct answers from the codes (a),
(b), (c) and (d) defined as follows :
(a) Statement I is true, Statement II is also true; Statement II
is the correct explanation of Statement I.
28. Statement I If a > 0 and b2 - 4ac < 0, then the value of
the integral [-------—------ will be of the type
ax2+bx + c
... -i x + A
p. tan
+ C, where A, B, C, |1 are constants.
~B~
(b) Statement I is true, Statement II is also true; Statement II
is not the correct explanation of Statement I.
(c) Statement I is true, Statement II is false.
(d) Statement I is false, Statement II is true.
26. Statement I
If y is a function of x such that
y(x-y)2 = x, then f= [log(x-y)2 -1]
J x-3y 2
dx
, .
„ , „
Statement II J ----- — = log (x - 3y) + C
x-3y
Statement II If a > 0, b2 - 4ac < 0, then ax2 + bx + C
can be written as sum of two squares.
/
■ ___ 1
■ dx = tan-1 (x2) + C
29. Statement I
J + x4J
Statement II
1
,
_i
- dx = tan x + C
1 -----1 + x2
J 2^ xd(cot 1 x) = 2^- -+c
X
30. Statement I
ln2
27. Statement I Integral of an even function is not always
an odd function.
d . _
Statement II — (ax + C) = ax Ina
dx
Statement II Integral of an odd function is an even
function.
g Indefinite Integral Exercise 4:
Passage Based Questions
Passage I
(Q. Nos. 31 to 33)
X + ^/l + X 2 xl5
=----- dx is equal to
■Jl + x2
32.
Let us consider the integral of the followingforms
f(x\, Jmx2 +nx+ p)17 2
Case I If m> Q then put
2 + nx + C -u + xjni
Case II If p > Q then put \mx 2 + nx + C = ux±y[p
(x + 71 4- X'.2x16
(a)
(b)
(C)
Case III Ifquadratic equation mx22 +nx + p = Ohas real roots
a and ft there put
31. If I = $
nix2 + nx + p = (x-a) wor(x-P)w
dx
,
T
,
to evaluate I, one of the most
x-y]9x 2 +4x4-6
proper substitution could be
(d)
10
______ 1
—+ c
—+C
15 (71 + X.2 + x)
____ 15
----- + C
(7i+*2 -x)
(x+ 71 + x.22x!5
)1
15
---- —+ C
33. To evaluate J----------- -^X
(x-l)7~x2 +3x-2
suitable substitution could be
(a) yjvx2 + 4x + 6 = u ± 3x
(a) J- x2 + 3x - 2 = u
(b) ^9xz + 4x + 6 = 3u ± x
(c) x = i
(d) 9x2 + 4x + 6 = |
----- one of the most
(b) 7~ x2 + 3x - 2 = (ux -J2)
(c)
x2 + 3x - 2 = u (1 - x)
(d)
x2 + 3x - 2 = u (x + 2)
Chap 01
Passage II
65
(h) 14,2 = | (~ sin5 x cos3 x + 8 f4> 4)
(Q. Nos. 34 to 36)
= Jsin"xcos m xdx. Then, we can relate In< m with
(c) h, 2 = | (sin5 x cos3 x - 8Z4 4)
(d) h, 2 = ~ (sin5 x cos3 x + 6L 4)
3
each of the following:
(ii) /n + 2,m
(0 In-2.m
Indefinite Integral
Passage HI
(iv) In,m + 2
(Q. Nos. 37 to 38)
(v) ^n-2,m + 2
(Vi) /n + 2,m-2
Suppose we want to establish a relation between I„t m and
Km-2 .then we get
P(x) = sin"n +1
+ 1 xcos"”1 x
...(i)
InI„tm andIn,m-2
exponent ofcosxis mandm-2
respectively, the minimum of the two is m-2, adding 1 to the
minimum we get m-2+l = m-l. Now, choose the exponent
m-\ofcosx in P(x) Similarly, choose the exponent ofsin xfor
P(x)= (n//)sin" xcos"1 x- (m- l)sin" + 2 xcos”1-2 x
Now, differentiating both the sides ofEq. (i), we get
= (n + l)sin" xcos”1 x-(m- l)sin" x(l-cos2 x)cos m - 2 x
= (n + l)sin" xcosm x- (m- l)sin" xcos"1-2 x
+ (m- l)sin" xcos" x
m x- (m-1)sin" xcos"1-2 x
= (n + m)sin"n xcos'"
If f: R -> (0, oo) be a differentiable function /(x) satisfying
f(x + y)- f(x - y) = fix)- {f(y) - f(y)~ y}, V x, ye R,
(/(y) * fir y) for all y e R) and f (0) = 201Q
Now, answer the following questions.
37. Which of the following is true for /(x)
(a) f(x) is one-one and into
(b) {/(*)} is non-periodic, where {■} denotes fractional part
of x
(c) fix) = 4 has only two solutions.
(d) fix) = /“’(x) has only one solution.
38. let gix) = loge(sin x\ and j figix)) cos x dx = h(x) + c,
(where c is constant of integration), then h
is equal
to
(a)0
Now, integrating both the sides, we get
sin"+1 xcos"1-1 x=(n + m)Inm -{m-l)/n>m_2
I
(d)^T
(c)l
Similarly, we can establish the other relations.
Passage IV
34. The relation between /42 and Z2>2 is
(Q. Nos. 39 to 41)
(a) I4.2 = ~ (“sin3 xcos3 x + 3 I2i2)
6
Let f'.R-^Rbea Junction as
/(x) = (x - l)(x + 2)(x - 3)(x - 6) -100 Z/g(x) is a
(b) I4,2 = - (sin3x cos3x + 3I2,2)
6
polynomial of degree < 3 such that |
(c) I4,2 = 7 (sin3 x cos3 x - 3f2> 2)
any logarithm junction and g(- 2) = 10 Then
6
39. The equation /(x) = 0 has
(d) If 2 = - (- sin3 x cos3 x + 2I2.2)
4
35. The relation between /4 2 andl6>2 is
(a) If 2 = - (sin5 x cos3 x + 8 I6_ 2)
5
(b) If. 2 = | (- sin5 x cos3 x + 8 I6- 2)
(b) -100
(d) -68
(a) -136
(c)-84
41. J/(x) dx, equals
(d)I4.2 = - (sin5 x cos3 x + 8Z6 2)
6
(a) I4,2 = ” (sin5x cos3x + 8 I4>4)
(a) all four distinct roots
(b) three distinct real roots
(c) two real and two imaginary
(d) all four imaginary roots
40. The minimum value of f(x) is
(c) h. 2 = - (sin5 x cos3 x - 82)
36. The relation between If 2 and I
dx does not contain
is
(a) tan'
x-2
2 ,
(c) tan-1 (x) + c
(b) tan
x —11
+c
1 J
(d) None of these
66
Textbook of Integral Calculus
g Indefinite Integral Exercise 5:
u Matching Type Questions
42. Match the following :
Column II
Column I
sinx - cosx
dx, where y < x <
| sinx - cosx |
(A)
If I = J
then I is equal to
(P)
sinx
(B)
lfl (X3 4- 1) (x3 4- 2) dx= -3 /II X3 4- 2 + C, then f(x) is equal to
(q)
x+C
(C)
If J sin
(0
ln|x|
• (s)
sin- x
x3 4- 1
x2
(x) — x-xf-\x)-2yll-x2 + 2x + C, then /(x) is equal to
2
x • cos-lx dx =
dx
= /(/(x)) + C, then /(x) is equal to
x/(x)
(D)
43. Match the following :
Column II
Column I
(A)
(B)
IfJ
( •>
2 A
X 4- COS X
1
IfJ ]x +y[x2+2 dx = y (x 4- Jx2 4- 2)3/2
— dx = A
x2
(D)
------- .
,then
(x 4- i]x2 4- 2)
B i
4 -x 4- 4 -^2-x-x',2 A
2 - x - x2
- sin
- 4- —f= log
472
x
x
>
•j2-x-x.2:
(C)
(P) 4 = 1
2 j
. -i '
cosecx .
cosec x dx = A cot x + B-------- then
secx
/
„2
sin2x
,fl sin4x 4- cos4x dx = Bcot-1 (tan2x), then
2x4-1
L then
3
(q)
B = -l
(r)
B=2
(s)
A=-l
g Indefinite Integral Exercise 6:
Single Integer Answer Type Questions
44. If [------- (2--3) dx------- = C - —, where /(x) is of the form of ax2 +bx + c, then (a + b 4- c) equals to
J x(x 4-1) (x 4-2) (x 4-3) 4-1
/(x)
45. Let F(x) be the primitive of 3x + 2
7x ~ 9
w.r.t. x. If F(10) = 60, then the sum of digits of the value of F(13), is
46. Let u(x)and v(x) are differentiable function such that
, (u(x)\
4- q ,
,
,
=7. If ^-X- = p and ----- = q, then -p—
- has the value
v(x)
V (x)
p-q
equal to
2
r
x-e
x —1
4- C, then find 24 A.
dx = 6A In
47. If j —L-ln
(x2 -1)
48. If j
x+1
<X4-1>
ex(2-x2)
(1 - x) -Jl-x 2
dx = pex
1 4-X
J-X
X
4- C, then 2(X 4-|1) is equal to
Chap 01 Indefinite Integral
49. If j cos x - sin x +1 - x dx = In {f(x)} + g(x) + C, where
e x + sin x + x
C is the constant of integrating and /(x) is positive, then
/(x) + g(x).
-------- - — is equal to
ex+sinx
dx
50. Suppose A = j
xz +6x4-25
If 12(A + B)= X-tan
and B = J
dx
x2 -6x-27
x —9
x+3
+ C, then
+ p.ln
x+3
4
the value of (X +11) is ....
+ cos 9x .
sin 4x
51. If J cos 6x- ---------------ax =
- sin x + C, then the
1 - 2 cos 5x----------- k
value of k is
tanx
52. The value of J
2
67
,
1
dx = x —= tan
1 + tan x + tan x
2 tan x +1
+ C, then the value of A is.
53. Jsin5/2 x cos3 xdx = 2sinA/2 x J. - — sin2 x +C,
B C
then the value of (A + B) — C is equal to
54. IfJ(x2010 + X804 +x“2)(2x,“’ +5x 402 +10)1/402
i'“2 dx
5X804 + lOx402 )a/402. Then(a-400)is
= —(2x 2010
2010 + 5X
10a
equal to
55. If Jexx3 ++xx2- *(3x4 +2x3 + 2x) dx = /i(x) + c Then the
value of h (l)/i(-1), is
Indefinite Integral Exercise 7:
Subjective Type Questions
56. Evaluate e (xsinx + co,x)dx
x'4 cos33 x - xsin x + cos x
2
2
x cos x
64. Evaluate j--------- —---------, a G N.
J (sin x + asecx)2
57. Evaluate J Jx + -Jx2 + 2 dx.
65. Evaluate j
dx
x - y[x2 + 2x + 4
58. Evaluate J _________ dx_________
(•^(x-a)2 -P2)(ax+ b)
66. Evaluate J
dx
X
59. Evaluate J
3
60. Evaluate J
x2~
dx.
67. Evaluate j
x4+l
dx.
x6+l
sin3 (0/2)d0
cos 0 / 2 -Jcos3 0 + cos2 0 + cos 0
61. Evaluate J
1 + ^x2 +2x + 2
(2 sin 0 + sin 20) d0
(cos 0-1) -^cos 0 + cos2 0 + cos3 0
62. Connect J x ra-1(a + bx”)p dxwith
X8 dx
Jxm-n “1 (a + bxn )p dx and evaluate J (l-x
3)173
63. Evaluate | cosec 2 x In (cos x + Jcos 2x) dx.
dx
,
68. Evaluate J ------------dx.
(l-x3)V3
69. Evaluate J
sjl + x 2 \15
=---- dx.
1 + x2
70. Ify2 -ax 2 + 2bx + c, and u„ = f — dx, prove that
J y
(n +1) a un + j + (2n +1) bun + nc yn , = xny and deduce
that aux = y-bu0; 2a2u2 = y(ax-3b)-(ac -3b2)u0.
68
Textbook of Integral Calculus
g Indefinite Integrals Exercise 8:
Questions Asked in Previous 10 Years' Exams
(i) JEE Advanced & IIT-JEE
2 x
7tJ (sec xsec
dx equals to (for some arbitrary
+ tan x)9/2
[Only One Correct Option 2012]
constant K)
[11,
< ------- (sec x + tan x) > + K
(a)
J
(sec x+ tan x)n/2 [11 7
x2l
_______1_______ [11,
4------- (sec x + tan x) > + K
(b)
J
(sec x + tan x)”/2 [11 7
-1
I 1 1 .
.21
< ---- 1- - (sec x + tan x) + K
(c)
(sec x + tan x)11/2 111 7
J
-1
W --------- 1----- 575 < — + - (sec x + tan x)2 > + K
J
(sec x + tan x)‘ 111 7
72. If 7 = J
ex
~dx,J = \ ____ e~x
— dx. Then,
J e-4x + e 2x + 1
e4x + e2x + 1
for an arbitrary constant c, the value of J -1 equals
[Only One Correct Option 2008]
e*x - e2x + 1
e2x
. . 1,
1
+C (b)-log
(a) -log
e4x+e2x + l
2
e2x
.. 1, e2x-ex + l
+C
(c)-log
e2x + ex + 1
2
+C
e*x + e2x + 1
+C
e'
e x -e2x + I
(ii) JEE Main & AIEEE
2x12 + 5x 9
73. The integral j
(x5 + x3 +1) 3
(a)
- x5
(x5 + X3 + I)2
+C
(b)
+C
(d)
X5
(c)
2(x5 + x3 + I)2
76. If j /(x) dx = \|/(x), then J x5f(x3) dx is equal to
dx is equal to
[2016 JEE Main]
x10
2(x5 + x3 + l)z
+C
-x10
+C
2(x5 + x3 + l)2
where, C is an arbitrary constant.
74. The integral j —
X* + Ip + C
I x4 J
[2015 JEE Main]
£
x4 T2
+ 1V +c
1
1 + x ——|e
xj
x+—
x dx is equal to
1
X+C
i
(c)(x+ 1) e
[2014 JEE Main]
1
(b) xe
«+c
1
(d) -xe + x + c
77. If the integral f $
X
J tan x - 2
| sin x - 2 cos x | + k, then a is equal to
(a) -1
(b) -2
(c) 1
(d) 2
1
<b)(x'+1)<+C
(c)—(x4 +1)« +c
(a)(x-l)e
(c) |x3y(x3) - J x2\|/(x3) dx + C
3 equals
i
75. The integral
(b) | x3y(x3) -3 j x\(x3) dx + C
(d) | [x3V(x3) - J x3y(x3) dx] + C
x2 (x4 +1)<
(a)
[2013 JEE Main]
(a) | [x3V(x3) - J x2y(x3)dx] + C
78. The value of V2 j sin x dx .
----- 7------- x 1S
• I x—
71 I
sm
k
4J
n
(a) x + log cos x---4
[2012A1EEE]
[2012AIEEE]
n
+ C (b) x + log sin x----- + C
4,
n
n
(c) x - log sin x----- + C (d)x-log cos x —
4.
4
+C
I-
Answers
Exercise for Session 1
lJ(x+l)
1.
-(x + 1) 3/2 + 1JI2 + C
3
2
2
3
i
2.--+-=---------2tan *x+ c
x 3X3 5x?
3
. if? ,
6. tan x- tan" x + c
5. - — + tan x + c
H3 .
1 (.
_
a2
7. — Ibx - 2a log |dx + a| + c
a + bx
6- I
4x
2V
8. ----------- + c
9. — + c
4
1 + log, 2
x"*’
10. ------ +------- + c
11.- - cos4x+ C
a + 1 log a
8 ,
1
______ _ 1 .
12. In | sec 3x | - In | sec 2 x| - In | sec x | + e
3
‘ 2
2
3
1
14. — sin x + — sin 3x + C
13. - sin 3x + 2 sin x + C
4
12
3
15. - — cos 2x + -J- cos 6x + C
64
192
1. tan x- secx+ C
+C
It
3
6. 2(sin x+ xcosa) + C
5. x + C
8. tan x - cot x + C
7. sec x - cosec x + C
9. (sin x+ cos x) sin (cosx- sin x) + C
11. - 72 cos
10. tan x - cot x - 3x + C
+C
13. - (x - sin x) + C
2
15.-4z + C
+C
14. - 2cosx+ C
42
Exercise for Session 3
, 1,
1. —tan
48
<4 J
2. —tan
36
+C
4x4 - 5
1
+C
3- —log
160
4x4+ 5
< 2 .
4. - sin
3
A log lx3 + Ja6 + /|+C
5. ■
3
7.
1
log, 2
log |2X +
(x +
x
16. log tan - + 1 + C
2
17.-log tan - + —
2
.2
6
4
I
+C
18. -log |sin x- cosxj + -cos 2x+ -sin 2x+ C
4
8
8
2 .
(cos372*) + C
20.---- sin
+C
19. sin
3
I 3 /
12 z,
„___ x
5
21.
(3 sin x - 2 cosx) log (3 cosx+ 2 sin x)+ C
13'
' 13 ■'
2x-3Y|
2-:_2)3/2--rfx-^74 + Sx-x2 25 .
22. --(4+3x-?);
+ — sm ------- +C
5 )\
3
2[l
2)
4
if£1
23. T^x2 + x + 1 + 2 log
+ x+ 1
1 - x + /? + x + 1
+C
./4rf + C
I3 J
fx3'2) + C
1. xV-2(x/-?)+C
2. - x2 cos x + 2(xsin x+ cos x) + C
3. x(log x) - x + C
4. x(log x)2 - 2(xlog x - x) + C
5. xtan x -
10.xlog(? + l)-2x+ 2tan"’x+C
13. xlog (log x)----- — + C
logx
J
15.
x2 + 1
14.-e^tan x+ C
2
17.
16. e1- EHZ + C
Vl-x
{acos (bx + c) + Asin (bx + c)} + C
—
u )
i 1 n A. I MMM ~
I
.....
1
t '*
18. sec xtan x + A- log
|sec x + tan x|' + C
_
“ '
2
2
19. - 2(- Txcos 7x + sin 7x) + C
20. x(sin~lx)2 - 2(- sin"1*-^1-x2 + x) + C
x- -log (l + x2)-(l-x)tan"1(l-x)+ — logfl + (1 + x2)} + C
2
2
3/2
23. --fl + 4
3l x22)
I 46
4
12. e*tan x + C
22. a ■ -tan
a
10. - 3 - 2x - x2 - 4 sin -1/X-1 + C
2x-l
a2 + b2
21. xtan
x—1
+C
76
9. Ilog lx2 + 2x+ 2| + tan-1(x+ 1) + C
6. x(sec"‘ x) - log |x + -Jx2 - 1| + C
9. - xcot - + C
2
11. f ■ log (secx)+ C
1 mam
2 4-3/
r 1 1
+C
6. — log
12
2 - 3e*
log |1 + x2! + C
_ x2
-1 i(x-tan-1x) + C 8. --(1 + logx) + C
7. —tan -1 ■
2
2
„3/2
-251 + C
11. —4 log I4X2 - 4x+ 17| + -1 tan"
81
6
2(x+l)
Exercise for Session 4
8. - 8-^5 + 2x- x2 - 3 sin
I
+ V*2 + x+ 1 + C
15.
24. (c)
2. sin 2x+ C
. 180 .
„
4.---- SHUT + C
3. -
8
Xx-b
- 6 log
Exercise for Session 2
12.
ll—Z + C
13. y]a-x(x- b) - (a - b) tan
4. — - x+ tan"’x+ c
3
3. logx+ 2tan"'x+ c
- cos 4x
3 +C
3 2 + -x
12. -J2X* + 3x - ~^= log x+ -lx
2
472
4V
2
4*
24.
+C
cosx
2cosx-sinx
I
S) 9( l+?
+c
2.
- - log 12cosx- sin x| + C
5 5
x
25. e,sm X(x - sec x) + C
70
Textbook of Integral Calculus
1.
C
t. - log |x — 1| — 4 log |x — 2|+ - log '|x- 3|+
2
2
4108
56. e(x“x+cMx)-
1 4- sin x
4-C
2 4- sin x
57. l(x+
3
12
1
40. (c)
43. A -4 p, q; B -4 p,r; C -4 r, D -4 q
48. (3)
46. (1)
47.(1)
49.(1)
55.(1)
54. (3)
52. (3) 53. (3)
2x-l
2. - log |1 4- x| -- log |1 - x 4- ?| 4- ?= tan
3
“
73
43
6
, 1 i
x"
+C
+C
3. - log
x2* 3
n
Z+ 1
5. log
34. (a)
31. (a) 32.32.(d)(d)
33.(c)
37. (b) 38.38.(d)(d)
39.(c)
42. A ->q; B -»r; C p; D -♦ r
Exercise for Session 5
1
x---------xcosx
I +c
-2
1
35. (a)
41. (a)
36. (b)
44. (5)
50. (4)
45. (6)
51- (4)
a
58. tan - = t
2
7x4- 7? 4-2
6. — log |1 - cosx|— log |1 4- cosx|4- — log |3 4- 2 cosx| 4- C
10
2
5
_ 1 . I1 4- sin x + ------ J------ 4-C
7. - log 4
1 - sin x 2(1 4- sin x)
go. tan-1 (cos0 4- secO 4- 1) 1/2 4-C
59. 2 (1 4- x.!/3)3/2
1'3)- + q
4cos 0 4- sec 0 4- 1 - 73
r
+C
61. - -2 log
.JcosO 4- sec 0 4- 1 4- 73
3
8. - - log |1 4- tan x| 4- - log |tan2x - tan x 4- 1|
62.
6
3
1
2tan x- 1
4- —= tan
73
75
+C
64.
2 log x 4- 1
4-C
9. log
3 log x 4- 2
10. -
(l-?):
8
20
40
63. - cot x log (cos x 4- ^cos 2x) - cot x - x 4- ■Jcot2x- 1 4- C
1
(4a2 - 1)3/2
2a sin
2a sin 2x 4- l')
2a 4- sin 2x J
65. 2 In | J? 4- 2x4- 4 - x|------ ,__ ——g------------
2(7? 4- 2x4- 4 — (x 4- 1))
Exercise for Session 6
lx* 4- ? 4- 1
1
_
■
x2
, 2 t .
2. —= tan
+ C,
73
/
- - In -J? 4- 2x 4- 4 - X-14-C
2 2
\
v
3. 2? - 3r2 4- 6r - 6 log 114- r 14- C, where, t = (x 4-I)1'6.
4.
x-b
(a 4- b) ^x4- a
7
(X4- 2)4-^? 4- 2x4- 2)
2
.
-—tan
3
i,,34-x
67. tan
6. (a)
7.(d)
1
68. - log
3
A
X
(?) + c
1.
--log (I-?)273
o
1 ,
—= tan
Chapter Exercises
3. (b)
9.(b)
15. (a)
21. (a,c)
27. (c)
6.(b)
4.(c)
5. (b)
10. (d)
11. (a)
12. (a)
16. (c)
17. (b)
18. (b)
22. (a, d) 23. (a, c) 24. (b,c,d)
28. (a)
29. (d) 30. (d)
+r
66. log (x + 1 + -J? + 2x+ 2) +
1/7
5. 42 sec A (7tan x sin ,4 4- cos /<) 4- C
8. (c)
9.(d)
9.
(d)
10. (c)
2. (a)
1. (b)
7. (b)
8. (d)
13. (c) 14. (b)
19. (b) 20. (d)
25. (a, b) 26. (c)
f 2a sin 2x4-1)
2a 4- sin 2x J
---------- ------- + C
(2a + sin 2x)
tan"1 x 4- log |x| - - log |1 4- X2) 4- C
2
x
!.• -I
L
y
73
69.
2+c
(X4- 714-?)
74. (d)
15
75. (b)
76. (c)
2(1-?) -x
71. (c)
72. (c)
77. (d)
78. (b)
4Tx
73. (b)
1/3
s
5. F(x) = J
Solutions
3x + 2
dx. Let x - 9 = t2
Vx-9
dx-2t dt
f 3 (t2 + 9) + 2 „ V
F(x) = J
I
f
J
= 2 J(29 + 3t2) dt = 2 [29t + t3]
I
*2
1. Letf = -------;-------------r«x
(1 + x )1 + (1 + x2)
F(x) = 2 [29 yJx-9 + (x -9)3/2] + C
F(10) = 60 = 2 [29 + 1] + C => C = 0
Given,
Put x = tan 0 => dx = sec2 d0
F(x) = 2 [29 Jx-9 + (x -9)3'2]
F(13) = 2 [29 X 2 + 4X2] = 132
Putting, x = tan0,1 = j
1 + sec 0
6. We have, J(xx)x (2x log, x + x) dx
= j (sec 0 - 1) dO
;■
= jxx (2 xlog,x+ x)dx
= log (sec 0 + tan0) - 0 + C
•
= jl-d(xx2) = xx2 +C=(xx)x + C
f(x) = log,(x + ^/x2 + 1) - tan-1 x + C
7. We have, j x log x (log x - 1) dx
/(0) = 0 => C = 0
/(I) = loge(l + 42) - tan-1 i
=>
= j logx(xlogx - x) dx
= 10ge (1 + 72) - y
= j(x logx - x) d (x logx - x)
4
2. We have, j /(x) dx = /(x)
7-{/(x)} = /(x)
dx
=>
__ (x logx-x)
-+C
2
'
=* -i-d{/(x)}
= dx
/(x)
log {/(x)} = x + log C => /(x) = Cex
1
1
X3
7
M 12 - —2 + —
{/(x)}2 = C2e2x
1
J
C2e2x
2
x2
J{/(x)}2dx = JC2e2x dx=------
X
4+C
2x2
x
+C
r'w=/(x)
2f(x)f'(x) = 2f(x)f(x)
In this case, we have
||x',|dx = j|x|n dx = jx"dx
[v|x| = x]
=>
LT(x)}2 = ^-{/(x)}2
dx
Now, /(0) = 2 and f*(0) = 3. Therefore, from Eq. (i), we get
{/'(0»2 = {/(0))! + c
In this case, we have | x | = * x
Jlx'1 |dx = J|x|n dx
= j (- x)n dx = - J x" dx
dx
{r(x)}2={/(x)}2+c
[v x£0=> |x| = x]
Case II When x £ 0
9 = 4 + C =s C=5
(T(x))2 = (/(x))2 + s
r(x)=75+i/(x»2
[v n in odd]
x" + 1 . n (-X)nx „ |x|"x „
=-------- + c =----------+ C =
+C
n+1
n+1
|x|"x
J|x"|dx = L-1— + C
n+1
V
9. We have,
4. We have the following cases :
Case I When x > 0
n+1
l2x* -2x2 + 1
or
= l[x!F(x!)-jF(x2)<((x!)]
Jxfx + C
X
r+C
~ => 1
-1 vz
2
2
jx3y(x2)dx = |j?f(x2)d(x2)
Hence,
2
1
2 + 4 -Z
n
2
Let
3. We have, J /(x) dx = F(x)
x'***
n+ 1
dx
-j—J-------- -
=>
= fdx
iog|r(x)+7s+Wx))2|=x+c,
(0
72
Textbook of Integral Calculus
/(0)=2 => log|24-3| = Q
Cj = log 5
(x2 + 1)
=b 11 +- cos
cos (x2 + 1)
log I J(x) + ^5 + {/(x)}21 = x + log5
= j tan GW d x2 + r = log sec
=> log
/(x)+75 + {/(x)}2
5
I 2
=x
2
7
dx
7
7
X2 4- 1
\ 2
4-C
sin (4x) dx
sin (4x - 2x) dx
15. f sin (2x)
cos (2x) cos (4x) -J sin (2x) cos (4x)
/(x) 4--^5 4-{/(x)}2 = 5ex
=> 75 + {/(x)}2+f(x)=5e’ and ^5 4-{/(x)}2 - f(x) = 5e
2/(x) = 5 (ex — e-x)
=>
x tan
„ r cos2xdx 1 ,, .
„
= 2---------------- (log | sec 2x - tan 2x |)
J cos 4x
2
dx - - (log | sec 2x - tan 2x |)
=2J (1-2cos2x
2
sin2 2x)
f(x) = 5-(ex-e~x)
£
/(4) = |(e’-e.-*) => f(4) = 5(e8-l)
=>
2e*
1 4- a/2 sin 2x
2
1 .
2y/2 2X1 Og 1 - y/2 sin 2x
: 2
-x e
-L-dx=J — •x*-xV
dx
x-1
x-1
.3
x2 + 4 Inx
10. We have, J
»x2
= j x3 e.xx22 dx = | J te'dt, [where t = x2]
1 + -J2 sin 2x
10g
1 - \/2sin2x
J [sin x
= -(t-l)ef +C = -(x2-l)ex2 +C
2
2
x
2
tanx
= —7~ + h
sin x
tanx „
p+q
rsin*(p4-q)x
<b = f{(p + q) *}
J
x
—+c
17. We have, f----------------- -—— * ------------------- dx
J (cos x 4- 3 cos x 4-1) tan-1 (sec x 4- cosx)
sin’x
2
= /—-2------------------------------------------ dx
J (cos x 4- 3 4- sec x) tan (sec x 4- tanx)
sin30 = | [tan 90 - tan30]
cos 90
=
1log (sec 1270) - log (sec 0) [ + C
-p
J 1 4- (sec X4- cosx)
cos x
f --------T-------------------------1
1
X-------------------------------- J tan- (sec x 4- cosx) 1 + (sec x 4- cosx)
(tanxsec x -sinx) dx
= f----- ------------------ d | tan-I(sec x 4- cos x)|
J tan- (sec x 4- cos x)
1 , ffiec 270
+C
= -log
sec 0
j2 sin (x2 4-1) - 2 sin (x2 4-1) cos (x2 4-1)
dx
J 2 sin (x2 4-1) 4- 2 sin (x2 4-1) cos (x2 4-1)
sinx(l - cos2x)
1
x----- ------------------dx
tan- (sec x 4- tanx)
sin0
sin 30
sin 90 d0 = | j(tan 270 - tan0) d0
+------- 4cos30 cos 90 cos 270
2 sin (x2 4-1)4- sin2 (x2 4-1)
1
_r
sin90 = | [tan 270 - tan90]
cos 270
2 sin (x2 4-1) -sin2(x2 4-1)
=> f(x) = tanx
sm x
sin0 = | [tan 30 - tan0]
13. On solving,
cos 30
14. We have, J x
dx = Ii + I2
sec x .
tan x „c tanx-cosx
dx
-— dx = —+’J
/.=J —
sin7x
sin8x
sm x . *
Now,
sin* (p 4- q) x fa = f{(P + q) x}
(p + q)x
dx
_ rsec2 x
tan2 x sec2x - sec2x + 1) dx
dx = /(x) => j
sin7 x,
J sin x
tan’x
--------- tanx 4- x 4- C
3
1 _
. .. .
A = B = -1 and _f(x) = x + C
3
sin* x
- - log | sec 2x - tan 2x | + C
2
7 '
16. J 1-7 cos2 x
sin7 x cos2 x
11. Jtan*xdx =
- | log | sec 2x - tan 2x | 4- C
= log, | tan-1 (sec x 4- cos x) | 4- C
18. We have,
f(x) = x [cosx |, — < x < 71
2
f(x)
x cosx [v cosx < 0 for x e (k /2, tc )]
Hence, required primitive is given by
j /(x) dx = - j x cosx dx 4- C = - x sin x - cosx 4- C
!
Chap 01 Indefinite Integral
!
19. Wehave, /(x)=(2x + l)|sinx|,n <x<2n
=>
Then,
I
/(x) = -(2x+l)sinx
Hence, required primitive is given by
- J(2x + l)sin xdx = - [~(2x + 1) cosx + 2sinx] + C
■
I
n
x2 - sinx cosx -2
2 - cosx
■
9J z
sinx-x2 2 - cosx
0
0
2x-l
/(*) = x2 -sinx
0
cosx -2
1 -2x
=>
I
o
2 - cosx
4
36
35
3
3
x + — log (9e2x - 4) - - Iog3 + C
2
36
2
ss — —
x2-sinx cosx-2
/(y)=(-D3 sinx - x2
4
= || log (9e2x - 4) = j log (e2') + C
[Interchanging rows and columns]
=>
1 dz
18 z + 4
9
2z + 8
+ 3>dz
J z (z + 4) [ 9
1 _ 1
dz
♦
z z+4
3
(2 3'
— — + — logz — log (z + 4) + C
19 4,
4
l-2x
0
0
2x —1
z+4
------ + 6
9
= 1 f 2z + 35 dz=l f2(z + 4) + 27
dz
9 J z (z + 4)
9 * z (z + 4)
= (2x + 1) cosx-2 sinx + C
0
20. We have, /(x) = sin x - x2
0
1 -2x
2x-l
0
25. J tan3x(sec2x- 1) dx = j tan33 xsec22 x dx - J tan’x dx
A = j tan3 xsec2xdx =
[Taking (-1) common from each column]
fM = -f(x)'
2
_ tan x
- In | sec x| + C2
2
J/(x)dx = 0
dx
21 J (x + l)(x-2)
tan4 x
+q
4
I2 = J tanx (sec2 x - 1) dx
/(x) = 0
=>
73
1
1
--------- +---------- dx
3(x +1) 3(x - 2)
26. The Statement II is false since while writing
f dx
, .
, _
J---- — = log (x-3y) + C,
J x-3y
= -1 log (x + 1) + | Iog(x - 2) + C
we are assuming that y is a constant. We will know prove the
A = --,B = 3
3
Statement I. From the given relation (x - y)2 = — and
y
j
2 log (x - y) = log x-logy.
x+y
Also, —
---------To prove the integral relation it is
dx
x-3y
A + B = 0 => - = —3-'=-i
B
1
/
22.
3
•
1
dx
X2 + 4
1
1
-1
=> - tan x — tan — = k tan x + I tan
3
6
2
1
kx 2 + 1
sufficient to show that — RHS = —- ---dx
x-3y
x
2
Now,
k = - and I = —
3
6
„22
x2
1
•2x---- dx
2
1 + x2
x3
=—log (i + x2y-J-~2---- fa
X2 + 1
23. I = j x log (1 + x2) dx = log (1 + x2)------ J
1
°
x2
/=
(tfx) =
24. I = j 4e2 + 6~ dx
■
fx2 + f log (1 + X2)- x2 + 1
I 2 >
x2 +1
2
, Y(x) = -
2
+C
_ 1 [log*-logy -logy = |[logx-3 logy]
2
2
4
1
— RHS =
dx
4
'1 + x2>
.
2
4e2x + 6
dx, put 9e2x - 4 = z => 18e2xdx = dz
9e2x - 4
yj
[log(x-y) - logy]
=
2 ,
9ex - 4e‘
'■J
1
RHS =-log --1
2
[y
1
X
3 dy
y dx
1 1
4 x
/
x+y
1
x-3yj x-3y
Thus, Statement I is true. Hence, choice (c) is correct
27. Let g(x) = f(x) + /(- x)
Assuming,
J /(x) dx = F(x) + C
74
Textbook of Integral Calculus
j g (x) dx = J {/(x) + /(- x)} dx
u
I = f— du= fa14,du = lu15
=>
J u
= j/(x)dx +j/(-x)dx
Similarly, integral of an odd function is not always an even
function. ,
Hence, Statement I is true and Statement II is false.
28. If a > 0 and b2 - 4ac < 0, then
2
dx
ax2 + bx + c
=>
where k2 =
=J-a
We can use case IH
=> Putting, 7~ x2 + 3x - 2 = u (x - 2)
(x -1) u or u (1 - x)
dP
4a
2
k/yja
f
X-----X,
2_
x.2
. —*
.2 .
x+7
2
2J
= 5 sin4 x cos2 x - 8sin6 x cos2 x
(
1
1 i
+---- log
2 2
2
K
■P = 5^4.2 “8/6>2
dx
I.
I “2
n
2 V2
\
>
1
X+--1
X
X+- + 1
X
J
36. Let P = sin5 xcos3x
+C
dP
. 4
.
6
,
— =5 sin xcos x-3sin xcos x
dx
= 5 sin4 xcos4 x - 3 sin4 x (1 - cos2 x) cos2 x
.’. Statement I is false.
30. Since,
cot
= 8sin4 x cos4 x - 3sin4 x cos2 x
k
x =---- tan
2
X,
tan'
d(cot-,x) = -j2',tan
d(tan-1 x)
“3^4,2
P
= 1(-P + 8J,.,)
d(cot-1 x) = - d (tan-1 x)
r
2
6
X.
X-----X
I,11
=---- 7= tan
4
. .
dP
— = 5 sin x cos x - 3 sin x cos x
dx
= 5 sin4 x (1 - sin2 x) cos2 x - 3sin6 x cos2 x
dx-- f
I +2
2
l(-p+2
■<
2
4
35. Let P = sin5 xcos3 x
1 j
1+~
\___ x.2 .J
2
P = 3^2,2
x+ A
+C
B
Thus, choice (a) is correct
=> / = M
2J
2
4
= 3sin xcos x-6sin xcos x
- > 0. which will have an answer of the type
a k/-Ja
29. / =j dx
x4 + 1
4
= 3 sin2 x (1 - sin2 x) cos2 x - 3sin4 x cos2 x
4ac - b,2
-------tantan
1--------4^- + C or g tan'
Thus, j 2
2
— = 3 sin x cos x - 3 sin x cos x
dx
b_
I + k2
2a
b_
-i X + 2a
4J
p=-2<0
Also, - x2 + 3x - 2 = - (x -1) (x - 2)
34. Let P = sin3 x cos3 x
2
1
1
m = -l < 0
or
4ac - b2
b_
I +
2a
4a
dx
ax2 + bx + c = a
15
33. Here,
which may be an odd function, if C + C = 0.
15
= —(x+ 71 + X2)15 + C
= F(x) + C+{- Ft-xJ + C'}
= F(x) - F(-x) + C + C
J
f(x+h)-f(x) [ /(xx-h)-f(x)
/>-» o
h
-h
37. Here, 2/* (x) = lim
2tan”1 X
In 2
-+c
= lim
h-t 0
:. Statement I is false. Statement II is true.
31. As m = 9 > 0, hence, we can substitute
:. 2f(0) Um
' f(x + h) - f(x - h)\
h
<
-h
h
0
•(0
J
j
7?x2 + 4x + 6 = u ± 3x
h-»0
32. Here, as per notations given, we can substitute
71 + x2 = (u-x)
As m = 1 > 0 and p = 1 > 0
h
,(ii)
Now by given relation, we have
and /(0) = i
/(/i) -/(-h) =
-h
Chap 01 Indefinite Integral
K
3rt ---- ------ .
42. (A) If— <x<—, then
sin x > cos x
8 ’
4
(x)
From Eqs. (i) and (ii), we have ------ = 2010
fix)
=>
/(x) = e201Ot,/(0) = l
r sinx - cosx .
r ,
„
---------------- dx= ldx = x+C
J | sinx - cosx|
J
:.
{/(x)} is non-periodic.
1
= -[3x.2‘
(B) J (x3 + x1)2dx
3
J
x
3
+1
(x3 + 2)
38. Here, j f(g(x)) cos x dx = j /(log (sin x)) .cos x dx
= je20101og(Iinx) cosxdx
= J(sinx)2010 . cos x dx
(C)Jsin xcos
(sinx)2011 + C
xdx = j —sin ‘x-(sin-1x)2 dx
~ (x sin-1 x + -jl- x2) - {x (sin-1 x)2
2011
(sinx)2011
2011
1
2011
x2 - x} + C (by parts)
+ sin’
=> sin x — x-xsm x-2>/l-x2|t y Ji -x2 +2x+C
2
.*. /“’(x) = sin-1 x, /(x) = sinx
Sol. (Q.Nos. 39 to 41)
Here, /(x) = (x -1) (x + 2) (x - 3) (x - 6) -100
J xln|x|
= (x2 - 4x + 3) (x2 - 4x -12) -100
= ln|ln|x|| + C
/(x) = ln|x|
a
= (x2 - 4x)2 - 9 (x2 - 4x) -136
( 22
43.
= (x2-4x + 8) (x2-4x-17)
X
• cosec2 x dx
(A)I = J k X 1++COS
x2
39. :. f(x) = Q=* (x2 - 4x + 8) (x2 - 4x + 17 = 0
=J
D<0
D>0
\2 + 1 -sin2x'
• cosec2x dx
1 + x2
Equation has two distinct and two imaginary roots.
= {((x-2)2-21} {(x-2)2+ 4}
=>
V (/(x)U=(-21)(4) = -84
_________ gjx)_________
(x2 - 4x -17) (x2 - 4x + 8)
Ax + B
--------- +
x2 -4x-17
2
2x = t - t
Cx+D
x2 - 4x + 8
l' z
=>
[given]
g(-2) = jD(4 + 8 -17) = -10
J /(»)
2
x2 - 4x + 8
,
1 t3/2
1
2
2
;=FT+ _11112 + k
x-2
+ C = tan'
2
x-2
2
1
3
x2 + 2).3/2
....
2-+k
^x + -^x.22 + 2
and B = 2
A=1
(c)i=J x22^2- x- -x x-2 x=.2: dx
dx
2
dx=2j
J x2 - 4x + 8
(x-2)!+(2)!
= 2 • - tan'
2
2 )
i
=> 2 dx = 11 + — I dt
\ t 7
=~ -2 J(fkVt + 4? J dt
D
gfx)=
.2
x2 - 4x + 8
(x2 - 4x -17) (x2 - 4x + 8)
g(x)_
2(x2-4x —17)
/(x) (x2 - 4x -17) (x2 - 4x + 8)
^x2 + 2 - x = -2
t
Put yjx2 + 2 + X-t =»
g(x) = D(x2 - 4x -17)
=>
dx=-cotx+cot lx+k
A = 1, B = -1
Clearly, A, B and C must be zero.
A
1 + x 2,
(B) I = J Jx + /x2 + 2 dx
which occurs at x = 2
f(x)
)
1
cosec,2 x —
40. f(x) = (x2 - 4x - 17) (x2 - 4x + 8)
41. v
1
dx
x3 + 2
x3 + 1
= -ln
+C
3
x3 + 2
= J (sinx)2010 . cos x dx
/Kx) =
75
+c
-J ^2 -dxx - x: -J- ^2-dxx- x‘
.2
.2
dx
x- X2
76
Textbook of Integral Calculus
+2J-
-f
t
46. u(x) = 7v(x) => i/(x) = 7V*(x)
-dt
t
dt
1
1
?
t ■?
p=7
= - sin'
= -sin
3
2 J ^2t2 - t-1
2x4-1
1 -J^t2 -t-1
T
2
3
= - sin'
g=0
p4-q_74-0^i
Now,
2 fit 2 - t-1
p-q
47. Let t = In
dx
x2 -1
(4 - x) 4- ^2 - X - x.2:
4x
48. Z=Jex
-> fI2x+3 1 + K
1
(1 - x) t/1 - x.2
.
,
I = ex
2
2 tanx-sec x ,
2
------------------ dx, put tan x = t
tan4 x 4-1
=> - cot(tan2 x) 4- C
dt
= - cot-1(t) + C
t2 4-1
B = -1
=>
/(x) 4- g(x) = ex 4- sin x .
a = 1, b = 3, c = 1 => a 4- b 4- c = 5
/(x) 4- g(x)
3x4-2
ex 4- sin x
dx.
=1
1
x+3
1
50. 12 -tan 1------ + — In x-9 = 3 tan'
4
4
2-6
X4-3
x-9 = t2 =>dx = 2t dt
3(t2 4- 9) 4- 2
F(x) = 2[29jx-9 4- (x - 9)3/2] 4- C
X 4-p = 4
51.
3
«
15
2 cos — x cos - x
2
2
1-2 I 2 cos2 —-1
=>
£(10) = 60 = 2(29 4-1] 4- C
C=0
F(x) = 2[29yJx-9 4- (x -9)3/2]
£(13) =2(29x2 4- 4X2]
= 4 x 33 = 132
Hence, sum of digits = 14-34-2=6
X4-3A , lx-9|
------ 4- In-----4 J
X4-3
X=3,p = l
■2tldt
t
= 2 J(29 4- 3t2) dt = 2[29t 4-t3]
Given,
I 4-C
f(x) = ex 4- sin x 4- x and g(x) = - x
dt
1
=C-yly=Cf
J t(t 4- 2) 4-1 =f (t + 1)2
x2 4- 3x 4- 1
F(x) =
1-X
1/2
= In (ex 4- sin x 4- x) - x 4- C
Put x2 4- 3x = t => (2x 4- 3) dx = dt
Let
1 4- X
r (ex 4- cos x 4-1) - (ex 4- sin x 4- x)
dx
J
ex 4- sin x 4- x
x
j.
4-'x-9
1 4- X '
, . ,
------- 4-C => I = ex
VI -X
(
33
2(|1 4- X) = 2 1 4- - = 2 X - = 3
k
2J
2
=>
44. [------------------------------------ dx
J (x2 4- 3x) (x2 4- 3x 4- 2) 4-1
45. F(x) = |
4-C
6A = — => 24A = 1
4
_____ \
1
t 1 4- X
dx
1 -X
(1 - x) 71-x2
/
As
s*n2y
fa, dividing N' and Lf by cos4 X
(D) J —
sin x 4- cos x
=> 2 tanx-sec2x dx = dt= |
t dt = - t2 4- C
4
2
-sin'
'=J
x+ 1
4
— + 2J2 08
X
2
/=! In x-1
1
fit r~5----------t-- 4-J2t -t-1 4-K
+—
2^2 log
0/0
° I
A
4J *
1
dt
J 2
3
1
x-1
t 1
1
1
———+———
2 2
16 16
^2 - x - x.2:
■^2 - x - x2;
7-0
dt
+ 2^
2x4-1
v(x)J
v(x)
+ 1 r___ dt
_1 f (4t-l)dt
4x) Y = 0
«(x)
Again,
1
put X = 2x4-1
(given)
I
2
.
n(
3 5x „
5x
3x
2 4 cos------ 3 cos — cos—
k
2
_2^
2
5x
3-4 cos2
2
»
5x
3x
= -2 cos — cos —
2
2
= -(cos 4x4- cos x)
sin 4x
4
- sin x 4- C
77
Chap 01 Indefinite Integral
sin x
52. I = j
tan x
dx = j
2
1 + tan x + tan x
= x2eI’ + ?_1 + C = /i(x) + C
COS X
sin x
—+„2
cos" X <COS X
dx
r sin 2x dx = fdx-2 j
J 2 + sin 2x
2 + sin 2x
1
dx
,x’ + x2-l
A(x) = x2-e
=>
/Kl)./<-!) = ?-e- = 1
56. I=J ^(x sin x + cos x)
sec2x
,
= X-2j -------- 2------------------- dX
2 sec x + 2 tan x
x4 cos3 x - x sin x + cos x
r dx
x2 cos2 x
= j ^xtinx+cwx).^ cosx)dx
_ f e(xrinx+cMx).£[
Let t = tan x, dt = sec2 x dx
r
—J
2 tan x + 1
1
I = x —tan
V3
dt
'3
^(x sin x + co» x) .
=>
x=
~7
7/2
p2-2
or dx
2p
2p2
= 1(x+7?72)3/2 -2
11/2
2 . i
.W2_£fU/2 + C=2sin7/2
x----- sm J1/2x + C
11
11
7
1
- — sin2 x + C
J 11
A=7,B=7,C = 11
58. I = J
__
dx________
(ax+b) ^(x-a)2-p2
Put (x - a) = P sec 0 => dx = P sec 0 tan 0 d0
f________ d9________
I
J a (a cos 0 + p) + b cos 0
-f
54. Letl=j(;■x2010 + x804 + x.<02)(2x16O8 + 5xW2 + 10)|’1/402dx
4O2O(x200’ + x803 + X401) dx = dt
^1/420+1
I
dt = —
4020 1/402 + 1
J 4020
•J
I
( aa + b
oft
aa + b
—(i)
__ 1
• cosec a1 log
Then, I =
(aa + b)
a
cot — +1
2
+
fl
I
j;
cot----- 1
.
2
aB
e
1
v
where —-— = cos a , tan - = t
2
aa + b
k403/402
1
f
4020 403/402
= J_(2x2010 + 5X804 + ^402)403/402
4030
Again, if
4*
aa + b
a
/=^__ cot a1 tan 1 t tan—
2
aa + b
:. a-400 =3
55. Lete■*; 3 + x* -1(3x4 + 2x3 + 2x) dx
= Jx2-e*’ + x2-
1
dQ_____
if
=t
1r
de
—
—1—
(“ + f,) cose + f flft
_ f(x2009 + x803 + x401)-(2 x2010 + 5X804 + lO402)1740^
Put 2x2010 + 5x804 + 10*
where sec a1 =
,xs + x -l.(2x)dx
3,
i
n
Applying by parts in first internal, we get
I = x/ + ? - *-J 2x e? + l2-1dx + J exi-xi~X2x)dx
dp
■* (aa + b) cos 0 + aP
+ x508 + x:4O1)-(2x1608 + 5 X402 + 10)(’,/4O2dx
=>
1
■jx+y/x.22 + 2
(A + B) - C = (7 + 7) -11 =3
.402
?:
(p2 + 2) dp
2p2
.11/2
= 2 sin7/2 x
=>
1
Pm (P2 + 2) dp=l1 j^df,+\p-^
= [t5,2dt-[t9,2dt = -——+ c
J
1
Let x + yjx2 + 2 = p or x2 + 2 = p2 + x2 - 2px
= | t2t,/2(l -12) dt = j [ts/2 -t9,2]dt
J
dx
-J Jx +Jx•2: + 2 dx
= j sin22x sin1/2 x (1 - sin2 x) cos xdx
■
X
x----------- + C
+C
53. jsin'i512 x cos3 x dx = J sin22.x sin1/2 x cos2 x cos x dx
.7/2
dxl X cos X
X COS X
57. I
1
----- --r
J
dx
= X-~!
2 J t2 + t + 1
i
59. f=J
e
aft
=> tan — = t
aa + b
2
■i
(•
J
X
-dx= j x
Vx2
2
1
1
m = — ,n = - ,p = 3
3
2
■
j
78 . Textbook of Integral Calculus
:.
m+1
------- = 1 le, integer
n
Let us make the substitution,
lx-“ dx = 2tdt
l + x,,’ = t2
3
I = 6 J t2 dt = 2t3 + C = 2 (1 + x1/3)?/2 + C
Hence,
60. I=j
u-V3
2 ,
't2 + 3 -^3
= "*?3 °g
? + 3 + -Ji
2 ,
sin3 (0/2)
—
.
-
■ au
cos 0/2 7cos
I
3 0 + cos2 0 + cos 0
2 sin 0 / 2 • cos 0 / 2 • 2 sin2 0 / 2
2 cos2 0/2 -Jcos3 0 + cos2 0 + cos 0
sin 0 (1 - cos 0)
(1 + cos 0) -Jcos3 0 + cos2 0 + cos 0
2 ,
= --log
dQ
a-i)
Jx2 + 1 / Xz + 1 -y/3
7cos0 + sec 0 + 1 - ^3
+C
yjcos 0 +~sec 0 + 1 + 73
62. Let
^-'(a + b^dx
dO
cos 0 = t=^-sin0d0 = dt
2 J a+1) /t3 + t2 + t
+c
="3 og •Jx2 + 1 / x2 + 1 + -J3 + C
„
-
■
4J
Put
2 ,
= "^3 °g u + V3 + c
(a+ b^ydx
and
P = xA + ,(a + bx")g + 1
Let
dt
where X and p are the smaller indices of x and (a + bx").
Here, X = m - n -1, |1 = p
_____
4J a+1)2 7^
4J
.2
p = xm~n (a + bxn)p + 1
dt
Differentiating w.r.t. x, we have
(i-i/t2)
dt
(t + | + 2) Jt + 1 + |
dx
= xm~n (p + l)(a + bx?)p (n bx"-1)
+ (a + bx/’)p+1 (m - n) xm-"“1
Put
t + 1 + - = u2
t
= nb(p +l)xm"1(a + bxn)p
=> (1 -
dt = 2udu
t2
2u du _ r du
= tan 1 (u) + C
(u
2
+ l)-u J u,2 +- 1
2J
+ (m — ri) x'n~ n"’ (a + bx*)p {a + bx"}
/=iJ
= nb(p + 1) x"1"1 (a + bxJ')p
1/2
= tan-1 t+l + i
t,
= tan"1 (cos 0 + sec 0 + 1)1/2 + C
61. /=J
(2 sin 0 + sin 20) d0
(cos 0-1)7cos 0 + cos2 0 + cos3 0
Put cos 0 = x2
=*
= b(np + m) x/n~i (a + bx^)p + a(m-n) xm~n~1 (a + bxn )pInt
egrating both the sides w.r.t. x, we get
.-. xm~n(a + bxn)p +1 = b (np + m) IM.l+a(m- n)
I
_xw~"(a + bxn)p*1
m 1
b(np
b (np + m)
- sin 0 d0 = 2x dx
(1 4-1 / X2) dx
=4j (1 / x - x) 7(1 / x - x)2 + 3
1
---- X = t
Put
x2
- 11 dx — dt
)
= -4j
a(m-n)
b (np + m)
Hence Proved.
Again, J
x8
(1 - x3)I/3
dx- j x9-1 (1 - x3) 1/3 dx
Here, m=9, b = -l, n = 3, p = -1 / 3, a = 1
.(/(l-x’H
6.
-8
8
X3 (1 - x3)2/3
(i)
3
(herem=6)
h = --------------- + — 12
-5--- 5
dt
tjt2 + 3
Again, put t2 + 3 = u',2
2t dt= 2u du
-udu
+ a (m-n)
p-b(np + m)
or
2) _ 2x dx
= 2f (1+x
(1-x2) 7’x2 + x4 + x6
=>
+ a(m-n) xm~n~' (a + bx/')p + b(m-n) xm-1 (a + b/)p
f
Z = 4j u(u2—3)~~ J
du
u2-3
(l-x3)2/3
2
Hence, /8=- -x6(l - x3)273- — x3(l - x3)273
8
20____
63. Let I = Jcosec2 x In (cos x + jcos 2x) dx
40
(l-x3 )273+C
- Jcosec 2 x - In {sin x(cot x + -Jcot2 x -1)} dx
Chap 01 Indefinite Integral
= jcosec 2 x In (sin x) dx
79
dx
65. I = j
x-yjx2 + 2x+ 4
+ J cosec2 x • In (cot x + -^cot2 x -1) dx
Put Jx2 + 2x+ 4 = t + x
In secortd integral put cot x = 1
x2 + 2x + 4 = t2 + x2 + 2tx
cosec2 x dx = dt
.’.
2x - 2tx = t2 - 4
1 = jcosec2 x-In (sin x) dx - j ln(t + Jt2-1) dt
t2-4 l(t2-4)
x =---------2-2t 2 (1-t)
In first integral (integrating by parts taking cosec2 x as second
integral) and in second integral (integration by parts taking
unity as second function).
dx = --
2[ (1-t)2 J
We have, (In sin x) (- cot x) - j cot x (- cot x) dx
-ln(t + 7^2 -l)t + j
t2 - 2t + 4
t dt
t2 - 2t + 4 .
------------ ;dt
2J -t(l-t)2
4^1
= - cot x (In sin x) - cot x - x -1 In (t + Jt2 -1) + yjt2 -1 + C
3
3
1 f 4
= -| - + (l-t) + (l-t)2. dt
2J t
= - cot x(lnsin x)-cot x-x-cot x {In(cot x + Vcot2x-l)}
+ Vcot2x-1 + C
3
= - 4 log 11|-3 log 11 -t | +
(1-t)
2
= - cot x • In (cos x + yjcos 2x) - cot x - x + -^cot2 x - 1 + C
.
cos* x dx
64. I = J----------—-------- 2 a 6 N = j
,2
(a
+
sin x cos x)2
J (sin x + a sec x)2
J
cos* x dx___________ _
J" (a2 + sin2 x ■ cos
■2 x + 2a sin x • cos x)
= 2 log | yjx2 + 2x + 4 - x | - - log
2
3
| Jx2 + 2x+4-l-x|------- ,
----------2 (Jx2 + 2x + 4 - x -1)
_ p______ 4 cos2 x dx______
e 1 + cos 2x ,
■ = 2 1 --------------- -- dx
J (2a + sin 2x)
J 4a2 + sin2 2x + 4a sin 2x
1
dx + 2 j
(2a + sin 2x)
= 21.-
cos 2x
dx
(2a + sin 2x)2
1
66. Z = j
(0
(2a + sin 2x)
dx
1 + Jx2 + 2x + 2
yjx2 + 2x + 2 = t - x, squaring both the sides, we get
dx
x2 + 2x + 2 = t2 + x2 - 2tx
(2a + sin 2x)2
2x + 2tx = t2 - 2
we know
*
t2-2
X~2(l + t)
du__________
Ii
(4a2-1)
74a2 -1 71 - u2
(2a - u)
.
t2 + 2t + 2 ,
dx =----------- — dt
2 (1 + t)2
=>
(2a ~ u)
n sin 2x + 1
1
du
u = 2a---------------= —=------ .3/2
: J
■Jl-u'2
2a + sin 2x (4a2 -1)
1 + yjx2 + 2x + 2 = 1 + t -
du _ (4a2 -1) cos 2x
dx
=>
2au -1
1
[2a sin-1 u + 71~ «2] = A and sin 2x =
2a - u
(4a2 -I)3'2
1
2a sin”1
(4a2 —1)3/2
t2 + 4t + 4
2(1 + 0
2(1 + t)
2a sin 2x + 1
(t2 + 4t + 4) • 2-(1 + t)I2‘
2a sin 2x + 1
2a + sin 2x
,
r (t2 + 2t + 2)
J (1 + t) (t + 2)2
Using partial fractions, we get
dt
I t +1
J (t + 2)2
/=(—-2f
2a + sin 2x
= ln| t + 11 +
+ . 11-
t2-2
2 (1 + t) (t2 + 2t + 2)
(2a + sin 2x)2
I=
Q
+ 2x + 4 - x | - - log
2
11 - a/x2 + 2x+ 4 + x | +--------- ,
3
■ -----2 (1 - Jx2 + 2x + 4 + x)
.2
------------------------------ 9 1
i ■ ii■
= 2 log |
------- -------- + C
(2a + sin 2x)
-1— + C
(t + 2)
y—' + C
x2 + 2x + 2
t
80
Textbook of Integral Calculus
67.
Ih 4^=)
x6 + 1
y
£'
(x2 + I)2 -2x2
dx
J (x2 + l)(x4 -x2 + 1)
x
71 + *2 >
x2 dx
(x3)2 + 1
(• (1 + 1 / x2) dx
-2J
x2 dx
(1 + 1 / x2) dx
2
J
I ■/ (x -1 / x)2 + 1 ‘ (x3)2 + 1
” ■* (x2 + 1 / x2 - 1)
a
(2ax + 2b)
H. + 1 =7- jx'1
2a J
■Jax2 + 2bx + c
9
= tan-1 (x - 1 / x) - - tan-1 (x3) + C
i
68. Z=J- dx
J (l -x3)1/3
dt
dt ’
~ f
'H t2 (1 -1 /1.3x1/3
3)1
t(t3-l)1/3
u2 du
=-J (1 + u3) • u
4J
_1 r du
"3 J u + 1
r
u du
J (1 + u) (1 - u + u2)
=> (n + 1) a Un + 1 + (2n + 1) bq, + nc- g,_, = x" y
Now, putting, n = 0 in both the sides, we get
aul=y-bu1)
= | log I u + 11 - i1 log I u2 - U + 11 - i1
A
6 6
Z 2
1
45/2
2u -1
tan-1
43
2a u2 + 3bui + cuq = xy
y-bu<>
+ cuo = xy
2au2 + 3b
a
+C
43
=> 2a2u2 + 3by - 3b2Uo + acuo = axy
71. Plan Integration by Substitution
i.e.
Z=|/{g(x)}Y(x)dx
Put
g(x)=t =>/(x)dx = dt
7=J/(t)dt
(1-x3)2'3 -d-x3),/3 + x2
—7= tan-1 -
x2
2(1 -x3)1/3 -x
4Tx
Description of Situation Generally, students gets confused
after substitution, i.e. sec x+ tanx=t.
+C
Now, for sec x, we should use
*
f yi + x.2)*5
2)1
•jl + X2
Put
r.
$
J-dx
(x + 71 + X2) = t
[from Eq. (ii))
=> 2a2u2 = y (ax - 3b) + (3b2 - ac) Uq
43
o
X
-(ii)
Putting n = 1 in Eq. (i), we get
1
1
.3
.1/3
+ 1|
= -log|(P-l)
1,3 + l|-ilog|(f.33-l).2/3 _{f3
3
6
2 (t3 - 1)1/3 -1
+C
—tan-1 ■
-Ifog
-0)
aUf + bu$ = x°y
du
1 r
6-1 ^^-du-lj
u2 - u + 1
2 J (u-1/2)2 +3/4
43
x71-1 (ax2 + 2bx + c) ,
b
dx--u„
-Jax2 + 2bx + c
a
aun + l=xp y-n[aun + i+2bun + cu„_1]-bun
lr u+1
,
—---------- du
3 J u -u+ 1
1 (• l/2(2u-1) + 3/2
du
3J
(u2 - u + 1)
(1 - x3)1'3 + x
n
a^ + 1 =x^y-n j
(using partial fractions)
1 r du
-3J u + 1
b
a
= - xn y - — f x"-1 ■ Jax2 +.2bx + c dx- — u„
a
aJ
a
Again, put t3 - 1 = u3 => 3t2 dt = 3u2 du
du
1+u
b r
----- dx
a -Jax2 + 2bx + c
Un+! = — [x" ■ 2-Jax2 + 2bx + c
2a
- J nx"-1 • 2 -J(ax2 + bx + c) dx] - - 14
J
a
Put x = l/t, dx = -1/12 dt
“ i
I15
—+ C
x" + 1
dx
■Jax2 + 2bx + c
(2ax + 2b) dx
yjax2 + 2bx + c
1 r
2«
2
= tan-1 (t) - - tan-1 (u) + C
69. Z=J
15
__1_ f x* (2ax + 2b) - 2bxn
dx
2«
.Jax2 + 2bx + c
2 f du
= Jf—
t! + l 3 J u2 + 1
= 310g
15
y
4
y
tdx
= = dt
-Jl + x,2‘
dr.J
70. vh, + i=J
In first integral put x -1 / x = t and in second integral put
x3 «= u
i
dx = dt or
sec2x-tan2x = 1
=> (secx-tanx)(secx+tanx) = 1
1
secx-tanx = t
Chap 01 Indefinite Integral
=>
sec2 dx
(sec x+tan x)9/2
Here,
81
(- 2x"3 - 5x“ 6) dx = dt
=>
(2x“3 + 5x-‘)dx = -dt
Put secx+tanx = t
=>
(sec x tan x + sec2 x) dx = dt
sec x • t dx = dt
dt
j
sec xdx — —
t
1
1
secx-tanx = - => secx = -| t + -1
2
t
t.
-3 + 1
74.
sec x- sec xdx
(sec x+tan x)9/2
1
1Y dt
t+- |2 __ ty t
=>
£.9/2
1
dx
x
dx
x2(x4 + 1)<
I dt
X5I
1+^‘
X
2
2
2 7t7/z + llt11/2 ■+K
1
f —Pdt
,
r .
1 1A 4
f
/ = J—r- = -| dt = -t + C = - 1 + —
J t3
J
I x44 J
+_______ 1_______
______ . 1
+K
7(secx+tanx)7/z ll(secx+tanx)11/2
1 I
1. + x - e
xj
75.
-1
[ 1
1 .
x2
— + - (sec x + tan x) + K
(sec x + tan x)11/2 ] 11 7
_1_
1
Xdx= je
Xdx +
1
---- -dx
,2x + e4x
I
Xdx + xe
= fe
(e3x-ex)
|e
x )
-
d
=fe
••• 7-1-J 1 + u2 + u
u,
Put u + - = t => 1
u
■du
1+
1
u,2
X-Je* Xdx
+ u2
76. Given, j f(x) dx = y(x)
-1
|du = dt
u )
+C
2
I = J x5/(x3)dx
Put
x3=t
=>
x dx = —
3
2,
dt
(i)
dtt
[Integration by parts]
.9
2x12 + 5x9
dx = j
dx
1 (x5 + x3 + l)3
x15 (1 + x~2 + x-5)3
(1 + X-2 + X-5)3
Let
7-(0<<4 dt
u2 - u + 1
1,
.
= -kjg
+ C = -1 log
2 ,
,
2x~3 + 5x~6
i
x+-
x -/«•• x dx
1
1,
t-1
=fJ—
t!-1 = -log t +1 + C
>
i
X
x+= xe■ X+C
du
73. Let I = J•
Xdx+xe
Xdx
x+-
-
£
Xdx = e
(u2-l)
2
x+1
J dx
ex = u => exdx = du
= J u+ -1
Xdx
X-J—(x)e
••• J-I=j + e2x + e4x dx
Put
+c
i
x| 1
Xdx + xe
e3x
ex
—dx and J = f
e4x + e2x + 1
J
72. Since,
xs
Put 1 + -i- = t4 => —^dx = 4t3dt
X
x
^=-t3dt
£.13/2 I
^9/2
2
I
+c=±
+c=^4_
+C
2t2
2( + 3 + 1)2
= |[tY(t)-J Y(t)dt]
= | [x3y(x3)- 3 j x^x3) dx] + C [from Eq. (i)]
dx
= | x3y(x3) -1 x2y(x3) dx + C
Now, put l + x”2 + x“5 = t
82
Textbook of Integral Calculus
_ jr (sin x - 2 cos x) + 2 (cos x + 2 sin x)
**
(sin x - 2 cos x)
5 tan x ,
----------- dx
tan x-2
To find The value of a, if
r 5 tan x
dx = x + a log | sin x - 2 cos x | 4- fc...(i)
J tan x-2
sin x - 2 cos x
sin x - 2 cos x
=> 1= jldx + 2 j
.(czYZrinz)^
J (sin x-2 cos x)
d (sin x - 2 cos x)
(sin x - 2 cos x)
5 tan x ,
Now, let us assume that I = j ----------dx
tan x-2
=> I = x + 2 log | (sin x - 2 cos x) | + k
Multiplying by cos x in numerator and denominator, we get
where, k is the constant of integration.
/=f
5sinx
dx
J sin x - 2 cos x
This special integration requires special substitution of type
= X(ZX) + B I —
l^xJ
=> Let 5 sin x = A (sin x - 2 cos x) + B (cos x + 2 sin x)
=> 0 cos x + 5 sin x = (A + 2B)sin x+ (B -2A) cos x
.(ii)
Now, by comparing the value of I in Eqs. (i) and (ii), we get
a=2.
sinx
78. Let I = 42 f
dx
• (
n
sin x---\
4.
Putx---- = t
4
=> dx = dt
Comparing the coefficients of sin x and cos x, we get
A + 2B=5 and B-2A = Q
n +• 1'1 dt
sin | —
4
)
sin t
Solving the above two equations in A and B, we get
r- r 1
1
,
= V2 —f= cot t + —f= dt
A = 1 and B = 2
=> 5 sin x = (sin x - 2 cos x) + 2 (cos x + 2 sin x)
5 sin x
,
=>/ = (----------------- dx
J d
sin x - 2 cos x
JLv2
42
= 1 + log | sin 11 + C
n
= x + log sin x---- + C
4,
!
CHAPTER
Definite Integral
Learning Part
Session 1
• Integration Basics
• Geometrical Interpretation of Definite Integral
• Evaluation of Definite Integrals by Substitution
■ Session 2
• Properties of Definite Integral
Session 3
• Applications of Piecewise Function Property
Session 4
• Applications of Even-Odd Property and Half the Integral Limit Property
Session 5
• Applications of Periodic Functions and Newton-Leibnitz's Formula
Session 6
• Integration as Limit of a Sum
• Applications of Inequality
• Gamma Function
• Beta Function
• Walli's Formula
Practice Part
• JEE Type Examples
• Chapter Exercise
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Session *1
•bF
Integration Basics, Geometrical Interpretation
of Definite Integral, Evaluation of Definite
Integrals by Substitution
Integration Basics
1 rn/2
= —Jq
What is Definite Integral ?
Let f be a function of x defined in the closed interval [a, b]
and 0 be another function, such that 0' (x) = /(x) for all x
in the domain of f, then
n
(1-2 cos 2x + cos 2x) dx
= —1
4
1-2 cos 2x +
_2
( 3-4 cos 2x + cos 4x dx
~ 4 Jo
1 + cos 4x
2
2
£ ft*) dx =[0(x) + c]* = 0(b) - 0(a)
71/ 2
ir
4 . „
sin 4x
= — 3x---- sm 2x +--------8
2
4
is called the definite integral of the function f(x) over the
interval [a, b], a and b are called the limits of integration, a
being the lower limit and b be the upper limit.
0
. o I «
2 (3it------n2 sin. it + —1 sm
2tt - 0
8 \ 2
4
J
S.i „ d 371
1 f 3n
------ 0 + 0 =—
8 2
.16
Remark
In definite integrals constant of integration is never present
1 'd
Working Rules
I Example 2 The value of j1 —^tan
-Ll dx
o
To evaluate definite integral j*/(x) dx.
dx
(b) 7t / 4
(a)n/2
(c) -n/2
2 dx is
X
(d) None of these
1. First evaluate the indefinite integral j/(x) dx and
Sol. Let I = f — tan'1—] dx
XJ
suppose the result is g(x).
2. Next find g(b) and g(a).
3. Finally, the value of the definite integral is obtained
by subtracting g(a) from g(b).
d. .-1 X
d (
.
-1
Here, — tan -’1 = —(cot x) =
dxl
X,
dx
1 + x2
I
Thus,
f(x) dx = [ g(x)]„ = g(b) - g(a)
(it Tt')
= —| 1
=
<4 4 J
I Example 1 Evaluate
(i) Jof— dx
3+4x
sin4 xdx
n
2
Hence, (c) is the correct answer.
’
1
J
Sol. (i) Here, I = f
, -------- dx =
Jo 3+ 4x
' I" (3 -h 4y) ~|1
4
. o
i (ln7
=A
..ln3] = lin(L)
4 *
(ii) Let I =
1
1 ndx
dx
L -2
J-1 1l + x2‘
= -(tan-1x)^j = ~[tan-1(l)- tan-,(-l)]
Remark
Note that [1 , — tan
2
x
tan
2V : tan-1(1) — tan-1(—1)
xj-1
- IE2E)=_
~4 I 4J"?
sin4 x dx
= - p2 (2 sin2 x)2 dx = - P72 (1 - cos 2x)2 dx
4 Jo
4 J°
is incorrect, because tan-11 — ] is not a anti-derivative (primitive)
vJ
d ftan-1-l^on the interval [-1, 1].
of —
dx V
xJ
Chap 02 Definite Integral
I Example 3 If ln = £ (log x)n dx, then ln+ nln_} is equal to
85
eb
i.e. J /(x) dx = Area (OLA) - Area (AQM) - Area (AfRB)
a
(a) e
(c) e -1
(b) e
+ Area (BSCD)
(d) None of these
Remark
Sol. We have,
=
eb
(logx)'fdx = £(logx)" -Idx
J f(x) dx, represents algebraic sum of areas means that area of
•a
function y=f(x) is asked between a to b.
=s
In = [x• (logx)" K - I*n-(logx)"-1 • -• xdx
J1
x
=
= (e - 0) -
(loge x)n-Idx = e - n ■ In^
••• fn+”-fn-i = e
Hence, (b) is the correct answer.
=> Area bounded =
| f(x) | dx and not been represented
by j* f(x) dx. e.g. If someone asks for the area of y = x 3
between -1 to 1, then y = x3 could be plotted as
I Example 4 All the values of'd for which
ji2{o2 + (4-4o)x+4x3}dx<12 are given by
(a) a=3
(b)o<4
(c) 0 <o <3
(d) None of these
Sol. We have,
{a2 + (4 - 4a)x + 4x3 }dx < 12
=>
[azx + (2-2a)x2+x4]2< 12
f
■1
-rfflli
]J]F’ 0
a2 +3(2-2d) + 15 <12
=>
a2 -6a + 9 ^0
-------- ►
1
i
=> a2 (2 -1)+(2 - 2a)(4 -1) + (24 -14) < 12
=>
y = x3
Figure 2.2
r° -x3 dx +
Area = J
x3 dx=2
(a-3)2 <0
a=3
or using above definition, area =/_*, I x3 |cbr=2£* x3dx
Hence, (a) is the correct answer.
x^
=2
Geometrical Interpretation
of Definite Integral
4
1
= —1
0
2
But, if we integrate x3 between -1 to 1.
fl
3
If /(x) > 0 for all x e [a, b], then £ /(x) dx is numerically
=>
equal to the area bounded by the curve y = f(x), the
Thus, students are adviced to make difference between
area and definite integral.
X-axis and the straight lines x = a and x = b i.e.
f(x) dx
J
. _
x dx = 0 which does not represent the area.
.
.
r3..................................
,
In general, j* /(x) dx represents the algebraic sum of the
I Example 5 Evaluate j | (x -1) (x - 2) | dx.
areas of the figures bounded by the curve y = /(x), the
X-axis and the straight lines x = a and x = b. The areas
above X-axis are taken with plus sign and the areas below
X-axis are taken with minus sign,
Sol. Let
x=a
fM
I = J* |(x - l)(x - 2) | dx
We know,
| (x - l)(x -2)| =
(x-l)(x-2), x<lorx>2
-(x-l)(x-2), l<x<2 •
x=b
+
\i
+
+
1
+
4-------
2
Using number line rule,
I
I
I
I
Figure 2.1
/ = J’ |(x-l)(x-2)|dx
86
Textbook of Integral Calculus
= J’ (x-l)(x-2)dx-|i2 (x-l)(x-2)dx
dx
• 2
Sol. Let
2
4 + x2
+ £ (x-l)(x-2)dx
x
•2
Ji
Jo
.
3x2 o T
x‘
+ j* (x2 - 3x + 2) dx
2
x 3 3x2 o
---------- + 2x
2
3
-----------+2x
3
Jo
2
+
Ji
3x2 o
J ------+2x
3
2
^3
3
2
+
2
27
8 12 J 11
------+ 6-- +------ 4 = —
2
3 2
6
3
27
7t
4
1 71
2 4
3
3
-—-+2
3 2
.j
)] =2 tan~1(1)"tan-1H)
= -tan-1
2
= f1 (x2-3x + 2)dx-[2 (x2-3x + 2)dx
i=5
=>
4
On the other hand; if x = 1 /t, then
fl/2
dt
dx
J1/2
t2 (4 + 1/t2)
4 + x2
'-L\
dt
= -fJ-1/2
V2 4t2 +1
1/2
= - - tan 1 (21)
1.2
J-1/2
Evaluation of Definite
Integrals by Substitution
Sometimes, the indefinite integral may need substitution,
say x = 0(t)- Then, in that case don’t forget to change the
limits of integration a and b corresponding to the new
variable t.The substitution x = 0(t) is not valid, if it is not
continuous in the interval [a, b].
I Example 6 Show that
dx
p/2
a2 cos2 x + b2 sin2 x
Sol. Let
a2 cos2 x + b2 sin2 x
sec2 x dx
fx = n/2
Jx = 0
a2 + b2 tan2 x
(divide numerator and denominator by cos22 x)
Put
tan x = t => sec2 x dx = dt
'=f
lt= 0
_1_
1
—
=7
+1
ab
2
5-o
2
I Example 7 Evaluate J_*
4
71
1.
1
r. I =---- , when x = 4
t
In above two results, I = - n / 4 is wrong. Since, the
integrand ——- > 0 and therefore the definite integral of
4+x
this function cannot be negative.
Since, x = 1 /1 is discontinuous at t = 0, then substitution is
not valid.
(v I = 7t / 4)
.
I Example 8 Evaluate j^x- sin-1 X=dx.
TbTx 2
Put sin-1 x = 0, then x = sin0 => dx = cos0d0
We find the new limits of integration t = tan x => t = 0
when x = 0 and t = 00 when x = n/2.
b2 j°
71
8
Jl-x
a2+b2t2
dt
71
8
Sol. Let I = [1/2x. sin” * dx
Jo
2
dt
= co
.
n
Remark
It is important that the substitution must be continuous in the
interval of integration.
dx
1 = f = K'2
J* = 0
= -^-; a,b>0.
2ab
= - -1 tan"1 (1) -1 - - tan”1 (- 1)
2
A 2
1
a/b
_i bt ']*’
tan 1 —
a o
Also when x = 0, then 0=0 and when x = -,
2
i=r.
sin0 •
Jo
7t
2ab
dx
4 + x2
_ 71
~ 6
then 0 = sin
0
■ cos0d0 = J^6 0 -sin0d8
71-sin2 0
= (-0 -cos0)f6 + Jo[”/6cos 0 d0,
using integeration by parts.
directly as well as by
the substitution x = 1/f. Examine as to why the answer
don’t valid?
.
=(-0cos0)?6 +(sin0)^6
71
71
«
■
71
A
= —cos—+ 0 + sin---- 0 =
6
6
6
-Jin 11
12
2
87
Chap 02 Definite Integral
Now, g'(x) = ^ = 71
dy
I Example 9 For any n > 1, evaluate the integral
r—4=-dx.
dt
When y = 0 Le. j
° (x+Vx2+1)"
,4
= 0, then x = 2
1 + t4
Sol. Let/= r
dx
Jo
Therefore, g'(0) = V1 +16 = 717
Hence, (c) is the correct answer.
Put
=> 71 + x2 = t - x
=t
t2-l
1. t —1
2
t
2t
dx = 1 1 + -^-jdt
d.
t2)
2
n o
then lim Y— is equal to
n
1 if
1
1 = I —’*“1 1 + ~
k=l2
t2)
2J>
tl-n
t 1
------ +
2 1-n -(n + l)
1
1
-2n
n
2|_1 —n 2
n2-l
1
I Example 12 Let an =f^/2 (l-sint)n sin 2t,
=> X ------------ or x = -
=>l + x2 =(t-x)2
2
+t
)dt
1
1-n
i
0-
(a) 1/2
(c)4/3
(b) 1
(d) 3/2
Sol. We have, o. - j*22 (l-sint)"sin2I<l/
n+l
Let 1-sint = u =>-cost dt = du
= 2^ un(l-u)du =■ 2^ undu
-fJo un + ,du
“n
=2
I Example 10 The value of
X2 4-2X-1
fe 1
Jo
x2 -2
2
(d)(Ve)'2"2
(C) 0
Jo
2
-* e
x2 - 2
dx + j * x log x • e
=2y
2 dx
n..An
=1
Put x + 1 = t in first integral
= 2(1)-
t2 - 2
x2 - 2
dt + J * x log xe 2
dx
t2 - 2
- + t ■ log t
A
Hence, (d) is the correct answer,
-x
dt
I Example 11 Let/(x) = J2X
2 Vi+F
and g be the
inverse of f. Then, the value of g'(0) is
(a) 1
(b) 17
(c) 717
(d) None of these
n + lj .-An
fx , ,
1 .
Ji
4
(a)7e
(b) e
,
.
,
<2'
•X
Sol. Let I = £ t\ntdt= Int —
2
2
X
h
2 ,
x2.
1 V
ifx i ■rdt
= —Inx —
2
2 2 h
2* t
x2 Inx
2
g
2
(d)e-1
(c)e2
2
/(x)=0
+... = 2-1 = 1
[ tlntdt = - is
dx
X
n+2J
. O(i iH 11
3 J <2 4 J V3 5
x2lnx
__ 1
7i+x4
1
1
1
1 1
n + lj 2^<n n + 2j
I Example 13 The value of x > 1 satisfying the equation
dt= logt-e 2
= (^-2
/'(*) =
1
Hence, (a) is the correct answer.
t
So/. Here,
n + 2y
lim 7^ = 2
(x + 1)
= fe-
n+l
"->0"n = in
x2 + 2x-l
Sol. Let I = T
1 A
1
1
Therefore, — = 2
n(n + l) n(n + 2),
n
----- — dx+ fe xlogx-e 2 dx is equal to
(x+1)
(a)^)”2*" (b)(^)e2-’
1
■7[x2’11=7
4
4
1
— x 2 =0 => [21nx-l] = 0
4
iInx = -1 => x = Ve
r
2
Hence, (a) is the correct answer.
(as x > 1)
88
Textbook of Integral Calculus
I Example 14 If lim
1 f” x2 + ox+1
- ’
1+x*
n. •
I Example 15 If the value of definite integral
x-o’[l°8°X1 dx, where 0 > 1 and [x] denotes the
— ox is
•tan
XJ
7t2
e-1
greatest integer, is —, then the value of ‘o’ equals to
equal to —, where k e N, then k equals to
A
(a) 4
1 + x4
Jo
(d) 32
(c) 16
(b) 8
01 t
r f- xz +ax + l
Sol. Let I =
--------- ----- tan
—
IxJ
(a)Ve
(b) e
Put loga x = t
= In a
xdx
1 + x4
(i-Uldx+af
1 + x4
n 71 + 071 _ 7l2
4L2V2 4 J [8-5/2
7tz + n2a
Z = lim a 8^2 16
=» a‘ = x
I = lna-Jfl (a1 -a-^• at) dt = Ina-J (af^d.af)jf
j _ 7t p (x2 +l) + ax
dx
4-1° 1 + x4
4 Jo
(d)e-1
Sol. Let I = f x-a’^^dx
dx
Put x = 1/t and adding, we get
[using tan”’(l/x)+ cot-1 x = n/2]
_n r°°
(c)
Ina a2t
= ;(“2-D
2 Ina 0 2
it
. 2a
[as {t} = t, if t e (0,1)]
-(a2 -1) = —- => a = Ve
16
= lim
(a^-af)dt = Ina-Jo a2t dt
2
71
2
_2
71
Aliter
n2
16
16
_(8>/2)a
2
x G (1, a)
loga x G (0,1) => [logax] = 0
I = T xdx = -(a2 -1) = —- => a = Ve
2
2
Hence, (a) is the correct answer.
=>
k = 16
Hence, (c) is the correct answer.
Exercise for Session 1
2. f72
2
I. [’"cosW
Jo
o
J.
Jo
f it/ 2 1
Jo
v1+cosxdx
4.
5. f Vx -Vx-1
rf*
•j
9.
Jo
dx
1 + cosx
sm2x-cosxdx
6. Jo logxdx
p* (sinx + cos x)^,
Jo 9 + 16sin2x
8. JJ»7(x-a)(b-x)
6-,-- 1—_ dx.b >a
tb x-a,
J------ dx
Mb -x
10. J”74-7tanxdx
Jo
13. If f(x) is a function satisfying
1
n
— ] + x^(x) = 0 for all non-zero x, then
\x J
nn
14. The value of J ]J(n + r)
0 V=1
(a)n
f(x)dx is equal to
X
1
dx equals to
= ,x+k
1
/
(b)n!
(c)(n+1)!
15. The true set of values of ‘a’ for which the inequality J°(3"2x
(a) [0-1]
3
f
-sinx
12. Jof4e sinx xcos°x
\dx
2
COS^X
11. J*cos2x.log (sinx)dx
(b) [-00, -1]
(d)n-nl
-2-3~* )dx £0 is true, is
(c) [Q 00]
(d)[-o,-1]u[1oo]
Session 2
Properties of Definite Integral
Properties of Definite Integrals
fb
{* b
ja
Ja
Property I. f- f(x)dx =
This property says that when integrating from 0 to a, we
will get the same result whether we use the function /(x)
or f(a - x). The justification for this property will become
clear from the figures below :
f(t)dt
i.e. The integration is independent of the change of
variable.
Proof Let <|)(x) be a primitive of /(x), then
dx
y4
dt
Therefore, f f(x) dx = [<t>(x)J ba = 0(b) - 0(a)
—(i)
and
-(ii)
Ja
Ja
J aa
Property IL jb f(x) dx = -
a
As x progresses from 0 to a, the variable a - x progresses
from a to 0. Thus, whether we use x or a - x, the entire
interval [0, a] is still covered.
From Eqs. (i) and (ii), we have
J* f(x)dx = f* /(t) dt
Ja
0
y+ •
f(x) dx
i.e. if the limits of definite integral are interchanged, then
its value changes by minus sign only.
Proof Let <|)(x) be a primitive of /(x), then
f f(x) dx = 0(b) - 0(a)
Ja
0
~ U/(x) dx=-[0(a) - 0(b)] = 0(b) - 0(a)
and
P /(x)dx = -[a f(x)dx
Ja
jo
Property III. f“f(x) dx =
a
~ x) dx (King’s property)
Proof On RHS put (a - x) = t, so that dx = -dt
Also, when x = 0, then t = a and when x = a, then t = 0
r .f(a -*)<&=-[ ° f(0 dt = f “ f(t) dt = [a y(x) dx
*u
ja
Jo
i/0
f(a-x)
----------- *-x
The function f(a - x) can be obtained from the function
/(x) by first flipping /(x) along the y-axis and then
shifting it right by a units. Notice that in the interval [0, a J
/(x) and f(a - x) describe precisely the same area.
There are two ways to look at the justification of this
property, as described in the figures on the left and right
respectively.
I Example 16 Show that
(i)
/(sinx)dx=|^/2/(cosx)dx
a
J* Kfl ~x)dx= J* /(x) dx
Remark
(ii) j”/2 f(tanx)dx=j*Q/2 /(cotx)dx
(iii) J^/2f(sin2x)sinxdx=j”/2V2/(cos2x)-cosxdx
This property is useful to evaluate a definite integral without first .
’ finding the corresponding indefinite integrals which may be
difficult or sometimes impossible to find.
Geometrically £ f(x) dx =^f (a-x)dx
= J ”/2/(Si n 2x) • cos x dx
(iv) Jo" x/(sinx)dx=y
/(sinx)dx
[1ITJEE 1982]
90
Textbook of Integral Calculus
Sol. (i) We know,
f n/2
r n/2
_
It
------ X
Jo /W *)<*-/<> f sin 2
dx
using J “ f(x) dx = J' f(a - x) dx]
f n/2
= J0
I Example 17 If f and g are continuous functions
satisfying f (x) = f(a - x) and g (x) + g (o - x) = 2, then
show that fo° f(x)g (x)dx = J® /(x)dx.
Sol. Let I = Jo° f(x) g(x) dx = £‘ f(a - x) g(a - x) dx
f(cosx)dx
. =£a /(x)-[2-g(x)]dx
••• O(sin x)dx ~ I02f{cos x) fa
using £a /(x) dx = £‘ /(a - x) dx
71
(ii)ff/(Unx)dx=Jf;[ tan -----X
dx
2
1
ra
fusing | “ f(x) dx =
/(x) = /(a ~ x) and g(a - x) + g(x) = 2
(givep
•• £° /(x) • g(x) dx = 2 £a /(x) dx - £a /(x) • g(x) dr
f(a -x)dx
r n/2
r n/2
,
"J, /(>“ *> * - J o
r n/2
(iii) We know, I = Jq
=»
/(cot x) dx
7t
>
2
J.
• I ---71 x I dx
j
• sin
12
J
a
using f0° f(x) dx = J" /(a - x) dx
f n/2
/(sin (ti - 2x)) cos x dx
=J
r n/2
I = Jo
V /(x) g(x) dx = £“ /(x) dx
/(sin 2x) cos x dx
C) I*"/2
dx
sin X-cos X ,
(iii) £n/4 log(1 + tanx)dx (iv) n/2 -------dx
1+sinxcosx
...(u)
r n/2
,
r n/2
Then, I =
= v2 J
/(sin 2x) sin x + — dx
4>
f” a)
21 =
fi.e. x = — - 0
I
4
/(cos 20) cos 0 d0
= 2^2^ /(cos 20) cos 0 d0 (since it is even)
/(cos 20) cos 0 d0
I = 4ijo
n
I = j x /(sin x) dx
Jsin x
■ —------ ■
(i)
21 = fW/2
^Sin X
0
-^sin x + ^cos x
=>
I=j
n
^sin
...(ii)
log (tan x) dx
Then,
- rtan (n--x
I = Jof «/2 log
21 = £
e n/2
fn/2
= Jo
/. j ” x /(sin x) dx = — j * /(sin x) dx
0
2
...F
(ii) Let I =
- Jo
£ /(sin x) dx
i
i
dx
X “
log (cot x) dx
Adding Eqs. (i) and (ii), we get
Adding Eqs. (i) and (ii), we get
21 = £” 7t /(sin x) dx => 7 =
. Idx
= [x]oK/2=--0 => 2Z = 2
2
f n/2
(tt - x) /(sin x) dx
dl
^sin x + ^cos x
0
I = Jo
~fa
Jo
0
^sin x + ^cos x
^COS X
r n/2
fit/n/2
Replacing x by (7t - x), we get
1 ~ i;
dx +
f n/2 ^/sin X + ^COS X
i
K
dx
^sin x + ^cos x
Adding Eqs. (i) and (ii), we get
/(cos 20)cos0 d0
= V2 j
7cos (7t/2 - x)
dr
^cos (7t/2 - x) + ^sin (ti/2 - x)
e n/2
Jo
7C ) .
Put x + - = - - 0
4 12
)
(iv) Let
Jo
/(sin 2x)(sin x + cos x)dx
r~ f n/2
(ii) J^72 log(tanx)dx
1+^/tanx
» , (i)
,.x .Let7=f
. . r^
_ rn/2
y/COS X
So/,
”722 — dx
dx ...(i)
J°
j +.- ^tan x
7cosx +Vsinx
Adding Eqs. (i) and (ii), we get
27 = £
fa
I Example 18 Evaluate
/(sin 2x) sin x dx
sin 2 ----- x
fa = 2 io
or 2 V
= f^2 /(cot x) dx
log (tan x) dx + £"/ log (cot x) dx
(log tan x + log cot x ) dx
f n/2
log (tan x • cot x) dx =
=> 2/ = 0 => 7 = 0
log (1) dx
I
Chap 02 Definite Integral
(iii) Let I =
log(l + tan x)dx
r n/4 ,
=j
’•
,
...(i)
1 + tanx + l-tanx
1 + tan x
2
J + tan x
= £”'* log (2) dx - Jo
(King’s property)
Also, when x = a, then t = b, and when x = b
•’’V
71/4
Then, I = [
Jo
fb
fb
Geometrically £/(x)dx=ja f(a + b-x)dx
[using Eq. (i)]
7t
...(i)
I
i n
\2
,
■
(n
1
(n
1 + sin---- x cos;---- x
I2
U )
-------X
2____
- cos; — x
1 + sin x cos x
21 = 0 =>
dx +
a+t>-x ;
I x
I ------- !
dx
a
...(ii)
Adding Eqs. (i) and (ii), we get
sin x - cos x
fb
log (1 + tan x) dx
1 + cos x • sin x
Jo'
fb
•••£ /(x)dx=£ f(a + b-x)dx
cos x - sin x ,
1= Jof71/2 -------------------dx
2/ = f n/2
dt^= " V f(a + b~ dt
+b~
= £ f(a + b-t) dt = £ f(a+b-x)dx
n.
n
=> 21 = — log 2 => Z = — log 2
4
8 ~
n/2 sin x - cos x .
(iv) Let
------- .---------- dx
* 1 + sin x cos x
sin
fa = //
dx
dx
=> / = (log2)(x)V-I
n/2
f(a + b - x) dx
Proof Put x = a + b-t =$dx = -dt
tan n/4 - tan x
1 +-------------------------- dx
1 + tan (n / 4) • tan x
/
fb
,
log [1 + tan (n/4 - x))dx
=r ><*
-r 4
rb
Property IV. £ /(x) dx = J
91
e n/2
cos x - sin x
Jo
1 + sin x cos x
b
x
As the variable x varies from a to b, the variable a + b-x
varies from b to a. Thus, whether we use x or a + b - x,
the entire interval [a, b] is covered in both the cases and
the areas will be the same.
dx =>
1=0
I Example 19 The value of £ log (cot a+ tan x) dx,
Sol. Let I = |
a
' cos a sin x
---- 4--------sin a cos x
-r
( cos (a - x) '
----------- ( dx
sin a cos x ,
= £ log [cos (a - x)] dx - J
dx
log (sin a) dx
log(cosx)dx
= £ logcos(x)dx-£alog(sina)dx-Jo log (cos x)dx
using |0V(x) dx = J f(a - x) dx to first integral
dx = - a log (sin a)
Hence, (b) is the correct answer.
+-X
I Example 20 Evaluate ["/3 —- •
•k/6 1 + J tan x
_ . _ . r f*73
dx
fJt/3
Jcosx-dx
Sol. Let I = I
------ == =
- y ----- ==
14.Jta
n
x
Jn/6
-^COSX+ySinX
1 + ^tan
=1
,
= - log (sin a)
b
The graph of f(a + b-x) can be obtained from the graph
of /(x) by first flipping the graph of /(x) along the y-axis
and then shifting it (a + b) units towards the right; the
areas described by /(x) and /(a + b - x) in the interval
[a, ft] are precisely the same.
log (cot a + tan x) dx
= [" log
-------------------- 1
I
(b) -a log (sin a)
(d) None of these
(a) a log (sin a)
(c) - a log (cos a)
a+b-x ;
!_x
where ae(0,n/2) is equal to
_
then, I =
rn/3
-Jcos (n/2 - x)
JlU6
yjcos (n/2 - x) + ^sin (n/2 - x)
(i)
dx
(v a + b = n/2)
1= Jn/6
f71/3
7sin x
yjsin x + Jcos x
dx
•••(ii)
92
Textbook of Integral Calculus
' Integrating both sides, we get
Adding Eqs. (i) and (ii), we get
c nJ3
21 = I
,
7t
It
71
3
6
6
1 dx =-------
Jk/6
I Example 21 Prove that
But
[b____ ZW----- dx = b- a
Jo f(x)+/(o + b-x)
Sol. Let
(0
'=r
/(q + h - x)
...(ii)
f(a + b- x) + /(x)
Aliter /(x)-f z(x)-/(-x)/z(x) = 0
~[/(~x)*/(x)] = 0
dx
=>
f(a + b- x)_____
dx
f(a + b - x) + f(a + b-(a + b- x))
*b
then, I = J
=>
/(x)-/(-x) = 9
f (x) + f(a + b - x)
Ja
/2(0) = C /. C = 9
2
,= r‘---------
■
ln/(x) = -/(-x) + C
ln[/(x)./(-x)] = C
f(x)-f(-x) = C
/(0) = 3
Integrating both sides, we get f(x)-f(-x) = Constant
Hence, (b) is the correct answer.
'51
I Example 24 j
Adding Eqs. (i) and (ii), we get
•b f(a + b- x) + f(x) dx
2I-J’« f(a + b - x) + /(x)
J-513+/(x)
has the value equal to
(t>)34
(d)0
'51
dx
f51
fa
Sol. Let I = J
'-51 3‘+/(x) ~ J-513 + /(-x)
(a) 17
(c) 102
21 = f6 1 dx = (b - a) => I = -—Jo
dx
ft*
2
I Example 22 Solve
using ^f(x)dx =j f(a+.b-x)dx
' - sin4 t
‘cos4 t
= pi
7/(cos2t-z) + 7f(z)
.
f - sin4 I
Sol. We have, I - | .
J f(z) dz
v■
.
1
^//(cos 2t — z) + ylf(z)
*cos' r
Jf(cos 2t - z) dz
Jf(cos2t-zj + Jf(z)
6+y(x)+/(-x)
dx
J-5i[3 + /(x)][3 + /(-x)]
(i)
f51
6 + f (x) +/(-x)
J-51 9+V(x) + /(-x)]+/(x)./(-x)
••(ii)
pi 6 + f(x) + /(-x)
J-5il8 + 3[f(x) + /(-x)]
using J6 f(x) dx = £ f(a + b- x)dx
,
2-51
dx =----=3’f
J-Sl
3
Adding Eqs. (i) and (ii), we get
27 = fr<,dz => 2l = (z)-“4'
•cos4 t
.
cos4 t
I = -i(sin4 t + cos* t)=-|(l-2sin2 t cos21)
2
= — —1 1 - - sin2 2t | = - - + - sin2 2t
2
‘
2' 4
2
Directions (Ex 23-25) Let the function f satisfies
1 = 21 = 17
3
Hence, (a) is the correct answer.
I Example 25 Number of roots of /(x) = 0 in [-2,2] is
(a) 0
(b) 1
(c) 2
(d)4
Sol. Let x = a be the root of /(x) = 0.
f(x)f'(-x) = f(-x) f'(x) for all x and f(0) = 3.
Z(a) = 0
/(x)-/(-x) = 9
I Example 23 The value of f(x)-f(-x) for all x is
(a) 4
(b) 9
(c) 12
(d) 16
Sol. Given, f(x)-/z(-x) = f(-x)-f'(x)
f'(x) f'(-x)
f(x)
/(-X)
.
(impossiEe
Put x = a, then
0=9
Therefore, /(x) has no root but /(0) = 3.
/. f(x) > 0, V x G R as f is continuous possible function
/(x) = 3e“x.
Hence, (a) is the correct answer.
Chap 02 Definite Integral
93
Exercise for Session 2
1. The value of J^4 log (1 + tan 0) d6 is equal to
(c)5 log 2
(b)-log 2
(a)^log2
(d) None of these
o
2. For any integer n, the value of j" e co^x • cos3 (2n + 1) x • dx is equal to
(b)1
(a)0
3. The value of £
Vx
(c)-1
(d) None of these
(c) 1/4
(d) None of these
dx is equal to
^5-x +
(a) 1/2
(b) 1/3
4. The value of [2---------------- —... „.
is equal to
Jo (17 + 8x -4x2)(e6(1 > + 1)
(a)-
1 log 2-J2~1
8VT1
2+ V21
(b)-
(c)-
2+ JTi
1
2-V2~1
log
- log
V21 - 2
8VT1
2+ V21
(d) None of these
2 + J21
1
log
V21-2
8V21
5.
5. Iff
If f is
is an
an odd
odd function,
function, then
then the
the value
value of
of J[a* ------- f(sinx) — dx is equal to
/(cos x) +f(sin2 x)
(b) f (cos x) + f (sin x)
(a) 0
(c) 1
(d) None of these
2
6. If [x] stands for the greatest integar function, then £1°
(a)1
(b)2
2
is
(c)3
(d)4
(c)-#sin a
... sin a
(d)------
7. The value of T------ —-----—(° < a < n) 's
J°1+cosa-sin a
(a)-Asm a
(b)-Asin a
a
8. If f, g, h be continuous functions on [0, a] such that f(a - x) = f(x), g(a - x) = -g(x) and 3h(x) = 5 +4h(a -x),
then the value of j*f(x) g(x)-h(x)dx is
(c)a
(b) 1
(a)0
9. If2f(x)+f(-x) = — sin fx—— then the value of [* f(x)dxis
x
v
xJ
JVe
(a)0
(b)e
(c)1/e
10. Prove that jxf(sinx)dx =
(d)2a
(d)e + 1/e
J7(sinx)dx.
n
)j
x2sin2xsin —cosx dx
11. Evaluate J”
2
(2x-n)
J
‘
.
12. Numberof positive continuous function f(x) defined on [0,1] for which jV(x)dx =1,^xf(x)dx =2andj\2f(x)dx=4.
'1 -----6 dx
J-----13. Let L = f
and1/,', = f1—------- dx. Then — is equal to
1 Joi + X •
2’
°e*3(2~x3)
I2
(a) —
e
14. iff(x) =
(a)1
(b)e
(c)3e
(d)J3e
eX / - r x\ • g{x(1 -x)}dx and l2 = j'(a^g{x^-x)}dx, then the value of p is
(b)-3
(c)-1
(d)2
Session 3
J
Mt.'W'XCW-e
Applications of Piecewise Function Property
Applications of Piecewise
Function Property
Proof Let us consider the function /(x) on [a, b] when
/(a + x) = f(b - x), then / is even symmetric about the
Property V (a), f * /(x) dx = f ‘ f(x) dx + j b f(x) dx,
a
Ja
Jc
c
Ja
mid-point x = - — on the interval [a, b] when
2
/(a + x) = - f(b - x), then / is odd symmetric about the
where c <-> R
mid-point x -
Proof Let 0(x) be primitive of/(x), then
f /(x) dx = 0(b) - 0(a)
Ja
„.(i)
a+b
2
of the interval [a, bj
0, if/ is an odd function
2/(x)dx, if / is an even function
Using [a f(x) dx =
J— a
and
fC f(x)dx + [b f(x)dx = [0(c) - 0(a)] + [0(b) - 0(c)]
Jcc
Ja
if/(a + x) = -/(b-x)
0,
= 0(b)-0(a)
...(ii)
=> £b/(^)dx = ' 2[V /(x) dx, if/(a + x) =/(b - x)
From Eqs. (i) and (ii), we get
Ja
jJ f(x) dx = JJ /(x) dx + £* /(x) dx
a
x2,for0<x<1
c
Generalisation Property V(a) can be generalised into the
following form
f(x)dx = JJ’ /(x)dx + £z /(x) dx +... + J^ f(x)dx
where,’
a <c, <c, <...<c„
iz
n i
n , <c„
n <b
Property V (b).
I Example 26 Given function, /(x)
' *
2
Evaluate J /(x)dx.
Sol. Here,
/(x) dx =
/(x)dx
/(x) dx +
1
Jo2/(x)* = [0 x2 dx + jj2 Vx dx
Jo° f(x) dx = j^2 /(x) dx + J“/2 /(a - x) dx
i
x3_
x3/2
+
3 o
3/2
Proof As we know,
■
J“ Z(X) dx = J”'2 f(x) dx + J°2 f(X) dx
x^
Put x = a -1 =£ dx = - dt in the second integral also,
when x = a /2, then t = a 12 and when x = a, then t = 0.
- i
3 J0
2
!
JI
2
2
+
-x
3
X
1
1-0 + |[2^-1]
Jo JW dx = J"72 /(x) dx + Ja°/2 /(a -1) (-dt)
3
= 1 + 4^_2 = 4V2_1 = 1
= J"72 /(x) dx + f*72 /(a -1) dt
3
12 /(a - x) dx
JJa /(x) dx = JJ72 /(x) dx + Joaa/2
f* f(x)dx =
da
if/(a + x) = -/(b - x)
Sol. By definition, | a - b | =
a +b
2La 2 f(x)dx’ ^f(a + x)=f(b-x)
3
3
3
3
I Example 27 Evaluate the integral I = Jq2 11 - x | dx.
Property V (c).
0,
3
Then,
|l-x| =
b - a, if a < b
a~ b, if a > b
(1 — x), 0<x<l
(x - 1), 1 < x < 2
:
Chap 02 Definite Integral
Jo2 |l-x|dx = Jol(l-x)dx + J]2 (x-l)dx
x2
X--------2
1
+
2
0
I Example 30 Evaluate J {x} dx, where (x) denotes
2
X2
---------X
95
the fractional part of x.
So/. £2 {x} dx = Jq2 (x - [x]) dx = £2 x dx - £2 [x] dx
Jl
_ ( x 2 V - Qo* to dx + |2 [x] dx
1
1— |-(0-0H +
2
'o
1 =1
= - + |o + -|
2
2
-1(4—0)- f1 Odx + f2 1 dx
'o
Ji
2
= 2 - (x)f = 2 - (1) = 1
I Example 28 Evaluate
(i)
| cos x | dx = J^72 | cos x | dx + j
So/, (i)
Remark
(ii) Jq2 |x2 +2x-3|dx
|cosx|dx
In above example, for greatest integer less than or equal to*, it is
compulsory to break it at integral limits.
J cos x | dx
= j^2 (cos x) dx -(cos x) dx
I Example 31 Evaluate Jq9 {Vxfdx, where (x| denotes
the fractional part of x.
So/.Jo’(V7^=f
= [sin x] o2 ~ [sin x]
= (l-0)-(0-l) = 2
(ii) £2
= f (x-)^-/; (VJjax
| x.22 + 2x - 31 dx
1
r2
= Jo I x22 + 2x - 31 dx + J
=i(x!'2):-j;’[^]jx
I x2 + 2x - 31 dx .. (i)
We have, x2 + 2x - 3 = (x + 3) (x - 1)
4(27H0’
.’. x2 + 2x - 3 > 0 for x < - 3 or x > 1
J
As
for -3<x<l
andx2+2x-3<0
[Vx] dx => 0 < x < 9 and 0 £ Vx < 3
Thus, it should be divided into three parts
So, | x2 + 2x - 31
(x2 + 2x - 3), for x < - 3 or x > 1
- (x2 + 2x - 3), for - 3 < x < 1
0<Vx£l, l<Vx<2
Le.
= 18- J/ [Vx]dx + j^ [Vxjdx + j^ [Vx]dx]
- (x2 + 2x - 3) dx + |2 (x2 + 2x - 3) dx
i
x3
= - — + x2-3x
3
+
o
2
X3
„
x
/ = 2(9)-Jo’ [^]dx
Eq. (i) becomes
/=
x -[Vx] dx
0 dx + | 4 1 dx + £’ 2 dx^
2
= 18-
i
= 18-[O + (x)*+(2x)’ = 18 - (3 + 10) = 5
— + x - 3x
3
=4
I Example 29 Evaluate j*
(x - [x]) dx, where [.]
denotes the greatest integral part of x.
Sot Let ; = £ (x-[x])dx=J_‘i x-dx- J ' [x]dx
I Example 32 If for a real number y,[y] is the
greatest integer less than or equal to y, then find the
value of the integral
Sol. We know,
[2 sin x] dx.
- 1 <sin x < 1 as XG [7t/2,37t/2]
- 2 < 2sin x < 2
2 sin x must be divided (or broken) at x = 5it / 6,7t, 7ti / 6.
As 2 sin x = + 1,0, -1 at these points.
=>
’,21,
1
fto
2
kJ-i
= -(l-1)- [
fl
-ldx+| Odx
= 0 + (x)!.1 -0=1
Jo
)
e 3iU2
JnJ 2
[2sinx]dx = £"‘ [2sinx]dx + fs*4 |[2 sin x] dx
♦r^xiax+c; [2 sin x]dx
96
Textbook of Integral Calculus
= f5X/6
* K
( TKjb
1 dx + f
0 dx +
(JSnJb
Jn
Js
3n 7n
_(5n _ 7t
« (7^
+ 0----- It 'I-2A
6
2
I6
JI 2
6
Jjt/2
r 3n/2
711/6
(-2)dx
xxlxx
7t
2
I Example 33 The value of Jo100 [tan 1 x] dx is equal
1
O
-2
x
2
to (where [.] denotes the greatest integer function)
(a) tan 1-100
(b) n/2-tan 1
(c) 100 - tan 1
(d) None of these
r 100
Sol. Let I ~ J
From the above graph, we need | min ({x}, {- x)) shown as
t
[tan"1 x] dx
where [tan-1 x] is shown as
V2
2
n/2-
lihb, xlflfl TlTlKXffff
zflffl
-2
Ithh
2
1
O
' tan 1
-tan 1
1
J_22 fM dx = 4 J1 /(x) dx = 4 X |1 x i x | = i
2
Hence, (b) is the correct answer.
-n/2--2 '
Lt
Li
2
I Example 35 The value of J2 (xCx2]
+[x2]x) dx is
1-—tan 1
equal to where [.] denotes the greatest integer function
0
(a) y+73+(2'/5-2'/5)+rL(9-3'/5)
4
log 3
tan 1
•—►
1 9-3'/’)
(b) X^3+^ + -L(2'/5-2'/5) + X(
—2
4
flOO
Jo
[tan
,
x] dx is shown as
0
4
■III
tan 1
' Iog3'
’
(c) -+—+—I—(2^-2^)+—(9-371)
4Y
1-
Iog2
I"'
3
3
Iog2
log 3
(d) None of the above
So/. Let 1 = ]^ (x[x2l + [x2]x)dx
100
= 1^ (x + ljdx + j^ (x2+2x)dx + J^ (x3+3x)dr
I72 x3 2X I71 x4 3*
x2
—+x + —+
. + —+
108 3jjJ
2
log 2
4
3
1
2
= 1 x (100 - tan 1)
Hence, (c) is the correct answer.
I Example 34 The value of
j 22 min {x - [x], - x - [- x]} dx is equal to (where [.]
-41
=-+^+- 2
4
3
1
-i-(32-3'/5)
log 3
log 2
Hence, (b) is the correct answer.
denotes the greatest integer function)
(a) 1/2
(b) 1
(c) 3/2
(d) 2
Sol. Let /(x) = min (x - [x], - x - [- x]) = min ({x}» {-x})
Graphically, {x} and {-x} could be plotted as;
I Example 36 The value of J
[| sin x | +1 cos x |] dx e
equal to
(b)n
I \ 3ft
(C)y
(d) 2%
97
Chap 02 Definite Integral
Z(t) = e'-l, for t >0
Sol. Let /(x) = [| sin x | +1 cos x |]
As,
| sin x | £ sin2 x and | cos x | > cos2 x
Therefore,
| sin x | +1 cos x | > 1
Again, for 0 < x < 1,
and
| sin x | + | cos x | <i yjl2 + 12
=>
1 < | sin x | +1 cos x | < 5/2
(corresponding to x > 1)
f(x) = ex -1, for x > 0
Z>x) = l
(vx = e‘)
f'(t) = 1, for t < 0
/(t) = t + C
Thus, [| sin x [ +1 cos x |] = 1
/(0) = 0 + C => C=0 =>f(t) = t, for t<0
£2’1 [| sin x | +1 cos x |] dx =
1 dx = 2n
/(x) = x, for x < 0
Hence, (d) is the correct answer.
Hence, (d) is the correct answer.
I Example 37 The value of the definite integral
£ sin|2x-a|dx, where a g [0,it], is
1 + cosa
(c)
2
(b) cosa
(a)l
(d)
1-cosa
2~
dt
sin|t|dt, where 2x-a = t => dx =—
2
Sol.
2 J-a
-sintdt +
o
1
-cost
2
-a
lp-a
2 Jo
I Example 39 The integral
(|cos 11sin t+ |sin 11cost) dt has the value
equal to
(a)0
(c)1/V2
Sol. Let
enJ2
rn
/ = J 2sint cost dt + J
[(-sintcost) + (sint cost)]dt
sin t dt
zero
f5K/4
n-a
1
-cost
2
o
=
fK/2
JnJ 4
,
sin 2t dt -
+ JJn
f5nJ4
These two integrals cancels
= -(1 - cosa) + -(1 + cosa) = 1
2
2
Hence, (a) is the correct answer.
=> Zero.
I Example 40 The value of £2/ (x)dx, where
Hence, (a) is the correct answer.
I Example 38 Let f be a continuous functions
satisfying
for 0 < X < 1
1
f'(lnx)
ex-l
for
(-2 sin t cos t) dt
sin 2t dt
Jn
= -[l-cosa]--[-cosa-1]
2
2
= -(l-cosa)+-(l + cosa) = l
2
2
x>1
and /(0) = 0, then /(x) can be defined as
(a)/(x)=
(b) 1/2
(d) 1
/(x) = -
0, when x =
n
n+1
, n = 1,2,3... .
3 is equal to
1, elsewhere
(b)2
(d) None of these
(a) 1
(c) 3
Sol. Here, J//Wdx = f+ f
+
ldx + ...
n
1 , if x<0
1 , if x<0
(b)/(x) =
1-ex, if x>0
ex -1, if x>0
21 dx .
n
x, if x<0
(C)/(x)=
ex, if x>0
x , if x^O
(d)/(x)=
ex-1, if x>0
Sol. Hlnx) = [1’ f°r 0<x-1
lx,
for
x>1
3
+... +
n
=------ +... + 1, as n —>00
n+1
Put
Inx = t => x = e‘
We take, limit n —> 00
For x > 1;
f'(t) = e* for t > 0
We have,
Integrating y(t) = e■f + C; /(0) = e°+C=> C = -l
[given, /(0) = 0]
5
■(M-O
2
f(x) dx = 1 + 1 = 2
Hence, (b) is the correct answer.
n
n +1
n-lx
• +... + 1
n /•<
98
Textbook of Integral Calculus
Exercise for Session 3
1. The value of J 3 {| x - 21 + [x]} dx, where [.] denotes the greatest Integer function, is equal to
(b) 6
(a) 5
<c) 7
(d) None of these
(c)3
(d) None of these
(| x | +1 x -11) dx is equal to
2. The value of J
(b)6
(a) 9
3. Let f(x) = x - [x], for every real number x (where, [x] is integral part of x). Then, the value of j1 f(x) dx is equal
to
(a) 0
4.
(d) None of these
(c) 2
(b) 1
2
The value of Jq [x + [x + [x]]] dx (where, [.] denotes the greatest integer function) is equal to
(a) 2
(c) -3
(b) 3
(d) None of these
2X
5. The value of J[x]
1 -r-j
dx is equal to (where, [.] denotes the greatest integer function)
)
2*
(a)-^L
(b)-(c)-W(c)
(b)
6. The value of J
{x) dx (where, {.} denotes fractional part of x) is equal to
'
log 2
(b)i
<a>|
(d) None of these
4 log 2
2 log 2
M
(d) None of these
7. The value of J 4 {x}[x 1 dx (where, [.] and {.} denote the greatest integer and fractional part of x) is equal to
(a) —
1
•
(b)13
12
12
(0 —
k 12
(d) —
12
8. The value of Jo* [t + 1]3 dt (where, [.] denotes the greatest integer function of x) is equal to
(c)
2
[X] ([X] +
(a)
[X] ([X]+ 1)
2
9. The value of
+ ([x] + ^3
(b)
[x]([x]+ p
2
3
+ (M+1)3«
3
10k
+ ([x] + D2 W
(d) None of these
[tan-1 x]dx (where, [■] denotes the greatest integer function of x) is equal to
(a)tan1
(c) 10n - tan 1
(b) 10n
(d) None of these
10. If f(x) = min{|x -1|,|x|,|x +1|, then the value of
f(x)dx is equal to
(a)1
(d)5
o
<c>7
4
11. The value of j~[2e-x ]dx (where, [•] denotes the greatest integer function of x) is equal to
(a) 1
(c)0
'
(b)loge2
(d)e
■
*
Chap 02
10ft
12. The value of J
99
([sec-1x]+[cot-1x])dx (where, [•] denotes the greatest integer function) is equal to
(b)1(ht-sec1
(d) None of these
(a) (sec 1)-10n
(c) re - sec 1
13. The value off
Definite Integral
ft/2
J-n/2
[cot-1 x }Jx (where, [•] denotes greatest integer function) is equal to
(b) it + cot2
(a) 71+ C0t1
(c) 7t + COt1+ COt2
(d) cot1+ cot2
14. The value of ^(tan'Xx -[x])+ tan'i""2(x -[x]))dx (where, [■] denotes greatest integer function) is equal to
(a) -n
(b)-L
n-1
(d)
W-A;
n(n +1)
(c)
W-A
n(n-1)
2
15. The value of J [x2 - x + 1]dx (where, [•] denotes the greatest integer function) is equal to
(a)
5+^
2
m u
b)
(c)
(c)
.
2
(d) 5~^
2
2
16. Evaluate J*[xn]dx, (where, [•] denotes the greatest integer function).
-X.................................... 1.
A
X
nJ
.
>
.. .
,
17. Prove that Jq [x]dx = x[x]-^[x]([x]+1), where [•] denotes the greatest integer function.
18. \ff(n) =
Jow
(where, [■] and {■} denotes greatest integer and fractional part of x and n eN). Then, the value
off(4)is ...
K I
_
19. If f(n) = j"*[cos t)dt, where x g (I 2rm, 2rm + —
Y,neN and [•] denotes the greatest integer function. Then, the value
of f(-
Vtt.
is...
20. If J*[x]dx =
\dx, x einteger (where, [•] and {■} denotes the greatest integer and fractional parts respectively,
then the value of 4{x} is equal to ...
I
Session 4
Applications of Even-Odd Property and
Half the Integral Limit Property
Applications of Even-Odd
Property and Half the Integral
Limit Property
I Example 42 Evaluate
Sol. Let
/(x) = x3 sin4 x, then
/(- x) = (- x)3 sin4(-x) = - x3 [sin (- x)]4
Property VI.
f_°fl
dx
.x3 sin4 xdx.
= - x3 (- sin x)4 = - x3 sin4 x = - /(x)
2 f /(x) dx, if /(x) is an even function
—
So, /(x) is an odd function.
J0
if f(x) is an odd function
0,
Proof We know,
Jt/4
Hence,
J - Jt/4 f(x) dx = 0
r n/4
i.e.
x3 sin4 x dx = 0
J /(x)dx=| /(xjdx+j /(x) dx, if a < c < b
a
c
ra
r0
ra
f0
r0
I Example 43 Evaluate J'"//i S'n2 X^X
dx
f_a fw fa=fwdx+Jo
Now, J
Sol. Let /(x) = sin2 x, then
y(x)dx=J f(-t) (- dt), where t = - x
= /;/(-1)dr
= -f
=
aa
/(- x) = sin2 (- x) = [sin (- x)]2 = (- sin x)2
= sin2 x = /(x)
So, /(x) is an even function, hence
f_* 2 (Sin2 dx ~ 2 f2 (sin2 dx
J(- x) dx (using properties I and II)
- 2 Jof71/2 1 C2°S 2X dx
| J(x) dx, if J(x) is even
n/2
sin 2x
2
o
...(ii)
f fl
X-------------
-J f(x) dx, if /(x) is odd
a
r n!2
2
dx- fw is even
0,
if f(x) is odd
I Example 41 Evaluate J1 (x3 + 5x + sinx)dx.
Sol. Let,
I Example 44 The value of jj log
(c)-1
(b)1
(a)^
I
2-x
2+x
/(x) = log
Now, /(- x) = log
Hence, J * (x3 + 5x + sinx)dx = 0
2+x
2-x J
= - log
/(x) dx =
2jy(*)d*J(x) is even
0, /(x) is odd
dx is equal to
(d)0
/(-x) =-x3 - 5x - sinx =-/(x)
So, f(x) is an odd function).
'2-Xs
<2+x,
Sol. Let
/(x) = x3+5x+ sinx
using
2
J-tt/2 sin2 x dx = —
From Eqs. (i) and (ii), we get
ra f(x) dx = 2 fo
71
f(- x) = - log
= log
f—T1
k 2+x ,
f 2~x
t 2+x
2-x
2+x
= -fM
Chap 02 Definite Integral
i.e. /(x) is an odd function.
/(x)dx = J_’i log 2-x dx =0
So,
I Example 46
k 2+x
if/(x) is odd
ra f(x)dx =
dx' fx is even
2V
r
— cosx
x sin (2x) • sin
(2x - n)
, 18
(a)n
(b)£8
/ i
=>/(x)is odd.
/'(x)is even.
=>y"(x) is odd.
Thus, /(x) + f"(x) is odd function, let
0(x) = (x2+!)•(/( x) + /"(x))
=>
4>(-x) = -0(x)
i.e. 0(x)is odd.
8
it
— COS X
2
n
Jo
dx
(■ n/2
—(i)
(2x - 7t)
J-n/2 0(x)dx = 0
Hence, (d) is the correct answer.
(7t - x) ■ sin 2 (7t - x) • sin I — cos (7t - x)
x2
f= JoT
dx
2(tt - x) - 7t
7t
(7t - x) - sin (2x) - sin
1= Jof"
— COS X
(ii)
n
2/=r
Jo
. r
r
• sin-1 (2x-Jt-x2)dx is equal to
x2
dx
(b)4(V21) (c)4(V2 + 1) (d) None of these
(a) 4^2
Sol Let I = J' -4. x
Adding Eqs. (i) and (ii), we get
(2x - 7t) sin 2x • sin
I Example 47 The value of
2
- (2x - it)
Jo
1 f" n •
2 sin x cos x • sin
2 Jo
— cos x ax
2
J
(put cos x = t, then - sin x dx = dt)
= -L'tsm —1 dt
2
Jo
2 J
I 7t
COS — t
wl
7t
cos -t
2
u
t-
it
_ 11
2
2
'0
o+l.
71
Put x = sin 0 => dx = cos 0 d0
nJ2
sin 0
.
I=2J
’
• sin
Jo
■Jl -sin2 0
<2
8
7t
7t2
2
= 2 JoI
. n
(2 sin 0 cos 0) • cos 0 00
• n/2
sin 0 -20 d0 +2 [
'n/4
20,
0 < 0 < 71 / 4
It - 20, 7t/4 < 0 <71/2
using sin-1 (sin 20) =
dt
=4
~
Jo
0-sm0dO+27t
f X/2
Jx/4
„
(7t-20)sin0 dQ
„
sin 0 d0-4
f X/2
Jn/4
0sin0 d©
= 4(a(-cos0))"’l-4p l-(-cosO)d0+27C(-cos0)^2
- 4 {0 (- cos 0)}J5 + 4 P2 (- cos 0) dO
f . (it
7t
sin — t
=2
using J0 f(x) dx = 2 P Z(x) dx, if f(- x) = /(x)
(using by parts)
dt
•sin-1(2x yll-x2)dx
Jl-x2
r n/4
= 2 P t-sin
=2
X
= 2 [’
dx
(2x - 71)
.. I = -
• sin-1 (2x 71 - x 2) dx
1-x2
7t
— cos x
2
2XCOS2 X/2
=>/(-x) = -f(x)
(C)^
71
x sin (2x) ■ sin
Sol. Let I = f
1
A
2___ J dx is equal to
Jo
X2
Sol. As, f(x) =
x2
2x cos22 x/2
e
sec x sin x + x3
2
x + tan x
COS X
I Example 45 The value of
e
sec x sinx + x3 ,
1
2
x + tan x
(x2+1)[/(x)+/"(x)]dx is equal to
(c) 2
(d) None of these
then value of
(b)-l'
(a) 1
(b)
-1
Hence, (d) is the correct answer.
71
COS X
lff(x) = X2
101
/o
Hence, (c) is the correct answer.
•K/4
= -^= + 2y[2 + y/2lt-^=-4+2y/2 = 4^2-4
V2
V2
= 4(72-1)
Hence, (b) is the correct answer.
102
Textbook of Integral Calculus
it dx
-r 1 +(itcos-2x)dx
-J”
=r
(it - x)
1 + cos2 x
I Example 48 Suppose the function
gn(x) = x2n+1 + anx+bn(neN) satisfies the equation
J1 (px+q)gn(x)dx = 0 for all linear functions (px+q),
Jo
Jo
I = 7t J”
Jo
then
(b) bn =0 ;on =-
(c)an=0;bn=- —
3
2
1 + COS X
dx
1 + cos2 X
dx
„ f n/2
dx
21 = it J n ---------— = 271
----------—
3
Jo
..h
2^ fWdx, f(2a- x)= f(x)
f 1n/2
I = it |
Jo'
Equating the odd component to be zero and integrating,
we get
tan x = t => sec2 x dx = dt
Put
Also, when x = 0,then t = 0 and when x = it 12, then t = °°
Hence,
/(2a -
I = 1t
if f(2a-x) = -f(x)
0,
(i)
/(x) dx; putting x = 2a -1, so
Consider the integral
f0
Jo
2 +12
it
”-0
X 00
o
7t2
~2^2
2
I Example 50 Prove that
Jo"”/2 log (sin x) dx = JJJo"/2 log (cos x) dx = - y log 2
J = J0*/2 log(sinx)dx
So/. Let
that dx - - dt
Also, when x = a,then t = a and when x = 2a, then t = 0.
Z
7t I tan-1 -U
yji I
V2
dt
Ji
Proof We know,
f(x) dx = jo“ /(x) dx + p /(x) dx
sec2 x + 1
r n/2
sec2 x
I=n I
--------- □— <dx
Jo
2 + tan x
Property VII (a).
[
A
sec2 x
-;--------«X
~2
"
(dividing numerator and denominator by cos 2 x)
2p + ~~~+ 2bnq = 0 for all p, q
2n+3
3
Therefore, bn = 0 and a* = 2n +3
Hence, (b) is the correct answer.
*
1 + cos X
/(2a - x) = - f(x)
0,
using j^20 f(x)dx =
Sol. We have, J1 (px + q)(x 2"+ 1+ anx + bn)dx = 0
r20 f(x) dx=|2 r
Jo
J°
1 + cos2 X
2n+3
3
= 2n + 3;’ b"n =- 2n+3
I(d)
2n+3
t* 2q
x dx
t n/2
(a)an=bn=0
••• Ja
Jo
f n/2
Then,
I = Jo
•(0
log sin (tc / 2 - x) dx
= fo"^ log (cos x) dx ...(ii)
a0
f(x)dx=Jfl /(2a-t)(-dt) = -Ja f(2a-t)dt
Adding eqs. (i) and (ii), we get
= Jo° f(2a -1) dt = Jo“ f(2a - x) dx
Ja/(x)dx, if /(2a-x) = /(x)
=.
...(u)
-Jo f(x)dx, if /(2a-x) = -/(x)
From Eqs. (i) and (ii), we have
f(x) dx
dx*^2a ~x^=■
2V
0
, if /(2a-x) = -/(x)
log sin x dx + J^”7
21 = Jo
r n/2
= Jo
log cos x dx
(log sin x + log cos x) dx
f n/21 / ■
\j
f n/2i C 2 sin x cos x>
dx
= Jo log (sm x cos x) dx = J^ log
2
/
f n/2
f sin 2x^ ,
= ]A log -------- dx
I 2 J
J
f n/2
= Jo
f n/2
log (sin 2x) dx - Jq
(log 2) dx
= Jo”/2 log sin 2x dx ~(log2)(x)0’n/2
t/2
I Example 49 Evaluate Jo" ---- ^-dx.
r n/2
1 + cos X
Sol. Let
x
dx
/ = J" 1 + cos'
X
Jo
2
Let
it
21 = J
log (sin 2x) dx-----log 2
2
7j = Jo
log (sin 2x) dx
.(in)
I
<
I
Chap 02 Definite Integral
J
A'J/logsin t ~
1
f 7t/2
log sin i dt (putting2x = t)
2/
I
f(x)dx =
= jo
71
X
4
2
= x (log 2)
-2
a
2 fo
~X^ =
dx’
7t
x
—+—
4 2
2/
0,f(2a-x) = -/(x)
e n/2
-2/
log(sint)Jt
= 2'2Jo
using
X
7t
—+ —
4 2
-2/
x
2
7t
4
x
2
7t
2 J
\ 4
log (cos 2z) dz
= x(log2) + /(x)
f(x)-2f[ 5 + ^ 1 + 2/
or
log (sin x) dx
103
4
= - x log 2
Hence, (a) is the correct answer.
:. Eq. (iii) becomes 21 = I---- log 2
2
1
I4
log sin x dx = -
Hence,
/
I Example 52 If J"
log 2.
Remark
forfo"
Students are advised to learn
C«/2
jo
,n/2
(1 + sin x)2
4
I
(a)-x log 2
(b) | log 2
(c) ~ log 2
(d) None of these
71
„
x A
f\ +—
4 2 I
f Jt/4
j
it
"A 4
(b) A-2it + n2
(c) 2n-A-it2
(d) None of these
2f
Io
2
x2•cos x .
------------- -dx
(1 + sin x)z
71
x
4
2
Tt
dx
1 + sin x
k
+2
o
- ----- dx
IJo ’ ----(1 + sin x)
B-A = -it2 +2K
where,
log (cos t) dt ...(ii)
(i)
x dx
(ti - x) dx
-J
’
1 + sin x
1 + sin x
K= Jor
K = n Jop ——------ K,
1 +sin x
Q,/(2a-x)=~/(x)
pa
using Jo f(x)dx = 2 /oa ^x)
“ x) = ^(x)’
= 2j--’%(cost)if-2j;— log (cos t) dt
It
Put t = — - z in first integral and t = — + z in second
integral, we get
„ f x/2,
(it
,
r x/2
= - 2 Jq log cos I — - z I dz - 2Jo log cos
= -2J
=r
B-A =
log (cos t) dt ...(i)
log (cos t) dt
r 7t/4 - x/2
.
~2f
(1 + sin x)2
Using by parts,
log (cos t) dt
rn/4 - x/2
7t
X
—
4- —
4 2
Jo
X2
X
'°8 <cos
x2 (2 cos2 x/2 - 1)
Jo
(-77/4 +x/2
= -Jo
(1 + sin x)2
B-A = [
1-71/4 +x/2
Jo
B= Jor« 2x2 cos2 x/2 dx
is equal to
log (cost) dt -JW(
f rt/4 ,
dx is equal to
(a) A+27t-7t2
Sol. Let
x
2
+ y j + 2/ (■ n
= -J0
dx = A, then the value
2x2-cos2 x/2
I Example 51 If /(x) = -1x log (cos t) dt, then the
Sol. Here,
J + sin x.
it
Iog(cosx)dx = --log2
log(sinx)dx = jo
value of /(x) - 2/
2
X
log
t n/2
=> 2K = 27t f
Jo
711 J
= 2lt [
Jo
Thus,
log (cos 2z) dz
dx
1 + COS X
71/2
2 X ,
n I
Xx
- sec — dx = 2ti (tan —
2
2
2 0
K = it
=>
= x log 2 - 2 Jo
e nJ2
= 271 [
Jo
1 + sin x
f n/2 1
z + — dz
4)
-(cos2 z - sin2 z) | dz
= o2jfox/2 (log2)dx-2jfo x/2 log (cos 2z) dz
dx
B-A = -it2+2K
B = A - 7t2 + 271
Hence, (a) is the correct answer.
= 27t
104
Textbook of Integral Calculus
As described in the figure above, the new variable t is
given by,
I Example 53 The value of
(cos-1 x-sin-1 Jl-x2)dx(o>0) (where,
J-a
r
Jo
’
cos
b' -a'
(x-a).
b-a ;
t -af +
x dx = A) is
(a) na - A
(b) 7to + 2A
(c) na - 2A
(d) na + A
Thus,
Sol. Let I = [“ (cos "’ x-sin”1 71“ x2)dx
dt = -——dx
b-a
(as a > 0)
J- a
b-a u
b -a J
= (° (cos'1 x-sin"1 ^l “ x2)dx
a + —----- (t-a )
J- a
+ J (cos"1 x - sin"1 71 - x2) dx
= J ° cos”1 x dx + A - 2 £ sin”1 71 “ x2 dx
= -f
Ja
7
The modified integral has the limits (a' to b'). A particular
case of this property is modifying the arbitrary integration
limits (a to b) to (0 to 1) i.e. a' = 0 and b' -1. For this case,
Property VTI (b)
|b f(x) dx = (b-a)
dx=U>~a)
7 =f
(n - cos”1 x) dx + A - 2A
= na - f ° cos"1 x dx - A
Jo
= na - A - A = na - 2A
Hence, (c) is the correct answer.
b-a'
dt
yb'-a')
+(*“«)0 dt
I Example 54 Evaluate
9 x--
J.;5e<»’>!dx+3fv2f e <
e(x + 5)'
dx.
Sol. Here, we know je** dx cannot be evaluated by indefinite
f[(b - a) x + a] dx
integral.
e(x + 5)2
Let
Sometimes, it is convenient to change the limits of
integration into some other limits. For example, suppose
we have to add two definite integrals I} and I2; the limits
of integration of these integrals are different, if we could
somehow change the limits of I2 into those of Ij or
vice-versa, or infact change the limits of both and I2
into a third (common) set of limits, the addition could be
accomplished easily.
Ij = - £* e(x "1)2 dx
Again, let
•••(*)
f2L = fJl/32/3 e9('x~2,3^dx
I 2-~ r‘ e 9
l3 3
Suppose that I = £ f(x) dx. We need to change the limits
23 _1
~2
1-2
3
3
io
1
- f1 e(x“,)2dr = i(/,)
3 Jo
33
17
(a to b) to (a' to b^. As x varies from a to b, we need a new
variable t (in terms of x) which varies from a' to b'.
b'
dx = (-5 + 4)Jo* e,[(-S
1 + 4)x-4 + 5]2 fa
where, I = /] + 3/2 =
c
+3
i2
dx
...(ii)
_A Y=A-A=o
3 J
e(x + 5)2
f5 e<* + 5>2dx+3f
2/3 e’(-2/3)2dr = 0
Jl/3
J-4
Property VII (c) If f(t) is an odd function, then
<Xx) = j f(t) dt is an even function.
a'
Proof We have,
a
x) = £ * f(t) dt
■>x
As x varies from a to b, t varies from a to b, t varies from a'
to b'.
t -a' b' -a
Thus,
x —a
b-a
=>
«-x)=f /«*+£’/ fwdt
<X-x)=0+[’xf(t)dt
J— a
f(t) is an odd function, then J
f(t) dt=O
Chap 02 Definite Integral
<t>(- x) = - fX f(-y) dy, where t = -y
=>
Ja
=>
105
(X-x) =-JQX/(y) dy
[v f is even function => f{- y) = f(y)]
<l>(-*) = £x f(y)dy
[v f is an odd function, then /(- y) = - f(y)]
=>
0(-x) = -<>(x)
Hence, 0(x) is an odd function.
<K-*) = Jafx f(0 dt
=>
0(- *) = <l>(x)
Hence, <f>(x) =
J
x
-X
Ja
'a
J(t) dt is an even function, if f(t) is odd.
Remark
If f(t) is an even function, then for non-zero ‘d, j * Kt) dt is not
I Example 55 if/(x>=r log "Izl" dt, then discuss
necessarily an odd function. It will be an odd function, if
fd f(t) dt =0, because, if «■*) = f \f(t)dt, it is an odd function.
JO
Ja t
$(-x) = -0(x)
whether even or odd?
So/. Let
4>(t) = log
0(- t) = 10g
=>
1-t >
l + t'
f(t)dt = -fc Kt)dt
= - log
1-t '
f ° Kt)dt- f ’ K-y)dy = - [ * Kt)dt - f * Kt)dt
0(— t) = - <|)(t), i.e. <|>(f) is odd function
Proof We have, 0(- x) = j X f(t) dt = where t = -y
x
/(- y) dy,
JO
JO
[where,y = -tin second integral of LHS=*/(-y) = /(y)]
2£°/Wt=Jo' Ky)dy-\^ Kt)dt
2£ Kt)dt=O
Property VII (d) If /(t) is an even function, then
/(t) dt is an odd function.
JO
Ja
♦W-J.’ Io8(tT7 dt is an even function.
<Xx) =
f(t) dt
£° f(t) dt + jo-' Kt) dt=-£° Kt) dt -
= -0(0
=>
-C Kt) dt=Q => f” Kt) dt=Q
or
0(x) = j* Kt)dt is an odd function, [when f(t) is even.]
Only, if
Kt)dt = 0
Jo
Jo
106
Textbook of Integral Calculus
Exercise for Session 4
.7. Let f: R —> R and g : R
R be continous functions, then the value of J
equal to
(a)-1
(b)0
n/2
[f(x) + f(-x)][g{x)-g(-x)]dx'is
(c) 1
(d) None of these
(c)0
(d) None of these
2. The value of j 1 (x|x|) dx is equal to
(a)1
x2 + sinx
dx is equal to
1+ x2
3. The value of j 1
(b) n - 2
(a)2-n
(d) None of these
4. If f(x) is an odd function, then the value of J
a
f(sinx)
dx is equal to
f(cosx) + f(sin2x)
(b)f(cosx) + f(sinx)
(a)0
/ 2x
cos
J+ x2
5. The value of J
+ tan
' 2x
J-x 2
1 + ex
(b)4
(a)|
6. The value of J "
2
COS X
(c)vk
2V3
(b)an
[x] + ioge
<”4
1 4- X
1-X
dx is equal to (where, [•] denotes greatest integer function)
(c)1
(d) 2 log (J
(c)^
(d)-2E
1
n/2
(a)0
e
sinx + .J dx is equal to
(b)1
9. If [•] denotes greatest integer function, then the value of J
n/2
-n/2
(b)2
(a) >■
70. The equation J
2
77. The value of J 2
(a) 1
a| sinx | +
(c)0
dx is
(c) a and b
sin2x
dx, where [•] denotes greatest integer function, is
x
1
+—
n
2
(b)0
(c) 4 - sin4
72. Let f(x) be a continuous function such that j
n+1
m
(a) 9
([>]•
(d)-2E
b sinx
+ c> dx = 0, where a, b, c are constants gives a relation between
1 + cos2 x
(b) a and c
(a) a, b and c
(d)-^=
(c)2n
(b)0
J-n/2
— is equal to
[IITJEE 2001]
(a) 7t
8. The value of f
dx
dx, where a > 0, is
1 + ax
7. The integral j
(d) None of these
(c)1
(b)-27
(d) b and c
(d) None of these
A
f(x)dx =n3,n e Z. Then, the value of J
(c)-9
(d)27
f(x)dx is
Chap 02 Definite Integral
1
ldx = a. Then, J t3 f(t)dt is equal to
ex +1
and[o y3
ex -1
ex -1
x(b)a
(c)2a
(d) None of these
(a)O
2
14. Let f: R ->R be a continuous function given by f(x + y) = f(x) + f(y) for all x, y e R. if J f(x) dx = a, then
13. Letf(x) =
2
J 2 f(x)dx is equal to
(a) 2a
(b) a
(c)0
(d) None of these x
(c)3
(d)4
2
15. The value of J 2 fx]|dx is equal to
(b)2
(a) 1
i
■ Directions (Q. Nos. 16 to 17)
= |l;H>lx|^1andg(x)=/(x + l) + /‘(x-l)foraUx€2?.
Let f(x)
0,|x|>l
16. The graph for g(x) is given by
,y
y
jo. D
(0,1)
(b)
■4-^X
(a) —-3+■
3
1
0
-3
-2
-3
1
-2
2
y
y
(C)
1
O
-+
1
0
2
3
■x
-3
-2
-1
(d) —I-------- k-------►
0
2
3
X
3
17. The value of |
g(x)dx is
(b)3
(a) 2
(c)4
(d)5
(c)-n
(d)0
■ Direction (Q. Nos. 18 to 20)
Lei Z.-J’
8t?.X.
J-k (1 + rc )sinx
dx;n=0,L2,
18. The value of ln+2 - ln is
to
(b) n
(a) rwt
10
19. The value of X /2m+i is equal to
/n = 1
(b)5re
(d) None of these
(a)0
(c)10n
10
20.
The value of y l2m is equal to
m=1
(a)o
(c) 10n
(b)5n
(d) None of these
107
Session 5
Applications of Periodic Functions and
Newton- Leibnitz's Formula
Applications of Periodic Functions
and Newton-Leibnitz’s Formula
Sol. Note that | cos x | is a periodic with period it.
If f(x) is a periodic function with period T, then the area
under f(x) for n periods would be n times the area under
J(x) for one period, i.e.
/(x) dx = n
| cos x | dx.
I Example 56 Evaluate
( cos x | dx
I=4J
Let
using property, J”7 f(x)dx = n^ f(x)dx
f(x) dx
cos x dx - [
cos x dx [
cos x dx - I
Ju
Jn/2
J
=4
f
'0
=4
sin x
Now, consider the periodic function J(x) = sin x as an
example. The period of sin x is 271.
n
xJt/2
Jo
y-
2n
i
n
x
2n a+2n
2n
1 ‘
n/2
I Example 57 Prove that j"^5 e x-[x] dx = 25(e-1).
■H
0 <a!
a
= 4 {1 + 1} = 8
sin x
Sol. Since, x - [x] is a periodic function with period one.
Therefore,
ex ~1x1 has period one
Z=f25X1 ex-^dx = 25f1ex-tx]dx
Jo
Jo
.
using property, JJ7 f(x)dx = n^ f(x)dx
Figure 2.6
= 25
ta+Zn
Suppose we intend to calculate
ex"° dx = 25(ex)10 =25(e’ -e°)
sin x dx as depicted
= 25(e-l)
above. Notice that the darkly shaded area in the interval
[2ji, a + 2ti ] can precisely cover the area marked as
Thus,
Ja
sin x dx =
r2n
Jo
sin x dx
This will hold true for every periodic function, i.e.
JO
[sin x + cos x] dx is
equal to
(a) -mt
(b) mt
(c) - 2mt
(d) None of these
(where [.] denotes the greatest integer function)
Sol. Let I = J
[°+Tf(x)dx = fT/(x)dx
Ja
I Example 58 The value of J
[sin x + cos x] dx
7t
We know, sin x + cos x = V2 sin
This also implies that
and
and
r+nT dx=fT f^dx=nf0T
Cn
nr^dx=ib
afMdx
•(*)
4
(where T is the period of /(x))
dx
sin x + cos x -
f*+"T f(x) dx = £ f(x)dx + njJ /(x) dx
r 2n
jo
1,
0<x<7C/2
0,
rt/2<x<3ir/4
-1.
37t / 4 < X < 7t
-2,
7t < x < 3n / 2
-1,
3ti/2< x <7tc/4
0,
771/4 <x <5 2n
p n/2
[sin x+cos x]dx=jQ
f• 3n/4
(1) dx + J
'n/2
(0)dx
109
Chap 02 Definite Integral
++C! <*2) *+*> *+C(0) *
371
T7C*0
(
3tin(3n
> ,
= - + (0) - 7t---------- 2------ 71—1
\22 J
{
4 J 12 2
J
771
4
3tt
2
+o
* Srt
I Example 61 The value of J
equal to
(a)^
(b)^2
(c)^
c:
3"
(d)7t2
= -7t
Since, sin x + cos x has the period 2tc.
[sin x + cos x] dx = n £ * [sin x + cos x] dx
So, I =
Sol. Let f = J 2n c°t”1 (tan x) dx
= - rm
=> I = 7 (* cot”1
Jo
Hence, (a) is the correct answer.
-
.
I Example 59 The value of Jf(x) dx; where
f(x) = minimum ({x +1}, {x -1}), V x g R and {.} denotes
fractional part of x, is equal to
7C
------X
cot
x
As we know, cot”1 (cot x) =
7t
----- X
2
/(x) = minimum ({x + 1), {x - 1}) = {x}
f5 f(x)dx = J_55 {x} dx = (5 - (- 5)) £’ {x} dx
=7
71
X2
— X-------
= 10^ (x-[x])dx = io£l xdx
< X2 V
N,
where n is a positive integer and 0 < V < it.
So/. Jof'B+V | sin x | dx =
=
fV
[IITJEE 1994, 2004]
,
r rar+V , .
. ,
| sin x | dx + Jy
| sin x | dx
sin x dx + n
rn
■
| sin x | dx
f(x)dx = n
using property, J"
IT
/
_2>
71
8
3tc
x2
2
2
+ — x-------
+
V
<37t2
2
”
>15/2
foil
-- JV2
2
4
8
7n2
2
[2x2-3]dx ••g(x)dx = 0, then g(x) = 0
when x g (0,2) has (where, [.] denotes the greatest inte­
ger function)
(b) atleast one real root
(a) exactly one real root
(d) None of these
(c) no real root
Sol. As, 1 < 2x2 - 3 < 2, V x G
y/sTi)
f(x)dx Le.
[2x2 - 3] dx > a v x G (0,2)
Li
|sinx|dx = nj^ |sinx|dx^
=» g(x) = 0 should have atleast one root in (0, 2).
[vg'(x)*0]
+nf’ sin x dx
o
>lt/2
_2 ,_2
_2
71
371
7t
-------------- 4. —
\v
= - COS X
Hence, (b) is the correct answer.
Jo
/
V
= (- cos V + 1) + n - cos x
\
2o
= - (cos V) +1 + n (1 + 1) = (2n + 1) - cos V
| sin x | dx = (2n + 1) - cos V, where n is a positive
integer and0< V<7t.
xX/2
I Example 62 Let g (x) be a continuous and
differentiable function such that
+ V | sin x | dx = (2n+1) - cos V,
rV
( _2
71
J
Hence, (b) is the correct answer.
=5
Hence, (c) is the correct answer.
I Example 60 Show
=7
71
rn ( 7t +---71 x Adx.
inJ2 I 2
,
dx +
2>o
2
[as {x} is periodic with period *1’]
= 10
, 0< x <n/2
n + x, 7t / 2 < X < 71
Sol. We know, {x + 1} = (x - 1} = {x}
Thus,
(i)
fW dx = (m-n) jj f (x) dx
”
(d) 6
(c) 5
(b) 4
dx
2
-J
(a) 3
cot 1 (tan x) dx is
'-2n
I Example 63 The value of x satisfying
f2[x + 14]
X
»{x}
dx = J
[x+14] dx is equal to
2
(where, [.] and {.} denotes the greatest integer and
fractional part of x)
(a) [-14,-13)
(b) (0,1)
(d) None of these
(c)(-15,-14]
Jo
110
Textbook of Integral Calculus
So/. Given,
f2[x + 14]
<■28 + 2[x]
Jo
f28
x
2
r• 28 + 2[x]
X
'28
2
I hr+I
Jo
J
dx +
using £nT /(x)dx = n
r a + nT
f2[xj
= 2x • cos x4
/(x) dx = j*0"T/(x) dx-, where T is period of f(x)
. (i) If the functions §(x) and ig(x) are defined on [a, b]
and are differentiable at a point x e (a, b) and f(x, t)
is continuous, then
d ’ p(x)
f(x, t) dt
dx . -W*)
dv(x)~
, j-(x.wM)
= Lxi
dx
d*M
(ii) If the functions (|>(x) and lg(x) are defined on [a, b]
and differentiable at a point x e (a, b) and /(f) is
continuous on [0 (a), <j> (b)], then
d
d
j dx
dx
r(” f(t)dt ]=^-{v(x)i/(V(x))
-^-{<>(x))ftt>(x))
dx
I Example 64 Find the derivative of the following with
respect to x.
dx
2
2
d £
2_
- COS
X
d
dx
rd?
11/x
x2
cos t2 dt
1
1
= —f= COS X + — cos
X2
x2
2^7
I Example 66 If ~ (|ye -t2
rx2
dt + I
Jo
f2
dxtJo
dt + f
x2
Jo
sin2 t dt = 0
d 2)
-y,22 f d / J -e~° ^-(0)]+sin2(x2)U(x
J —(y)f
ax
dx
lax
dx
, fd
-sin20 <— (0) ■ =0
[dx
2 x'
„2
e~y — + 2xsin2x2=0 => — = - 2x ey2 sin‘
dx
dx
I Example 67 Find the points of maxima/minima of
rx2 t2 - 5t+ 4
dt.
0 2+
x2 t2 ------ St +—
4 dt,
Jo
2 + e'
, A
2
2 4- ex2
= cos X
using Leibnitiz’s rule, —
dx
f* ™f(t )A = 4- (v(x))/(V(x)) - ~ («x)) + /(«x»
dx dx
x2
/
.-.f'(x) = x4 - 5x2 4- 4 ■2x-0
- cos 0 • — (0)
dx
1
sin2 tdt =0,
find^.
dx
Sol. We know, — [y
dx{ x
1
1
1
2>/x
= —COS X + —- • COS
So/. Let/(x)=f
cost2dt
cos t dt
d
I dx
X
cos t2 dt .
J
=> e
•’•y-(/(x))= cos(x)^ -^-(x)
= 2x cos x4
I Example 65 Evaluate y- (
dx
*r
(cost2)dt
J
-T-(fW)
= cos
dx
Property IX. Leibnitz’s Rule for the Differentiation
under the Integral Sign
Z(x) = jo
f x2
d
dx
So/. Let/(x) = J^ cos t2 dt
14 + [x] = (14 + [x]) {x}
(14 + [x])(l-{x}) = 0
[x] = -14
xe[- 14,-13)
Hence, (a) is the correct answer.
So/. (i) Let
d
= cos (x2)2 • l^-(x2)l - cos (0)2U(0)
' dx'.....................
dx
(dx
J
[dx
d
dr = (14 + [x]) {x}
/J /(x) and
(i) £ cost dt
= COS X
(ii) Let /(x) = Jo cos t2 dt
dx = (14 + [x]) {x}
Jo
cos t dt
'0
r x2
dx = Jo{X> (14 + [x]) dx
X 1 .
14 [2
d
dx
dx = jj*1 [x 4-14] dx
|
Jo
(x - 1) (x 4- 1) (x - 2) (x 4- 2) • 2x
2 + ex2
+
-2
0
+
1
4-
2
Chap 02 Definite Integral
From the wavy curve, it is clear that f'(x) changes its
sign at x = ± 2 ± 1 0 and hence the points of maxima are
-1,1 (as sign changes from + ve to - ve) and of the
minima are -2, 0, 2 (as sign changes from - ve to + ve).
J3 - sin2 x-A (x) _ 3 _ sin2 ZE d . n ]
dy
— — + cos y ■ —
dx
V
3 dx k 3
dx
- cos (0) A (o) = 0
dx
I Example 68 Find ~ f f, —1— dt .
r
dx^J*2 logt J
so/. Af
rf — dt
dxljx2
1__
•og x3 dx
logt
d (
-^3 - sin2 x + cos y
4^’)-
1
logx2 dx
3x2
2x
3 log x
2 logx
X
K is
(a) 10
prove that—+K2y = Kf(x).
dx
sin x
Now, J
sin K^x ~ dt + T
dx
4 2esin x*
x
Hence, (c) is the correct answer.
I Example 72 The function
rx
(
1A
J(x) = Jo log| sjnf 11 sin — I dt, where x e (0,27t), then
AL {/(t) cos K(x-t)} dt
dx
J(x) strictly increases in the interval
+!/(x) • cos K(x - x) • A(
d x)l - {/(o)cos K(x - 0) • A (0)}
dx
~K -K joX /(t)sinK(x-t)df+ f(x)-0
/(t)sin K(x - t)dt + Kf(x)
^Az = -K2y + Kf(x) => ^~ + K2y = Kf(x)
dx
I Example 70 If j*
,L
i
sin (
—dt
t
K = 16
=>
Again, differentiating both the sides w.r.t. x, we get
dx2
sin (x2)
= (F(0l6 = F(16)-F(l)
=>y:- = K|o f(t) cos K(x - t) dt
= - K2 J
*=1
4 e
(V x2=t)
= K J X f(t) cos K(x - t) dt + 0 ~ 0
dx
•(>)
X
• {f (x) sin K (x - x)} - A (o) {/(0) sin K(x - 0)}
dx
I
(d) 18
f------ dx = F(x)
f(t) sin {K(x - t)} dt
Differentiating w.r.t x, we get
dx2
cos y
esinT
d
T-(f(x)) =
dx '
x
Sol. We have,
J
Az
dx
(c)16
(b) 14
j
?
dx = Jo dx
^3 - sin2 x
dx
x
2esin>2
x dx = F(K) - F(1), then the possible value of
Ji
I Example 69 If y = j * f (t) sin {K(x -1)} dt, then
Sol. We have, y =
dy
sin x
(x2 - x)
— dt =
logt J logx
d2y
=0
I Example 71 Let4(F(x)) = e ----- ,X>0. If
1
dxljx2
Ill
-^3 — sin2t dt+cos t dt = 0,
dy
then evaluate —.
dx
Sol. Differentiating w.r.t. x, we have
d
f J3 - sin2 t dt 1 + — [y (cos t) dt =
0
J dxJ<>
dx 'n/3 y
, . ( it
Sit
1 6
6
a
(c)
(b)[ —,2ti |
6
71 lit
Sit lit 'I
(d)
6 ’T
)
T’T J
Sol. Here,/'(x) = logjsinx| sin x + - > 0
2J
sin x + - < 1 and
2
sin x + - > o
2j
0<sinx + -<i
2
-l/2<sinx<l/2
XG
(Sit 7lt
6 ’ 6
as x e (0,2it)
Hence, (d) is the correct answer.
112
Textbook of Integral Calculus
R and F(x) =
I Example 73 Let f: (0, °°)
F"(x) = f'(x) => F"(2) = f'(2)
From Eq. (i), f'(2) = ^256+1 = 4257
tf(t)dt.
12
If F(x2) = x4+ x5,then S f(r ) is equal to
Hence, (c) is the correct answer.
r =1
(a) 216
Sol. Here,
(b) 219
■ (c)221
(d) 223
F(x2) = x*+ x!=f/ff(r)A
Property X. Let a function f(x,a) be continuous for
a < x < b and c < a < d, then for any a e [c, d],
I Example 76 Evaluate 1(b) = j
f(x2) = 2 + ;X =>/(r
+ |r
=> f(r21)) =
= 22+-r
=>
2\ «
5 ,
\ n (5n -F13)
/(r2) = 2n + -n(n + l) = -^—---r=l
4
2
Jo db
db
Hence, (b) is the correct answer.
= fo1’
•
I Example 74 A function f(x) satisfies
f(x) = sinx+j* /'(t)(2sint-sin2 t)dt, then f(x) is
(dt
1
j
t
Also, f(0) = 0, hence C = -1
...
1
„ V-1 +sinx
f(x) =----------- 1 =------------- !
1-sinx
1-sinx
Hence, (b) is the correct answer.
—-— + C
1-sinx
I Example 77 Prove that
sinx
1-sinx
Now,
no=
(d)
rn/2
dx
7t(o2 + b2)
Jo
(a2 sin2 x+b2 cos2 x) 2
4a3 b3
a2 sin2 x + b2 cos2 x
1
F(x) = f f«)dt =» F'(x) = f(x)
Then, Z = p
2
Jo
sec2 x
dt
dx= Jof"
a2 tan2 x + b2
a2 t2 + b2
(where, t = tan x)
15^17
t8■2t
-2t 2x11
2V1 +18
,2------------- —
1 dx
M2
Sol. Let
U
Sol. Here,
(when b = 0) ...(ii)
1(0) = 0
1(0) = log (1) + C
Z(0) = C
From Eqs. (ii) and (iii),C = 0
=>
1(b) = log (b + 1)
yll+U 4
—du, then the value of F"(2) equals to
J17
bi x dx = Jo‘(x‘)Ifc =
In x
? + 1>0
Also, from Eq. (i),
I Example 75 If F(x) = Jx f(t)dt, where
(C) V257
dx+0-0
Integrating both the sides w.r.t. b, we get
1(b) = log (b + 1) + C
I(b)=C ^Adx
Given,
In x
cosx
(1-sinx)2
■x
(b)-^
dx;b>0.
b +1
dx
cosx
.
Integrating, we get f(x) = j ----------r dx
(1-sinx)2
7
(a)
4-/l7
In x
-i (/((,» = -2-
=> (1 + sin2 x-2sinx)/'(x) = cosx
n \
>
1
1
[i-o] =
“ b+1
h+1
x
... sinx
,. .,1-cosx
1-cosx
... tanx
(b)
(c)--------(d) —
(b) -------(a) -------1-sinx
1-sinx
cosx
1-sinx
Sol. Differentiating both the sides w.r.t. x, we get
f'(x) = cosx + /'(x)(2sinx-sin2x)
Put 1-sin x = t,
1 XbP -1
x b -1« \
12(60 + 13)
Z /(r2) =
4
r=l
cosx
l + sin2x-2sinx
da
xb -1 ,
Sol. We have, 1(b) = jji ---dx
» In x
12
=>
dx
if /(a) = [ f(x, a) dx, then
= ["
Jo
da
Ja1
Differentiating w.r.t x, we get
x2 f(x2)- 2x = 4x3 + 5x4
/=±
ab
68
(0
f x/2
Thus, I
Jo
*-0
2
71
2ab
1
-dx = — = fM ...(I)
2ab
a2 sin2 x + b2 cos 2 x
Chap 02 Definite Integral
Differentiating both the sides w.r.t. ‘a’, we get
j-n/2
0
=>
f'(x)=—^=
dx - - n
dx = (a2 sin2 x + b2 cos2 x)2
2a2b
2^1 + x •
-2asin2x
sin2 x
r n/2
Jo
I
0
Hence, (a) is the correct answer.
2
cos x
C=-n
/(x) = n 71 + x - n = n (^l + x - 1)
=>
Similarly, by differentiating Eq. (i) w.r.t. ’b’, we get
r’t/2
/(x) = n 71 + x + C
where, f(0) = 0 =>
, = ~—
n ~•
dx
4a3b
(a2 sin2 x + b2 cos2 x)2
.
n
dx =
4ab3
(a2 sin2 x + b2 cos2 x)2
...(iii)
ru */1 •;
I Example 79 Let /(x) be a continuous functions for
allallx,x,such
x/(t)suchthat
that (/(x))
(/(x))2 2==j£7
(t) •
dt and
Adding Eqs. (ii) and (iii), we get
(sin2 x + cos2 x) dx
r n/2
Jo
(a2 sin2 x + b2 cos2 x)2
(• n/2
dx
(a2 sin2 x + b2 cos2 x).2‘
Jo
n
n
—7- +
4a3b 4ab3
n
TaW
(a2 + b2)
I Example 78 The value of
rit/2 log (1 + xsin2 0)
d0; x > 0 is equal to
’ -k
sin2 G
(a)nQT+x - 1)
(b) n (^1 + x - 2)
(c) 7k (^1 + x -1)
(d) None of these
/(0) = 0, then
(a)/0=lo8l
5
4
(C)/W = 2
(d) None of these
Sol. Here,
2 sec2 t .
----------- dt,
4 + tan t
2f(x).f'(x) = f(x)-
=>
1
sin2 0 „ A .
----------- ---------- — 40+0-0
l + xsin20 sin20
(using Newton-Leibnitz’s formula)
f
40
j* n/2
cosec2 0 40
l + xsin2 0
0
cot2 0 + (1 + x)
\n/2
i= ■ tan 1 cot 0
+x
71 + x zo
< Vi
f'(x) =
2 sec2 x
4 + tan x
sec2 x
4 + tan x
Integrating both the sides, we get
As above integral is function of x. Thus, differentiating both
the sides w.r.t. ‘ x’, we get
Jo
<b< = 3
4
On differentiating both the sides w.r.t. x, we get
Sol. Given, f(x) = J^2 log(l + xsin2 0) dd;x>0
sin2 0
r n/2
113
n.
2yjl + X
J 4 + tan x
Since,
dx = log (4 + tan x) + C
/(0) = 0
0 = log (4) + C
C = - log 4
/(x) = log (4 + tan x) - log (4)
log (4 + 1)-log (4) = log4
Hence, (a) is the correct answer.
114
Textbook of Integral Calculus
Exercise for Session 5
1. The value of J1° sgn (x -[x])dx is equal to (where, [•] denotes the greatest integer function)
(c)11
(b) 10
(a) 9
(d)12
2. The value of £ (x -[x])dx (where, [•] denotes the greatest integer function)
(b)
(a) [x]
•nn-n/4
3. The value of j
'-n/4
(c) x[x]
(d) None of these
(c) ~^n
(d) N°ne of these
|sin x + qos x\dx is equal to
(a)2V2?
(b) V2n
4. Letf(x) = x -[x]for every real number x (where, [•] denotes greatest integer function), then |1 f(x)dx is equal to
(b)2
(a) 1
5.
=x+
(d)2
(c)0
tf(t) dt,then the value off(1) is
(a) J
(b)0
(b)0
(d)-21
(c)1
(c)1
2
x
6. The least value of the function <KX) = J5n/4(3sinf + 4cosf)dt on the interval
d
*
is
(b)-2V3 + - + -l
2 V2.
(a)V3+2
3 1
(c) - + -y=
(d) None of these
7. The points of extremum of 0(x) = J* eH2/2,(1-t2)dt are
(a)x = 1,-1
(b)x = -1,2
(c)x = 2,1
(d)x = -2,1
8. If f (x) is periodic function, with period T, then
(a)j“f(x)dx = j“‘7(x)dx
(b) j‘ f (x)dx =
(c)j“r<x)dx = j‘y(x)dx
(d) j“ f (x)dx = [^Tf(x)dx
d
e”nx
-4 2e810x2
dx
(a) 4
x
x
Let-j-(F(x))
9. Let
—(F(x)) = -—, x>0. If r
10. Let f:(0,oo)
W45
(b)8
f(x)c/x
dx = F(k)-F(1), then one of the possible value of k is
(c) 16
(d)32
R and F(x) = f* f(t)dt. If F(x2) = x2(1 + x), then f(4) is equal to
(b)7
(c)4
(d)2
11. Let T > 0 be a fixed real number. Suppose f is continuous function such that f(x + T) = f(x) for all x e R. If
I = J7 f(x)dx, then the value of f3+3Tf(2x)dx is
(a)|/
12. Let f(x) =
(a)±1
(b)2f
(c)3f
(d)&
V2-t2dt, then the real roots of the equation x2-f'(x) = 0 are
(b)±-^
(c)±J
(d)±V2
Chap 02 Definite Integral
13. Let f(x) be an odd continuous function which is periodic with period 2. If g(x) =
(a) g(x) is an odd function
(c) g(2n) = 0 for all n e N
>x
115
f(0 dt, then
(b) g(n) = 0 for all n e N
(d) g(x) is non-periodic
•X
14. Letf(x)be a function defined byf(x)=J* t(t2-3t + 2)dt, 1<x <3.Then, the range off(xJis
<b)['T
4
4
(a)[Q2J
fcosx
2|
I
.
COS 1(f)
15. The value of ....
lim
----------- -rfis
x->o 2x-sm2x
<b>-:
(a)0
76
(d) None of these
4
(0H
(d)l
If fx bfcos4f~asin4f^ _ asin4x
for all x * 0, then a and b are given by
Jo
f2
x
(a)a = -J,b = 1
(b) a = 2,b - 2
(c)a = -1b =4
(d)a = Zb =4
17. If f(x) =
{f(t)}~'dt and £{f(t)}“1 = V2, then
(a)f(x) = V2?
(b)f(x) = V2to^7
(c) f (x) = V3x-1
(d) None
of these
•X
18. Let f be a real valued function defined on the interval (-1,1) such that e~xf(x) = 2 + J* 7f4 + 1 dt, for all x e (-11)
and letf-1 be the inverse off. Then, (f"1)'(2) is equal to
[ht jee 2010]
(b) 1/3
(d) 1/e
(a) 1
(c)1/2
■ Directions (Q. Nos. 19 to 20) Consider the function defined on [0,1]-> R
r(x)= sin x - x cos x ifx # 0 and f(0) = 0.
[IITJEE 2012]
19. jo f(x)dx is equal to
(a) 1-sin (1)
(b) sin(1)-1
20. lim ~
f(x) dx is equal to
(a) 1/3
(b) 1/6
(c) sin (1)
(d)-sin (1)
(0)1/12
(d) 1/24
Session 6
Integration as Limit of a Sum, Applications of Inequality
Gamma Function, Beta Function, Walli's Formula
Some Important Results
to Remember
Integration as Limit of a Sum
Applications of Inequality
and Gamma Integrals
(i) Z r = n(n-H)
r=l
2
n
An alternative way of describing J f(x) dx is that the
r
definite integral j* f(x) dx is a limiting case of
r=l
summation of an infinite series, provided /(x) is
<iii) X r3
"-1
b
continuous on [a, b], i.e. [ f(x)dx= lim h S
Ja
n —» oo
2 _ n (n +1) (2n +1)
r=l
(iv) In GP, sum of n terms, S„ = •
J
(iv) The lower and the upper limit of integration are
r
limiting values of — for the first and the last term of r
respectively.
Some particular cases of the above are
(a) lim
"1
or lim
n —y OO
Pn
(b) lim
n~^oar = \
n
n-1
i
( r
r_Tj
=1
n
[n
s -fF Jo* f(x) dx
1 (r = f3 f(x)dx
~f - Ja
«
where, a = lim — = 0 (as r = 1)
n —n
and
0= lim - = p(asr = pn)
n —♦<» n
’
|r|<l
_ sin n 012
•cos[a +(n -1) 0/2]
sin0 / 2
(iii) Replace—by x and—by (dx) and lim £by the sign of f.
1
r=l
(v) sin a + sin(a + 0) + sin(a +2 0) +...+sin [a + (n -1) 0]
= smiPZ2.sin[0l + (n_1)P/2]
sin 0 / 2
(vi) cos a +cos(a+0)4-cos(a+2 0) +...+cos [a + (n -1)0]
i.e. lim X - f\ ~ ]•
n
n
n
an,
(1 - r)
i f r
(i) Express the given series in the form £ — / — •
n“
n
(ii) Then, the limit is its sum when n —»°°,
n->°°
«(rn~l)
|r|>l
(r-1) ’
a(l-rn)
Method to Express the Infinite
Series as Definite Integral
n
n2 (n + 1)2
4
r=1
f{a + rh), where h = -—- ■ The converse is also true, ie, if
n
we have an infinite series of the above form, it can be
expressed as definite integral.
n
6
7t2
,
, 1
1
1
1
1
(vu) 1-- +----- +---- +.. . oo =----12
22 32 42 52 62
7C2
(viii) 1 + — + — + —+ — + — +.. . oo = ■—6
22 32 42 52 62
(ix) 1 + — + — + ...«> = 2L_
32 52
8
,'111
(x)
4--- 1- + ... oo =-22 42 62
24
gie -e'*
e#i0 +e -i0
—, and sin 0 = —
2
2i
-e
e
e -e-e
eG + e
(xii) cos h0 =
— and sin h0 = —
2
2
(xi) cos 0 =
Remark
The method of evaluate the integral, as limit of the sum of an
infinite series is known as integration by first principles.
Chap 02 Definite Integral
n—1
I Example 80 Evaluate the following:
(i) lim
n—> oo
fl
2
3
n-1
V nz
n
rr
n
—+—+ —+••• +
(ii) lim
n—> <»
/
n
(iii) lim
77?
n—> oo
np+1
n—> oo
2~
1
r = 0 •^4n2 - r2
Sol. Let S, = £
4s
n -»00
n
00
n
dx
V
= sm
=> S = lim
n x-
n
1
E i(£n
as n —> 00 =>---- >0
n
"V"
r.l
= f1 x dx = Jo
(ii) Let Sn =
1
1
1 + ... + —
1 = y
v
+—
n+r
n+1 n+2
2n
=s
=s n [i + -
n —» oo
I
n
1
n
1
Hence, S = lim Sn = lim - £
n-»~ n /T'
F = ,i i + L
n
_ f1 dx = [log (1 + x)]^ = log 2
~ Jo 1 + X
1
= Ern 2 + (r/n)
r=l
:--V 2+dxx = [ln|2+ x|]J
= log 6 - log 2 = log 3
S = ln3
I Example 83 Evaluate J4 (ax2+ bx+c) dx from the
. first principle.
n
n
n
(iii) LetS„= 2
2 + n2 4- 92 + ”• +--- 2
2n2
n2 +12 n +z
4-1 3
n
n
As x —> 1, r —»0 and x —> 4^ r —> n
Sol. Let x = 1 + rh, where h =
1
r = 1 n(l + r2/n2)
-2
r=1
n
J (ar2 + bx + c) dx = lim
1
S„ = lim — Y
Hence, S = lim
]
** —r w
1+7
I
rp
"
3
f
I
3rf
3r)
+ b 1 + — +c
I
nJ
nJ
r=o n
p
3r ) + c
afl + ^r- + — +i>fl + —
n—
( n2 nJ '
n
r = n0 n
= 3 lim
X 1
r=l n
1 A
p
n"4“ nr = l
1
P+1 o
= or
3 lim —1 a I n +
n -»oo n
S = lim Sn = lim — V
xf + 1
A
r=0
• = lim £ - a 1 + —
n
dx
= tan-1 x =
4
1 + x2
.0
r \ t
1P +2P + ... + np
(1V) Let Sn =---------—------n
h [a(l + rh)2 + b(l + rh) + c]
"
I
X h f (1 + rh)
r=0
n
= lim
r2
n ->« n r = 1
= Jorx p dx =
-J
S= lim S,
r=1
-r
4n
1
r = 1 2n + r
1
nI -4 oo
1
1'
1
+-------+ ... + —
k2n +1 2n + 2
6n y
4n
r — 1
n
6
I Example 82 Evaluate
' 1
l
1)
1
lim
+... + — •
I —> 00
6n)
/2n +1 2n + 2
Sol. Let S„ =
2
-i i
—
2 Jo
n
6
S = sin
■ -1'
r=2
r=2
? - (r/n)2
r=0
. -i x
n-1
>. -1
= x -n2
n
,P>0
3
n
1
Hence, S = lim Sn = lim — Y
(1) Snn —
-• + —
■— + —- + ... + ■
^2
x2
n”
r=0 n
74 -(r/n)2
3/4 - X2
2
1
= z- 4 -(r/n)
2
1
71 r = 0
Sol. Let
1
n~
as n -»oo.
r=0
n
n
+ —2--- -+... + —n2 +22
2n2 7
(1P +2P +... + np)
(iv) lim
1
I Example 81 Evaluate S = £ ■. ■ - -
1
1
1
+---+ ... + —
n+1 n+2
2n
f
117
1
P+1
I
9n (n + l)(2n + 1)
6n2
6n (n + 1)
2n
3n (n + 1)
+b n+
+ cn
2n
= 3 [a (1 + 3 + 3) + h(l + 3/2) + c]= 21a + — b + 3c
2
118
Textbook of Integral Calculus
I Example 84 The value of
y/n
( . n , 2n .
. (n
v — 1) tc
is equal to
lim sin — • sin — sin — ...sin-?
n 7
k 2n
2n
2n
(d) None of these
(a> I
Sol. Let A =
271
7t
371
lim sin —-sin —-sin — ... sin
2n
2n
2n
2(n -1) 7C
2n
n -> 00
;. log A = lim
n —> «>
= lim
n —>«
271
7t
— log I sin — • sin — ... sin
2n
2n
n "V
_1
n
2 <n - i)
2(n -1) n
2n
(b)-l + V5
1
^77
(C)-1 + V2
(d) 1+V2
2
2n
Sol. We have, lim —»V -;
n—n r=\yn2+r 2
r
1
f2
X =Jx = (a/x2 + 1)o =75-1
= lim —J n
n->-n ,=1
£ log sin —
2n
Hence, (b) is the correct answer.
(?){tf(x)dx
using lim E — f
n
n
•a
TtX
= f" log (sin t)- — dt
putting— = t
2
Jo
2-2 r n/2
equal to
(a) 1+V5
= is
h-t-n
m
r=l
= J0 log sin
=—j
l/n
1 2/1
I Example 86 The value of lim — V
log(smt)dt
71
using jo2a f(x) dx = 2
f(x) dx, if /(2a - x) = /(x)
Applications of Inequality
Sometimes you are asked to prove inequalities involving
definite integrals or to estimate the upper and lower
boundary values of defmite integral, where the exact value
of the definite integral is difficult to find. Under these
’ circumstances, we use the following types
Type I. If /(x) is defined on [a, b], then
_4
71
n i
o
using£r72 log (sin x)dx =
~l‘°g2
fb
fb
£ f{x)dx <£ |/(x)|dx.
log 2
= - 2 log 2
Equality sign holds, where /(x) is entirely of the same
sign on [a, b].
log A = log(1/4) => A = 1/4
Hence, (c) is the correct answer.
I Example 85 The interval [0, 4] is divided into n
equal sub-intervals by the points x0,x1,x2,...,xn_1,xn
whereO = xo <x, <x2 <x3 <...<xn = 4.
I Example 87 Estimate the absolute value of the
. ri9 sinx ,
lnte8ra Jio ~dx1+ X
n
.
11’
Sol. To find, I = f' I’ sin x dx
< J10
I
Ji1,0 1 + x8
If 8x = x/-x/_1 for/ = 1,2,3,...,n, then lim yx(8xis
8x->0“T
equal to
(a) 4
Sol.
(b)8
(d) 16
(c)y
Since,
limSxtXj + x 2
The inequality
8x —»0
4 4
sin x
dx
(i)
1 + x8
(using type I)
| sin x | < 1 for x > 10
8
12
= lim----- + - + — +.
n
n->~n n n
.+ 4-n
n
sin x
<
1 + x8
Also,
1
...(u)
10<x£19 => 1 + x8 >10”
1
1
1
­
------ r <
<—
—Tr Ol
or
r- <10"8
1 + x8 10
11 + x 8 1 .
io8
From Eqs. (ii) and (iii),
f 19
u Xl x2 x3
’fix’fix'
+
"-♦oonz
2
Hence, (b) is the correct answer.
1 + x8
sin x
1 + x8
= lim -^-(1 + 2 + 3+...+ n) = lim ~
n-»«nz
sin x
=8
fi9
J*o
,(iii)
< 10’8 is fulfilled.
.
f19
dx < J10 10"8 dx
sin x
1 + x8 dx < (19 - 10). 10"8 < 10" 7
(v the true value of integral ~ 10"8)
Chap 02 Definite Integral
f8
I Example 88 The minimum odd value of ‘ d (a > 1) for
,. , r 19 sin x , 1 .
which Jio -—- dx < -, is equal to
(a) 1
Sol. Let I = [ 1’ sin x dx
Ji,I0
1 + x“
r 19
<f
”
J10
dx
J10
Jl°
f(x) = 0, VxG[4,8]
=>
/(6) = 0
Type IV. For a given function/(x) continuous on[a, b], if
sin x
v sin x < 1 => --------<
1 + xfl
dx ■<”
dx
1 + xa
=>
.a
/
,a
/(x)dx = 0, which shows /(x) can’t have any value
greater than zero in the sub-interval [4, 8].
=> f(x) is constant in the sub-interval (4, 8] and has to be
zero at all points, x G (4,8].
(d)9
(c)5
(b)3
119
you are able to find two continuous functions j\ (x) and/2 (x)
on[a, b]suchthat /j(x) < /(x) </2(x), V x e[a, b],then
/(x)dx<£* /2(x)dx.
£*
1 +10°
(*.* 10 < x < 19 => 10“ + 1 < 1 + x° < 19a + 1)
f 11
dx
I Example 91 Prove that - < Jq
’ /zi-x2-x33
9__
1 + 10°
.
9
< - => 1 + 10“ > 81 orl0a>80i.e. a = 2,3, 4,5,...
’’1 + 10°
9
Sol. Since, 4 - x21 4 - x 2 — x3J | 4 — 2x2 > 1, V x G [0,1]
■J 4 - x 2
.'. Minimum odd value of ‘a’ is 3.
Hence, (b) is the correct answer.
l-J4 ~ x - x3 174 - 2x2 > 1, V x G [0,1]
2
_____ 1_
"777^ 1/4-x!
Type n.
f f(x) g(x) dx
7t
= <—=■
b
g2(x)dx
,
£
dx
=<
.3
- X
1
V4 - 2x 2
. V X G [0.1]
dx^
4 - x2
where f 2(x) and g2(x) are integrable on [a, b].
-x3
i
dx.V xg[0, 1]
^4 - 2x2
I Example 89 Prove that J1 7(1+xM1 + x3) ^x c3™0*
=>
exceed ^15/8.
. -1 x
sm —
2
Sol. £* -J(l + x)(l + x3)dx < ^fo'(i+x)ax]Qo1(i + x5)ax
6
____ dx
^4-x2 -x3
75
dx
^4 - x2 - x 3
4V2
sin
-i x ]
Type V. If m and Af be global minimum and global
maximum of f(x) respectively in [a, b], then
m(b-a)<£ f(x)dx<M(b-a).
Type HI. If/(x) lg(x) on [a, b],then
£ f(x) dx > £b g(x) dx. In particular, if /(x) > 0,
Proof We have, m < f(x) < M for all x 6 [a, b]
£ m dx < £ /(x) dx < £ M dx
=>
m(b-a)<£ /(x) dx
then£ /*(x)dx|o.
I Example 90 If /(x) is a continuous function such
that /(x)| 0, V x e [2,10] and j*8 f(x) dx = 0, then find
M (b - a)
I Example 92 Prove that 4 < £3 ^3+x5 dx <2V30.
Sol. Since, the function f(x) = -^3 + x3 increases
f(6).
monotonically on the interval [1, 3].
Sol. /(x) is above the X-axis or on the X-axis for all x G [2,10].
.*.
M = Maximum value of -^3 + x3 = -^3 + 33 = 730
and
m = Minimum value of -^3 + x3 = ^3 + I3 = 2
If /(x) is greater than zero at any sub-interval of [4, 8],
then £ f(x)dx must be greater than zero. But
Textbook of Integral Calculus
120
m = 2, M = ^30, b - a = 2
(ii) If n e N, then T(n+ !)=(»)!
(iii) f(l / 2) = Vn
ra
22SJ, *3 + x3
Hence,
r fm + 1 r n + 1
™ j;
n/2
4<
I Example 93 Prove that 1 <
Sol, For 0
x
m
2
2
x ■ cos'* x dx = —
2T
e*22dx < e.
m + n+2^
2
(v) FnF(l-n) = —-—,0<n<l
sin nn
1, we have e° < ex < e1
e°(l-0)< Jo‘ ex2 dx < e1 (1 - 0)
J' ex2dx £ e
=*
sin
1
(vi) FmFm + -=-^-T2m
22j 2 2m -I
l
n-1
Jo
...r
(vii) rir
n
Gamma Function
n-1 A _(2Jt) 2
n
7
nm
I Example 95 Evaluate J* e -x x3 dx.
If n is a positive number, then the improper integral
f e~ x x" ~1 dx is defined as Gamma function and is
Jo
denoted by Tn.
Di = [
Jo
i.e.
Sol. By definition of Gamma function,
e~x x"’1 dx, where xE Q+.
f
Jo
e"xx3dr=f
Jo
f
Jo
e~x x3 dx = 6
I Example 96 Evaluate r1
I Example 94 Evaluate
(0 n
(ii) T2
I e ~x dx
Sol. (i) n=fe "X.x1-1dx= IfUm
—4 co *0
e-xx4-1dx=F4
, 1
logx
n-1
dx.
.n -1
dx
log — I
So/. Let I =
b
= Um
= Um (~e~b
- e
Put log — = t => dx = - e
x
o
=0+1=1
tn-1(-e’f)dt = Jo“ e
...
f e~x • x dx
(ii) r2 = Jo"e— x 2~'dx = i,Um
—> 00 *0
,
i
log-
tn-ldt
n-1
dx = rn
X
= Um
dt
-xe~x -e~x
JO
= Um [(- b e~b -«-*)-(0-1)]
= Um
Um
'7 + 1
1
■7
+1
= Um
-(6 + D + 1
> OO
eb
Beta Function
The beta function is
B(m, n) = j° x m -1 (l-x) n“1 dx,
(using L’Hospital’s rule)
where m, n > 0
=1
Properties of Gamma Function
Gamma function has following properties :
(i) n = l,ro = «’ and F(n + l) = nFn
e.g. T5 = 4T4 = 4 X3 T3 = 4 X3 X2T2
=4x3x2xifl=4x3x2xl
Properties of Beta Function
(i) B (m, n) = B (n, m), where m, n > 0
(ii) B(m,n) =
n wherem, n >0
T(m + n).
(iii) B(m>n) = jo°°
xm “i
----dx
m+n
J
Chap 02 Definite Integral
I Example 97 Evaluate j1 x6 ^(1-x2) dx.
Sol. Let,
A sin x dx = Jof cos"x dx
Jo
n-1
=----n
n -1
n
/ = Jof x 6 -J(l _ x2) dx
Put let x2 = t
=> 2x dx = dt
2 Jo
I = -B
2
n-3
n-2
n-3
n-2
I Example 98 Evaluate
5/2 7(1 ~ 0 dt
7 3
2* 2
r[- + (2
5 3 1 r~ 1
n ■1 2 2 2
2
4-3-2-1
2
n-5
n-4
n-5
n-4
n/2
3.2. ZE, n is even.
4 2 2
4
9
------ 1, n is odd.
5 3
sin4 x-cos6 xdx.
Sol. Using Gamma function, we have
1 T7/2F3/2
2
121
n
. 4
r(5/2)r(7/2) r(5/2)r(7/2)
6
sin x • cos x dx = —7----------- - - -----------------^44-6 + 2)
2f6
’
J°
I 2 J
2
V 5 3 i
3 1
-x-x-xf(l/2)
-x-xr(i/2)
A 2 2 2________
2 2
2x5!
571
256
3n
512
I Example 99 The value of Jo e -ȴ dx is equal to
ft °°
Walli's Formula
An easy way to evaluate J”7 sinm x • cos" x dx, where
m,ne I+.
n/22 sinm x • cos" x dx = £
We have, j"^
sin
(b)^2o
(d) None of these
(a) —
2o
(c)“r=
sin" x-cosmx dx
(m - l)(m - 3)... (1 or 2) .(n - l)(n-3)... (1 or 2) 7t
Sol. Let I = f
Jo
e ~a2x2 dx
Jo
2aylt
(m + n)(m + n - 2)... (1 or 2) 2
* A^lfe-r1'2 dt
2a
when both m and n G even integer.
(m - l)(m - 3)... (1 or 2)-(n - l)(n - 3)... (1 or 2)
(put a2 x2 = t =>2a2x dx = dt)
_ i r “ ri/2-i
(m + n)(m + n - 2)... (1 or 2)
2a J”
•e~‘-dt
(using £ e - X xn-1dx = rn)
when either of m or n G odd integer.
Remark
If n be a positive integer, then
1 t^I
1 ,
= — I - = — -yJK
2a
2
2a
Hence, (a) is the correct answer.
9
(using T - = Vn)
2
122
Textbook of Integral Calculus
Exercise for Session 6
n/2
1. The value of f(x) = Jq'
(a)-l(Vu7-1)
log (1 + x sin2 0)
d&, x Z 0 is equal to
sin2 0
(b) Vn C/1+ x -1)
(c) n (^1 + x - 1)
(d) None of these
(c) log 2
(d) None of these
7t
1 n
2. The value of lim - y
f r vis equal to
+r>
(b) log 4 - 1
(a) 1 - log 2
3. The value of lim
n —♦
— {(n + 1)(n + 2)(n + 3)...(n + n)}Vn is equal to
n
(a)4e
(d) None of these
(c)e
<b)7
4
4. If m, n e N, then the value of J* b° (x -a)m (b - x)n dx is equal to
(b -a)m+n + 1-m In I
(m + n + 1) I
(a)
(b - a)m+n-m In I
(m + n) I
(b)
(c)
(b-a)m-rnl
mI
(d) None of these
2n4 +1
5. The value of lim
n I 'isn5 +1
is equal to
2
(a)e
(b)—
e
6. The value of lim n
n —>»
|5
(c)
1
+ 3n2 + 16n + 16 + ‘” +
3n2 +8n +4
(d) None of these
n terms is equal to
(b) J1 '09(1)
3
(d) None of these
1 r
it
,3?l+... + nsin3™ l
7. The value of lim —=■ sin3 — + 2 sin
is equal to
4n
4n J
«->«nz
4n
J?
(a)3£(52-15n)
(b)-?j(52-1&>)
9jcz
9re2
(c)-L(15n-15)
9rcz
8. The value of f(k) = jo
I
(d) None of these
log (sin2 0 + k2 cos2 0) d0 is equal to
(a) n log (1 + K) - it log 2 (b)n log 2- log (1+ K)
(c) log (1+K)-7tlog2
(d) None of these
9. Ifn?,n eN.then/^n = J1 xm (1 - x)n dx is equal to
(a)
m In!
(m + n + 2) I
(b)
2m In I
(m + n + 1) I
(c)
m In I
(m + n + 1) I
(d) None of these
10. The value of/(n)^" sin2 n0 de is (V n 6 N)
sin2 0
(a) rm
(b)y
(C)?4
(d) None of these
JEE Type Solved Examples:
Single Option Correct Type Questions
*Ex-1 'fTfMdx=^ and f(x) is an even function,
• Ex. 3
The true set of values of ‘a’for which the inequal­
ity £ (3”2x -2 • 3"x) dx > 0 is true, is
then P f I x-— \dx is equal to
Jo
\
(a) [0,1]
xJ
(b)2
(a)4
(d) None of these
(c)ic
(c)[0,oo)
(b) (-00,-1]
(d) (-«>,-l]w[0,°o)
-2)dx>0
So/. We have, | 3 (3
Put 3"x = t =>3”x In 3 dx=-dt
So/. Here, ^f(x)dx=^
ln3j’ (t-2)dt>0 = — ~2t
2
i
Putx=t-t
=>
(q-2fl
(
11
I
t 2J
A
- ----- 2-3-"
I 2
dx = 1 + — dt
>0
--2|>0
.2
J
3
=>
■rXf4)"+P(‘4) ±d,
<4>+<
3"a>31 =>a<l or 3‘a<3°=>a>0
Thus, a e(-°°,-l]u[0,oo)
•Wy)
■rt }*-jXy’i dy
f
1
3'2fl-4x3"fl +3>0 => (3"*-3)(3"a-l)>0
put t=—
y.
as /(x) is even
y)'
Hence, (d) is the correct answer.
• Ex. 4
71,3 value of the definite integral
max(sinx, sin-1(sinx))dx equals to (where, n e I)
1
t
1
=1X'"7WX
t
(b)n(7t2-4)
(b)
n(it2-4)
(a)
1
n(7t2~8)
x.
(c)
• Ex. 2 The value of
( n
\f n
v=1
/
n
1 n<*+') e
(a)n
4
f2nlt
Sol. Let I = I
Jo
1
V=1
n(7t2-2)
(d)
x+k
max (sin x, sin" (sinx))dx
Y|
dx equals to
J
(b) n!
......L..
(c) (n +1)!
(d)n-n!
n
■--J--...
So/. The given integrand is perfect coefficient of PJ (x + r)
1
1= n<x+r)
1 x
ilt-X
0
rcl
n
4
i
—n/2 i
3n/2
it
2n
=(n + l)!-n!=n-n!
.... ________
0
Aliter (x+l)(x + 2)(x + 3)...(x+n)=e‘
So that, when x = 0,then t = Inn! when x=1, thent =ln (n +1)!
+ ln(x+ n)] = t
(ln(x+l) + ln(x + 2) +
( 1
1
1
dx = dt
------- F------- +... +
(x+n)J
\(x +1) (x + 2)
Therefore,
I
pin (n+1)1
Jinn!
l^T”'
,ln(n+1)1 _einni =(n +1)> -n! = n-n!
Hence, (d) is the correct answer.
i -nJ2
j
' fH/2
fit
xdx+l
=n L
Jlt/2
JO
W2
TT2
1 (
2
_2
q_2
i
,
f2it
(n-x)dx+l (sinx)dx
Jx
TT2^
it
n i
= n — 4.------- !t!2 -— -2
8
2 2^
'
71
7t
Jit
=n --- +------------- 2
8
2 8
4J J
_n(n2-8)
4
Hence, (c) is the correct answer.
*x
124
Textbook of Integral Calculus
Applying Leibnitz theorem, we get
(I1 - 22-33. .n njn2 is equal to
• Ex. 5 lim n
o»4.
ve
e
a,
Vi-y'<x))! -r<x)
i-<r(x))2=/2(x)
(d)V7
■ 4»1)
Sol. Let L=limn 21
</'(x))2=i-f2(x) =>r(x)=±Vi-/2(x)
2
y--±71~y2> wherey-f(x)
dx
<iy = = ±dx
•(l’-22-33...n")"2
lfn + 1
lnn +
X k In k
InL = lim—
n—2
n
n k=l
1 n
if n + 1
= lim-n—»<»> 2
n
lnn + —
2
Integrating both the sides, we get sin-1(y) = ± x + C
1 yx », k
lfn + 1
= lim--|------|lnn +
n->“> 2
/(0) = 0 =>C = 0
(fclnfc-fclnn + fclnn)
n k=l
n
n k=i
Inn yx
n
n
y = ±sinx
y=sinx = /(x), given f(x)^0 forxe[0,1]
=>
Its known that, sinx < x, V xe R+
Jt=1
Hence, (c) is the correct answer.
S
1
—
= [ x lnxdx = -- => L = e *
The value of lim — fx Nn(1 + f) dt
x-» o
t4+4
PIT JEE 2010]
• Ex. 8
n i
Hence, (c) is the correct answer.
r!7«2
n
Jt2
x2
4
(a)-/(2)
71
71
(b)l
(a)0
(c)±
64
Sol, lim ± f ~?g(1!t) dt
X-+0X3J0
is equal to
4+?
Using L’Hospital’s rule,
16
2
(b)-/(2)
1 □ • IM1 1
<-1 and
sin 2
13 3. >
•
1 ( n +1Y
fi
.
1 ( n +1
=— ------ In n + xlnxdx + - ------ Inn
n J
Jo
2< n .
• Ex. 6 lim
1
<2
sin
ifn+lY
1A k. k Inn n(n + l)
= lim- - ----- Inn + - \ - In - + — ----- -------2\ n J
nrf.
n n
2
k=i n
PIT JEE 2007]
xlog(l + x)
lim
x-»0
(d) 4/(2)
n 1
4+x4
3x2
3x
x-»o
.. I log(l + x)_i
using lim
rsec
sec z x
x
£ /(»■*< = lim /(sec2x) • 2 sec2 x tan x - 0
Sol. lim
n
4
K
X->4
x
16
1
1
4+x4 "12
= 1^1^20
X
"x->0
Hence, (b) is the correct answer.
2x
(applying L’Hospital rule)
^/(2)-4 ^8/(2)
71/2
7t
4
• Ex. 9 The value(s) o/£ x4(1-x)
—dx is (are)
1 + x2
PIT JEE 2010]
22
/LX
3
(a)— -it
Hence, (a) is the correct answer.
7
• Ex. 7 Let f be a non-negative function defined on the
interval [0,1]. //J* ^-(f'(t))2dx = J* f(t)dt,Q < x < 1 and
2
(b)
—
(b)-=105
(d)-- —
(c)0
2
15
_fix4(l-x)4
Sol. Let I Jo------- T~dx
Jo 1 + x2
fi(x4-l)(l-x)4+(l-x)4
/(0) = 0, then
(a) f\
(c)/
PIT JEE 2009]
1
, x-l
< - and /|
>; >»'GJ
1
3
1
1
2
3
<2and/t3j<3 (d)^
1
□ rl
> - and f
2
1
dx
(1 + x2)
;
=f1(x2-l)(l-x)Mx+f,<— x2-2x)'-dx
Jo
Jo
(1 + x2)
=£ (x2-1)(1-x)4+(1 + x2)-4x+
□ r\
> - and f\
2
1
Sol. Given, ^l-(f'(t))zdt = J/(t)dt,0<x< 1
1
>3 '
Jo
3
4x2
(1 + x2)
= £1 (x2-l)(l-x)4 + (1 + x2)~ 4x+4-
4
1-x2
dx
dx
Chap 02 Definite Integral
4
dx
1 + x2
22
n
—
— o =—n
7
.4
0/(xcos8)dx
= [’ x6-4x5 + 5x4 + 4-
(c) sinO
1 4 5 4 ,
- ------ +------ + 47 6 5 3
(d) tanO J’"10
J°\
1
Sol. Let, T =
Hence, (a) is the correct answer.
pin 0 tan 8
fCOt6
•'snO
f(x)dx
f(xtan0)dx.Put xtan0=zsin0
/ =j^/(zsin0)cos0dz
f (xtan0)c/x 0 # —, n G I j is equal to
• Ex. 10 If £
=>
dx=cos8dz
(a) -cos0 |)tane/(xsin0)c/x
ftanO ,
= - cos 0jj
(b) -tan0 [Sin0/(x)</x
125
fUnfl
f(z sin 0) dz = - cos 0^
/(xsin 0) dx
Hence, (a) is correct option.
j cose
JEE Type Solved Examples:
More than One Correct Option Type Questions
• Ex. 11 Let f(x) be an even function which is mapped
Sol. Here, min [| x- n|,| x -(n +1 )|} can be shown as
from (-it, 7t). Then, the value of^ (f(t) dt + [/(x)]^ dx
can be (where, [•] denotes greatest integerfunction)
(a) 0
(b) it
(c) 2it
(d) ‘47t
Sol. As, f(x) is an even function, then | f(x) dx is an odd function.
[as 1 < f(x) < 3]
Also, [/(x)] = 1,2
-£ (pw ■*)*+£, r/wi *
fit
fZl+ 1
A=f
or
fit
J-K
1,11
=-xlx-=2
2 4
[vsince [_f(x)] = 1 or 2]
(>n+ 2
-r
idx
J-n
(min{|x-n|,|x-(n+ l)|})dx
Now, A, = |
Jn+l
2dx=2it or 4n
(|x-n|-|x-(n+ l)|)dx
=|(|r|-|r-l|)dr
Hence, options (c) and (d) are correct.
rt
• Ex. 12 LetM^r 1(min{|x-n|,|x-(n+1)|})c/x,
[putx=n + l => <ix=dt]
.t
aid A=£*I!(|x-(n+'l)|-|x-(n+3)l)dx,
^2=f++l20X-n|-|X-(n+1)D</X’
[put x=n +t => dx = dt]
=/’((4-f)-(3-0)dt=j ’ldI=l
A3 =^23(|x-(n+4)|-|x-(n+3)|)dx
and
1
9
Also, gfx) = A, + A2 + A3=- + l + l= —
4
4
g(x)=A1 +A2 +A3, then
(a) A, + A2 + A3 = 9
9
(b) 4, + A2 + A3 = 4
(c)£«(x) = ^
n=l
100
(d) £ S(x) = 300
n=1
.
100
X £<x) = g(l)+g(2) + 5(3) + ...+ g(100)
11=1
9 9
=-+ -+
4 4
9 900
=---4
4
I
126
Textbook of Integral Calculus
JEE Type Solved Examples:
Statement I and II Type Questions
■ Directions (Q. Nos. 13 to 15) For the following questions,
choose the correct answers from the codes (a), (6), (c) and
(d) defined as follows.
(a) Statement I is true, Statement II is also true; Statement II
is the correct explanation of Statement I.
/(-x)=-sinx+ xcosx+ C
On adding, /(x) + f(-x) = constant which need not to be zero.
2dt-,g(x) = 4i
For Statement I /(x)
.2
f(-x)=r
x4T+t2dr,t=-y
'o
(b) Statement I is true, Statement II is also true; Statement II
is not the correct explanation of Statement I.
(c) Statement I is true, Statement II is false.
/(x) + /(-x)=0
=> /is odd and g is obviously even.
Hence, (c) is the correct answer.
(d) Statement I is false, Statement II is true.
• Ex. 13 Statement I (f/(x)=J0 (x f(t)+V)dt,then
£ /(x)dx = 12
• Ex. 15 Given, f(x) = sin3 x andP(x) is a quadratic
Statement II /(x) =3x+1
Sol. Let |*/(t)dt=fc, so /(x) = xk +1
polynomial with leading coefficient unity.
Now,
P(x) ■- f "(x) dx vanishes.
Statement I j
j’(kt + l)dt = fc => ^ + l=k, sok=2
Statement II
/(x)=2x + l
r2"
f(x)dx vanishes.
Sol. P(x)=ax2 + bx + c, f(x)=sin3 x
j’/(x)dx=12
*=ffr(x).r(x)dx
Hence, (c) is the correct answer.
i
Also
Using
• Ex. 14 Statement I
The function f(x)
1 + t2c/t is
P(x) • /'(x)l?1 - J* Pz(x) • f'(x)dx
zero
an oddfunction and g(x) = /'(*) is an even function.
Statement II For a differentiable function /(x) if f'(x) is
an even function, then f(x) is an odd function.
Sol. If /(x) is of odd, then f'(x) is even but converse is not true.
e.g. If f'(x) = xsin x, then /(x) = sin x - x cos x + C
n
II
I
= - [Hx)-y(x)]*-jV(x)-/(x)<ix
f2n
a
=Jo ? (x)’/(x)^x=2j0 sin3xdx = 0
Hence, (a) is the correct answer.
JEE Type Solved Examples:
Passage Based Questions
Passage I
(Q. Nos. 16 to 18)
Suppose we define the definite integral using the following formula £ f(x)dx = —-(f(a) + f(b)),for more accurate result
force (a.b), F(c) =
+ /(c))+^(/(i) + /(c)).
2
When c =
fl+^,thenj^ /(x)rfx = ^——
(/(a) + /(6) + 2/(c)).
[IIT JEE 2007]
I
Chap 02 Definite Integral
r/2 sin
. xdx. is. equal/4to
Sol.
• Ex. 16 j
127
F'(c)=(b-aY(c) +f(a)~ f(b)
i
F\c) = f\c)(b-a)<0
(a)|(1+V2)
(b) 5(1+72)
F'(c) = 0
4
f \
71
(0—7=
8V2
Sol.
f'(c)
(d)—-7=
4v2
/
n
f«/2
It «
T"0
Jo
sinxdx = —----4
0+ —
2
2
sin(0) + sin^^ + 2sin
b-a
Hence, (a) is the correct answer.
=?W)
8
Passage II
(Q. Nos. 19 to 20)
x.
Hence, (a) is the correct answer.
cos (a + P) -sin(a + P) cos2P
Let /(a,P) = sina
cosa
sinP .
-cos a
sina
cosP
• Ex. 17 If f(x) is a polynomial and if
f /(x)dx-^—
(/(f)+ /(*))
lim----------------- ?—
= 0 for alia, then the
(t-a) 3
degree off(x) can atmost be
(a)1
(0 3
• Ex. 19 The value of
/
7t
/ = 12 e3H /(0,0) + / ^,p
Sol. Applying L‘ Hospital’s rule,
/(<■))lim------- --------------- —
(t~a)
f'(t)
2 ----- = 0
(b)0
(c)2(2ex,2-1)
(d) None of these
=>
rX/2 a(
"
/=/o ej/(0’°) +
= £ e^(2 + 2cosP + 2sin3)dp
3(t-a)2
Hence, = [e^(2 + 2sinp)]o/2
r<o-;(/(o+/w)-
no
(t-a)
lim-------- ------------------ 2 ------= 0
t-»a
12(t-a)
= 4eK/2-2 =2(2e’V2-1)
Hence, (c) is the correct answer.
nt/2
bn,
• Ex. 20 If I = I
/-►a 12
(t-a)
HO
lim-------- ------------------ 2 —=o
e-»a
12(t-a)
lim ?
= 0 /*(a)= 0 f°r
a-
fit/2
J-rt/2
/•JV2
=2
(d)0
(b)2
(d) None of these
cos.2
\
X,
“
JJ
cos2 p(cosp + sinP) dp
fK/2
that a <c <b and (c,f(c)) is the point on the curve for which
F(c) is maximum, then f'(c) is equal to
2b-a
I 2
J-it/2
=
2f(b)-f(a)
I
(c)2(2e’t/2 -1)
• Ex. 18 Iff "(x) < 0,V x e (a, b) and c is a point such
(c)
r 7i
(a)^2
f(a) is atmost of degree 1.Hence, (a) is the correct answer.
(b) 2(/(b)-/(a))
(b)
b-a
(
cos 2 P /(0,P) + / 0.--P
[/] is
Sol. Again, I =
(a)/(b)-/(a)
b-a
dp/s
Sol. Here, /(a,P)=2cosP i.e. independent of a
/(>)-;(/«)+/(<■))HO
lim----- 2—;-------- -- 2 ---- = o
‘-♦a
ATT
(a)e’''2
(t-a)
=>
--P
Jo
(b)2
(d)4
x\
,(3n n
J-n/2
fR/2
Jo
COS
cos2Psinpdp
cos;3pdp + 0
I = ® =>(11=2
Hence, (b) is the correct answer.
.
dp, then
7
128
Textbook of Integral Calculus
JEE Type Solved Examples:
Matching Type Questions
• Ex. 21 Match the following.
Column I
jt cosx
— 1 —x
is not defined at
sinx2
x = 0. The value of f(0\ so that f is continuous at
x = 0,is
(A) The function /(x)=
(B) The value of the definite integral ^“7=—i
is
equal to a + 6 In 2, where a and b are integers, then
(a + b) is equal to
„fnsec20-tan0
(C) Given, e"Jo------ -g------ dQ=1, then the value of
Column 11
(p)”-T~
Putnx~t => dx = -dt
n
1 fl
(q)
0
tan
Use L’Hospital’s rule, lim---sin'
(r)
(s)
Column I
i dx
1-11 + x2
1
exco1x_1_x
/(*) =
(s) ~2
i
is an reven function.
1 + x2
z-2Jor* 1 +dxx2 = [2(tan-1 x)]o = 7t2
(B) Put x = u6 => dx =6u5 du
"3 +1-1
•1 u
=6f
— - - du=5-6 In2
J u+1
=>a + fe=5-6 = -l
(C)e"^e-e(sec2 0 - tan 0) d0 = 1
0» £ r?=[sm xlo=7
r • -I
<o £
dx _ f3
1-x2” J2
-en | e* (sec21+tant)dt = l
(usep(rtx) + /'to)=<f(x)]
11
dx
x2-l
dx
(D)LetJ = [2
(D)
Jl/n+l
lim zl/n
Jl/n + l
x j
tan (nx)dx
sin-1 (nx) dx
- -In
2
x-1
13
2
[putx=sec0 => dx=sec0tan0d8]
Xylx^-l
[ef tant]^" =1 =>-en [-e"" tann] = l =>tann = +1
71
1,2 1/3
= — ln- = -ln 2 3 2 \2
Put - 0 = t => dd = -dt
-i z
j
71
J-i 777
= —1
2
r1/n
i
dx
H
(A)
exco*x(-xcosx-sinx-sinx) + ex COiX(-xsinx + cosx)
=>
71
Sol. (A) -»(s), (B) -+ (s), (C) -> (q), (D) -> (r)
= lim----------------------------------------------x-»o
2
•i 6us du
J u3 + u2
(r)
£
XCttX
•(-xsinx+ cosx)-l
= lim------x-*o
2x
2
dx
•’xTx2 -1
x2
(applying L’Hospital’s twice)
1
(A) {log
|
dx
(c> r•3'217?
Sol. (A) -> (r), (B) -> (p), (C) -» (s), (D) -» (r)
[IITJEE2006i
, . 1, f2A
r dx
"
X
2
2
(P> 2 °8I 3 J
(B) £777
equal to
ex cos x -1 - x
(A) lim
= lim
2 sinx2
x -------2
7C
n
n+1
Column II
^7
n-fh
tH1
7t
n+1J_7
=A=1
• Ex. 22 Match the following
bn = f", sin-1 (nx) dx, then lim — has the value
1----
x
n
1/2
/H-1
J
0r
-form
0
sin-1 (t) dt
n Inhi +1
tan n is equal to
(D) LetG„ = J"| tan-I(nx)dr and
tan'1 (t) dt
lim n^LLL
=r
•«/3 sec0tan0d0
Jo•)
__ pR/3
Jo
sec 0 tan 0
ld0 = 3
Chap 02 Definite Integral
129
JEE Type Solved Examples:
Single Integer Answer Type Questions
• Ex. 23 Let f-.R —> Rbea continuous function which
satisfies f(x) = £ f(t)dt. Then, the value o//(ln5) is.
• Ex. 25 Let /(x) be a differentiable function satisfying
fM + /f x + j = 1,Vx e R andg(x) = J* f(t)dt. Ifglf) = 1,
PIT JEE 2009]
oo
Sol. (0) From given integral equation, /(0) = 0
Also, differentiating the given integral equation w.r.t. x, we get
Sol. (6) Here, /(x) +
f(x)
=> ec = 0, a contradiction
/(0) = 0
X (g(x + k2) - g(x + k)
U=i
f'M_1 => f(x)=e'ex
f(x) * 0, then
is.
n
n=2
/'(x) = /(x)
If
8
then the value of^
J •
X+2 = 1
Replace x by ^x +
, we get /x+il + /(x+l)=l
/(x) = 0,V xeR => /(ln5) = 0
On subtracting, J(x) = f(x +1)
• Ex. 24 For any real number x, let[x] denotes the largest
integer less than or equal to x. Let f be a real valuedfunc­
tion defined on the interval[-10,10] by
/(x)=-
Also, g(x+l2)=j'+1 /(t)dt=Jo7W<fr + f*+l f(t)dt
Since, /(x +1)=/(x)
^x+l!)=fo7(r)dt+l!.£/(()dt
x-[x],
if[x] is odd
1+[x]-x, if [x] is even
7C2
...(i)
gtx + 21)=[’f(l)dt + 22J‘ f(t)dt
10
Then, the value of -^-|io/(x)cos7txdx is.
PIT JEE 2010]
x-[x],
if [x] is odd
Sol. (4) We have,/(x) =
1 + [x] - x, if [x] is even
g(x + k!)=joy(t)dn-e-j'of(t)dt
...(ii)
and s(x+*)=J’/(t)df+t-f’y(t)*
/(x) and cos Ttx are both periodic with period 2 and are both
even.
n
n
fc=l
fcwl
...(iii)
Thus,£ «x + V)-^x + i)) = £
J w /(x) cos Ttx dx=2 J*°/(x) cos TCX dx
fWdt
= X (k2-k)-g(l). givengfx)=j7(0^
k=l
= 10 J _f(x) cos Ttx dx
- ( n(n + !)(2n+ 1)
k
6
Jo 7(x) COS Tlx dx = £ (l-x)cOSTtxdx = -Jo ucosTtudu
_n(n-l)(n + 1)
3
• \ AAA /
-10-9
-2-1012
9
10
• • s-X (gtx + k^-glx + k)) -s
8
riO
,40
J io (x)cos,lx^x=!“20Jo ucosnudu=—
8x3
n=2 (n-l)n(n +1)
'
n=«2
k=2
= 12jp f(n + l)-(n-l)
n-2
J /(x)cosTtxdx=| (x-1)cosnxdx = -ucosnudu
,
= 12
(n-l)n(n + l)
1
(n-l)n
1
n(n + l),
f 1
1_
1
+
+
Jx2 2x3, <2x3 3X4 f
...+
It2 fio
— J io /(x)cosTtxdx = 4
n(n + 1)
xl
2
= 12
1
1
2 n(n+l)
=6
1
k(n-l)n
1
n(n + l)
130
Textbook of Integral Calculus
Subjective Type Questions
• Ex. 26 Find the error in steps to evaluate the following
integral.
f
Jo
-r
=f
dx
rn
sec2 xdx
_ r« sec2 xdx
l + 2sin2x Jo sec2x + 2tan2x Jo
Jo l + 3tan2x
Remark
Students are advised to check continuity of anti-derivatives
before substitution of integral limits.
| sin x | dx = 8 and^a+b
8 ant
| cos x | dx = 9,
• Ex. 27
= -U [tan-1 (-Ji tan x)] J = 0
then find the value of^ x sin x dx.
V3
Sol. Here, the Newton-Leibnitz’s formula for evaluating the defi­
nite integrals is not applicable because the anti-derivative,
Sol. We know, cI b |sin x| dx represents the area under the curve
•fl
/(x) = -U [tan-1 (fl tan x)]
V3
has a discontinuity at the point x = — which lies in the interval
from x=a to x-b.
We also know, area from x = a to x=a + it is 2.
f |sinx|dx=8
•fl
It
rI n ,
3 tan---- h
<2
LHL = lim -U tan'
h—♦ 0 ^3
,
at x = —
2
871
b-a = —
2
= lim -4= tan-1 (-?3 cot h)
h-*oj3
a fl + b
Jo
Similarly,
|cosx|dx=9
,
1
-i/x
n
= lim -7= tan («») = —7=
A-0V3
2d3
-.(i)
„
9it
a+b-0=—
2
—(i)
.(ii)
From Eqs. (i) and (ii), a = 7t / 4, b = Ylit 14
Also,
/-
RHL = lim -L- tan
/>-»o V3
I n
71
i
V3 tan — + h
<2
tt
at x = —
2
Hence,
Cb
Ja
xsinxdx=
fl7n/4
Jx/4
xsinxdx
f i7n/4
= [-xcosx]’„7/74 + [
= lim -7= tan-1 (- -73 cot h)
h-»o V3
i-
1
*
-1/
= lim
tan
*-♦0 V3
\
Jlt/4
n
(-«>) =----- —
2V3
1771
177t 7t
It . . i17m/4
=--------cos------ + — cos — + [sm x]^
4
4
4
4
47t
(ii)
From Eqs. (i) and (ii), LHL # RHL at x = it 12
=“7s
V2
=> Anti-derivative f(x) is discontinuous at x = 7t / 2.
[ xsinxdx=-2fin
So, the correct solution for above integral;
«
dx
'■J.’ • 1 + 2sin2 x
Using, £* /(x)dx=-
f*
Ja
'a
sec2 xdx
l + 3tanzx
(iii)
• Ex. 28 Evaluate
/(2a-x) = -/(x)
0,
cos x dx
cos<cos" x)
Jcos (cos 1 a)
dx.
sin (sin 1 x)
2f‘f(x)dx, f(2a-x) = f(x)
Sol. We know, cos (cos-1 x) and sin (sin-1 x) could be plotted as
We know, if f(x) =
sec2 x
l + 3tan2x
rlU2
.'. Eq. (iii) reduces to I =2 £
■« dt
=2'3^'0 i + t2
/(7t-x) = f(x)
sin (sin-1 x) and cos (cos-1 x) are identical functions.
sec2 xdx
/an(dn-1P) COSMOS-1 x)
- rin(an-1 P)
dx=fjc
ldx=(P-a)
Jcos(co«_| a) sin (sin-1 x)
'co»(coi-1 a)
1 + 3 tan2 x
(put 5/3 tanx = t => J3sec2xdx=dt)
y=cos(cos'1x)
1
y=sin(sin~1x)
1
•2
=4(tan-,(t)]7
V3
0
.
2 f 7t
j It
2 ,
= -7= (tan 00 - tan 0) = —j=----- 0 =
V3U2
J
V3
it
dx
1 + 2sin2 x y/3
f
n
I
x
1
x
131
Chap 02 Definite Integral
-J(a2+b2)/2
• Ex. 30 Evaluate |.------------
x-a ,
------- dx.
fs P-x
• Ex. 29 Evaluate
x-a .
- -----dx
P-x
Sol. Let
Put
X
JV(3a2 +62)/4
2 -a2)(b2-x2)
J(a2+b2)n
x
Sol. LetZ = [
— dx '
+
/(x2-a2)(b2-x-2)
x = acos2t + Psin2t
x-a = acos2 t + Psin2 t-a=P [sin2(t) + a(cos2t-l)]
2xdx = [2a2 cos t(-sin t) + 2b2 sin t(cos t)] dt
=>
Similarly, p - x=(P - a) cos21
where,
dx=2(P-a)sintcostdt
x = a=>sin2t = 0 => t = 0
and
x=P=>cos2t = 0 => t = n/2
(va<p)
Eq. (i) reduces to
For
xdx = |(i>2 -a2)sin2 tdt
2 a2+ b2
x.2z=-------- = a :2cos2t+b2sin2t,
2
a2 + h2=2(l -sin2 t)a2 +2&2sin21
or (a2 + 62)=2a2 + 2(fe2-a2)sin2t
z=rJon/2 ——* • 2(P - a)sin t cos t dt
(P - a) cos t
-2(P-a)sintcosrdt
-J,> —
cost
For
sin2t
sin f
2 = +----- aste[0,Jt/2]
cost
cos t
=(P~a)f 2sin2 tdt=(P~a) f
Jo
Jo
x/2
sin2t
=(P-a) t2
o
(l-cos2t)dt
2
1
n
sin t = - => cos2t = 0 => t = —
2
4
3a2 + b2
x! = --------- = a 2 cos2t + b2 sin2t,
4
3a2 + b2 = 4a2 + 4(b2 -a2)sin21
It
1
_. 1
sin 2 t = - => cos2t = - => t = —
4
2
6
.'. Eq. (i) reduces to
• X/4 1
'k/6
7t
1
0--sin
0 j =(p-a) —
2
2
—--sinrt
2 2
= (P~a)
2
(■ X/4
Jn/6
(b2 -a2)sin2t
dt
J(b2 - a2) sin2 t(b2-a2) cos21
/ \ */4
it n
4 6
dt = t
x/6
Z=j(p-a)
e8ecx
put ^/P - X = t
or P - x = t2
=> -dx=2tdt
x-a .
------ dx
P-x
When x = a, thenp-a = t2 => A/p-a=t
Sol. Let / = [
Jo
7(p-t2)-a (-2t)dt
I
4JJ ,
----------- -dx.
cos x (1 - sin x)
■ I x+ —
e**1 sin
I
4,
cosx(l -sinx)
dx
if*4 £,8ecx (sin x + cos x) (1 + sin x) dx
Ji
cos x(l -sin2 x)
/P -a
t
t°
= -2 | ,__ 7(P - a) - r2 di
J^P -a
-a
=2p
Jo
it
12
f
sin x + —
lt/4
• Ex. 31 Evaluate P
Jo
e1
When x=P,thenO = t2 => 0 = t
0
...(i)
x2=a2cos2t+b2sin2t
Put
=psin2 t-asin2t =b (x-a)=(P~a)sin21
Aliter Let
dy
-J-f*74 £,8ecx (sinx +- cos x)(l + sin x)
Ji J®
COS X’COSX
VWP^a)2-*2 dt
e«cx
-a
/
•(sec x tan x+sec x) (sec x+ tanx)dx
=2
(P-g) . 1
sin
2
=2
(P-a)dn-VM'
o+®^>
sin-1(l) • -• 0 + --------sin (0)
2
2
,Becx’[secxtanx(secx+ tanx+l) + sec2 x]dx
2
^*ecx • sec x tan x (sec x + tan x +1)
pH
2
2
*
8*cx • [sec x tan x (sec x + tan x) + sec2x
+ secxtanx]dx
jt/4
e»ecx
•sec2xdx
132
Textbook of Integral Calculus
4
Applying integration by parts,
7t
I =-7= {(sec x+ tanx+ l)esccx}0’lM
V2
1
2 \
fM4
f n/4 esecx
secx j
2 x
<
—7=1 (secxtanx+sec
sec x)-e
xye
42 0
dx + -7=J
v2
<
7t
•sec2 xdx
8
V2
x2 (sin 2x - cos 2x) dx
(14-sin 2x) cos2 x
r n/4
Jo
8 Jo
4 Jo
t2 (sin t - cos t)
(1 4- sin t) (14- cos t)
7=i4 fJo’
x2 cos X
Jo
(1 4- sin x)2
a
/(x) dx =
a
m2
>
(1 4- sin x)
Jo
dx
-1
—-—dx
•(i)
14-sinx
a
using,
/(x) dx=£ f(a - x) dx
/=-tc24-2 [n
’ ^Ldx
■(ii)
Jo1
•••(ii)
14- sin x
Adding Eqs. (i) and (ii), we get
71
2I=-2n2 + 2 [«’ ----------dx
Jo> 14- sin x
TT2 A
7t
. .
= -27t2 4-271 /»
[’ 1-sinx .
Jo -5
.
(sin t - cos t)
_ 1 r n/2 \
4 Jo
T 2x-
f(a - x) dx
------------------ dt
(1 + cos t)(l + sin t)
7t t - —
n
^0
n
Adding Eqs. (i) and (ii), we get
21
-1
.
(put 2x = t)
•(cost-sin t)
(
dx=Jo x2 {(14-sinx)-2 cosx}dx
j=fx2 (1 + sinx)~P
2
—-t
2
n/2
Sol. Let I = f"
dx.
I -L.
...(0
using,
71
X2 COS X
Applying integration by parts,
t2 (sin t - cos t)
dt
(1 4-sin t) (cos21/2)
1 ,n/2
4
16
• Ex. 33 Evaluate f*
Jo
V2
- «« r- ,
rn,A x22 (sin 2x-cos 2x) dx
• Ex. 32 Evaluate
■— ---------------------- -—
Jo
(1 4- sin 2x) cos2 x
Sol. Let I = I
7t2
1
=----- 1- 7t log —j= 8
8
8
V22
42.
04
V2
V2
V2
k
7t2
71 + 4 log -7= • -
(Ji + 2) e42 - fie —(e72 - e) = 4= (1 + Ji) e42 —
=
o
0
7t2 1
7t2 It .
— --7tlog2 = — --log 2
— (1 + 1) e}—J=[e',secx j n/4
V2
{(V2 4-14-1)
=
n/2
t
= — • (n - 0) + 4 log cos 8
2
(1 + cost)(l + sint)
cos2 x
J°
dt
...(iii)
=-27t2 4- 2ti [tan x - sec x]on
sin t - cos .t
------------------------- dt = O
(1 + cost) (1 + sin t)
= - 2712 4-27t [ 0 - (-1 -1)] = - 27C 2 4-47t
/ = -7t2 4-27t
= °’^2a “
as
=“
• Ex. 34 Compute the following integrals.
.'. Eq. (iii) becomes
1=18 Jof"'2 7it(sint-cost)dt
(i) f”/(x"+x"")lnx—=o
Q
J°
x
(1 + cos t) (14- sin t)
(ii) Jo" /(x
(,-nn 4-x~
■ n)lnx dx =0
1+x 2
_7t j.n/2 t {(1 4- sin t) -(1 + cos t)} dt
8 Jo
(1 4- cos t)(l 4-sin t)
_1t r M2
8 Jo
_7t t M2
~ 8 Jo
_2n p M2
8 Jo
Sol. (i) Let
p n/2
tdt _7t fM2
t dt
14- cost 8 Jo
J°
14-sint
tdt
14- cost
7t p
r M2
t dt
1 + cost
7t2 rn/2
dx=e‘ dt or — = dt
x
(n!2-t)dt
14- cost
J°
8 Jo
16 Jo
Also,
and
dt
1 + cos t
8
2t-tan2. 0
n/2
X = °° => t = oo
0
= j J^21 (sec21 / 2) dt~~^
n/2
X = 0 => t = -<»
f~/(x" + x“")lnx — =f" /(en'+e-n,).tdt = 0
2
it
t = lnx => x=e‘
(sec21 / 2) dt
t
-I>’ l-2tan-dt
2
7t2
16
x
J—
(•/ integrand is an odd function of t)
n/2
2tan2. 0
(ii) Jo"/(^ + x-n)lnx. dx =£"/(eB, + e-"').t. e ■dt
l + e2*
1 + x2
■
Chap 02 Definite Integral
2
te‘
Now, if h (t) = f(e"‘ + e~nt) •
1 J>) _l(/(x))2
1 + e'
2
e'
Then, h (-1) = f(e~n‘ + ent) - (-1) • ——
l + e*
e +e
h(-O=-h(t)
Thus, integrand is an odd function and hence
n[r*>
l+x
A
K sin xdx and J = J e~a cos xdx
I =-e~a cosx-sj
...(0
J = e"“ sin x + sZ
(ii)
=2 ±72 or
=>
2
In x + constant
or ex
where, Cj and c2 are constants of integration.
But g is continuous on [0, <»), then c2 x1-'^2 is ruled out
sinx-s cosx
1 + s2
Hence,
^x)=q xU+V2
1
Also,
g(0) = c1 =1 => g(x) = ?+72
• Ex. 37 Letlrt =fn/4 tann x dx (n > 1 and is an integer).
s
Show that
Sol. (i) Given, IH = J
f cosxdx = lim f e ffcosxdx=lim A—= 0
Jo
j_»o Jo
j—♦ o $* + i
C tUi
2
1 +
\2
.
f
tan
x-tan2 xdx
,
r
xdx-I
Jo
tan
-(
'
(n-i,
,
xdx
1
n-1
o
(ii) For 0<x<n/4,we have 0<tan" x<tan’
Z(X)=V ^dt
So that;
f'(x)-g(x)
0<4<4-2
f. + fn+2<2/,</B + /JI-2
1
(i)
-[£ mo]
104
= J>
dt - Ia_2, where t = tan x
Jo
and positive in (0, <») satisfying g(Q) = 1 and
.
.
1
</nn <-------2(n+l)
2(n-1)
tan" x dx = |
=Jo sec
1 x-tan
• Ex. 36 Find a function g:R—>R, continuous in [0,00)
=>
1
(ii)
n-1
s
fx
gtfdt
Io
Jo
n
1
sin xdx = lim i e~a sin xdx = lim
=1
2
o Jo
j-»o5* + i
-|2
-41
s
s2
smx—----- sinx- --------r COS X
1 + s2
s2 + l
f“e-“cosxdx =
. r
Z(x)
g(x) = /,(x)=CiX1+,/2 or
=>
sz + l
=>
-2
t2 - 4t + 2 = 0, where t = —^X/(x)
j[x) = cx
Thus, f e-orsinxdx = 1
Jo
s2z +. l
Sol. Let
xgfxf
=4
In /(x) =2 ±
cos x + s sin x
1 + s2
XX
x2
I fM )
Subtracting Eqs. (i) and (ii), we get
x
[using Eq. (i)]
x
I xg(x)
Using integer by parts, we get
and
x
IW*)]2 X2 = 2x /(x) (g(x)) - (/(x))2
(b)Jo cosxdx = 0
Sol. Let us first evaluate;
Jo
2
1 [/'Ml1 ~ T2-f(l)<'(x>-Wx»i
2
(a) Jo sinxdx = 1
Now, [
x
Differentiating both the sides w.r.t x, we get
• Ex. 35 Show that
and
2
2
i/JcrWA _12 Cf(x))
x
j f(x" + x"")lnx-
and
133
or
1
----- <21. <-----n+1
n-1
1
_
2(n + l)
1
2(n-l)
x
134
Textbook of Integral Calculus
Sol. Toprove; sin2Kx=2sinx[cosx + cos3x + cos5x +
...+ cos(2K-l)x]
• Ex. 38 lfUn = £n -—C°- n - dx, where n is positive
1 - COS X
Taking RHS, [2sinxcosx + 2sinxcos3x + 2sinxcos5x +
integer or zero, then show thatUn + 2 + Un=2Un + r
sin2 9
Jo
So/.
... + 2 sin x cos (2K -1) x]
1
— =—
nrt.
Hence, deduce that f”72 S--
=(sin 2x) + (sin 4x - sin 2x) + [sin 6x - sin 4x) +
...+(sin2Kx-sin(2K-2)xJ
=sin2Kx
2
-[1 - cos(n + 1) x]
tu.-iu.-f [1 - cos(n + 2)1 x]- cos
x
-f
cos (n +1) x - cos (n + 2) x &
Jo
sin2K x-cot xdx=
fnl2
. x
2 sin n + x-sm —
l
2
2
---------- dx
2sin2 —
2
=£
(1 + cos2x)dx + £
[cos2K x+cos(2K-2)x]dr
(n/2
(0
(cos2nx)dx = Q,W nel and n#0
Jo
r n/2
• =>
Jo
fk
f /2
1
sin2Kx-cot xdx =
JoJo
ldx+ 0 = —
2
...(ii)
1
• Ex. 40 Evaluate
2x
• sin 1
1 + x2
2J
. x
sin —
2
n 2cos(n +l)x-sinx/2 ,
-------------------------- dx
>
sinx/2
2J
Sol. Let
'l-j*
dx
f
1
1 + x2
J + x2 J
k 1+X
dx
2 tan- x,
if-l£x£l
it -2 tan-1 x, if x>l
r V5
,
.t
2x
1 1
. ------ rsin
1 + x2
J + x2
1
l
r i 2 tan- x
J°
=> <4. <4
(
U}-U0 = it
(common difference)
=2
r V3
f V3 tan-1 x
1
2 A K/4
t2
Now,
1 + x2
it2
_ fK/2
n Jo sin22 0 ^ = Jo1>’
=— + n
n/2 1 -COS2/J0 „
16
--------- tro
l-cos20
£ fw 1-cosnx .
1
~ 2 Jo :" dx = -rm
1 - COS X
2 .
(put tan-1 x=t)
5-tan-11}-2
n
it
3
4
7t2
7C^
9
16
—
12 L
-7 IT*
= — 71
72
• Ex. 41 Prove that £* e Xt ■e-'2 dt = ex2/4
X
Jo e
Sol. Letl = f’e,xt
: • e~‘ dt. Put t = X + z => dt = - dz
Jo
• Ex. 39 Prove thatfor any positive integer K,
sin2Kx
sinx
2
2
,2
=2[cosx+cos3x+...+cos(2K-1)x]
It
Hence, prove that £n/2 sin2Kx-cotxdx=—
2
•••
dx
( 2 \ nl3
+ it {tan
Un=U0 + njt=mt
Un=HK
dx
1+x
=2f0n4 tdt + n^x)f3-2^ tdt
« 1-cosx ,
dx = n
M.>’ -------1 - COS X
2x
•sin
Idx
1 + x2
i tan- x ,
Jo
1 - COS X
2
-2
=2J>^dxx+n
+ it J
J. 777^.
dx = 0
dx.
2
\
2x
sin
2x
Now, sin'
n
=2£ cos(n + l)xdx =2 sin(n + l)x = 0
n+1
o
=> Un+2 + Un=2Un^
— 1
(cos4x+cos2x)dx
But we know that,
sin| n + - |x
k
2>
dx
. x
sin —
2
I
M.
•cosxdx
2cosx[cosx+ cos3x+...+ cos(2K-l)x]dx
r n/2
.
3
1A
sin n + - x-sm n + - x
Now,
sinx
= Jo
From Eqs. (i) and (ii), we get
<4+2 + 11.-2 If,
sin2Kx
•+...+ £
- (
1
sin n + - x
k
2,
-dx
. x
sin —
2
W,
C n/2
Jo
3
3
tU.-tU.-f
r n/2
Jo
1 - cos x
I
=r
Now,
f—
2 >
> ■'e -fI —
2 >I dz
dz =-f’
2J-
-P/4
dt.
i
135
Chap 02 Definite Integral
1x
--Z
X
-
=-f e*
2J-x
Jo (1 + t2) + a(l -12)
2
=—r —
=-ex2/4 rx e"'2/4 dt^-ex2/4-2[3 e-'2/4 dt
2
J-x
2
Jo
dt_____
l-a Jo
Jie)'
t2+
/• a
using, J_“ f(x)dx = 2^“ f(x)dx,when/(-x) = f(x)
Jo
jxext'•x-,2dt = ex:■2/4
Jo
<■
2________ 1
l-a ^(1 + a)/(l - a)
=ex-1'4 fxe-,2/\dt
=>
Jo
2dt
(1 - a) t2 + (1 + a)
- f" ------- ----------- = f "
*2/4 dz
4 dz
2
tan
it
n_
it
? 2
e~,2/4dt
(integrating both the sides w.r.L ’a’)
/(a) = Ttsin"1 a + c
•Ex. 42 If f(x) = e«+j;,(e’ + te~x) f(t)dt,find f(x).
y(0)=fff log 0 + 0 cos x)dx = o
Jo
cos X
But
Sol. We can write f(x) = Aex + Be~x, where
c=0
/(a) = 7tsin-1 a
A = l + |o f(t)dt and B = Jo tf(t)dt
• Ex. 44 Evaluate Cn/2 cosec 0 tan
Jo
A=l+ [’ (Ae‘+ Be~‘)dt=l + (Ae‘-Be~‘)‘
Jo
A=J + A(e’-1)-B(e-1-1)
=>
t
Sol. Let
(2-e)i4 + (e-1 -1)B = 1
(i)
= A (te‘ - ef )q + B (- te~‘ -‘A1
i
cosec 0-tan" (c sin 0)d0
/(c) = Jo
B = A+B(l-2e-1)
f 11/2
A-2e-1B = 0
...(ii)
Jo
From Eqs. (i) and (ii), we get
4e-2e2
e-1
4-2e
4e-2e
e-1
4-2e
Hence,
cosec2 0d6
c2 + cosec2 0
/
1
— — —= =
tan-1
where, f(0) = A
Sol. Given,|a|<l
But
/(0) = Jo
=$
/(c)=^log(c + 7c2 + l)
Let
cosx
, =f
dx
dx
dx=
------------/'W=f0* ------------------Jo
cosx-(l + a cosx)
'o 1 + a cosx
cosec2 0d9
(c2 + l) + cot20
1V2
cot 0
Integrating both the sides, we get
71 de
/(c) = J "J
j log (c 4- 7c2 +1) + A
2^c2 + 1
• Ex. 43 lf\ a | < 1, show that
rjt log(1+acosx)(
dx=7tsin 1 a.
cosx
log (1 + a cos x) ,
------------------ dx
cosx
=r!
7c2+i 7fl
+1 I
£
(c sin 0) dQ.
z.,, v f^2
Q
(sin0)
/ (c) = J
cosec 0----- —— d0 + 0-0
1 + c2 sin 0
rlt/2
d9
Jo
1 + c2 sin2 0
B=£ t(Ae‘ + Be~‘)dt
=>
,
fn/2
cosec 9 tan-1 (0) d0=0
f«
(differentiating wit ‘a’ using Leibnitz rule)
f
x
dx
2
put tan — = t => — =--- : y,whenx = 0, t = 0
2
dt 1 + f
^and when x = 7t, t
/
2dt
1 + t2
h«)=j;
1+a
\l + t2 J
• Ex. 45 Evaluate [
Jt/2
Jo
sec 0 • tan 1 (a cos 0) dQ.
Sol. Given definite integral is a function of‘a’. Let its value be 1(a).
Then, 1(a) =
=>
f ^2
Jo
.
sec0-tan-1(acos0)dO
r^2
I (a) = J0 sec0.
f kJ2
= Jo
1
_ -cos0d9+O-O
1 + a cos,2 0
(using Leibnitz rule)
_
,2
sec2 0
1
d0 = f «/2
’
d9
Jo’ sec20+a2
1 + a2 cos2 0
136
Textbook of Integral Calculus
Jo
sec2 0
1 + tan2 6 +a
=r
dt
__ 1
,
t2 + (a2 + l)“A/7
>>n/2
1
7a2 +1
tan'
t
/(*) =
It
=> I'(a) = —
2 7<j2+i
2L-o
2
From Eqs. (ii) and (iii),
180
, , 80
a =— and b = —
119
119
Eq. (i) reduces to
(put tan 6=t)
80x2 + 180x
119
• Ex. 48 Prove that
Integrating both the sides w.r.t. ‘a’, we get
du-
Jo* <r f dt)du=I* _
f(a) = ^-log|a + 7a2 + ll + C
Sol. Here, applying integration by parts to
Since,
Z(0) = 0 + C
Ila) = — log | a + 71 + a2 |
2
• Ex. 46 Let f be a continuous function on [a, b]. Prove
that there exists a number x G [a, 6] such that
‘
£’ /(t)dt=[‘ fit) dt.
and
fit) dt
(rb
=> gla)’glf) = ~
jMf"
gib) = £* fit) dt
=(“ Jo"
“ Jo*
= X £* fit) dt -
uflu) du
= f lx-u)flu)du
\2
J
• Ex. 49 Evaluate f3,1/2
(log | sin x |) (cos (2nx)) dx, neN.
Jo
It implies that g(x) will become zero at least once in [a,b].
So/. Let /(n) = Jo
(log|sinx|)(cos2nx)dx
Hence, f fit) dt = f fit) dt for atleast one value of x G [a, b].
I
Jx
sin 2nx
/(n)= log|sinx|2n
• Ex. 47 If /(x) = x + £(xy2+x2y)(/(y))dy,>d/(x).
=X[1+JO* y2f^dyj
,2
0
...(i)
jo
' +ty5T
=1 + '
+---- a = l +
45 /o
\
sin(2n + 2)x’cosx ,
---------------------- dx
sinx
*
(sin(2n + 2) x -sin2nx) • cos x
••• A(n+l)-/i(n)=Jo
sinx
T /
a b
—+—
4 5
20a =20 +5a + 4b or 15a-4b =20
x
f M2
|>3n/2 2cos(2n + l)x-sinx-cosx
Jo
3 4
/o
sinx
f3rt/2
= Jq
(cos(2n + 2)x+cos2nx)dx
(H)
b = Jo’ y fly) dy = £’ ylay + by2)dy
\
sin2nx .
cot x---------ax
2n
sinx
a = l+f1 y2 f(y)dy =!+(’ y2(ay + by2)dy
f ay* * by* Y
■3n/2
-i:J
#3
fix) = ax + bx2 or f(y) = ay + by2
and
3x/2
Let /,(„)-J*3«/2
o sin(2nx)cosx dx
fix) is a quadratic expression.
jo
n
(using integration by parts)
rin\) = n0_ 1 If3,I/2 -----sin2nx-cosx
„ 1 T/
l(
---------- dx. = Q--I
iln).
...(i)
2n Jo
sin x
2
Sol. Given, fix) = x + x f * y2 f(y) dy + x2C y fly) dy
jo
»o
where,
du
f(t)dt] <0
Clearly, g(x) is continuous in [a, b] and g(a) • g(b) < 0.
Ja
’
i
i.e. taking T’ as second function and J fit) dt as first function,
we have
f(t) dt ■ du - [ f " f(t} dt] • (w)J - f * flu) ■ u du
Jo
[Jo
r°
Ju
Sol. Let £(x)= [' /(Odt-f f(t)dt,xe[a,b]
We have, g(a) = -
'
“n
X 3JI/2
' f sin(2n + 2)x sin(2n)x
I. (2n + 2) +
2n
,0
a b
=> b = - + 3 4
12b = 4a + 3b or 9b-4a = 0
=0+ 0
/,(«+l)=/,(n)=...=/,(!)
(iii)
.
2sinxcosx-cosx .
j (i) _ f3x72 sin2x-cosx dx= f3«/2 --------------------dx
1
J°
sin x
Jo
sinx
Chap 02 Definite Integral
pn/2
Jo
f 3it/2
,
2 cos xdx =
Jo
Again, I = [ (tx + 1 -x)" dx - [’ {(1 - x) + tx}" dx
(l + cos2x)dx
2
Jo
Jo
3n/2
“I x + Sin2x
137
=Jo* {"Co (1 - xf + "Ci (1 - x)"-1 (tx) + nC2 (1 - x)"-2
0
(tx)2 +...+ "Cr (1 - x)" ~r {txf + ... + "C„ (tx)"} dx
=>
Ji(l) = ^ = A(n) => f(n) = -^-/j(n)
2
2n
:.
rz \
371
I(n)=~
4n
[using Eq. (i)]
=Jo'o’ E nCr(\-x)n-rW dx
,c<'L'ki-xr'x'idx
■
...(ii)
r®0
• Ex. 50 Evaluate
e x sin”x dx, ifn is an even integer.
From Eqs. (i) and (ii),
Pi
n
Sol. Here, In
e
i
i (l-x)"~rxr dx =----- {t" + t"-1 + t"-2 + ... + t+l}
sin" x dx
0
r=o
= (- sin" x e"x)“ + n |
sin'
n +1
Equating coefficient of t* on both the sides, we get
xcosxe x dx
nCk f’ (l-x)""fc?dx = —
I
=>
n
Jo
ln =n[(-sin"-1 x-cosxe-x)^
+ (n-1) [ sin"-2xcos2xe-xdx-[
Jo
Jo
l. = n(n-l)f e-xsin"
-2xdx-n2 f
sin"-2
Jo
Jo
sin"xe-xdx]
monotonic and differentiable, prove that for any real
numbers a and tr,
e-xsin"xdx
(n2 + l)/n=n(n-l)/n_2
Jo
o
■
Replacing n by (n - 2), (n - 4).... 2, we get
In
(n-l)(n-3) (n-4)(n-5)
In—6
(n-2)2 + l (n-4)2 + l
... and so on.
J/(o)
f -1(x) exists.
Let
/-1(x) = t=>x = /(t) hrdx = f'(t)dt
As,
x = /(a)=>/-,{/(a)} = t=>t = a
.
and
<
rf(b)
c*
f; 2x{b-r*(x)}dx=fJu 2f(t)(b-t)f\t)dt
^2(2-1)
I2
I0
22 + l
Jf(a)
_n(n-l) (n-2)(n-3) (n-4)(n-5)
In
n2 + l (n-2)2 + l (n-3)2 + l
n/2
{/2(x)-/2(o)}dx = f^‘” 2x{b-f -1 (x)}dx
Sol. As, f(x) is differentiable and monotonic.
_n(n-l)
In-2
n2 + l n~2
_(n-2)(n-3)
(n-2)2 + 1
(l-x)"-t?dx =----- -----"Ct(n + 1)
• Ex. 52 Given a real valuedfunction f(x) which is
= n(n-l)Jn-2-n2^
In
n+i
Jo
= nf (sin"-1 x-cosx)(e-x)dx [where(-sin" xe-x)7 = 0]
Jo
n!
I(>~
n {l + (2r)2}
n!
n/2
7
=f (b-t){2f(t)f'(t)}dt
2(2-1) T
22 + l °
i
n
f {/(r))2dr
-.•J.-p-dx-l
=-(h-a){/(«))2+J‘ {f(t)}2dt
II 1 + 4r2
Ja
= [b (.{fit)}2-{f(a)}2)dt
Ja
• Ex. 51 Evaluate J1 (tx +1 - x)n dx, n e N and t is
= f‘ l/'2(x)-/2(a))<ix
*a
independent ofx. Hence, show
1
^xk0-x)n~kdx = -
• Ex. 53 Evaluate
C^n+1)
ri sin 0(cos2 0-cos2 n/5)(cos2 0-cos2 2tt/5)
------------------------------------------------------------- dv.
Jo
sin 5 6
Sol. Here, /=£’ (tx + l-x)" dx=f* ((t-<i)x+l)"dx
Jo
[(g-Dx+ir1
z10-l=(z2-l) z2-2cos
. (n + l)(t-l)
1
■
-(i)
2
271 I
z 2 -2 cos— z + 1
0
'
=—— {t" +t"-1 + t"-2 +... +t +1}
n+1 ‘
Sol. We know,
i
4n
x z 2 -2~ cos —
z+1
5
1
5J
2„
6n
z -2 cos — z + 1
5
138
Textbook of Integral Calculus
„
n 1
z -2 cos — + 5 z
1
z5
or
z5
I
2n 1 1
z-2cos— + —
5
z )
6n
1
z - 2 cos — + - J z -2 cos — + 5
Z
5
z /•
x
z = cos 0 + i sin 0
Put
1
/=-
0]
^+Hn|
a 2 + b2
Again,
a bl + b2I = ~
2
and
abl -a2] = a\n(b/a)
=> 2i sin 50 = (2i sin 0) (2 cos 0- 2 cos 7t ! 5) (2 cos 0 - 2 cos 271 / 5)
x (2 cos 0 -- 2 cos 4 7t / 5) (2 cos 0-2 cos 671 / 5)
sin 50 = 16 sin 0(cos‘ 0 - cos2 n / 5)(cos“ 0- cos2 2n / 5)
=>
sin 0(cos2 6 - cos2 rt / 5)(cos2 0 - cos2 2ti / 5)
1
sin 50
16
• • j:
.2
sin 0(cos 0- cos n / 5)(cos2 0 - cos2 2n / 5)
Aliter Convert a cos x + bsin x into a single cosine say
cos (x + 0) and put (x + <|>) = t.
® Ex. 56
dd
Inxdx
Evaluate
x2 + 2x + 4
sin50
= f* —d0 = —
Jo 16
16
nc
lim
• Ex. 54 Show that
= e-2.
n
.
= Inn Y —nCi
lim V —
K=0 n k(K + 3)
n
=2
K + 2dx
K +3
K=0
nC
■
lim
n—♦oo
Intdt
1
1 ,
putt =— =>at = —-dy
t2 + t+l k
y
y
Joo
dx
—+-+1 y
ly
y
1
>
K=0
x
1+n
12 =iero
lny»(+l)
-dy
/2=Jr° fit
k\
x2
w;
■”
x+2dx V
= lim £ nc
'-K ■
nK Jo
n —f °°
2Jo t2 + t+l
h
1
Intdt
' •«
In 2 + In t ,
ln2
dt
— ----------- dt = 2 Jo
t
2
+
t+\
J° 4(t" + f + l)
1
x -0 K + 3
’
Sol. I = f - ?n XC^X— (put x=2t =odx = 2dt to make coefficient of
Jo x2 + 2x+4
x2 and constant term same)
nK (/C+3)
i. -0
n—
Sol.
bn
1
On subtracting, we get J .a2 + b2[ 2
n
-r
Inydy
^2
dx
2
y2 + y + 1
Inydy
T
2-
y +y + i
oo
,n
= jJo‘x•2: • exdx
•: lim
1+n
=(x2-ex)o-f‘ 2x-exdx = e-2\
[
Jo
•*o
= ex
1
t+- I
2
eXdx
_2
=e-2{e-e+ 1} =e-2
• Ex. 55 Letl = \
Jo
and J =
•h
12
COSX
------------------- dx
acosx + bsinx
dt
Now, I] =£
7t
2
2
+
2
—f= tan
V3
7t
2n
“
j_ln2 271
~ 2 3^3
• Ex. 57 Find a function f, continuous for all x (and not
—(0
W->H
zero everywhere) such that f2 (x) =
f(t)sint
2 +cost
Sol. We have,
/2(x)=f
m™Ld,
J0 2+cost
(Note that /2(x) being an integral function of a continuous
bl - aJ = In [a cos x + bsin x]q/2
W-J’lng)
function is continuous and differentiable)
...(ii)
2/(x)/'(x)
_/(x)-sinx
2+ cosx
From Eqs. (i) and (ii),
2»_
t~
ait
cTl +abj = —
b2I-abj = b\n(b/a)
7t ln2
3^3
[Hint By putting x = 1 /1, we get I =-I, so I = 0]
Compute the values of I and J.
and
0
btxdx
Note For a > 0,1 = f
=0
Jo
-J ax2 + bx + a
,
al + bj = —
2
•n/2 bcosx-asinx .
-------------------- dx
J
acosx+tsinx
\/3
U
~ >/3 L 2
sinx
----------------;—dx, where a > 0 and b > 0.
acosx + bsinx
Sol. ‘:
U + (l/2)]2
Integrating, we get 2/(x) = C - ln(2 + cos x)
x = 0 => /(0) = 0 => C = hi 3
,, v 1,
3
/(x) = -ln
J' ' 2 2 +cosx
139
Chap 02 Definite Integral
• Ex. 58 Evaluate £° tan
ax-tan
ita
,2
X
parameter.
Sol. Let
or I =n71-a2 +C
z=r tan ax-x tan -dx
Jo
If a = 0 and I = 0, then C =-n
di _ (•«
bx
dx = [“ dx
da Jo (l+a2x‘2 )x ' Jo
J° l + a2x2
1 r°°
“7Jo
dx
—j
If
.22 ,
1
-joe
It
a
• Ex. 60 Evaluate C72 In (1+asinx dx
Jo
1-asinxJ sinx
2a
_
.It.
dZ = — —- => I = — Ina + C
J
2J a
2
When a = 1 and Z = 0 then C = 0
Sol. Let
Jo
X
2
7o-x
.2
-2axz
dZ _rl
da
Jo
Jo
2a dQ
a2sin20-l
pW/2
Jo
=r
2 sinx
dx
(l-a2sin2x) sinx
2sec2x
rlU2
Jo
------------2-------- 2------
flt/2
2sec2 x
dx
(l-a2)tan2x+1
,
1 + tan x-a tan x
2)
Jo
dx
rlt/2
Jo
Jo .
2a
2a dt
Jo (l-a2)t2 + l °r’(l-a2)
(put tan 0=1)
dt
Jo
t2 +
\2
1
<71-a22 >
Jo
t2+
=>
dl =
it da
Jl-a2
When a = 0 and 1 = 0, then C = 0
Z = nsin*1(a)
Hence,
I = 7tsin ra + C
\2
1
<71-a,22 >
2 =[tan-1(tVl-a2)]7 = it
Vl-a
■Jl-a 2
2a sec2 0 d0
a2 tan20-(l + tan20)
2a sec2 0 dQ
(1-a2) tan2 0+1
(put tanx = t)
2
dt
=r(l-a2dt
r
—
r
2)t2 + l (1-a2)
(put x=sin0)
(l-a!x2)'x27b^.2
rlU2
J°
dx(a2 <1).
In(l-a2x2)
.
—v r - —ddx, a <1
X27(l-z!)
Let
dx
I=J. In 1-asinx J sinx (|a|<l)
da
In (1-a2x2)
f
• Ex. 59 Evalute
1 + asinx
rnl2
di _ c^/2
I =—Ina
2
Hence,
l-a2-l
I = it
—-= — [a tan x]0 =—
It f da
e
So/.
2a =[tan'1 t71-a2]7 =- it-2a
2^1^a,2
■Jl^a
x
—dx, where a is a
g Definite Integral Exercise 1:
Single Option Correct Type Questions
(y2 — 4y + 5) sin (y — 2)dy
equation /(x) = minimum /(x) is x = 0 and /(x) = 0 has
root a and p, then J
-
x dx is equal to
(b) !(<.= -!,’)
4
(d) None of these
(ajlfp'+a*)
4
(C) o
3. f J(3 - 2sin 2 t)dt + £ costdt = 0, then
J»t/2 V
and y = 7t is
(a)^
(b)-V2
(c) - J5
(d) None of these
« sin"’(sin x)+cos”1 (cos x)
(1 + x2) 1+
k
' c - It
+frrctan
dx = log
(3atx=n
t2 +1
x - 2 = 0(0< a < tc), then the value of
7
xis
(a)± 1
(c)±
a
(b)±
2 sin a
! a
sin a
(d)±2.
'2 sin a
a
Isin a
V a
•x tdt
—, then f' (2) is equal to
2 l + t‘
and g(x) = £
9. Iff(x) = e
(a) 2/17
(c) 1
(b) 0
(d) Cannot be determined
10. If a, b and c are real numbers, then the value of
f(l + x2?
a
x2
t2 sin 2t ,
" J-3
2. Let f(x) = x2 +ax + b and the only solution of the
fp
x2
t2 + 2t cos a +1?
(b)2
(d) None of these
(a)0
(c) -2
dt
8. If x satisfies the equation J
is
(2yz-8y + l)
7
17 ])
(where, [■] denotes greatest integer
J + CK y
lim In
(1 + a sin bx)e,x dx equals
t -» 0
(a) abc
(b)^
. . be
(c) —
a
(d)i
r = 4n
function), then number of ways in which a - (2b + c)
distinct object can distributed among -— persons
c
equally, is
9'
12’
15’
(d)-^-x3!(6!)J
11. The value of lim y
n-> °° r = 1
■Jr (3-s/r + Wn)2
is equal to
So’
(aW
5. The value of the definite integral
12. Let /(x) = J t el dt and h(x) = /(I + g(x)) where g(x)is
--------- —--------- (a > 0) is
(l + x‘)(l + x22))
defined for all x, g' (x) exists for all x, and g(x) < 0 for
f
Jo
x > 0. If h' (1) = e and g' (1) = 1, then the possible values
{a)74
^72
(c) 7t
(d) Some function of a
6. The value of the definite integral
p3n/4
Jo
[(1 + x) sin x + (1 - x) cos x] dx is
3k
(a) 2 tan —
8
(b) 2 tan ~
(c) 2 tan
(d)0
7. Let Cn
to
(a)l
(c)-l
rl/n
Jl/n + 1
——dx, then lim n2 • Cn is equal
sin-1 (nx)
“
(b)0
which g(l) can take
(a)0
(c)-2
(b)-l
(d)-4
13. Let /(x) be a function satisfying f' (x) = /(x) with
f(fi) = 1 and g be the function satisfying
f(x) + g(x) = x2.
The value of the integral J1 /(x) g(x) dx is
o
(a)e- -e2--
2
2
(c)|(e-3)
(b) e - e2 - 3
(d)e-|e2-|
2
2
Chap 02 Definite Integrals
dt
M. Let/(x)=
141
[200S
\
21. The value of JI ft f(£
J k I Jo
x | sin tcx | dx ] is equal to
#0
where g(x)= P”* (1 +sin t2)dt. Also,h(x) = e-1 x| and
(a)-72008
(b) ft-72008
(c) 1004
(d) 2008
Jo
/(x)=x2sin —, if x^0and/(0) = 0, then/'
is equal to
x
n
22. lim
—, x > 0 is equal to
n2 + k2x
(a)Z'(O)
(b)R'(O")
(W)
(d) Um
(a) x tan"1 (x)
(c) tan~‘ <*)
xsinx
12 - J 2 /(sin x) dx, then — has the value equal to
12
(a) 1
(c) 2
23. Let a>0 and /(x) is monotonic increasing such that
/(0) = 0 and f(a) = b, then J /(x) dx + J /"’ (x)dx is
'o
equal to
(a) a + b
(b) 1/2
(d) 4
(b) ab + b
x*
24. Iff
r --------- cos
J-Uj3 1-x4
16. Let / be a positive function. Let
A = J*_t (*) f(x(l - x)) dx-, I2 = J*
/(x(l-x))dx,
-it
A
(b) 1/2
(d) 2
(c) ab + a
•
x4
1 + x2
-, then
1-x4
(d)l
dx is equal to
x
x
(b)l
17. Suppose that the quadratic function /(x) = ax2 + bx + c
(d) Cannot be evaluated
. 26. lim
X-+ o
f‘ (l + x)x dx |
o
4
(b)e
(a) 2 In 2
(a) A = /(-!) + /(I)
Jo
(c)2
(a)0
is non-negative on the interval [-1,1]. Then, the area
under the graph of / over the interval [-1,1] and the
X-axis is given by the formula
(d) ab
_j
‘k’ is equal to
(a) 7t
(b) 2it
2s- r 4*
where 2k -1 > 0. Then, — is
(a)fc
(c) 1
x2
X
15. For /(x) = x4 + j x |, let I, = J" /(cos x) dx and
0
(b) tan"1 (x)
(d) tan~‘(x)
1/X.
is equal to
44
(c)lne
, V .
(d)4
27. If g(x) is the inverse of /(x) and /(x) has domain
(b)A = /
x 6 [1,5], where /(I) = 2 and /(5) = 10, then the values of
(c) A =1 [/(-I) + 2/(0) + /(!)]
j /(x) dx + J
(d) A = i [/(-I)+ 4/(0)+ /(!)]
(a) 48
g(y) dy is equal to
(b) 64
(c) 71
(d) 52
28. The value of the definite integral
18. Let 1(a) = f
Jo
x
2
— + a sin x
a
r»/2
dx, where ‘a’ is positive real.
The value of ‘a’ for which 1(a) attains its minimum value,
(,)fl
(d)
Ji
19. The set of values of ‘a’ which satisfy the equation
Jo (* " log2 a)dt = log2
(a) a 6 R
(c) a < 2
/
20. lim xs
(b) a € R*
(d)a >2
1
(a);
(b)-£2
(c)-;11
(d)i
3
3
3
6
29. If/(x) = J (f(t))2dt,f: R-> R be differentiable function
and /(g(x)) is differentiable function at x = a, then
(a) g(x) must be differentiable at x = a
(b) g(x) may be non-differentiable at x = a
(c) g(x) may be discontinuous at x = a
(d) None of the above
30. The number of integral solutions of the equation
f- In t dt
, „ „
4-------------- 7t ln2 = 0;x>0,is
Jo
filx
J-Ux
sin x sin 2x sin 3x dx is equal to
|
dt is equal to
1 + e1
-1-
(a)0
(b)l
(c)2
(d)3
31. f’/B cos —71 — dx is equal to
2
(c)l
(d)0
(a)0
n
(b)l
(c)2
•(d) 3
142
Textbook of Integral Calculus
(ax2 - 5) dx = 0 and 5 +
32. If J
(bx + c)dx = 0, then
(a) ax2 - bx + c = 0 has atleast one root in (1, 2)
(b) ax2 - bx + c = 0 has atleast one root in (-2, - 1)
(c) ax2 + bx + c = 0 has atleast one root in (-2, - 1)
(d) None of the above
33. The value of
(Jx + -J12x - 36 + -Jx - yfl2x -36) dx
is equal to
(a)6>/3
(c) 12^3
34. If In = J
({x + 1}{x2 + 2} + {x2 + 3} {x3 + 4}) dx, (where,
{} denotes the fractional part), then
2
1
(a) - -
is equal to
1
(c) (d) None of these
'3
sin (2n -1) x
r«/2 sin 2 nX
KT
35. Let/. = [’" r
—,neN,
n Jo
sin x
sin x
then
(b) - -
(b) 4^3
(a)JB+1-I=/B
(d) None of these
(C) Jn +1
(b)J.+I-JB=/B+I
In = In
g Definite Integral Exercise 2:
More than One Option Correct Type Questions
36. If f(x) = [sin-1 (sin 2x)] (where, [] denotes the greatest
integer function), then
fx/2
(a) J
71
/(x) dx = — - sin- (sin 1)
(b) /(x) is periodic with period n
(c) lim /(x) = -1
(31
3
ill
(a) T 7 +/F =t
4
(b) the value of J equals to 1/2
f o\
. . [1 1
' \3
e x/2
sin x dx
Wj. (sin x + cos x)y has the same value as J
2
(d) None of the above
37. Which of the following definite integral(s) vanishes?
In (cot x) dx
(a) j
/(x) = x2 +x2
(b) j sin3 x dx
1 + cos 2x
C J«/'x(lnx),,s
dx
38. The equation 10x4 - 3x2 -1 = 0has
(a) atleast one root in (-1, 0)
(b) atleast one root in (0, 1)
(c) atleast two roots in (-1,1)
(d) no root in (-1,1)
39. Suppose I1 = j ’ cos (n sin2 x) dx;
c*/:
^I=Jo cos
tf(t)dt + x3 £ f(t)dt,
then which of the following hold(s) good?
30
(a) £ l/(<) dt = 1?
(b) /(I) + /(-I) = “
(c) J’, t/(t) dt > £ fit) dt (d) /(I) - /(-I) = “
43. Let f(x) and g(x) be differentiable functions such that
y(x) + £ g(t)dt = sin x (cos x- sin x) and
(/'(*))2 +(<?(*))2 =1. then /(x) and g(x) respectively,
can be
■>
r*12
sin x) dx and fj=J0o cos (re sin x) dx,
(a) - sin 2x, sin 2x
2
(b) --------,cos 2x
2
(b) I2 + /, = 0
(c) - sin 2x, - sin 2x
2
(d) - sin2 x, cos 2x
then
(a)A=0
(c)
42. Let /(x) is a real valued function defined by
+ I2 + /, = 0
(d)
COS 2x
(t sin at + bt 4- c)dt, where a, b, c are
44. Let f(x) = J
40. Let/(x) = j’i (l-|t|) cos (xt) dt, then which of the
following holds true?
(a) /(0) is not defined
(b) lim /(x) exists and is equal to 2
x-» 0
(c) lim /(x) exists and is equal to 1
x-* 0
(d) /(x) is continuous at x = 0
41. The function f is continuous and has the property
f(f(x)) = 1 - x for all x g [0,1] and J = Jo f(x) dx, then
non-zero real numbers, then lim
x-i 0
X
(a) independent of a
(b) independent of a and b, and has the value equals to c
(c) independent a, b andc
(d) dependent only on c
45. Let L = lim J
(a)n
n dx
where a G R, then L can be
1 + n2x2>
(b)n/2
i(c)0
(d)l
Chap 02 Definite Integral
143
g Definite Integral Exercise 3:
Passage Based Questions
Passage I
52. The number of solutions f(x) = 2ex is equal to
(Q. Nos. 46 to 48)
j;
Suppose lim —
x-» o
(a)0
(c)2
t2 dt
(b)l
(d) None of these
53. Hm (/(x))/(’x) is
i/p
— = /, where
bx - sin x
(b)6
(d) None of these
(a) 3
(c)l
peN,p> 2,a> 0,r> Oandb* 0.
54. The function /(x) is
46. If I exists and is non-zero, then
(a) b > 1
(b) 0 < b < 1
(c) b < 0
(d) b = 1
(a) increasing for all x
(c) decreasing for all x
Passage IV
47. If p = 3 and I = 1, then the value of ‘a’ is equal to
(a) 8
(c)6
(b) 3
(d)3/2
.
(b) non-monotonic
(d) None of these
(Q. Nos. 55 to 57)
/(x) = J (4/4 - at3) dt and g(x) is quadratic satisfying
48. If p = 2 and a = 9 and I exists, then the value of I is equal
g(0)+6=g'(0)-c=g"(c)+2Z> = Q j=A(x)and y = g(x)
intersect in 4 distinct points with abscissae x,; i = 1,2,3,4 such
to
(< •
that'S'— = %,a,b,ce R* andh(x)= f (x).
A
Passage II
(Q. Nos. 49 to 51)
Suppose /(x)and g(x)are two continuous functions defined
forO<x< 1. Given, f(x) = £ ex + 1 •• f(t)dtand
g(x) = Jo eX+< '8^dt+x49. The value of /(l) equals
(a) 0
(b)l
(c)e“
(d)e
55. Abscissae of point of intersection are in
(a)AP
(b) GP
(d) None of these
(c)HP
56. ‘a’ is equal to
(a) 6
(b) 8
(c) 20
(d) 12
57. 'c is equal to
(a) 25
(b) 25/2
(c) 25/4
(d) 25/8
Passage V
50. The value of g(0) -/(0) equals
(>)A
3-e1
(b)-Te 1-2
-
51. The value of
(c)-Ae -1
(Q. Nos. 58 to 60)
(d)0
(c)7
Passage III
(Q. Nos. 52 to 54)
We are given the curves y - J
f(j)dt,let us define^ as
= V (x) f2 (v(x)) - u (x) f2 (u(x)) and the equation of the
(dy\
tangent at (a, b)and y - b = —
(x - a).
equals
g(2) H
(a)0
=
f{.t}dt through the point
58. If y = J
t2 dt, then the equation of tangent at x = 1 is
(b) y = x -1
(d)y = x+l
(a) x + y = 1
(c)y = x
59. Ify= [
(lnf)dt, then lim —is equal to
x-0- dx
Jx1
y = /(x) where /(x) >Qand f(x) is differentiable,
VxeR through (0,1). Tangents drawn to both the curves at
the points with equal abscissae intersect on the same point on
the X-axis.
(a)0
(b) 1
(d)-l
(c) 2
60. If/(x) = J' e ,I/2 (1 - t2)dr,then — /(x)atx = lis
dx
(a)0
(c)2
(b)l
(d)-l
144
Textbook of Integral Calculus
Passage VI (Q. Nos. 61 to 62)
71 7t
Consider /:(0, °o)-> —, — defined as /(x) = tan
2’ 2/
f logfX
Jlogex)2+1;
[tan]dx is equal to (where, [•] denotes
62. The value of j
67. The about function can be classified as
(a) Injective but nor surjective
(b) Surjective but not bijective
(c) Neither injective nor surjective
(d) Both injective and surjective
the greatest integer function)
(b)£
(a)~2
(c)-l
2
(d)l
g Definite Integral Exercise 4:
Matching Type Questions
63. Let ^lim y £ (sin x +sin ax)2 dx = L, then
Column II
Column I
Column I
Column II
(P)
0
j: 1 dx + J 7“ In x dx is equal to
(B) For a = 1, the value of L is
(q)
1/2
(C) For a = - 1, the value of L is
(D) For a g R - {- 1,0,1}, the value of L is
(r)
1
(s)
2
J--------------
#l/e
i
(A) For a = 0, the value of L is
(q)
(B) The value of the definite integral
_i_
(C)
lim
<il»2a • 3* ...(n-iy-’-jfy
n
l
I +2+3+
+ ji
I
J
(r)
equals
(s) e
64. Let J(0 = £ (x + sin 0)2 dx and g(0) = £ (x + cos 0)2 dx
'0
66. Match the following.
where 0 e [0,2n) The quantity /(0) - g(0), V 0 in the
interval given in column I, is
Column I
Column II
Column I
(A)
.where
If/(x)=L'
Jo 71+?
Column II
(P)
"2
(q)
2
(r)
1
(s)
-1
JO
(A)
it 3n
(P)
negative
P
g(x) = J
4 ’T
3ti
(q)
positive
(C) F3Jt
(r)
non-negative
(B)
—, 7t
. 4
(B)
-
(1 + sin t )dt, then value of
If f(x) is a non-zero differentiable
function such that J /(/) dt = {/(x)}2,
(D)
7t'I
( 7tc _
KMT’2rl
(s)
non-positive
Vx g R, then /(2) is equal to
(C)
65. Match the following.
Column I
(A)
(1 + 2008 x200s) d™ dx equals
JO
*
pb
If I (2 + x - x ) dx is maximum, then
a + b is equal to
Column II
(P) e
(D)
If lim
sin 2x
| = 0, then 3a + b
"7“ + a + ~
x )
has the value
Chap 02 Definite Integral
145
g Definite Integral Exercise 5:
Single Integer Answer Type Questions
67. If f(x) = £^ and J y(x)dx = log
!0
4
n«l
68. The value of I = [
J-k/2
then the value of (m + n) is ...
cos x dx
(where, [•] denotes greatest integer function) is
1 + 2 [sin-1 (sin x)]
— —dG (n g N\ then the value of /(15) + /(13)
• 2
sin
2 0~
/(15)~/(a)
69. If f(x) = — [
KJo
70. Let f(x) = j e <1+ 0 dt and g(x) = (h / x), where h(x) is defined for all x g h. If g1'2 = e4 and h'(2) = 1. Then, absolute value
of sun for all possible value of h(2), is ...
sin x • log (sin x)dx = log
71. If = £
• -G
J
. Then, the value of Kis...
.*
@ Definite Integrals Exercise 6:
Questions Asked in Previous 10 Years' Exams
(i) JEE Advanced & IIT-JEE
2 COS X
72. The value of J
~K/2 1l + e
ex
.
dx is equal to
[Only One Correct Option 2016]
+2
4
(d)n2 + e'/2
(a)^--2
4
(c)rt2-e”t/2
73. The total number of distincts x e [0,1] for which
dt - 2x -1 is
IJ«l—
+ t4
[Integer Type 2016]
74. Let /(x)=7tan* x+7 tan6 x-3 tan4 x-3 tan2 xfor all
XG
I
7C
7C I
—, — . Then, the correct expression(s) is/are
2 2J
[More than One Correct Option 2015]
t« /I
1
(a) J x/(x)dx = J°
12
(b) £" Af(x)dx = 0
(c) £ x/(x) dx = |
(d) JoM7(x)dx = l
75. Let/(x) =
192x3 for all x g R with f R j = 0. If
2 + sin4 nx
m £ £ /(x) dx < M, then the possible values of m and M
axe
(a) m = 13, M = 24
(c)m = -H,A4 = 0
[More than One Correct Option 2015]
(b) m = —, M = 4
2
(d) m = 1, M = 12
76. The option(s) with the values of a and L that satisfy the
[ easin'1 at + cos4 at)dt
equation Jo* K
= £, is/are
£ e'(sin6 at + cos4 at)dt
e4' -1
(a)a=2,L = -----e -1
e4’ -1
(C)a = 4.L = 7—e -1
[More than One Correct Option 2015]
e“ +1
(b) a = 2, L, = ———
e' + 1
eAK---+ 1i
(d)a = 4,L = ^
e +1
■ Directions (Q. Nos. 77 to 78) Let F: R R be a thrice
differentiable function. Suppose that F(l) = 0, F(3) = -4
and F* (x) < 0 for all x e (1,3). Let f(x) = xF(x) for all x g R.
77. The correct statement(s) is/are
[Passage Based Questions 2015]
(a) /(l)<0
(b) /(2)<0
(c) f (x) * 0 for any x g (1,3)
(d) /(x) = 0 for some x g (1,3)
78. If £ x2 F'(x)dx = -12 and £ x5 F"(x) dx = 40, then the
correct expression(s) is/are
(a) 9/(3) + /(l) - 32 = 0
(b) £/(x) dx = 12
(c) 9/(3) - /(I) + 32 = 0
(d) £/(x) dx = -12
146
Textbook of Integral Calculus
79. Let f :/?-> 7? be a function defined by f(x)
[x], x<2
x>2
0,
where [x] denotes the greatest integer less than or equal
Xf(x2)
to x. If J = f —-— dx, then the value of (47 -1) is
-i 94.
2 + f(x
ZY•*■4-1
+ l)’1
[Integer Answer Type 2015]
x
12 + 9x2
80. If a = f(e 9x + 3 tan"1 x )
dx, where tan-1 x takes
Jo
1 + X2
C.
X
J^cos^-J dx
f172
o i
D.
(2 cosec x)17dx is equal to
[Only One Correct Option 2014]
(a) Jo
J
2(e“ + e'“) du
(b) £
I(e“+e-)”du
(c> J.
(e* -e‘“)”du
(d)
2(e“ -e’“)'6du
c
D
s
P
s
p
-•X/2
is
[0, 2] and is differentiable on (0,2) with J(0) = L
LetF(x) = f /•(Vt)dt,forxG[0,2].IfF'(x)= f'(xXV
83. Match the conditions/expressions in Column I with
statement in Column II.
[Matching Type 2014]
D.
(c)
(b) y-4
2
2
+4
(d) ~
2
[Integer Answer Type 2011]
Column II
(a) llog|
(b) llogl
j:
dx
1 + x2
P-
(22
3
(c) logj
(d) llogl
j:
dx
q-
1,
-log 3
2 A3.
Ct
O
4
88. Let f: [1, oo] -»[2, oo] be differentiable function such that
2Iog(|
/(l) = 2.If6|i /(x) = dt =3x/(x)-x3, V x>l then the
f 3 dx
J 2 1-X2
dx
x^~\
r.
n
3
s.
7t
value of /(2) is ....
89. The value(s) of j X
o
7
List I
B-
n + xJ
I
Column I
J’
P
s
xsinx2
, .
87. The value of f *3------ ---------------------—dx is
•'V’^sinx + sin(log6- x2)
<(a)\ y-n
22
84. Match List I with List II and select the correct answer
using codes given below the lists.
[Matching Type 2014]
A.
P
D
[Integer Answer Type 2014]
rt-x'l
r 22 .
x2 + log
log ------ cos xdx
w
c.
0
[Only One Correct Option 2012]
(a) 0
Jo
B.
s.
d2
22)5 • dx is
85. The value of jo4x33< y-y(l-x
86. The value of the integral J*X/2
x e (0,2), then F(2) equals [Only One Correct Option 2014]
(a) e: -1
(b) e* -1
(c) e— 1
(d) e*
C
s
B
r
r
A
(b) q
(d) q
82. Let f: [0,2] -> R be a function which is continuous on
A.
4
—Y equals
Codes
A B
(a) r q
(c) r q
4 )
\
I
r.
Jo cos 2xlog^—J
[Integer Answer Type 2015]
flogU+Jl)
r» 3x‘
J-U + e' dx equals
| is
only principal values, then value of (loge |1 + a |
81. The integral J
List II
List I
.2
p.
The number of points in the interval [—713, x/13 ] at
q.
8
1 + x2
- dx is (are)
[Only One Correct Option 2014]
(b) —
105
(d)
2
90. For a e R (the set of all real numbers), a
.
2
X^
15
List II
The number of polynomials /(x)with non-negative
integer coefficients of degree < 2, satisfying
/(0) = 0 and J /(x) dx = 1, is
which f(x) - sin(x2) + cos(x2) attains its maximum
value, is
(c) 0
[Integer Answer Type 2011]
-1,
,.
(l‘+2‘+...+ «-)
lim---------------- ----------------------- ----------------(n +1)4”’ [(na +1) + (na + 2) +... + (na + n)]
Then, a is equal to
(a) 5
/ \ ~15
(0 —
_1_
60
[More than One Correct Option 201 Jj
(b) 7
/JX -17
(d) —
Chap 02 Definite Integral
■ Directions (Q Nos. 91 to 92) Let /‘(x) = (l-x)2 sin2 x + x2,
2(t-l)
VxeTJand g(x) =
-Ini /“(i)dt Vx e (1, «>).
'i
t+1
[Passage Based Questions 2010]
91. Consider the statements
Q: There exists some x e R such that 2/( x) +1=2x( 1 + x).
Then,
(a) both P and Q are true (b) P is true and Q is false
(c) Pis false and Q is true (d) both Pand Q are false
92. Which of the following is true?
(a) g is increasing on (1, <»)
(b) g is decreasing on (1, <»)
(c) g is increasing on (1, 2) and decreasing on (2, «»)
(d) g is decreasing on (1, 2) and increasing on (2, «>)
93. For any real number x, let [x] denotes the largest
integer less than or equal to x. Let / be a real valued
function defined on the interval [-10, 10] by
x-[x[,
if/(x) is odd
1 + [x[-x,
2
94. Let/ be a non-negative function defined on the interval
[0,1], If £ £1 “(/ (0)2 dt = £ f(t) dt, 0 < x < 1 and
.0
/ (0) = 0, then
[Only One Option Correct 2009]
<a>and 41) >1 <b>4;)
P: There exists some x g R such that,
/(x) + 2x = 2(l +x2).
sin nx
llnteaer
[Integer Answer Tvoe
Type 20101
2010]
10 -l°
J
z/1
1
dx, n =0,1, 2
then
(1 + rc*) sinx
[More than One Correct 2009]
=10*
(b)
(a)
m *1
10
(0 S',. =0
(d)
n ■1
n
andT.=£- n
, for
2
n
2
+kn
+
k
2
" ^
k=0
k=aan
n 2+kn + k2
96. Lets. =X
n = 1,2,3,..., then
[More than One Correct Option 2008]
‘
f (x) cos nx dx is
Then, the value of — J
11
>- and f - > b. 3
2
(c) /(I) < -21 and. f\3J- < -31 /JX(d) f4I2J-1) > 2-1 andJ fcl\3/-31 I < 31
if /(x) is even
10
147
. .
3x3
It
(c) T' < 77T
3x3
3x3
(ii)JEE Main & AIEEE
97. The integral
Jxm
102. Statement I The value of the integral
——— is equal to
1 + cos x
[2017 JEE Main]
Jit 16
(b) — 2
(d) 4
(a)-l .
(c) 2
98. Let IB = j tan"x dx,{n > 1). I4 + Ib = a tan5 x + bx5 + C,
where C is a constant of integration, then the ordered
[2017 JEE Main]
pair (a, b) is equal to
(a)
99. lim
0
(b) -
5
)
(n + l)(n+2)...3n
wQ.-i)
(c)
1/n
(b)^
e
<c)7
dx is equal to
logx2 +log(36-12x + x2)
[2015 JEE Main]
(a) 2
(b)4
0
(a) it - 4
(c)4^3-4
Statement II £ /(x) dx = £ f(a + b - x) dx
(a) Statement I is true: Statement U is true: Statement II is a true
explanation for Statement I
(b) Statement I is true: Statement II is true; Statement II is not a
true explanation for Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
103. The intercepts on X-axis made by tangents to the curve,
X
y = £ 111 dt, x e R which are parallel to the line y = 2x,
'0
are equal to
(a)±l
(c) ±3
104. If g(x) = J cos At dt, then g(x + k) equals
[2012 AIEEE]
(a)^>
X
l + 4sin2---- 4sin—dxis equal to
2
[2013 JEE Main]
(b)±2
(d) ± 4
(d)6
(c)l
X
101. The integral f
is equal to n/6.
an x
’
(d) 3 log 3 - 2
logy2
100. The integral £
[2013 JEE Main]
is equal to
[2016 JEE Main]
, , 18
dx
r* /J
2
[2014 JEE Main]
(b) — - 4 - 4^3
3
(4) 4x6 - 4 - 7C/3
gto
(b) g(x) + g(n) (c)g(x)-g(K) (d)g(x) gfn)
105. The value of j > 8 log (1 4- x) dx is
-° 1 + x2
(a) £ log 2
0
(b)^lo62
(c) log 2
[2011 AIEEE]
(d) n log 2
r*:
148
Textbook of Integral Calculus
106. For x G [o, — j define f (x) = f * Vt sin t dt. Then, f has
°
2/
I
[2011 AIEEE]
(a) local minimum at n and 2rc
(b) local minimum at n and local maximum at 2jc
(c) local maximum at n and local minimum at 2tc
(d) local maximum at 7t and 2rc
= l,p'(x) = p'(l-x), for all xg [0, 1], p(0) = l
[2010 AIEEE]
'o
(a) V41
(c) 41
[cot x] dx, [ ] denotes the greatest integer function, is
- 0
Jo
[2009 AIEEE]
equal to
n
(d) -T
'
2
1
cosx ,
109. Let I = jSin X dx and J = J —— dx.
X
0
0 ■dx
(c) -1
f(x)
and p(l) = 41. Then, J p(x)dx equals
X
(b) 1
107. Let p(x)be a function defined on R such that
lim
108. [
(b) 21
(d) 42
[2008 AIEEE]
Then, which one of the following is true?
2
2
(b) I < - and J < 2
(a) I >- and J >2
3
2
2
(d) I >- and J <2
(c) I < - and J >2
Answers
Exercise for Session 1
1.2.1
8
3.2
7. — log, 3
20
-Llog(j2-1).
10. —7= + “7=
6. -1
2V2
13. 10!
V2
14. (d)
13. (c)
2
3
19. (a)
%/3 . 4V2
4
4.--— 5.—
3 ‘
8. n
--42]^
4
)
Exercise for Session 2
1. (c)
2. (a)
3. (a)
4.(0)
5. (a)
6. (c)
7. (b)
8.
8- (a)
9. (a)
11.-
12.0
13. (c)
7C
14. (d)
Exercise for Session 3
1. (c)
7. (b)
12. (b)
t
2. (a)
8- (d)
13.(c)
3. (b)
9.(c)
14. (b)
4. (b)
10. (b)
15. (d)
5- (a)
11. (b)
6.(d)
16. £(r-l)-(r1/"-(r-ir) + H9-^")
r■I
18. 3
19. 1
20.2
Exercise for Session 4
2. (C)
I- (b)
7. (a)
8- (C)
13. (c)
19. (c)
20. (a)
14- (c)
3. (c)
9.(c)
15. (d)
4. (a)
10. (b)
16. (c)
5- (c)
H. (b)
17. (a)
6. (a)
12- (b)
18. (d)
5. (a)
H. (c)
6.(b)
12. (a)
Exercise for Session 5
1. (c)
7. (a)
2.
(b)
2.(b)
8. (a)
3. (a)
9. (c)
4- (a)
10. (c)
15. (c)
16. (a)
17. (a)
18. (b)
4-(b)
10. (a)
5.(c)
6.(c)
Exercise for Session 6
9.-(b-a)
2
12.1 +
11.—
2
15. (d) •
14. (c)
20. (b)
1. (c)
7. (a)
2. (a)
8. (a)
3.(c)
9.(c)
Chapter Exercises
5-(a)
2.(c)
3. (a)
4. (a)
1. (a)
8.(d)
10. (a)
9. (a)
7. (d)
11. (c)
16. (d)
13. (d) 14. (c) ’ 15.(c)
17. (d)
21. (d)
22. (c)
23. (d)
19. (b) 20. (a)
28. (d)
25. (a) 26. (b)
27. (a)
29. (a)
33. (a)
34. (b)
31. (a) 32. (b)
35. (b)
40. (c,d)
39. (a,b,c)
37. (c,d) 38. (a)
44. (d)
42. (b,d) 43. (c)
45. (a,b,c) 46. (d)
48. (b) 49. (a)
52. (b)
50. (a)
51. (b)
54. (a) 55. (a)
57. (a)
56. (c)
58. (b)
62. (c)
60. (a) 61. (c)
63. (A) -4 (q), (B) -4 (s), (C) -4 (p), (D) -4 (r)
64. (A) -> (q), (B) -4 (r), ?C) -4 (s), (D) ->(p)
6. (a)
12. (c)
18. (a)
24. (a)
30. (c)
36. (a,b,c)
41. (a,b,d)
47. (a)
53. (c)
59. (a)
65. (A)—> (s), (B)-4 (p), (C)-> (q)
66. (A) -> (s), (B) -4 (r), (C) -4 (r), (D) -4 (q)
72. (a)
67. (8) 68. (0)
69.(3)
70. (2)
71.(2)
76. (a,c) 77. (a,b,c)78. (c,d)
73. (1) 74. (a,b) 75. (b)
79. (0) 80. (9) ’ 81. (a)
82. (b)
85. (2) 86. (b)
83. A—4S, B—>s, C—>p, EMr
84. (d)
92. (b)
87. (a) 88. (8/3) 89. (a)
90. (b,d) 91. (c)
98. (c)
97. (c)
95. (a,b,c) 96. (d)
93. (4) 94. (c)
99. (b) 100. (c)
101. (d) • 102. (d) 103. (a) 104. (c)
105. (d) 106. (c)
107. (b) 108.
108.(d)
(d) 109. (b)
= log (1 + rt2) + 2n(tan_| x)4 - [log (1 + 16) - log (1 + n2)]
17
= log(l + tc2) + 2n(tan-1 4 - tan-1 it) - log
1+ K2
Solutions
. (1 + n2)-(l + rt2)
= log -------- —------- - + 2rt tan
17
On comparing with log
(l + n2)2>|
a
(2y2-8y + l)
Put
=>
(z2 + l)sinz
dz is an odd function.
(2z2—7)
2. Hence, /(x) = x2 + ax - b
c
5. Put x = tan 0
h: 1 + (tan ey
r x/2
6. I =
=
Jo
=2
Jo
x(i(sin x - cos x) dx
x
v.
"
u
(sin x + cos x) dx
px/4
Jo
»Ji/c
J0
(sin x + cos x) dx
r4
(sin x + cos x) dx + 0 = 2 [- cos x + sin x];
1
7. We have, C, = J",
tan'1 (nx) ,
.
dx
sin' (nx)
c.=ir
(put nx = t)
n J— sin (t)
tan'
Now, L = lim n2 • C, = lim n J „
— dt(« x 0)
-^77 sin'
=0
L -^~
--J3 -2sin2x
dx
i
fJX/4
JX/4
=2 + l + ■+1 = 2 (72 + 1) = 2 tan —
8
V2 J2
Differentiating both the sides, we get
(•^3 -2sin2 x) • 1 + (cosy)
(cosey
~
T n
d0 => I = —
(sin 0)‘ + (cos 9)8
4
(3«/4
J0
-j(3 - 2sini2: t)dt+ j( costdt-Q
dy
(sin x + cos x) dx +
x3dx = 0
rr f(x)dx = 0, when/(-x) = f(x)j
3. Here, J
!
i.x/2
Jo
+ [x(—cos x -sin x)]’1'4 +
— = 0 or a = 0
2
y(x) = x2 - b = 0 has roots a and p.
a = -y/b and p = - 4b
J Vo
de__
f3x/4
-a
The solution of f(x) = minium/(x) is x = —.
2
Given, /(x)]
is at x = 0.
J cia
, we get
Thus, the number of ways to distribute 9 distinct bijective into
9!
3 persons equally is —y.
(3!)
I = 0, as | /(x) dx = 0, when f(-x) = - /(x)
=3
1 + CK
a — (2b + c) = 17 - 8 = 9 and -—- = 3
(z2 + l)sinz
dz
(2z2 -7)
’-y.
/(*)
J
c—k
+ Im tan
a = 17, b = 2 and c = 4
y -2 =z
dy -dz
and
1+ 4n
4 — 7t
(l + *7
+ 2k tan'117
1+ 4n
= log
1 (y2 - 4y + 5)sin (y - 2)dy
1. Let/=p
Jo
4— K
tan
L=
cosy
tan-1 t
sin’1 t
dt
0 .
- form
.0
T
n
n
tan'
n + 1 ( 1__
0------n {(n + I)2
sin'
n+1
Applying Leibnitz rule,
'^*/(x ,x)
4. Here,-4 £ x
-1
x2
4 => 0 £ — <1
17
L = lim
’x2l
— = 0 and sin-1 (sin x) is an odd function.
Let./=f (1+X2)X1
cos'1 (cos x) ,
------ 4---- -dx
J-4 (1+ x2)xl
== 0+2
0 + 2 [f’’ COS"<CO5X> dx
Jo
J" (l+x!)
7
8.
dx
4
K
t2 sin 2t
f1
dt
J® t2 + 2t cos a + 1
11
1=2
2
j: t2 + 1 dt = 0 as the integrand is an odd function.
Also,
2k
TC
1
t + cos a
tan'
sin a
sin a 0
a
2 sin a
1
2
150
Textbook of Integral Calculus
Thus, the given equation reduces to
x2-------------2 = 0 => x = ± 2
2 sin a
9. f(x)-e-■I(I) .g'(x) andg'(x) =
'sin a
a
N°w.r^)=
X
But
/*(x) = e,(x>-
X
=
-lash'(0+) = -l
I, = £ /[cos (tc - x)] dx = J /(- cos x) dx
o
= f /(cos x) dx
f
Jo
i
fK/2
rX/2
Jo0
Jo0
Zj = 2 (
tic
/(cos x) dx • 2 |
12
•
Aliter Let u = cos x => du = - sin x dx
f(u) = du .
dx
Vx(3>/x + 4)2
s = ly____ 1
y/l-U,21
^2/'^
Jo 71-u2
3 1
Put 3-Jx + 4 = t =>----- 7= dx = dt
_2 r10^_2riT
t2
3[.t -ho
dt
...(ii)
10.
*( ’ e1’ dt = /(I + g{x))
(given)
From Eqs. (i) and (ii), — = 2
^2
16. Given,
= J * x f(x (1 - x)) dx and
f2=J*t/(x(l-x))dx
Differentiating, we get h'(x) = e(1+x(x)) •g'(x)
h'(l) = e
0 7^
2 6
1
3 40 ”10
12. Given, /(x) = J , e'* dt,h(x) = f(l + g(x))-, g(x) < 0 for x > 0
Now,
•••(*)
T(0
’
2 1
3 [4
du
Similarly, with sin t = t,
2 Vx
Now, /i(x) = j
/(sin x) dx = 2/2
=> i=2
OR
3
I2 J
15. Clearly, f is an even function, hence
(1 + a sin bt /" .
= In lim --------------- -— = In e
11. Tr =
f
Hence,
[ (1 + a sin b x)e/x dx
= In lim —-------------------------
t-»o
/(7t/2)
■J1 + g2(K 12)'
(fJ-.M-.
10. lim In - [ (1 + a sin bx)c/’ dx 1
,t J“
J
I-+0
g'(x);
(x) = [1 + sin (cos2 x)] (- sin x)
e<(2)=e°=l
f(2) = e-M.S'(2) = e>= A
Hence,
i
14. Here, f(x) =
(given)
Using King’s property Z, = J*
(1 - x) /(x (1 - x)) dx
e
=>
=>
or
2A = I-* f(x^~x^dx = I2
(l + g(l))2=l
l + g(l) = ±l
g(l) = 0 (not possible)
g(l) = -2
13. Given, f(x) = /(x) => /(x) = Cex
^=2
A
17. A = J j (ax2 + bx + c) dx = 2 J (ax2 + c) dx
Since, /(0) = l
/.
1=/(O) = C
f(x) = ex and hence g(x) = x2 Thus, £ fW g(x) dx = £ (x2ex - e‘-)dx
Y4
o
= [x2ex]’0-2 [’ xex dxJ0
=(e-0)-2[(xe')l-(«')■. ]-l(e’-l)
=(e-0)-2[(e-0)-(e-l)]-l(e!-l)
= e--e2
2
3
2
-1 0
1
a
1
= 2 — + c = — [2a + 6c]
3
A=jl/(-i)+4y(o) + /(i)j
>x
151
Chap 02 Definite Integral
18. 1(a) =
■*(x2
0
.
7t’’
ita2
„
/ =(2008)’
I xsin xdx-lt
— + a2sin2x + 2xsinx dx
a
)
22.
T.=
a
7C
l(a) = —7 + —+ 2n =n —- + — + 27C
3a2
2
3a2
2
- it
2
K
a
VJa
72.
-ff;
______ 1______
n[l + (k/ n)2 x2]’
s=ly___ 1___
n
2lt
+ ~T= + 2n
V6
1 + (k I n)2 x2
V2
dt
1 + t2x2
23. Let y = f(x) => x = g(y) and dy = f (x) dx
=>
ftg
f
Also,
[/(a) I™ =2n + it2
2
t2
19.
=j:
1
-1/Xf _ tan-1 (x)
dt
= — tan (tx) =
x2 J® t2 +(l/x2)
x
x
1(a) is minimum when -i- = ~^= => a2 = it
y/3a
=> JI =2008
n
n2 + k2x2
0
— - log, a ■ t
2
0
I-.,.
ln(l + t2)
dt
1 + e~‘
•••(0
f l/x
24. Letl=fr1'/’
J-i/Vj
x4
cos
i-x4
.i/V5
x4
-1/75 1 - x4 cos
’
2x
dx
-G)
-2x
dx (using King’s property)
1-x4
2x '
',/75
x4
J-i/75 i-x4
=» /=I
7C - COS
1-x4 J
dx
-(ii)
On adding Eqs. (i) and (ii), we get
.\i4i X*
x4
dx
21 = it J-i/75 i-x4 dx => 21 =2k J(
0
1 -X4
X-4-
0 r
— form
.0
X
Using L’ Hospital’s rule,
x4 ln|l + ^|.
x J
I = lim
x-> 0
1
x2
25. Put In x = t or x = e' =^dx = e‘ dt
... I = J" f(e’ + e-) 4 e‘ dt = J" f(e' + e") t dt = 0
(as the function is odd)
Aliter I Put x = tan 0
= - lim x2 In [ 1 + —1~
3
-3
i .. , ( , i 1
= - lim In 1 + —
3—
I
x\
k=n
:.
,
ln(l + t2)dt
= lim 0-------- r------f
I
x22.
In tan 0
rX/2
,
sec 0 00
/. f tan 0 H----tan-—0j ---------tan 0
(1~ form)
r«/2
(
Jo
I
X2
In tan 0
tan 0 J sin 0 cos 0
1
f tan 0 +
=|
1+-‘H
x--3
0
Jo0
x-» ••
= lim -x2
dy,y =
+L
J0
Jo
On adding Eqs. (i) and (ii), we get
21 = j‘/X In (1 + t2) dt = 2 J•l/x
’/XIn (1 + t2) dt
1= lim xs f’/Xln(1 + t2)dt
W) = q
= f Cf(x) + xf (x)) dx = [xf(x)] o = a f(a) = ab
.(ii)
Hence,
f(0) = 0
=> x = f~' (y) = g(y) = £ f(x)dx + £ xf (x) dx
I = Jf'-l/x
/X ln(l + t2) e' dt
1= f',Xln(l + t2)dt
b
0
(using King’ property)
=>
f(a) = 8
1 ~ Jo
(2-2 log, a) = 2-2 log, a .
2 log, a = 2 log, a => a e R+
•l/x
ln(l + t2)
20. Consider I = f
dt
J-l
-l/x
1 + e'
=>
f(a)=by
(v a > 0)
= 2-log, (a2)
=>
• l/x
a
.
---------------- uu
Aliter II Put x = l/ t=>/=-l =>2/ = 0=>7 = 0
i \1/X
21. Put7tx = t => dx = —
26. lim
n
■*.
1
K f^OSX
I =------ [
1 f200»» .
11 sin 11 dt = — f
(i)
1 r 2004 k
=> I = —
(20087C -1) | sin 11 dt
It
On adding Eqs. (i) and (ii), we get
200871 /200«»
|
7t
Jo
0
l/l
,
-(ii)
= lim
2* 1 -1
= lim
X +1
i*i
= e‘-^
,
(1 4- x)
X +1
, .
11 sin 11 dt
it Jo
n it J°
11 (1 + x)1 dx I
1/1
(1“ form)
)
Um
=e
'•[ 1(IH) J
1 -1
.
| sin 11 dt = (2008) • [ | sin 11 dt
Jo
= ex^
1
J = e2
e
152
Textbook of Integral Calculus
27. Uty=f(x) =» x = f~'(y) = g(y)
32. J
(ax2 - 5) dx + J (tx + c) dx + 5
dy = f(x) dx
1=+r **
= J * (ax2 - 5-bx + c + 5)dx = 0
Hence, ax2 - bx + c = 0 has atleast one root in (-2, -1).
When y = 2, then x = 1
and when y = 1, then x = 5
33. I = J‘ ((7^3 +
X
7x^3)) dx =6^3
+
34. A=J’, (W + {x3}){x2}dr
y
1
KD = 2
5
f(5) = 10.
2
x3
= -2j‘ {x!|dx = -2x—
3 o
3
-«/2sin2 nx - sin2 (n -1) x
sin2 x
35- J.J,
l=J’(/(x)+xf(x))dx = |x/(x)|,!
_ rx'2sin(2n -1) x-sin x
dx = /„
J°
sin2x
= 5/(5) -/(l) =5 10-2 = 48
i.e. .
28. I = f sin x sin 2x sin 3x dx
-0
Jo
=
I-X-l=/x
(i)
Xm-X=/x+1
r«'J .
Jo
(n
7t
** x Isin- 2----71 x isin• 3
„ I----71 x L
sin ---—
dx
^2
2
2
J
36.
y
fX/2
- cos x sin 2x cos 3x dx
= J#
...(ii)
5 ..
2
ftIZ
- sin 2x (cos x cos 3x - sin 3x sin x) dx
.'. 21 = I
Jo
=
[K/2
<0
1--
- sin 2x cos (4x) dx
-3
2
X'-
fx/2
=-J
n
4
k
3re
0
>X
5
sin 2x (2 cos 2x -1) dx
Putcos2x = t => - sin 2x x 2 dx = dt
21 = f"'-(2t2 — 1) dt = - [—
2
2
3
1F2
2 b.
1_ ' 2
2
=------- + 1-- + 1 =2[ 3
3
’ ?
f X/2
37. (a) I = f
. 1=1
Jo
6
I =-
29. Here, r(x)=(/(x))’ >0;^-/(j>(x))L.
dr
g(x) - g(a)
= f (<?W) lim
x -a
It implies that g(x) must be differentiable at x = a.
x2
30. On putting t = — and then solving,
z
Inx . 27t In x
dt =--------.2 . *2
x
=>
=>
ln2
X
X
x = 2 and 4 i.e. two solutions.
xn
31. Put — = t
4/
_ 4n2 f*
COS y [t] dt = o
n J®
In (cot x) dx => I =
r X/2
Jo
In (tan x) dx
Jo
-0
In (cot x) dr =*I=-I=> 1 = 0
2X
(b) I = f sin3 r dx = - f sin3 x dx => I = 0
Jo
Jo
(c) Atz=i,f=r
t
-l/t(lnt)*'3
I = -1 or 1 = 0
l + cos2x
2
(d)
>
„
fX [1 + cos 2x
io V
dt
t(ln t)*'3
dr > 0
38. I = fo
1 (10x
4 -3x2 -1), dr = [2x5 --x3-x]‘=0
.
Jo
Since, /(x) is even, hence must have a root in (-1, 0).
39. We have, I, =
J, =
jr
n nx .
n r"•ifai
cos — — dx=—I cos y[t]dt
2 nJ
n J®
! r
f X/2
n
Ita*
o
O'
2
11 ?(-!)
’-(-I)- |(1)3-1
4
\3
2 [[3u
In x
_n
2
-t
fR/2
J0
fx/2
cos (n sin x) dx
Jo
cos (% cos x) dx
On adding, 2It = f
Jo
-0
<• x/2
=
=>
Jo
A=0
cos (n sin2 x) + cos (n cos2 x) dr
M
(n
n J
2 cos — • cos — cos 2x dx = 0
\2J
2
(i)
Chap 02 Definite Integral
Now,
Z2 =
pX 12
Jo
f X/2
X
1
21 =fJo
cos {tc (1 - cos 2x)} dx
2
| 1
.
1
-7= sinx + -r=cosx
v2
V2
f IC/2
=-
dx
1 ,x/2
----------------- - dx
J (sin x + cos x)
cos (n cos 2x) dx
Jo
= -1 £ cos (n cos t) dt
2 t*'2
=--J
(put2x = t)
2 Jo
<4
_ 1 f*'2 cosec2 [ — + x j dx = - 1
2 Jo
\4
J
2
fK/2
X/2
cos (n sin t) dt
Jo
42. We have, f(x) = x2 + ax2 + bx>
where,
a = J‘( t/(t) dt and b = £ /(t) dt
i
Now,
a = j'] t[(a + 1) t2 + bt']dt
0
=>
a = 2b f t*dt = —
Again,
b=
=>
b=2 [' (a + 1) t2 dt
=>
^2(0 + 1)
3
40. /(x) = 2 f (1 -1) cos (xt) dt
v
- -
J *■
_ -
= 2r(1-t)^
X
1
+ — f sin xt dt = 2 0---- - cos xt
x2
x Jo
0
1 - cos x
x2
/(x) = 2
(x*0)
If x = 0, then/(x) = J
(1 -| t1) dt-2 £ (1 -t) dt = 1
option (c) is correct.
Hence,
/(x) =
2
1 - cos X
X2
.
|, ifx*O
, ifx = O
1
41. Given,f(/(x)) = -x+l
yw(x)))=-/w+i
Z(l-x) = -/(x) + l
/(x) + /(l-x) = l
Hence,
Jo
JO
2J=f‘dx = l => J = i
1_
1
3
+ F 11 -- | = 1 =>F
=1
+F
4
4j
4.
4.
..
fX/2
sin x
Now I =
Jo (sin x + cos x):
. (n
sin---- x
U
•X/2
'=1.
0
fsin f— - xl + cos f— - x
fn
=> z = Jof
2
3
5a
2(a + 1)
2 "
3
2
a=3
11
2
—a=6
3
£ tf(t)dt = ± and j‘/(t)dt = ^
=>
30
/(l) + /(-l) = 2(a + l) = ^
and
20 <
/(l)-/(-l)=2& = ^
2
Put x = i in Eq. (i),
F
5
2
-(ii)
/(I) = (a + l)O
/(-l) = (a+l)-&
2j=f’(/(x) + /(l-x))dx
Jo
Jo0
/(x) =(a + 1) x2 + far’
Now, J = f /(x) dx = f /(I - x) dx (using King’s property)
=>
7(0^ = ^ ((a + l)t2 + bt')dt
4 and> b. = —
io
a=—
11
11
Replacing x by /(x), we get
=>
-(i)
5
J°
From Eqs. (i) and (ii),
f is continuous at x = 0.
option (d) is correct.
Jo
»(H"
=>!=2
...(ii)
I2 + I3 = 0
Hence, Z, + I2 + I3 = 0
•'0
2
dx
sin2 f — + x
cos (n cos t) dt = I3
I22 + I3=°
Ij = -
153
COS X
.
----------------- - dx
J (cos x + sin x)
ydx
43. Given, Cf(x))2 + (g(x))2 = 1
/(x) + J g(t) dt = sin x (cos x - sin x)
Differentiating both the sides, we get
f (x) + g(x) = cos 2x - sin 2x
Squaring both the sides of Eq. (i), we get
Cf (x))2 + (g(x))2 = 2f (x) • g(x) = 1 - sin 4x
=> 1 + 2/*(x) • g(x) = 1 - sin 4x
2/*(x) g(x) = - sin 4x
sin 4x
Now, substituting g(x) =---------- in Eq. (i), we get
2/* (x)
(0
154
Textbook of Integral Calculus
f(x) - S*n 4 - = cos 2x - sin 2x
J
2f(x)
47. I = lim
Put
f(x)=t
=> 2t2 - 2 (cos 2x - sin 2x) t - sin 4x = 0
2
=---= 2 lim----- *-o(a + xr)1" a"'
2 (cos 2x - sin 2x) ± -^4(1 - sin 4x) + 8 sin 4x
“
4
Taking + ve sign, 2t = cos 2x - sin 2x + cos 2x + sin 2x
=>
t = cos 2x
Taking - ve sign, t = - sin 2x
/(x) = e’ (’ e‘ -f(t)dt
JOo
/(my)
f(x) = Aex .
=>
f(t) = Ae' .
where, A=[‘
cos 2x
2
A = f ef - Ae' dt; A = A ( e2t dt
=>
Jo
Now, A r [ e^dt -1 j = 0 => A = 0, as j e2' dt * 0
Hence,
J(x) = | sin 2x or f(x) = cos 2x -1
2
If f (x) = cos 2x, then gfx) = - sin 2x
If f(x) = - sin 2x, then g(x) = cos 2x
B = B f’ e2' dt + f e' • t dt
But
[' e* dt = -(e2 -1) and [ te' dt = l
Jo
Jo
2
Jo0
cos at
=>
3B = Be2 + 2
0
a
COSafdt+|ct|* (using I.B.P.)
J
a
♦j:
- x cos ax
1 .
+ — sin ax + ex
a2
cosax sin ax
+------- + c
= lim 2 x^0
a
a ■ ax
n dx
45. Consider I = J
1
\
n
I n
>
,
L - lim---- tan
■
an
.1
if a < 0
if a > 0
x~»0 b - cos x
=>
B=
2
3-e2
2
2
ex + x=>g(Q) =
3-e2
3-e2
/(0) = 0
S(0)-/(0) =
2 -0=-^
3-e2
3-e,2
3-e2
g(Q)_ 2
6
g(2) 3-e2
Using L’Hospital’s rule
For existence of limit, lim denominator = 0
b -1 =
b=1
1
3
Solutions (Q. Nos. 52 to 54)
We have the equations of the tangents to the curve
y = J f(t)dt and y = f(x) at arbitrary points on them are
’'-f,
x!
--------------- = lim----------------
bx - sin x
Also,
Jt
V. f K
n/2, ifa = 0
0,
t’dt
J<0
2e2
6
St i(2) = ----- 7 + 2 =
3-e2
3-e2
n,
r
From Eq. (ii), g(x) =
= -n(tan nx)0= — tan an
n
\2 2
n’fx’+l
r
=>
2B = B(e2 -1) + 2
a
-0
B = ^(e2-1) + 1
n
1 1
= 2 - - + - + c =2c
a a
46. lim
Jo
=❖
=2
r
Jo
= 2 [ (t sin at + c) dt
=2
x
g(x) = Be* + x
.(H)
g(t) = Be' +t
B = [' e‘ g{t) dt => B = [' e‘ (Be' + t) dt
even
Jo0
lim
g(x) = e' £ e' g(t) dt + x
50. Again,
(tsinat + Irt + ,c )dt
even
0
/(x) = 0 => f(l) = 0
where,
i.e. /(x) = - sin 2x and g(x) = - sin 2x
2
cos 2x -1
and g(x) = cos 2x
=> /(*) =
2
x-aO
Jo
|_Jo
C, =0 and C2 = -1/2
~X
.(i)
JO
0
/(0) = 0
44. Consider f(x) = J
•>
2
2
48. If, p = 2 and a = 9, then I = —
9./2-3
_f(x) = -sin2x
/(x) = - sin 2x + Q or /(x) =
2
=>
2
■
If p = 3 and I = 1, then 1 = — => a = 8
49. Here,
=> 2t = (cos 2x - sin 2x) ± -Jl + sin 4x
f (x) = cos 2x or
/
a‘"
4t = 2 (cos 2x - sin 2x) ± ^4(1 - sin 4x) + 8 sin 4x
Since,
x2
x2
1
(a + xr),/p ‘ (1 - cos x) x2
•••(>)
and
Y-/(x) = /’(x)(X-x)
-09
As Eqs. (i) and (ii) intersect at the same point on the X-axis
Putting Y = 0 and equating x-coordinates, we have
155
Chap 02 Definite Integral
Also, f(x) has y = 0 as asymptotes.
f(x) can be shown as
£ f(t)dt
f(x)
x--—--X-
/(x)
f(x)
=>
n
£/w* /(x)
e, tan
2 ’
=>
(Hi)
>x
0
As, £(0) = 1 => J 0 f(t) dt = c x 1 =*c = ±
K
J
~2
/(t) dt = - f(x); differentiating both the sides and
integrating and using boundary conditions, we get f(x) = e2x;
y = 2ex is tangent to y - e2x.
:. Number of solutions = 1.
Cearly, f(x) is increasing for all x.
(oo° form)
Jim (e2x)‘‘” =1
Clearly, f(x) is neither injective nor subjective, also graph for
[/(x)] can be shown, as
Solutions (Q. Nos. 55 to 57)
0
We have, g(x) = g(0) + xg'(O)+ — g* (0) = - bx2 + ex - 6
2
ft(x) = g(x) = 4x4 - ax3 + bxz - ex + 6 = 0 has 4 distinct real
roots. Using Descarte’s rule of signs.
Given biquadratic equation has 4 distinct positive roots.
Let the roots be jq, x2, x3 and x4.
12 3
4
—+—+—+—
24
jq x2 x3 x4 >«
Now,
Xj x2 x, x4
4
2
2>2=> —=
x, x2
4
3
-o
62. j;[/(x)]dx=j0‘ -l dx+ [ 0dx=-(x)- = -l
63. (A) For a = 0,1 = j sin2 x dx
rt 1 - cos 2x ,
= 1 ------------ dx
Jo
2
r
x 1 . „
-------sm 2x
L2 4
jo
=k
xs x4
2 3 4
-L. . —. — = fc4
X, x2 x3 x4
=>
1
1 sin2T
L =----lim----------2 r-“ 4 T
— = k* => k = 2
3/2
=>
1 3
Roots are -, 1, - and 2. Also, a = 20 and c = 25
2 2
58. At x = 1, y = 0, — = 2x • (x4)2 -(x2)2 = 1
dx
:. Equation of the tangent is y = x -1.
59.
dx
y(x)=p’'!(i-t!)<ft
/'(l) = e”’-0 = 0
61. Here, f(x) = tan
(C) For a = - 1, f 0dx = 0=>L = 0
J0
•
(D) For a
0, -1,1,
0
l-cos2x 1 - cos2ax
+ cos(a - l)x - cos(a + l)x dr
----------- +
2
2
x—>0*
f(x) = [«'■'■ (l-x’)]’
Now,
■>
rT
(B) For a = 1, J 4 sin x dx =$ L = 2
T
lta&= 64 lim x3 (In x)2 - 27 lim x2 (In x)2 = 0
60. We have,
1
2
Jo
= 64x’(lnx)2 -27x2 (In x).2:
x-» 0’
T 1
—- —sin2T
2 4
I = f (sin2 x + sin2 ax + 2 sin x • sin ax) dx
= 4x’(ln x4)2 -3x2 (In x1)2
»-to‘ dx
0
0
iQgrX
logex+l
,0 < x < *>
attains minimum at x = - and maximum at x = e.
e
-iT
1
1 .
sin(a-l)x sin(a + l)x
x — sin 2x----- sm 2ax-------------------------------4
4a
a-1
a + ln
T ,
1
.'. L = lim---- lim —
t--T
r-“T
sin (a -1) x
- sin 2x - — sin 2ax +
a-1
4
4a
=> L = 1
sin (a + 1) x
a+1
1T
156
Textbook of Integral Calculus
64. v
/(0) =
~(x + sin 0)*
(1 + sin 0)* - sin* 0
3~~
3
and g(0) =
(x + cos 0)3
. 3 1
o
_ (1 + cos 0)’ — cos’ 0
••• f
3
-1
1 4- 3 sin 0 + 3 sin2 0
, ... 1 + 3 cos 0 + 3 cos2 0
(B) We have, J*/(x) dx = {/(x)}2
/(0) = —-------- - ------------and g(0) =------------- - -----------Differentiating both the sides, we get
Now, /(0) - g(0) = (sin 0 - cos 0) + (sin2 0 - cos2 0)
/(x)=2/(x) r(x) =>
and /(0) - g(0) = (sin 0 - cos 0) (1 + sin 0 + cos 0)
1
2
Now verify all matchings.
Integrating both the sides, f(x) = - x + C
.
r(x)=i
65. (A) Let a = 2008, then
where, /(0) = 0 => C = 0
/(x) = ^ =>/(2) = l
I = [’ (1 + ax’) ex* dx
Jo0
I = f (ex* + ax* ex‘) dx
(note -.ax? = ax- x*-1)
Jo
I = [’ (ex* + ex‘ • x - ax*-1) dx
A
(C) Maximum when a = -1, b =2 =>a + b = 1
sin2x
b „ ,
hm —— + a + — = 0, then
(D)If
Jo
x-to
(x)) dx, where £(x) = ex‘
= f’ tf(x) + X
X
x-»0
Hence, J = [xex ]'o =e
For limit to exist 2 + b = 0=>b = — 2
/ = A+/2
Consider I2 = Jc,,e 1— In x dx
sin 2x + ax’ - 2x
hm----------- ;--------- = 0
x-*0
Put^/-ln x = t=>-lnx = t2=>x = e-’’
II
.. ;.-£ dx = -2t e-1’ dt
1 ■ (-2fe-’) dt = [ft-1 ] I - j’ e-’dt = 1 - [’ e~‘ dt
V
JO
■ ■ v.—
I
Q
n
, 1
dt-- = e
°
e
Note that, if /(x) = e-x’, then f~'(x) = J- In x
I2 = f1 e-x dx + 1 - f11 e
Hence,
g
Jo
Jo
(J 0'
n
n
General term of In L
r
t
In n2
n
"
1
= 0-1 —
2 2
n
(r
x2 • In x
2
-Ifx
’lax
2 Jo
x
£
0
4
4-cosx-isinx
Jo
Jo0
4sinx
1,
=;log
17 -0cosx
1
(5'
Ci
1
(
(I)
= log I- = log —
\n J
m + n =8
cosxdx
r*/2
68. Here, f = J
1 + 2 [sin-1 (sin x)]
f-»
4
cosx ,
r 0 cosxdx
’J..,,—“H--» -1
fl
f»*/2 cosx ,
+ J cosxdx + J
3
-- dx
X/2
/(*)
+ s’w
and S' (x) = (1 + sin (cos2 x)) (- sin x)
„
Hence,
1
4
4sinx
4sinx
d =----------------- : =>/(*) =
17 -0cosx
17 -0cosx
and f f(x)dx = f
r
A L = e-,/<
.
66. (A) We have, _f(x) =
3a + b = 2
sinnx sinx sin2x
67. Let B = £
=----- + ——H
4"
4
42
x
2x
and A = 1 + cos— + cos— + .
4
42
e‘2x
1
A A + iB = 1 + —
4
+-+
Sum = —- V - ln|n
n
n
Jo0
4
Using left hand rule and solving, we get a = —
Thus, B imaginary part of
£ + 2- In 2 + 3 • In 3
In L- f x In x dx =
JO
x
4
In L = lim -1n
1 • In
x
.. sin 2x + ax’ + bx n
hm----------- r--------- = 0
Jo
(B)
x3
„. . (1 + sin (cos2 x)) (- sin x)
f(x) = ---------- ,
, ~---------
If-"'2
J
= -3J-1 cosxdx+ Jo[ cosxdx + f cosxdx + -f cosxdx
Jo
3J1
= -J
=0
cost(-dt) + j cos(t)-(-dt)
cosxdx+ ^ f
3J1
cosxdx
Chap 02 Definite Integral
i [“ ‘2 sin’(n + 1)9 - sin’ n9
W. Here, /(x + 1) - f(x) = —
------------------------- - ---------------------- UU
7tJo
sin’9
sin(2n
+1)9
-sin9
_ 1 f*/2
dd
n Jo
sin’9
= /2 sin (2n + 1)9 de
sin’9
.x/2
r«'2sin2n9- cos9
cos2n9d9
I®
sin
9
It
if-t’
•K?)
2|_ 3
2 [3
=-[[12i-ikfi.ikfi.ik...'
3J <4 5/ <6 7J
= log,2 -1 = log I - I
\ej
K=2
•»« x’cosx .
72. Let
-------
(i)
-nl2 1 j. e*
[" J* /(x)dx = £/ (a + b - x) dxj
sinO
7C
r2|
L
1
cos9 + cos30 + ...+ cos (2n -1)9 • cos9d0 + 0 j
On adding Eqs. (i) and (ii), we get
f*/2
21 = (
J-K/l
• +...(2cos(2n-l)cos9)}de
-r
>X/2
l d9, as j
n Jo
cos2n9de = 0
J-xl2
/(n + l)-y(n)=l
z
Z/
2
Z \
z
n = 2,/(3)-/(2) = i1
If
2 ..
n
3
=> /(3) = - and so on
2
z
15 + 3
/05) + y(3) _ 2__ 2 _o
Hence,
15 _9
Z(15)-/(9)
2 2
70. Here, g(x) = J^<X) e:1(,*°’dt
•*.
/(x) = /i,(x)e(,+M,>)
=>
g'(x) = A'(x)-«
x2 cosx-(l) dx
when y(-x)=/(x)
f x/2 ,
21 = 2j°
x cosxdx
. 21 =2 [x’(sinx) -(2x)(- cosx)+ (2)(-sinx)] J/2
21=2 —~2
4
1 = ^-2
4
73. Letf(x)=f'——
Jo 1 +
dt
/(2) =/i'(2)-e(,+*(x>)’
■
=* e4 = i.e(1 **<’»’, given/(2) = e4 andh'(2) = 1
(1 + W=4
l + h(2)=2,-2
h(2) = -3,l
Absolute sum for all possible values of h(2) = |-3 +11 = 2
*
.
t*'2
1 t*12
71. Let / = J sinx log (sinx)dx = -J sin x-log (sin x)dx
=>
=*f& =
Put cosx = t =$ -sinx dx = dt
■■■
.2
2
2
dr
dt
o<-^ <-1
2
JX
iog(i-?)dt
If*
2Jo
> 0, for all x € [0,1]
J0 1 + t
Because,
sinx-log (1-cos x)dx
x’
.’./(x) is increasing.
At x = 0, y(0) = 0 and at x = 1,
2 ®
o
1
= “Jo
dx
Using integration by parts, we get
/(") = “
=>
1
1-r—+
x2 cosx
~2
11 [
2
1
nr = „l./(2)-/(l)
... ... = ;2 as/(l) = i =>/(2) = ;
If
--(ii)
i + e'x
-</2
7t Jo
=>
x’ cos(- x)
H.
1 r«/i
=—
{(2 cos9 cos9) + (2 cos30 cos9)
1
= -|
1
1
— + —...
10 21
fl
11
------ + —+ —+
12X3 20 42
21
10
1 i
i.e. cos9 + cos39 + cos50 +...+ cos(2n -1)9 - 1
2
••/(n+l)-/(n) = ^
i -i
iii
i
= — —f- — 4- —
1J’
Using, cos9 + cos30 + cos50 +....+ cos (2n -1)9
. n-(29)
sin——_ ___ 2 • cos(0 + (n -1)9)
3
2
e 7
t>
10 + 21
157
t2
dt<('-dt
J»2
z
Textbook of Integral Calculus
158
1 \
33
192 x‘-—
S/(x)-/(0)<24x'-^
12
16/
2L
3
16x4-1 S/(x)£24x4 —
Thus, f(x) can be plotted as
Yf
1/2
/(X)
o
. .
. .. . 1
Again, integrating between the limits - to 1, we get
2
J’i(16x--l)dxSji'i/(x)dx£J‘! 24x*--}dx
2J
-x
1
y = f(x) and y = 2x -1 can be shown as
Y
16x5
------- x
=>
(ID
L 5
•*
3
----- X
l/2
2 1/2
6>
y = 2x-1
10
1/2
y = f(x)
2.6 < J* /(*)<(* ^3.9
1/2
X'-
0
76. Let/, = £ e'(si;in6 at + cos6 at)dt
■X
1
r2K
2n
= £ e‘ (sin6 at + cos6 at)dt +J
e (sin e at + cos6 at)dt
3X
+ [ e (sin at + cos6at)dt+£ e'(sin6 at + cos6 at)dt
r
(0.-1)
From the graph, the total number of distinct solutions for
x e (0,1] = 1. [as they intersect only at one point]
74. Here, f(x) =7 tan* x +7tan6x-3tan4 x-3tanzx
••
fj — f2 + Z3 + /4 + f5I
Now,
I3 = j e'^in6 at + cos6 at)dt
Put
t = n+t=> dt = dt
I3 = f e* *' • (sin6 at + cos6 at)dt
for all xef—,—1
Jo
I 2 2)
•(ii)
/(x) = 7 tan6 x sec2 x-3 tan2 x sec2 x
= (7 tan6 x-3 tan2 x) sec2 x
Now,
•(>)
.2k
Now,
J2K
Put
x (7 tan6 x-3 tan2 x) sec2 xdx
f
xf(x)dx = f
Jo
Jo
1
I. = f e'(sin6 at + cos6 at)dt
t = 2n + t=>dt = dt
n
= J e‘+2* (sin6 at + cos6at)dt
= [x(tan7 x- tan3 x)] ’ '*
= e2’J2
1 (tan7 x-tan3 x)dx
-f
Jo
and
=Le(sin
Put
t = 3n + t
fH/4
= 0-£
tan x(tan x-l)dx
f»/4
=-
at + cos6 at)dt
=L’e“" (sin6 at + cos6 at)dt
tan x (tan x-1) sec x dx
Jo
.(iii)
(iv)
Puttanx=t =>sec2 xdx = dt
.’.
£
From Eqs. (i), (ii), (iii) and (iv), we get
/, = I2+eK -Ij + e2’ •I2+e3’1 -12 =(l+e" +e21t +e”)/2
x f(x)dx = -J t3(t2 -l)dt
?T 1_1
5•'Jo
Also,
fX/4
[
Jo
»«/4
/(x)dx=|
Jo
,
4
6
1
12
f e' (sin6 at + cos6 at)dt
L = Jq
f e' (sin6 at + cos6 at)dt
,
Jo
(7tan x-3 tan x) sec1 xdx
= (l + eK+e1K +e3,)=1'(—~1) ferae K
ex -1
= £(7t6 -3t2)dt = [t7 -t’]‘o = 0
77. According to the given data, F(x) <0, Vxe (1,3)
192 x3
f(x) = 2 4- sin4 Ttx
192x3
192x3
75. Here,
3
~J
We have,
=>
=>
2
1
On integrating between the limits - to x, we get
2
f
Jl/2
3
.IxSjr’/fxldxS^
Hx)4xsf
Jl/2J
Jl/2
2
Also,
Now,
dx
=>
f(x) = xF(x)
/*(x) = F(x) + xF/(x)
-.(i)
f(l) = F(l) + F(l)<0
[given F(l)=0 and F(r)<0]
/(2)=2F(2)<0
[using F(x) < 0,Vxe(l, 3)]
f (x) = F(x) + xF'(x) < 0
[using F(x) < 0, Vxe(1,3)]
fM<0
159
Chap 02 Definite Integral
3rc
=> logja + l|----- =9
4
j’x2F(x)dx=-12
78. Given,
(x2F(x)]’ - J 2x • F(x)dx = -12
9F(3)-F(1)-2j3/(x)dx = -12 [v xF(x) = /(x), given]
=>
81. Plan This type of question can be done using appropriate
substitution.
(2 cosec x)17 dx
Given, I = f
J* /<
-36-0-2jf(x)dx = -12
=>
r« « 217 (cosec x)'4 cosec x (cosec x + cot x) ,
_■
dx
Jtl4
(cosec x + cot x)
“• I
= -12
Jr
cosec x + cot x = t
(- cosec x - cot x - cosec2x) dx = dt
and
cosec x-cot x = 1/t
„
1
2 cosec x = t + t
r
1V4
Let
J xiF"(x)dx = 40
and
[x5r(x)]’-JJ3x2F(x)dx = 40
=>
[x2(xF'(x)]’-3x(-12) = 40
=>
{x2[/'(x)-F(x)]},j=4
=>9[f(3)-F(3)]-[r(l)-F(l)] = 4
=>
9[r(3)+4]-[r(l)-0] = 4
=>
9r(3)-r<D=-32
[x], x<2
79. Here,/(x)
0, x>2
and when t = V2 + l,e*=V2 + l
u = ln(-j2 + 1)
r®
/=-|
r
Jln(j2 +1)
ffa(V5+i)
= 2jJo
2 + /(x+ 1)
x/(x2)
----- ——-—dx +
2 + f(x + 1)
2 + /(x + 1)
+ [’ xJ(x’> dx
A + f(x + 1)
fi
r4i X’ 1
Jo
Ji 2 + 0
tr1
= J ( 0dx+ f Odx+f ---- dx+f 0dx+|
/
dxLJ*<x>
Put
9x + 3tan 'x = t
=>
f-A
(
1+x2
(x» 1/V(x)]- f W (X» 1/ ♦ (X)l
Lox
J
= l(2-l) = i
4
4
12 + 9x2
dx
1 + x2
=>
f(x) = Kext
Now,
/(0) = l
1 =K
/(x) = ep
Hence,
Jo
dx
dx = dt
Put x = tanO
J + SX/4 ,
=>
e'dt = [e']’ + ”/4 = e■’ + 5‘/4-l
3TC
Iog,|l + «I = ’+V
4
lax
J
[K=e‘]
F(2)=f4e'dt = [e,];=e4-1
83. (A) Let I = J*
a=JJo
J
In/(x) = x2 + c => f(x) = ezt
4/ = 1 => 4/ -1 = 0
Jo
/(0
fHx) dx = j 2xdx
J fix)
=>
and^3 <x <2=>3 < x2 < 4 => f(x2) = 0
CC=
(e* + e-)’* du
/(X)
4i<x<^Ji =>2<x2 <3=>/(x2) = 0,
80. Here,
e
W=2x
1<x2<2
=>[xj]=1
2 <x +1 < 1 + 5/2 =>/(x+1) = 0,
' =£
—
F(x) = 2x/(x)
F(x) = /(x)
2x/(x) = f(x)
0<x<l=>0<x2<l=> [x2] = 0,
^x .
.
—dx= —
‘ 2
-«U4 e*du
F(x)=\
‘' f(41) dt
Given,
Odx
V-1 < x < 0 => 0 < x2 < 1 => [x2] = 0,
=>
. .
2(e + e
82. Plan Newton-Leibnitz’s formula
Also,
I
>
Let t = e‘ => dt = e"diL Whent = Le" =l=>u = 0
' = J.
°-^2).-dx +
-I-2
j: xf^)—dx
+ /(x + l)
r
2
<
2
-dx
- 2+/(x + 1)
+
dt
t
t
217
=> dx = sec20d8
f X/4
I
=2J.
2
160
Textbook of Integral Calculus
(B) Let
Put
r’
dx
(D) Plan J‘ /(x)dx = 0
I=f , JoVl-x2
If f(-x) = -f(x), i.e. /(x) is an odd function.
x = sin0 => dx = cos0d0
1=
f Kit
Jo
Let /(x) = cos 2x log
IT
ld0 = 2
1+x
1 -x^
<c> J.’tS ■Ws)I
/(-x) = cos 2x log
= -f(x)
J + xj
Hence, f(x) is an odd function.
fl/2
So,
= [sec,x]I2 =--0 = <x7?=
-1
3
3
(A) -> (q); (B) -> (r); (C) -> (p); (D) -4(s)
•
84. (A) Plan
(p) A polynomial satisfying the given conditions is taken.
(q) The other conditions are also applied and the number of
polynomial is taken out.
f(x) = ax* + bx+ c
Let
85. Plan Integration by parts
J f(x) g(x) dx = /(x) j g(x) dx - J
Given, I = f 4xs-^-r(l - x2)5 dx
Jo i dx2
u
=r4x’Ai-x1)1
f(0) = 0 => c = 0
£/(x)dx = l =>
Now,
“+4.P—
— — 1 => 2a + 3b = 6
3 2
As a, b are non-negative integers.
So,
a = 0, b = 2 or a =3, b = 0
f(x) = 2xor f(x) = 3x2
(B) Plan Such type of questions are converted into only sine
or cosine expression and then the number of points of maxima
in given interval are obtained.
/(x) = sin (x2) + cos (x2)
-izRx’a-x’n-j.12x(i-x‘)'I’dr]
2
It
J
tt .
•
It
4
= 2mt => x2 = 2mc +
, for n = 0
x=±
=>
[9it _
x = ± .1—, for n = 1
V 4
So, /(x) attains maximum at 4 points in [—713, -J13].
/(-x) dx
(q) j* f(x) dx = 2 J‘ /(x) dx, if /(-x) = f(x), ile. f is an even
function.
h:
3x2
p2
dx and
3x2
=>
21
I=J
, 3xV)
3x2
dx
3x2 dx =>21=2^ 3x2 dx
i=[x’i;=»
= 12x
6
it - x
it + x
• x It
88. Z = J_
-tit
.
= 12 0 + - =2
6.
cos xdx
:. /=J
t* 12
2
^x cos xdx+0 =2j#
o
■>
(x2 cos x) dx
= 2{(x2 sin x)J/2 - f 2x-sin x dx}
*•
* 2
„
it
= 2 ——2 {(—x-cos x)J/2 — [ ' l (-cosx)dx)
44
*o
\2
^-2
= 2------ 2(sinx)’/2 = 2
4
4
87. Put x2 =t
—-4
2
xdx = dt/2
dt
j = f1”-------------- 2-------J1<*2sint + sin (log 6 -1)
(C) Plan
(p) £ f(x) dx =
0
‘ (1 -x2)6'1
fK It
/T"
.
= V2 cos x cos — + sin — sin (x ) = V2 COS x2
4
4
For maximum value, x2 -
= 0-0-12(0-0) +12^ 2x(l-x2)sdx
As, J‘a f(x) dx = 0, when /(- x) = - /(x)
= J2 -J=cos(x2) + 4=sin(x2)
V2
12 x2 —(1 - x2)’ dx
dx
= px3 X5(l-x2)4 (-
2 J.
=>
.V2
-f.
dx
L
.3
[/(x)] j gfx) dx] dr
0
dx
-0)
Using, J /(x) dx = j f(a + b - x) dx
_ 1 p10*3 sin (log 2 + log3 -1)
2 J10** sin (log2 + log3 -1) + sin
(Iog6-(log2 + log3-t))
-If10*3
sin (log 6 -1)
dt
2 J10*2 sin (log6 -1) + sin (t)
•I®*2
sin (log 6-t)
k>«2 sin (log6 -1) + sint
•Cu)
Chap 02 Definite Integral
lim ^1, where, — is replaced with x.
•-»“ n
n
E is replaced with integral
ri
..
1* + 2“ + ... + n*
1
Here, lim---------- ;------------------------------------------- = —
«-‘-(n +I)4 {(na + l) + (na+ 2) + ...+ (na + n)} 60
On adding Eqs. (i) and (ii), we get
t + sin (log 6 - t)
21 =12 Jk*2 sin
sin (log6 -1) + sint
2J = 1
i (10g3 - log 2)
2
1
4
7=;logb
L''
n(n + 1)
(n + I)4"1 • n*a +
2
Using Newton-Leibnitz formula.
Differentiating both sides
=>
6/(x)-1 - 0 = 3/(x) + 3xf (x) -3x2
1+1
n.
X
X2
=>
13
d x
=1
dx lx
Z(x)
1
3.
X
1
2
- = l + c => c = - andf(x) = x2--x
3
3
3
=>
1____ 1
l-(2a + 1) ~ 60
=>
= 1
2-[x4H],0
(2a + l)(a + l) 60
60
(2a + 1) (a + 1) = 120
2d2 + 3a + 1 - 120 = 0
2a2 + 3a-119 = 0
(2a + 17)(a-7) = 0
rd)=|
4
•|2u + 1 + k
n.
_1_
2
1
(2a + 1) (a + 1) ” 60
f(2) = 4-- = 3 3
Note Here, /(I) = 2, does not satisfy given function.
2
1
1+1
nJ
V /(!) = ;
j£-L2- = x+c
60
■(2na + n + 1)
(;])■-
On integrating both sides, we get
=>
60
2Z => lim
=> 3xf(x)-3f(x) =3x2 =>
xf(x)-Hx) Z , •
1
=> lim--------- p2---
88. Given, f(1) = 1 and 6 j f(t)dt = 3x f(x) -x’, Vx £1
=>
161
8
Forthat /(x) = x2--xand/(2) = 4-- = -
„ “17
89. Let Z=
r)<
dx
} 0 1 + x2
_p(x4-l)(l-x)4+(l-x)4^
■'«
(1 + x2)
• 2
91. Here, /(x) + 2x = (1 - x)2 sin2 x + x2 + 2x
x)2
dx
O’-ixi-xr^x+j;°i(1 +(1x+2-2
x 2)
= f‘](x2-l)(l-x)4+(l + x2)-4x +
4x2
Jo
= [ | (x2 —1)(1-x)4+(1 + x2)-4x+ 4*01
#1
Jo
x‘-4xs+ 5x4-4x2 + 4-
4
4
dx
1 + x2
dx
x7 4xs 5x5
4xs
.
. .
-----------+--------------+ 4x - 4 tan x
7
6
5
3
1
it J 22
_1 4.5
4 5 4 , Jit
-- +------ + 4 — 4----- 0 =----- K
~7 6 5 3
7
4
90. Converting infinite series into definite integral
Hn)
Le.
lim-^
n
•
•<*(■)
f \
lim- £ f - = f/(x)dx
•dx
>-(i)
where, P: /(x) + 2x = 2(1 + x)2
-(ii)
2(1 + x2) =(1 - x)2 sin2x + x2 + 2x
=>
(1 -x)2 sin2 x = x2 — 2x + 2
=>
(1 - x)2 sin2 x =(1 - x)2 + 1 => (1 -x)2 cos2x = -1
which is never possible.
P is false.
Again, let Q: h(x) = 2f(x) + 1 - 2x (1 + x)
h(0)=2/(0) +1-0 = 1
where,
h(l) =2f(l) + 1 - 4 = -3, as h(0) h(l) < 0
=> h(x) must have a solution.
Q is true.
-
92. Here,J(x) = (l-x)2- sin’x+ x2 >0, Vx. ,
and gfx) =
2(f -1) - log r /(t) dt
t+1
2(x-l)
-log x -/(x)
(x+1)
+ ve
For g'(x) to be increasing or decreasing,
g,(x) =
let
= 7(x77
"Iog r
+ 1)
...(i)
162
Textbook of Integral Calculus
4»'(x) =
1 _-(x-l)2
x x(x + I)2
4
(x + I)2
0'(x) < 0, for x > 1 => 0W < 0(1) =* 0(x) < 0
From Eqs. (i) and (ii), we get
g'(x) < 0 for x e (1, °°)
A g(x) is decreasing for x 6 (1, «»).
x-[x],
if [x] is odd.
93. Given, f(x)1 + [x] - x, if [x] is even.
...(ii)
nx sin nx .
----------------- dx
(1 + nx)sinx
On adding Eqs. (i) and (ii), we have
(■’ sinnx , „ r« sin nx ,
2I"=J x~“ dx=2
------- dx
smx
Jo sinx
sin nx .
.
[v/(x) = ■------- is an even function]
sinx
_
r" sinnx .
=>
I. =
------- dx
sinx
sin (n + 2)x-sinnx ,
Now,
♦ 2 “ Ai ~ J.
;
dx
Jo
smx
_ r« 2 cos(n + 1) x-sinx
■'°
sinx
10
j f(x) cosnx dx =2 j f(x) cosn xdx
-w
0
Y
\ AAA
-2-1O
-10 -9
J 1 >X
9 10
1 2
.(0
Using j /(x) dx = j f(b + a - x) dx, we get
J(x) and cosn x both are periodic with period 2 and both are
even.
10
.
r«
sin nx
I. ---------------- dx
•'-’(1 + nx)sinx
95. Given
= 2 [ cos(n + l)x dx = 2 sin (n(n++1)1)x‘ ’ 0
J0
2
= 10 J /(x) cosnx dx
/B+2=A„
-(iii)
A,
=
J
S
^
n
dx
=$
I\=n
and
I
2
=
0
Since,
J® sinx
1
2
From Eq. (iii) It = I3 = = ..= n and I2 = I4 = It = ...= 0
0
1
1
1
Now, j f(x) cosnxdx= j(l — x) cosnx dx = -|ucosnudu
0
0
2
I
0
2
10
1
and j/(x) cosnx dx = j(x-l) cosnx dx = -Jucosnudu
1
S I.2m =°
m-l
'
Correct options are (a), (b), (c).
10
1
40
jf(x) cosnx dx = -20 jucosnu du = —
-10
= 10n and
m«l
-o
1
10
=>
96. Given, S„ = V —----- n + kn + k
0
2 10
* *
— | /(x) cosnx dx = 4
to -10,A
94. Given
=f
X
J 0o
0
X
=i-fa*
0£x<1
\
Differentiating both sides w.r.t. x by using Leibnitz’s rule,
we get
7i -(/'«(’ =/(x) => y-(x)=±7i-(/(x)} !
_2_ n
73 13
sinx<x, V x>0
rl
• 1 <-1
sin
< -1 and. sm
3
3
2
J
1
f - < - and
2
1
<3
n
6J
dx =
2
1
x+2,
tan
V3
n
3x/3
c
’
x
3^3
Similarly, T > ~^=
3V3
sin-1 {/(x)} = ± x + c
x = 0 => sin-1 {/(0)} = c
c = sin-1(0) = 0
/(x) = ± sinx
but
/(x) a 0, V x e [0,1]
/(x) =sinx
As we know that,
-T
J 0 1 + X + X2
Put
=>
1
____ 1_
< lim > —
k k2
l+‘+
n n /
n
nJ 7
A1—1
-J -mr
=>
,
[v/(0) = 0]
97.
'=J,K/4 1 + dxCOS X
'-L 1 - dxcos x
•W
r3x /4
•••(ii)
Adding Eqs. (i) and (ii)
•3X/4
2
.3X/4
dx =$ I = f
*'4 sin2x
I = -(cot x)’7< =2
Jlx/4
cosec xdx
98. I, + /t=f( tan4 x + tan6 x) dx = J tan4 x sec2 x dx
1
.
= - tan x + c
5
a = -, b = 0
5
Chap 02 Definite Integral
n
x-a,
101. Plan Use the formula, |x —a | =
(n + 1)-(n + 2)... (3n)
99. (b) Let I = lim
163
x>a
-(x-a), x<a
to break given integral in two parts and then integrate
separately.
2/i
1
(n + 1) •(« + 2)... (n + 2n)"[n
= lim —
x | dx = J |l-2sin—|dx
l-2sin—
I
= lim
n+1
n+2
n + 2n
n
n
n
n
. . x
1 -2sin— dx- L
>
22)
Taking log on both sides, we get
log I = lim - log
2
1+n
1+1
n,
"-*“n
, (
1 I ,
21
. f, 2n
2n
log 1 + - + log 1 + - +...+ log 1 + —
n
n
n~
=> log I = lim —
«-*-n
h:
X /4
1+
f2
log I =
=>
=>
1
,
log I = log (1 + x) • x - j -------xdx
1+x
r
ifft
X
n
2[3
6.
12
/=- -
i-— dx
i+x
log I =2• log3 -[x-log 11
=>
=>
2I=j*'*dx => 2/= [x]Ji:>
.2
0
...(h)
On adding Eqs. (i) and (ii), we get
o
.
log 1 = [log(1 + x)-x]2 - f 2X+1-1
------------ dx
JOJ
1+x
log I = 2 • log 3 -
I n
tan---- x
K.2
•*
x /•
4. ^tanx
_ / tan y
'* 11 +
2
=>
lb
dx___
Jx/6 1 + Jcotx
_ r« n y] tan xdx
log (1 +x) dx
Jo
...(i)
dx
X /I
1 2"
=> log I = lim log 1 + n.
j
dx
■Jtanx
X /3
102. Let
X 1j
1 -2 sin— dx
2)
x1
x+ 4cos— •
(
x 3
= x + 4cos—
2. 0
2n
1+—
n.
■
n
.
Statement I is false.
But j f(x)dx = | f(a + b- x)dx is a true statement by
Io
property of definite integrals.
log Z =2-log 3-[2-log 3]
103. Given, y = J |t| dt
log I = 3 • log 3 - 2 => log I = log 27-2
l = e^”-2 = 27-e’2=^
e
= I x | • 1 - 0 = | x|
•b
100. Central Idea Apply the property j f(x)dx = J f(a + b-x)dx
[by Leibnitz’s rule]
v Tangent to the curve y = f 111 dt, x e R are parallel to the
J0
line y = 2x
and then add. Let
Slope of both are equal => x = ± 2
1_______ ___________ dx
Points,
2logx2 + log(36-12x+x2)
y= f
111 dt = ±2
Jo
=r___ _______ dx
Equation of tangent is
_ r<
For x intercept put y = 0, we get
•*2 2 log x+ log(6 — x)2
y - 2 = 2 (x - 2) and y + 2 = 2 (x + 2)
21ogxdx
0-2 = 2(x-2) and 0 + 2=2(x + 2)
•'22[logx+ log(6-x)]
_ H
log xdx
'2 [log x + log(6 — x)]
r«
log(6 - x)
&
j2log(6 - x) + logx
=>
[’•*
On adding Eqs. (i) and (ii), we get
21
= plogx4- log(6-x)^
=>
(i)
x=± 1
104. Given integral g(x) = [ cos 4t dt
J0
.(ii)
To find g(x + 7t) in terms of gfx) and g(n).
g(x)=l cos4tdt
= //<fl + b~ x)dx]
Jo
=>
s(x+jt)=r1
Jt «0
cos 4t dt
fX
fX* x
cos4tdt
= J cos4tdt + J
= g(x) + A
=>
=>
21 = jdx = [x]J
21=2 => I = 1
(say)
cos4tdt=| cos4tdt
Jo
(definite integral property)
164
Textbook of Integral Calculus
=>
g(x+ n) = g(x) + g(n)
But the value of I, is zero.
X
sin 4n
sin 4t
4
4 0
• f=J
I =8 |
-J 0
sin x = 0
x = n, 2n
f"(x) = Vx cos x + —7= sin x •
2Vx
At x = it,
Local minimum at x = 2it.
107. We have,
log (1 + tan 0) d0
,=8L 108 1 +
p X/4
0
p X/4
= 8J, 108
f X/4
= 8JJ 0 log
0
p(x) = -p(l~x) + C
1=-41 + G
=*
=>
C = 42
=>
...(i)
p(l - x)dx
1 - tan 0
1 +----------1 + tan 0
2
1 + tan 0
21 = j (p(x) +p(l-x))dx = j 42dx = 42
=>
1=21
log {1 + tan 0} + log
Jo
108. Let I = f
de
tan---- 0
<4
r X/4
=>
I = £ p(x)dx =
=>
[cot x] dx
-(i)
J 0
=> I = f
Jo0
de
[cot (n - x)] dx = [
Jo0
[- cot x] dx
...I■(ii)
On adding Eqs. (i) and (ii),
2/=Jo [cot x]dx+J# [-cot x] dx=
de
,(ii)
0
(-l)dr
0
[■•' [x] + [- x] = -1, if x £ z and 0, if x 6 z]
= [- x)o =-7t
Adding Eqs. (i) and (ii), we get
J 0
p'(x) = /(I - x), V x e [0,1], p(0) = 1, p(l) = 41
0
- 0
21 = 8]
f"(lt) = - Vrc < 0
:. Local maximum at x = it. At x = 2it, f"(2it) = -fin > 0
Now,
Using ^f(x)dx = ^f(a-x)dx, we get
0
=>
p(x) + p(l-x) = 42
8 log [1 + tan0]
■sec20d0
1 + tan2 0
f x/4
-X/4
...(i)
;in x - 0}
(using Newton-Leibnitz formula)
f(x) = Jx sin x = 0
.■
0
2 /
\
X
4 J
0=^
•• 4
i
J°
^“1 = 0
=>
g(x + re) = g(x) - g(it)
In my opinion, the examiner has made this question keeping
g(x) + g(it) as the only answer in his/her mind. However,
he/she did not realise that the value of the integral 7, is
actually zero. Hence, it does not matter whether you add to or
subtract from g(x).
_ r >8 log (1 + x)
dx
I
105.
(1 + x2)
x = tan 0 => dx = sec20d0
Put
=> tan 0 = 0
When x=0
0=0
x = 1 = tan0
When
X/4
_f(x) = [ -Jt sin t dt, where x e [ 0, — |
Here,
2
1 + tan 0
de
I = 4 f SMlog2d0 = 4-log2(0);M
2
109. Since, 1 = f ‘ ^dx < f *-£=dx,
Jo Vx
J 0
(7t
A
= 4 log2-l---- 0 l = K log2
\4
)
106. If 4>(x) and \g(x) are defined on [a, b] and differentiable for
every xand j[t) is continuous, then
7-[(*U7(t)dtl = f[V(x)J~\|/(x)- f [ 4>(x)]<Xx)
dxLJ*<x>
J
dx
dx
Jo Vx
because inxe(0,l),x>sinx.
I< j^Vxdx = |[x3,2]l0 => I<2
3
and
J 0
J <2
J 0o
X 2dx = 2
CHAPTER
Area of
Bounded Regions
Learning Part
Session 1
• Sketching of Some Common Curves
• Some More Curves which Occur Frequently in Mathematics in Standard Forms
• Asymptotes
• Areas of Curves Given by Cartesian Equations
Session 2
• Area Bounded by Two or More Curves
i
■
Practice Part
• J EE Type Examples
• Chapter Exercises
Arihant on Your Mobile!
Exercises with the @ symbol can be practised on your mobile. See inside cover page to activate for free.
Session 1
Sketching of Some Common Curves, Some More Curves
which Occur Frequently in Mathematics in Standard Forms,
Asymptotes, Areas of Curves Given by Cartesian Equations
Sketching of Some
Common Curves
For finding the area of a given region, we require the
knowledge of some standard curves.
(i) Straight Line
Every first degree equation in x, y represents a straight line.
So, the general equation of a line is ax + by + c = 0. To draw
a straight line find the points, where it meets with the
coordinate axes by putting y = 0 and x = 0 respectively in its
equation.
By joining these two points we get the sketch of the line.
Sometimes the equation of a line is given in the form
y =mx.This equation represents a line passing through
the origin and inclined at an angle tan-1 m with the
positive direction of X-axis. The equation of the form
x = a and y = b represents straight lines parallel to Y-axis
and X-axis, respectively.
Region Represented by a Linear Inequality
To find the region represented by linear inequality
ax + by < c and ax + by > c, we proceed as follows
(i) Convert the inequality into equality to obtain a
linear equation in x, y.
(ii) Draw the straight line represented by it.
(iii) The straight line obtained in (ii) divides the
XY-plane in two parts.
To determine the region represented by the inequality
choose some convenient points; e.g. origin or some points
on the coordinate axes.
If the coordinates of a point satisfy the inequality, then
region containing the points is the required region,
otherwise the region not containing the point is required
region.
I Example 1 Mark the region represented by
3x+4y<12.
Sol. Converting the inequality into equation, we get
3x + 4y = 12.
This line meets the coordinate axes at (4, 0) and (0, 3),
respectively. Join these points to obtain straight line
represented by 3x + 4y = 12.
ty
This straight line divides the plane in two parts. One part
contains the origin and the other does not contains the
origin. Clearly, (0,0) satisfy the inequality 3x + 4y < 12. So,
the region represented by 3x + 4y < 12 is region containing
the origin as shown in the figure.
(ii) Circle
The general equation of a circle is
x2 + y2 +2gx+2fy + c=0
The second degree equation in x, y
such that coeff. of x2 = coeff. of y2
and there is no term containing xy; it
always represents a circle. To draw a
sketch of a circle, we write the
equation instandard form
(x - h)2 +(y - k)2= r2, whose centre
is (h, k) and radius is r.
4
c r
(h,k).
o
Figure 3.1
Remarks
1. The inequality (x - a)2 + (y - b)2 < r2 represents the interior of
a circle.
2. The inequality (x - a)2 + (y - b)2 > r2 represents the exterior of
a circle (i.e. region lying outside the circle).
Chap 03 Area of Bounded Regions
(5) y2 = - 4a (x - h); a, h > 0
[iii) Parabola
Y
It is the locus of points such that its distance from a fixed
point is equal to its distance from a fixed straight line.
Taking the fixed straight line x = - a, a > 0 and fixed point
(g,0), we get the equation of parabola y2 = 4ax.
X
16; o)/(h.o)
Steps to Sketch the Curve
Figure 3.6
(i) It passes through (0,0).
(6)y2 = - 4a (x + h)\ a,h>Q
(ii) It is symmetrical about axis of X.
Iy
(iii) No part of the curve lies on the negative side of axis
ofX.
-V
__ (0,0)
7PT6)
(iv) Curve turns at (0, 0) which is called the vertex of the
curve.
(v) The curve extends to infinity. It is not a closed curve.
(1) y2 = 4ax (Standard equation of parabola)
>x
Figure 3.7
,2=4ax
y
(7) x2 = 4ay, a > 0
+y
v
9
. I
*X
(0,0)
t
%
9
9
*x
v (0.0)
Figure 3.2
(2) y2 = 4a (x - h); where a and h are positive.
Figure 3.8
(8) x2 = 4a (y + k); a > 0, k > 0
O__
(0. 0)
>X
ty
t
I
I
I
I
I
Figure 3.3
(3) y2 = 4a (x + h); where a and h are positive.
9
9
(0.0)/
Y4
X
9
v
>X
0
(0.-A)
Figure 3.9
(9) x2 = 4a (y - k); a, k > 0
Figure 3.4
(4) y2 = - 4ax; a > 0
aY
v___
(0.0)
v (0»
(0,0)
Figure 3.10
Figure 3.5
*X
167
168
Textbook of Integral Calculus
(10) x2 = ~4ay; a>0
kY
V
>X
Z'(o,o)’
Definition 2. An ellipse is the locus of a moving point
such that the sum of the its distances from two fixed
points is constant. The two fixed points are the two foci of
the ellipse. To plot the ellipse, we can use the
peg-and-thread method described earlier.
Stantard Equation
2
2
x +<-=1
y
—
2
a
bl2
Ifa<b
If a> b
Figure 3.11
(11) x2 = - 4a (y + k); a, k > 0
|Y
*X
(0.0)
(0, b) and (0, - b)
Foci
(0, be) and (0,-be)
Major axis
2a (along x-axis)
2b (along y-axis)
Minor axis
2b (along y-axis)
2a (along x-axis)
Directrices
a .
a
x = -*ndx = -~
e
e
b ,
b
y = - and y = —
t
e
Eccentricity e
i
Latus-rectum
2^
a
2a2
Focal distances of (x,y)
a±
ex
a±ex
b±ey
b7
TZ
V >■
b_____________
(x-q)2
Figure 3.12
(12) x2 =-4a(y -k);a,k>Q
Y
v (0, k)
(0,0)
X
\
t
!
(a, 0) and (- a, 0)
(ae,0)and (- ae,0)
And lastly, if the equation of the ellipse is
t
t
Vertices
!
%
Figure 3.13
(iv) Ellipse
Basics of Ellipse
Definition 1. An ellipse is the locus of a moving point
such that the ratio of its distance from a fixed point to its
distance from a fixed line is a constant less than unity.
This constant is termed the eccentricity of the ellipse. The
fixed point is the focus while the fixed line is the directrix.
The symmetrical nature of the ellipse ensures that there
will be two foci and two directrices.
(y-p)2
=1
a2
b2
instead of the usual standard form, we can use the
transformation X —> x - a and Y —» y - 0 (basically a
translation of the axes so the axes so that the origin of the
new system coincides with (a, 0). The equation then
becomes
X2 Y2
t+—7=i
a2 b2
We can now work on this form, use all the standard
formulae that we’d like to and obtain whatever it is that
we wish to obtain. The final result (in the x-y system) is
obtained using the reverse transformation x -> X +a and
y-*Y + 0.
x2 v2
(1) — + ^— = 1; a > 0, b>Q (Standard equation of the
ar b
ellipse)
-y
(0.b)
/
(-a.0) \
(0.0)
(0,-6)
Figure 3.14
y(a,0)
+X
Chap 03 Area of Bounded Regions
x2 y2
(2) — .+ ~ = 1; b > a > 0 (Conjugate ellipse)
a
b
Latus-Rectum
The chord(s) of the hyperbola passing through the focus F
(or F”) and perpendicular to the transverse axis. The
length of the latus-rectum can be evaluated by
substituting x - ± ae in the equation for the hyperbola :
YA
(0.b)
!
%
I
I
I
(-*.0)!
I
>x
(0, 0)
x2
j2
a2
a2(e2-l)
I
I
I
169
!
y2
e2
=1
=1
a2(e2-l)
(O.-b)
2 ■b4
. y2=a2(e2-l)2=a4
Figure 3.15
(3)<^
- + (y-fc)— = l;a>0,b>0 and a > b
y=k
2b2
Thus, the length of the latus-rectum is —
a
4 - - - ------------------- JJ.
CJ
0
>X
x=h
b2
We discussed in the unit of Ellipose that an ellipse with
centre at (a, p) instead of the origin and the major and
minor axis parallel to the coordinate axes will have the
equation
2
X2 Y2
(x-jx)
(y-P)
a"2
-+
=1 or 'F+F=1
(x-q)2
(y-p)2
a2
b2
b2
• a
b
where X —> x - a and T —> y - p.
The same holds true for a hyperbola. Any hyperbola with
centre at (a, P) and the transverse and conjugate axis
parallel to the coordinate axes will have the form
Figure 3.16
a2
y=±—
a
b2
YA
<4)
a2
2
2
a2
£
= 1, where a < b
YA
I
I
-I
b
y=k
or
>X
0
Figure 3.17 .
M Hyperbola
A hyperbola is the locus of a moving point such that the
difference of its distances from two fixed points is always
constant. The two fixed point are called the foci of the
hyperbola. Constrast this with the definition of the ellipse
where we had the sum of focal distances (instead of
difference) as constant. As in the case of the ellipse, we have
Focal distance of P(x, y)
I
a
dy =e(PF) = el x — = ex -a
\
e)
I
a i
d2 = e(PF ) = e x + - I = ex + a
I
eJ
a-2
y2
b2 -1
where X —> x - a and T —> y - p.
We can, using the definition of a hyperbola, write the
equation of any hyperbola with an arbitrary focus and
directrix, but we will rarely have the occassion to use it.
2
2
x
y
(1)------ — = 1 (Standard equation of hyperbola)
a2 b2
AY
(-a.0)
(0.0)
Figure 3.18
(a.0)
*X
Textbook of Integral Calculus
170
2
2
Some More Curves which Occur
Frequently in Mathematics in
Standard Forms
(2) —---- — = 1 (Conjugate hyperbola)
b1.2 a_2
1.
>x
«■
(0.0)
Y
y = -x
Figure 3.19
y=x
>x
(0,0).
(3) xy = c2 (Rectangular hyperbola)
Ya
Figure 3.23
(c.c)
Modulus function, y = | x |
>X
(0.0)
2.
2
Figure 3.20
X
■+
1
(x-h)2
(y-k)2
a2
b:
(4)-
2
3
= 1; a > b > 0
-2
Y
y=k
cl
f
(y.W
k
I
Figure 3.24
Greatest integer function y =[x]
\
3.
y
-x
0
y=e*
xy=x
x=h
Figure 3.21
(5)
(x-h)2
(y-k)2
a2
b2
(0.1)
y= Inx
/ (0.0)/(1.0)
*X
y
Figure 3.25
y=k
4.
(x,A)
bI
-X
0
+y
(0.1)
y = e"x2
+-X
!x = h
Figure 3.22
Figure 3.26
Chap 03 Area of Bounded Regions
5.
y
(1.1)
a<l
-a=1
------- >-X
a> 1
(0, 0)
Figure 3.27
aY
(1. 0)
(0. 0)
y=i~b
--------- ►x
171
(b) Symmetry about Y-axis If all powers of x in the
equation of the given curve are even, then it is symmetric
about Y-axis. e.g. x2 = 4ay is symmetric about Y-axis.
(c) Symmetry in opposite quadrants If by putting - x
for x and - y for y, the equation of curve remains same,
■then it is symmetric in opposite quadrants.
e.g. xy - c 2 , x 2 +y 2 =a 2
are’ symmetric m opposite
quadrants.
(d) Symmetric about the line y = x If the equation of a
given curve remains unaltered by interchanging x and y,
then it is symmetric about the line y - x which passes
through the origin and makes an angle of 45° with
positive direction of X-axis.
(ij) Origin and Tangents at the Origin
Its cartesian equation is x2/3 +
+y2/3
2'3 = a2/3
See whether the curve passes through origin or not. If the
point (0, 0) satisfies the equation of the curve, then it
passes through the origin and in such a case to find the
equations of the tangents at the origin, equate the lowest
degree term to zero. e.g. y2 = 4ax passes through the
origin. The lowest degree term in this equation is 4ax. '
Equating 4ax to zero, we get x = 0.
Its parametric equation is x = a cos3 t, y = a sin3 t
So, x =0 i.e. Y-axis is tangent at the origin to y2 = 4ax.
Figure 3.28
1
y=—
,a >0
z
a
X
7. The Astroid
and it could be plotted as
Figure 3.29
Curve Sketching
For the evaluation of area of bounded regions it is very
essential to know the rough sketch of the curves. The
following points are very useful to draw a rough sketch of
a curve.
(i) Symmetry
(a) Symmetry about X-axis If all powers of y in the
equation of the given curve are even, then it is symmetric
about X-axis, i.e. the shape of the curve above X-axis is
exactly identical to its shape below X-axis. e.g. y2 - 4ax is
symmetric about X-axis.
(iii) Points of Intersection of Curve
with the Coordinate Axes
By putting y = 0 in the equation of the given curve, find
points where the curve crosses the X-axis. Similarlv by
putting x = 0 in the equation of the given curve we < < u
find points where the curve crosses the Y-axis.
e.g. To find the points where the curve
xy2 = 4a2 (2a - x) meets X-axis, we put y = 0 in the
equation which gives 4a2 (2a - x) =0 or x =2a. So the
curve xy2 - 4a2 (2a - x), meets X-axis at (2a, 0). This
curve does not intersect Y-axis, because by putting x = 0
in the equation of the given curve get an absurd result.
(iv) Regions where the Curve
Does Not Exist
Determine the regions in which the curve does not exist.
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 values of y for which x is imaginary.
The curve does not exist for these values of x and y.
172
Textbook of Integral Calculus
So, the origin is a point of inflexion.
e.g. 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
a2y22 -x2 {a- x) does not exist for x > a as the values of y
are imaginary for x > a.
X
o
p
(v) Special Points
dy
Find the points at which — = 0. At these points the
dx
tangent to the curve is parallel to X-axis.
Find the points at which — =0. At these points the
dy
tangents to the curve is parallel to Y-axis.
(vi) Sign of dy/dx and Points of
Maxima and Minima
dy
Find the interval in which — > 0. In this interval, the
dx
function is monotonically increasing, find the interval in
dy
which — < 0. In this interval, the function is
dx
monotonically decreasing.
dy
_ d2y
Put — = 0 and check the sign of —- at the points so
dx
dx2
obtained to find the points of maxima and minima.
Keeping the above facts in mind and plotting some points
on the curve one can easily have a rough sketch of the
curve. Following examples will clear the procedure.
I Example 3 Sketch the curve y = x3 - 4x.
Sol. We note the following points about the curve
- ■ ;
(i) The equation of the curve remains same, if x is .
replaced by (- x) and y by (- y), so it is symmetric in
opposite quadrants.
Consequently, the curve in the first quadrant is
identical to the curve in third quadrant and the curve
in second quadrant is similar to the curve in fourth
quadrant.
(ii) The curve passes through the origin. Equating the
lowest degree term y + 4x to zero, we get y + 4x = 0
or y = - 4x. So, y = - 4x is tangent to the curve at the
origin.
(iii) Putting y = 0 in the equation of the curve, we obtain
x3 - 4x = 0 => x = 0, ± 2. So, the curve meets X-axis at
(0, 0), (2, 0), (-2, 0).
„it
Putting x = 0 in the equation of the curve, we get
y = 0. So, the curve meets Y-axis at (0, 0) only.
(iv) y = x3 - 4x => — = 3xz - 4
dx
I Example 2 Sketch the curve y = x3.
Sol. We observe the following points about the given curve
(i) The equation of the curve remains unchanged, if x
is replaced by - x and y by - y. So, it is symmetric
in opposite quadrants. Consequently, the shape of
the curve is similar in the first and the third
quadrants.
(ii) The curve passes through origin. Equating lowest
degree term y to zero, we get y = 0 i.e. X-axis is the
tangent at the origin.
(iii) Putting y = 0 in the equation of the curve, we get
x = 0. Similarly, when x = 0, we get y = 0. So, the curve
meets the coordinate axes at (0, 0) only.
(iv) y = x.3 =,^ = 3Z^=6xand4 = 6
dx
dx2
dx3
Clearly, — = 0 =
at the origin but
dx
dx
dx3
(v) As x increases from 0 to y also increases from 0 to
*>. Keeping all the above points in mind, we obtain a
sketch of the curve as shown in figure.
Now,
=>
— >0=>3x2-4>0
dx
2
2
x —y=
>0
Ji
• v3
2
2
=> x < —7= or x > -j= (using number line rule)
V3
and
dy
dx
n
2
V3
2
V3
So, the curve is decreasing in the interval
2
(- 2 / 5/3,2 / -73) and increasing for x> —
or x<--p
■ .2-
Ji
'3
2 .
.
-1
.
,
2 .
x = - -j= is a pomt of local maximum and x = —j= is
73
0.
point of local minimum.
2
When x =
, then y =
16
3^3
Chap 03 Area of Bounded Regions
When
2
x=-
16
3^3
, then y =
L 2, 16)
l/3 3/3<
173
The lowest degree term is x + y. Equating it to zero,
we get x + y = 0 as the equation of tangent at the
origin.
+Y
(iii) Putting y = 0 in the equation of curve, we get
x2 - x = 0 => x = 0,1. So, the curve crosses X-axis at
t
t
t
(0, 0) and (1, 0).
i
!
Ro)7
/
X
t
X
i
x
/(2.0)
0 X
Putting x = 0 in the equation of the curve, we obtain
y = 0. So, the curve meets Y-axis at (0, 0) only.
t
1Y
Keeping above points in mind, we sketch the curve as
shown in figure.
(0.0)
(1.0)
>x
I Example 4 Sketch the curve y = (x -1) (x - 2) (x - 3).
So/. We note the following points about the given curve
(i) The curve does not have any type of symmetry about
the coordinate axes and also in opposite quadrants.
(ii) The curve does not pass through the origin.
(iii) Putting y = 0 in the equation of the curve, we get
(x - 1)(x - 2) (x - 3) = 0 => x = 1,2,3. So, the curve
meets X-axis at (1, 0), (2, 0) and (3,0).
Putting x = 0 in the equation of the curve, we get y = - 6.
So, the curve crosses Y-axis at (0, - 6).
We observe that
and
(iv) y = x2 - x => — = 2x - 1 and —= 2
dx
dx2
Now,
dy
1
dx
2
1 d2y
x = ~, ~y>0
2 dx2
At
1
So, x = - is point of local minima.
2
x< 1
=> y<0
1< x<2
=> y>.0
2< x<3
=> y<0
x>3
=> y>0
(v) — >0=$»2x-l>0=>x>dx
2
So, the curve increases for all x > - and decreases for
2
y
0
all x < -. Keeping above points in mind, we sketch the
2
curve as shown in figure.
(1,/0) (2,0)\
/(3.0)
X
zZ(0.-6)
Clearly, y decreases as x decreases for all x < 1 and y
increases as x increases for x > 3.
Keeping all the above points in mind, we sketch the curve
as shown in figure.
I Example 5 Sketch the graph for y = x 2 — X.
I Example 6 Sketch the curve y = sin 2x.
So/. We note the following points about the curve
(i) The equation of the curve remains unchanged, if x is
replaced by (- x) and y by (- y), so it is symmetric in
opposite quadrants. Consequently, the shape of the
curve is similar in opposite quadrants.
(ii) The curve passes through origin.
(iii) Putting x = 0 in the equation of the curve, we get
y = 0. So, the curve crosses the X-axis at (0,0) only.
Putting y = 0 in the equation of the curve, we get
sin 2x = 0 => 2x = nrt, n e Z
So/. We note the following points about the curve
mt
x = —, n e Z
2
(i) The curve does not have any kind of symmetry.
=>
(ii) The curve passes through the origin and the tangent
at the origin is obtained by equating the lowest degree
term to zero.
So, the curve cuts the X-axis at the points
..., (- 7C, 0), (- 7C / 2,0), (0,0), (7C / 2,0), (n, 0),...
174
Textbook of Integral Calculus
^y =
- 4 sin 2x
dx2
(iv) y = sin 2x => — = 2 cos 2x and
dx
Now,
— = o => cos 2x = 0
dx
=>
2x = ±-,±—....
=>
x = ± — ,± —....
, 7t
, 37C
2
2
.>
371
4
4
3rc
atx =----4
dx
0 => sin 2x = 0 => 2x = nit, n e Z
nn
' •=> x = —, n e Z => x = ± 7t / 2 ,± it , ± 3tc 12 ,± 271,...
2
Clearly, ^-2-<0atx = ±7t/2,±37i/2,±57t/2,...
dx2
d2y
—^->0atx = ±K,±27t, ±3rc...
dx2
So, x = ±n/2,±3n/2,±511/2,... are the points of
local maximum and local maximum value at these
points is 1. Points x = ± 71, ± 2it, ± 37C, ... are points of
local minimum and the local minimum value at these
points is 0.
and
„
i “
7t 57t
d22yy „ \
Clearly, —Oat x=—,—....
X dx
4 4
dx 2
and
Now,
7it
4
, d 2y n 4
371 lit
and —4 >0 at x = —, —
dx2
4 4
(iv) y = sin2 x => ^ = sin 2x
5n
K
X =----- ,—
and at
4
4
Clearly,
> 0 when 0 < x <
7t 571 771
•So, the points x = —
4 4
’»
i
»
»•• •
4
and x = - —, - —.... are points of local maximum
4
4
and local maximum values at these points are 1.
7t 5n
,
3tt 77C
Similarly, x = - —,----- .... and x = —, —,...are
4
4
4 4
points of local minimum and local minimum value at
these points is (- 1).
(v) sin 2 (x + it) — sin 2x for all x. So, the periodicity of
the function is 7t. This means that the pattern of the
curve repeats at intervals of length 7t.
+Y
(-371/4,1)
(71/4,1)
(v) sin2 (it + x) = sin2 x for all x. So, the periodicity of the
function is it. This means that the shape of the curve
repeats at the interval of length it.
(-571/2,1) (-37T/2.1) (-ti/2,1) (tck/2,1)
/2.1)
(37t/2,1)
(3K/2,1) (57t/2,1)
(57r/2,1)
AAAAAA/?
(n,0) (2n,0)
(-271,0)
(57t/4,1)
(-71,0)
0
(3t:,0)
>X
(0.0)
(-rt/4,-1)
dy n ,
7t
— < 0 when — < x < it.
dx
2
So, the given curve is increasing in the interval
[0, rt / 2] and decreasing in [tu / 2,7t ].
and
(3tc/4,-1)
(771/4,-1)
Thus, keeping in mind, we sketch the curve as shown in
figure.
I Example 7 Sketch the curve y = sin2 x.
Sol. We note the following points about the curve
(i) The equation of the curve remains same, if x is
replaced by (- x). So, the curve is symmetric about
T-axis, i.e. the curve on the left side of T-axis is
identical to the curve on its right side.
(ii) The curve meets the coordinate axes at the same points
where y = sin x meets them.
(iii) y = sin2x =>^ = sin2x and ^y = 2 cos 2x
dx2
Keeping the above facts in mind, we sketch the curve as
shown in figure.
Asymptotes
The straight line AB is called the asymptote of curve
V =/(*)» if the distance MK from M a point on the curve
y = f(x) to the straight line AB tends to zero as M recedes
infinity.
In other words, the straight line AB meets the curve
y - f(x) at infinity (K is a point on AB). Thus,
1. If /(x) —> ± oo for x -» a, then the straight line x = a is
the asymptote of the curve y = f(x).
Chap 03 Area of Bounded Regions
2. If in the right hand member of the equation of the
curve y = f(x) it is possible to single out a linear part
so that the remaining part tends to zero as x —> ±
i.e. if y=f(x) = Kx + b + g(x) and g(x) —>0 for
x —> ± oot then the straight line y - Kx + b is the
asymptote of the curve.
f(x)
-
3. If there exist finite limits lim ------ = K and
x-> ±°° x
Thus, its graph is as shown in figure
+y
y = 1 (asymptote)
x2-1
ys77i
(-1.0)
(1.0)
lim [ /(x) - Kx] = b, then the straight line
-
----------
(0.-1)
X-»±oo
y - Kx + b is the asymptote of the curve.
175
1
I Example 9 Construct the graph for /(x) = x+ - ■
X
Methods to Sketch Curves
While constructing the graphs of functions, it is expedient
to follow the procedure given below
(1) Find the domain of definition of the function.
(2) Determine the odd-even nature of the function.
(3) Find the period of the function if its periodic.
(4) Find the asymptotes of the function.
Sol. The function is defined for all x except for x = 0.
It is an odd function for x 0.
It is not a periodic function.
For x —> 0 +, /(x)—> + <»; for x —> 0 -,/(x)—» - <»
For x —> - °°, /(x) ->-<»; for x —> /(x) -> °°
lim (/(x)-x) = 0
X —> ± “
(5) Check the behaviour of the function for x —> 0 ±
(6) Find the values of x, if possible for which /(x) -» 0.
(7) The interval of increase and decrease of the function
in its range. Hence, determine the greatest and the
least values of the function if any.
The straight lines x = 0 and y = x are the asymptotes of
the graph of the given function.
Now, consider /(x2) - /(xi)(for x2 > xj
. 2-xlx) +--------1
1
= {x
x2 x}
= (x2-xl) 1----- — < 0 for X] x2 g (0,1]
Remark
(5), (6) and (7) gives the points where the function cuts the
coordinate axes.
I Example 8 Construct the graph for f (x) =
x2 — 1
x2-1
x2 + 1
2
So/. Here, /(x) = ±—- = 1---x2+l
xz+l
(1) The function f(x) is well defined for all real x.
=> Domain of /(x) € R.
(2) f(- x) = /(x), so it is an even function.
(3) Since, algebraic —> non-periodic function.
/(x)—» 1 for x—>±°°
and
/(x)-»-l for x—>0±
It may be observed that /(x) < 1 for any x g R and
consequently its graph lies below the line y = 1 which
is asymptote to the graph of the given function.
22
Again,------- decreases for (0, 00) and increases for
x2 +1
(- 00 0), thus J(x) increases for (0, 00) and decreases
for (- °°,0) in its range.
(4) The greatest value —> 1 for x —> i 00 and the least
value is - 1 for x = 0.
*1 X2 J
L
and it is > 0 for xt x2 g [1, «>).
Thus, /(x) increases for x g [1,00) and decreases for
X G (0, 1],
Thus, the least value of the function is at x = 1 which is
/(l) = 2. Thus, its graph can be drawn as
Y
I
I
(1.2)/
(0.0)
*x-
•(-1.-2)
V
I
I Example 10 Construct the graph for f(x) =
Sol. The function is defined for all x except for x = 0. It is
neither even nor an odd
function. It is not a periodic function.
176
Textbook of Integral Calculus
Forx—>0 + _f(x)-> 0;forx—» 0-,
For x—>°o
/(x)->l
then J f(x) dx will equal Ai - A2 + A3 - A4 and not
for x—> - °°, /(x)->2
2
lim /(x) = |
/(x)->
Aj + A2 4* A3 +A^.
The straight line y = - is asymptote of the graph of the
2
given function.
If we need to evaluate Aj + A2 + A3 + A4 (the magnitude
of the bounded area), we will have to calculate
dx +| £f(x) dx | J‘/(x)
+|{
*f(x)dx
From this, it should also be obvious that
As x increases from (0, «>), — decreases from (0, «>) and e1,x
x
decreases from (0, o°).Thus, (1 + e1/x) decreases from (2, «>).
f(x) increases from ^0,
for x e (0, oo).
Similarly, J(x) increases from (1 / 2,1) for x —> (- «>, 0).
i.e. f (x) is an increasing function except for x = 0.
Thus, its graph can be drawn as shown in figure
Ya
(0.1)
Y= 1/2 (Asymptote)
(0. 0)
■X
£/(x)dx £j*|f(x)|dx
'a
(2) The area under the curve y = f(x) from x = a to x - b
is equal in magnitude but opposite in sign to the area
under the same curve from x = b to x = a, i.e.
J*f(x)dx = - £a/(x)dx
This property is obvious if you consider the
Newton-Leibnitz formula. If g(x) is the anti-derivative of
x(/), then J /(x) dx is g(b) - g(a) while | f(x) dx is
g(a)-g(b)
(3) The area under the curve y - f(x) from x = a to x = b
can be written as the sum of the area under the curve from
x = a to x = c and from x = c to x = b, that is
£ f(x) dx = £ /(x) dx + £ /(x) dx
Let us consider an example of this. Let c e (a, b)
y#
Areas of Curves
y=f(x)
(1) Suppose that /(x) < 0 on some interval [a, b]. Then,
the area under the curve y - f(x} from x = a to x = b will
be negative in sign, i.e.
a2
A.
This is obvious once you consider how the definite
integral was arrived at in the first place; as a limit of the
sum of the n rectangles (n -» <»). Thus, if /(x) < 0 in some
interval then the area of the rectangles in that interval will
also be negative.
This property means that for example, if f(x) has the
following form
a
c
+x
b
It is clear that the area under the curve from x = a to x = b,
A is A} + A2.
Note that c need not lie between a and b for this relation to
hold true. Suppose that c > b.
y+
y=/(x)
yt
A2
Ai
a
b
c
Chap 03 Area of Bounded Regions
177
Observe that area (rect AXYB) < J /(x) dx < area (rect DXYC)
Observe that A = J /(x) dx = (A + A]) - At
= £/(*) d*-£/(*) dx
= £/(*) dx+£/(x)dx
Analytically, this relation can be proved easily using the
Newton Leibnitz’s formula.
(6) Let us consider the integral of f{ (x) + f2(x) from x = a
to x = b. To evaluate the area under fx (x) + f2(x), we can
separately evaluate the area under (x) and the area under
f2(x) and.add the two area (algebraically).
Thus, £(/i(x) + f2(x)) dx = £ fx (x) dx + £ f2(x) dx
Now consider the integral of kf(x) from x = a to x = b. To
evaluate the area under kf(x), we can first evaluate the area
under /(x) and then multiply it by k, that is :
£ kf(x) dx = fc£ /(x) dx
This is because the curve of /(x) lies above the curve of •
g(x), or equivalently, the curve of /(x) - g(x) lies
(7) Consider the odd function /(x), i.e. /(x) = - f(- x). This
above the x-axis for [a, b]
measn that the graph of /(x) is symmetric about the origin.
y= W
(4) Let /(x) > g(x) on the interval [a, &]. Then,
£/(x) dx >£g(x) dx
y= gW
—a
a
a
>x
b
fa
This is an example where /(x) > g(x) > 0.
From the figure, it should be obvious that J /(x) dx =0,
£/(x)dx=Ai + A
because the area on the left side and that on the right
algebraically add to 0.
■2
£g(x)dx=Ai
while
Similarly, if /(x) was even, i.e. /(x) = /(- x)
Similarly, if /(x) < g(x) on the interval [a, b} then
y+
£/(x) dx <£g(x) dx
(5) For the interval [a, b], suppose m < f(x) < M.
That is, m is a lower-bound for /(x) while M is an
upper bound.
/(x) dx < M (b - a)
.
Then
+a
a
£ _f(x) dx =2£ /(x) dx because the graph is symmetrical
This is obvious once we consider the figure below:
Y|.
M
D
C
y=M
about the y-axis. If you recall the discussion in the unit on
functions, a function can also be even or odd about any
arbitrary point x = a. Let us suppose that /(x) is odd about
x = a, i.e. /(x) = - f(2a - x)
Y+
m
A!
-41
x=a
-IB
! Y
x=b
x=a
>x
17 8
Textbook of Integral Calculus
The points x and 2a - x lie equidistant from x = a at either
sides of it.
r2a
Suppose for example, that we need to calculate J
f(x) dx.
It is obvious that this will be 0, since we are considering
equal variation on either side of x = a, the area from x =0
to x = a and the area from x = a to x = 2a will add
algebraically to 0.
Similarly, if f(x) is even about x = a, i.e. /(x) = /(2a - x)
y+
Thus fav(b-a) = f/(x)dx => fav =~—
Ja b-a
£/(x)dx
This value is attained for at least one c e (a, b) (under the
constraint that / is continuous, of course).
I Example 11 Sketch the graph y = | x+11. Evaluate
J
| x+1| dx. What does the value of this integral
represents on the graph,
Sol. Here, y = |x + 11 = -
'(x + l),
if x2>-l
-(x + 1),if x<-l
which can be shown as
c.
.e
---------------------- 1------------------a
■>
(0.1)
3
then we have, for example
j^/(x)dx=2 j“/(x)dx
I
A
D
‘2
0
3
From the discussion, you will get a general idea as to how
to approach such issues regarding even/odd functions.
(8) Let us consider a function /(x) on [a, b]
I It
3
.’.J Jx + l|dx=j ’|x + l|dx + |2Jx + l|dx
=J ’ - (x+l)dx + J 2 | x +1| dx + j ^(x + l)dx
y,.
fib) -.................. -y
x2
—+x
2
+
x^
2
-2
:
=9
Representation of the value 9 of integral on graph.
.'.If2 | x +11 dx = 9 represents the area bounded by the curve
f(a)
a
b
We want to somehow define the “average” value that f(x)
takes on the interval [a, b], What would be an appropriate
way to define such an average? Let fav be the average
value that we are seeking. Let it be such that it is obtained
at some x = cG[a, b]
y,.
f(b)------------------------ -J
y = |x + 1|. X-axis and the lines x = - 4 and x = 2, i.e. if is
equal to the sum of the ares of AABD and ACE,
1
1
9 9
i.e.
;(3)(3) + i(3)(3) = £ + - = 9
2
3
2 2’
1
(v area of triangle = - x base x height)
2
_
I Example 12 Find the area of the ellipse
vz2
y2
a"
b2
= 1.
Sol. Using the symmetry of the figure; required area is given by
A = 4 (area OABO)
f„=f(c)
k
ta
x2 v2
= 4 J q y dx, where — 4=1
/(a)
a
c
b
We can measure fav by saying that the area under /(x)
from x = a to x = b should equal the area under the
average value from x = a to x = b. This seems to be the
only logical way to define the average (and this is how it
is actually defined!).
B (0,b)
0 (0,0)
A (a,0)
Chap 03 Area of Bounded Regions
2
2
1,2
n2
I— a2
a2
= 2 a 3a2----- log (2a + a-73) + — log a
2
2
^- = 1- —
a
a
2
, 2
y =7(. -x2)
=>
= 2 a2
In the first quadrant,
b
-x2
ya
a2 - x
Sol. Given curves are
-x2 dx
2+fisin-.ii.
2
a
1
2
0 + —sm -
a
a
J
'3 - a,22 log (2 + -73) sq units
I Example 14 Find the area common to the parabola
5x2 - y = 0 and 2x2 - y + 9 = 0.
A = 4 f - Ja2 - x2 dx = 4
Jo a
2 '
2a + aj3
= 2 a2j3 - — log
2
y = ±-/a2 -X2
a
= 4*
a
179
2
a
and
y = 5x2
...(i)
y. = 2x2 +9
-.(ii)
a
Y
x2=y/5
Jo
.L
2
Y2_y-9
2
(0.9)
- {0 + 0} =-------- sin (1)
a‘ "2
= 2ab
(0.0)
X
A = nab sq units
I Example 13 Find the area bounded by the hyperbola
x2 - y = a2 between the straight lines x = a and
x = 2o.
Sol. We use the symmetry of figure.
Required area, A = 2 f
Ja
2
-a =y
y dx, where x2 - y 2 = a 2
Remark
In such examples, figure is the most essential thing. Without
figure it just becomes difficult to judge whether y, to be
subtracted from y2 or otherwise.
Let us solve Eqs. (i) and (ii) simultaneously,
2
5x2 = 2x2 + 9
3x2=9=>x2 =3
=>
x = -V3
x = V3
or
In the usual notations, the required area is given by
A
(2a,0)
(a,0)
■ x = 2a
Let us take x = 0, which lies between
x = - J3 and k = V3
In the first quadrant; y = + \x 2 -a2
a
A = 2\2
a JS^7dx
a
=2
x
2
When x = 0 from Eq. (i) y = 0
a2
.
x2 - a,2 - — log(x +
________ -|2a
When x = 0 from Eq. (ii) y = 9
7^7)
Now,
n2
aJla.22- a2----- log (2a + Jia2- a2
2
9>0
Parabola Eq. (ii) is above parabola Eq. (i) between
Ja
I-----------------
=2
We have to find which curve is above and which is below
w.r.t. X-axis in order to decide yj and y2.
Take any point between x = - J3 and x = J3
I---------------A
a',
x = - -7'3
0- —loga >
and
x = j3.
180
Textbook of Integral Calculus
y of the curve (ii) is to be taken as and y of the curve (i)
is to be taken as y2.
. . f 75
Area(A) = J_7- {(2x2 + 9)-5x2}dx
•
i
= Jo’ (x3 - 3x 2+2x)dx + J2 (x3 -3x2 +2x)dx
..
( x4
V
------- X3,3 + X2
+
I4
_ [73
“J-V3 (9-3x2)dx
7o
2
f x4
--------- X
14
+ x2
7i
1-1+1 + (4-8 + 4)4
= 2 jo^(9-3x2)dx
,3
1-1+1
4
111
= — 4— = - sq unit
4 4 2
= 2[9x-xJ]/5
= 2 [9 73 -373]
I Example 16 Find the area between the curves
y = 2x4 - x2, the X-axis and the ordinates of two
minima of the curve.
Area = 1273 sq units
I Example 15 Find the area enclosed by
y = x (x -1) (x - 2) and X-axis.
Sol. The given curve is y = x (x - 1) (x - 2). It passes through
(0, 0), (1, 0) and (2, 0).
The sign scheme for y = x (x - 1) (x - 2) is as shown in
figure.
------------------- 1----------- ----------- 1------------------------1_
Sol. The given curve is y = 2x 4 -x2.
When y = 0, then x = 0,0, ± -4=
72
The sign scheme is as shown below
+
_1
~J2
+
1-2
0
From the sign scheme it is clear that the curve is + ve when
0<x<lorx>2, hence in these regions the curve lies
above X-axis while in the rest regions the curve lies below
X-axis.
Remark
Sometimes the discussion of monoton icity of function helps
us in sketching polynomials. In the present case, ^=3x2-6x + 2
dx
When — = 0 , then; x = 1 ±
dx
y[3
Sign scheme for — is
dx
+
-+-±
1+=L
J3
. (
.
1)
Thus, it is clear that the curve increases in - «>, 1 1 \
1 4
.
decreases in 11 —
, 1 + — and again increases in
—4—
---- F-
+
1
co
£
Therefore, it is clear that the curve cuts the X-axis at
1 n , 1
x - —f=, 0 and -?= •
yl2
V2
The curve is -ve in | —1L,0 j and 10,-4= | while positive in
\ *2)
I\ *vV22 j
the rest. Now, — = 8x3 - 2x. The sign scheme for — is as
dx
dx
below
+
+
0
1_
2
i.e. The curve decreases in (-*>,- 112) and (0,1/2) and
increases in the rest of portions. Also, the function possess
minimum at x = - 1 / 2 and 1/2 while maximum at x = 0.
Therefore, the graph of the curve is as shown below
y+
f 1 + -4= . °° Therefore, the graph of the curve is as below
I -73
y..
>x
1
73
1
W
>x
=
Required area = 2 j^2 12x4 - x2 | dx
=2
Hence, required area
fl
06
2
V3' ’ ‘ y/3)
/o
-►
1
f 2
x(x - l)(x -2)dx + J
x (x - 1)(x - 2) dx
r 1/2
r5 x3
-(2x4-x2)dx =-2 2—- —
Jo
7
=— sq unit
120
_
vi
Jo
Chap 03 Area of Bounded Regions
181
Exercise for Session 1
1. Draw a rough sketch of y = sin 2x and determine the area enclosed by the curve, X-axis and the lines x = n /4
and x =3n/4.
2. Find the area under the curve y = (x2 + 2)2 + 2x between the ordinates x = 0 and x = 2.
3. Find the area of the region bounded by the curve y = 2x - x2 and the X-axis.
4. Find the area bounded by the curve y2 = 2y - x and the Y-axis.
5. Find the area bounded by the curve y =4 - x2 and the line y =0 and y = 3.
6. Find the area bounded by x = at2 and y =2at between the ordinates corresponding to t -1 and t = 2.
7. Find the area of the parabola y2 =4ax and the latusrectum.
8. Find the area bounded by y = 1 + 2 sin2 x, X-axis, x = 0 and x = n.
9. Sketch the graph of y = 4x +1 in [0,4] and determine the area of the region enclosed by the curve, the axis of
X and the lines x =0, x =4.
10. Find the area of the region bounded by the curve xy -3x - 2y -10=0, X-axis and the lines x = 3, x = 4.
Session 2
Area Bounded by Two or More Curves
Area Bounded by Two
or More Curves
If confusion arises in such case evaluate
Ja
Area bounded by the curves y = f(x), y = g(x) and the
lines x = a and x-b.
Let the curves y = f(x) and y = g(x) be represented by AB
and CD, respectively. We assume that the two curves do
not intersect each other in the interval [a, b].
Thus,
shaded area = Area of curvilinear
trapezoid APQB - Area of curvilinear trapezoid CPQD
= fb f(x)dx —P g(x)dx= P {f(x) -g(x)} dx
y = f(x)
0
y=gw
A
B
\y = f^)
C
y=g(x)
D
------ >-x
Q
P
x=a
x=b
Figure 3.34
aY
O
Yf
a
j qa
Ja
I f(x) ~ &(x) I dx which gives the required area.
Area between two curves y = /(x), y = g(x) and the lines
x = a and x = b is always given by J {/(x) - g(x)} dx
provided f(x) > g(x) in [a, b]; the position of the graph is
immaterial. As shown in Fig. 3.34, Fig. 3.35, Fig. 3.36.
Y4
x=a
x-b
x=b
Figure 3.32
Now, consider the case when /(x) and g(x) intersect each
other in the interval [a, b].
First of all we should find the intersection point of
y = /(x) and y = ^x). For that we solve /(x) = g(x). Let
the root is x - c. (We consider only one intersection point
to illustrate the phenomenon).
Thus, required (shaded) area
0
Y = /(x)
t
y=g(*)
Figure 3.35
yi
y=W
o
->x
= JJ {/(*) -£<*)}dx+jjg(x) -f(x)}dx
Y+
y=gN
y=gW
x=b
x=a
y = f(x)
0
a
c
Figure 3.33
b
-+X
Figure 3.36
I Example 17 Sketch the curves and identify the
region bounded by x = 1 /2, x = 2, y = loge x and
y = 2X. Find the area of this region.
[IIT JEE I99y
Chap 03 Area of Bounded Regions
183
Now, we know that
Sol. The required area is the shaded portion in the following
figure.
/y = 2*
C £ 0 denotes the region, inside the circle C = 0.
P < 0 denotes the region, inside the parabola P = 0.
S < 0 denotes the region, which is negative side of the line
S = 0.
Required area = Area of curvilinear SOMRO
+ Area of trapezium MNSR - Area of curvilinear SONSO
y=logex
------- >X
= f2 Jtx dx + - (MR + NS) ■ MN
Jo
2
- (Area of square ONSG - Aiea of sector OSGO)
x=2
x=1/2
- »2\
1
= £2 V8xdx + |(4+3)-l-^3
:2
2
In the region - x <> 2; the curve y = 2X lies above as
2
compared to y = loge x.
Hence, required area = J^2 (2X - log x) dx
7t • 5
4 7
(9k l')
=-------- 1 sq units
I4
6j
I Example 19 Find the area of the region
{(x,y):0<y<x2 + 1,0<y <x + 1,0<x<2}.
T2
2X
-—- - (x log X - x)
1O
82
Jl/2
log
Sol. Let R = {(x,y):0<y £ x2 + l,0< y< x + l,0< x< 2}
2
5
31
— log 2 + - sq units
2
< log 2
= {(x,y):0^y< x2 + 1} n {(x,y):0<y< x + 1}
2J
I Example 18 Find the area ^iven by
y £ x2 + 1}
where, R{ = {(x, y): 0
x+y<6,x +y2 <6y and y2 <8x.
n{(x,y):O<x<5 2}
r'l R3
— Rj
R2={(x,y):0^y^x + l}
Sol. Let us consider the curves
and
P = y2 - 8x = 0
(i)
R3 = {(x,y):0^ x<2}
Thus, the sketch of Rlt R2 and R3 are
C = x2 + y2 = 6y
..y
Le.
x2 + (y - 3)2 - 9 = 0
...(ii)
5
and
S=x+y-6=0
...(iii)
4
y = X2 + 13
(1.2)
(6 - y)2 + yz - 6y = 0
i.e.
/xy=x+1
(2.3)
V
The intersection points of the curves (ii) and (iii) are given
by
\ 2
(0.1)>
y = 3,6
y#
P(0,6)
G
Ol
lllllllllliillllllilii
,fl(2,4)
0
2
*X
From the above figure,
<5(3,3)
Required area =
MN
1
( x3
>X
J
(x2 + 1) dx + J (x + 1) dx
V
+
0
' x±
\2
23
= — sq units
< 2
I Example 20 The area common to the region
determined by y > Vx and x2 + y2 < 2 has the value
Therefore, the points are (0, 6) and (3,3). The intersection
points of the curves (i) and (iii) are given by
y2=8(6-y),
i.e. y = 4, -12
Therefore, the point of intersection in 1st quadrant is (2,4).
(a) 7t sq units
(b) (2n - D sq units
1
(c) | - - -1 sq units
4 6
(d) None of these
7C
Textbook of Integral Calculus
184
Sol. The region formed by y £ Vx is the outer region of the
parabola y2 = x, when y £ 0 and x £ 0 and x2 + y2 < 2 is
the region inner to circle x2 + y2 = 2 shown as in figure.
Thus, required area
- 3 (1 + x) - ^(3 - x) (x - 1^
J___
2
f- 3 (1 + x) + ^(3- x)(x -1)>
I
y =fx
x.y 0
= -
2
dx
712 — (x —2)2 dx
x)(x-l)dx =
/
____________
- ■ -1 (x - 2) J- x2 + 4x -3 + - sin
2
2
x-2
xl 3
1
i
n
= — sq units
Now, to find the point of intersection put y2 = x in
x2 + y2 = 2.
and g(x) = {x}
2
x2 + x — 2 = 0
1,
(x + 2)(x —l) = 0
x = 1, as x > 0
(where, {.} denotes fractional part of x), then the area
bounded by f(x) and g(x) for xe [0,10] is
Required area = J * (^2- x 2 - Vx) dx
X-J2-X 2
2
3/2
-1 X_
— + sin
r= X
y/2 -----V3
2
n 2 | n 1
= — +------ -4 3 I 4 6
1
2
i
0
sq units
Hence, (c) is the correct answer.
(b) 5 sq units
. . 10
(c) — sq units
(d) None of these
{x}, x « z
}2, where both f(x)
1, xez
and g(x) are periodic with period T shown as
Sol. As, /(x) = -
1
Sol. Comparing ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get
a = 5, b = 2, h= 3, g = 7 / 2, f =3andc = 6
h2-ab = -l<0
xeZ
(a) | sq units
I Example 21 Find the area of the region enclosed by
the curve 5x2 + 6xy + 2y2 +7x+6y + 6 = 0.
=>
M, XgZ
I Example 22 lff(x) = <
Jw
zz:
0
12
3
♦
4
5
6
7
8
9
10
Thus, required area = 10 J * [-JW “ {*}2 ] dx
So, the above equation represents an ellipse.
= ioj‘ [(x)1/2 -x2]dx_
2y2 + 6 (1 + x) y + (5x2 + 7x + 6) = 0
- 3 (1 + x) ± ^(3 - x) (x - 1)
= 10
Clearly, the values of y are real for all x e [1,3} Thus, the
graph is as shown below
= 10
x3/2
4—
3/2
3
Jo
2.3 _r3; = —10 sq units
Y-‘
0
1
£
3
I Example 23 Find the area of the region bounded by
the curves y = x2, y = 12 - x21 and y = 2, which lies to
-31
-3 -
-6
Hence, (c) is the correct answer.
the right of the line x = 1.
ix:'(3.‘-6)
[iitjee 2002]
Sol. The region bounded by given curves on the right side of
x = 1 is shown as
•
Required
area = J
n
{x2 -(2-x2)}dx + jJL {4-x2}dr
Chap 03 Area of Bounded Regions
= ^(2^-2)iX + ^
I Example 25 Let /(x) -x2, g(x) = cos x and
a, p (a < p) be the roots of the equation
18x2 - 97tx + 7t2 =O.Then, the area bounded by the
curves y = fog(x), the ordinates‘x = a, x = p and the
X-axis is
(a)^(n-3) sq units
(b) y sq units
(4 - x2) dx
/
(
3
„ X
= 2-------- 2x
x3
+ 4x - —
185
2
/
(c)
(d) ~ sq units
sq units
So/. Here, y = fog(x) = f{g(x)} = (cos x),2: - cos2 X
0
y
1 12
18x2 — 97tx + 7t2 = 0
Also,
(3x - 7t) (6x - k) = 0
^-2^-g-2)+(£
{^2
>x
8-?
3
I
x = -,-(asa,p)
6 3
=>
3)
= [_4^+2»] = 20 -12>/P sq units
I
3J
3
A
7
11
I Example 24 The area enclosed by the curve
| y | = sin 2x, when x e [0, 2tc] is
(a) 1 sq unit'
(b) 2 sq units
(c) 3 sq units
a6 3
.*. Required area of curve
(d) 4 sq units
So/. As, we know y = sin 2x could be plotted as
1
0
=
A A A'T
”/2 V tV2"
f n/3
2
,
1 fK/3 /.
„ \ ,
cos xdx=-\
(l + cos2x)dx
J n/6
2 J K,(>
k/3
1
7t
7t
1 I . 27t
sin 2x
1
2
2
1
2
—+ 6 2 k ‘2
7t
2I
3
1(^3
. 27t
2tt 1
+ -sm----- sm — 2
3
6
6
ft
12
2 ,
Hence, (d) is the correct answer.
Thus, | y | = sin 2x is whenever positive, y can have both
positive and negative values, i.e. the curve is symmetric
about the axes.
371
sin 2x is positive only in 0 < x < —■ and K < x <
• Thus,
2
71
I Example 26 Find the area bounded by the curves
x2 + y2 = 25,4y =|4-x2|and x = 0 above the X-axis.
vSo/. The 1st curve is a circle of radius 5 with centre at (0, 0).
the curve consists of two loops one in 0, — and another
2
in
371 •'
7t,— .
2
■ ■’nW? •■•v’".
The 2nd curve is y =
4 — x2
4
4
which can be traced easily by graph transformation. ’*-«—■
y
->
x
o
Thus, required area = 4
=4
fJt/2
(sin 2x) dx
cos 2x
2
(2,0)
(4,0) 1(5,0)
n/2
= - 2 (cos 7t - cos 0)
0
= - 2 (- 1 - 1) = 4 sq units
Hence, (d) is the correct answer.
When the two curves intersect each other, then
Textbook of Integral Calculus
186
2\2
1-^- = 25 => x = ±4
x2 +
I Example 28 Let /(x) = max«sin x, cos x, i ►, then
4J
Hence, required area = 2 J *
^25 - x 2
dx
4
=2
2
x2dx-j 0
25
= 2 6 + — sin
2
1-
X2V
f 41
4 I
j2I
/
x2>|
1-— dx
determine the area of region bounded by the curves
y = f(x), X-axis, Y-axis and x = 2tc .
1 Graphically, /(x)
Sol. We have, f(x) = max J sin x, cos x, -I.
4 J
2
could be drawn as
&)-H]={25sin’,6H
1
1/2
I Example 27 Find the area enclosed by | x | +1 y | = 1.
cosx sinx
o
cosx z
1/2
I
T''
p/4 \ 5p/6 \P
>X
/ 5p/3 /2p
Sol. From the given equation, we have
|y| = l-|x|lo
[•■•lyllo]
Here, the graph is plotted between 0 to 2it and between the
points of intersection the maximum portion is included,
thus the shaded part is required area
-1<X<1
Value of f (x)
Interval
0< x< 7C/4
7t/4< x<5tc/6
5k/6< x<5n/3
5K/3< x<2k
i.e. for
for
for
for
cos x
sin x
1/2
COS X
Hence, required area
f 571/3 1
51t/6
r
cos x dx + f
sinxdx+I
-dx+ 5n/3 cosx dr
5n/6 2
J
Jo
J 71/4
I= f
,5n/6 . 1 ,
.
Therefore, the curve exists for x e [-1,1] only
and for - 1 < x < 1;
f
' -0
|x|-l
y = ± (1 -1 x |) i.e. y
,5n/3
= (sin x)J/4 -(cosx)k/4 +
+-(x)
7<x);5n/6 + (sin x)ifi/3
V2
-(|x|-l)
Thus, the required graph is as given in figure.
2
J3
1_
2
/2J
. 1 (5lt
571
3
6
+2
/
+ 0+—
2
I 2J
+ -fi + 5/3 j sq units
Required area = (V2)2 = 2 sq units
1
Exercise for Session 2
1. The area of the region bounded by y2 = 2x + 1 and x-y-1 = 0is
(a) 2/3
(b)4/3
(c)8/3
(d) 16/3
2. The area bounded by the curve y =2x - x2 and the straight line y = x is given by
(a) 9/2
(b) 43/6
(d) None ofthese
(c)35/6
3. The area bounded by the curve y = x |x|, X-axis and the ordinates x = -1,x=1is given by
(b) 1/3
(a) 0
(d) None ofthese
(c) 2/3
4. Area of the region bounded by the curves y=2x,y=2x-x2,x=0 and x = 2 is given by
(a)JL_4
(b)JL+4
(c) 3 log 2--
3
Iog2 3
3
Iog2
5. The area of the figure bounded by the cuves y = ex, y = e
(a)e+e
(b)e-e
(d) None of these
and the straight line x = 1 is
(c)e + --2
e
(d) None of these
Chap 03 Area of Bounded Regions
187
6. Area of the region bounded by the curve y2 =4x, Y-axis and the line y = 3 is
(b) 9/4
(a) 2
(c) 6V3
(d) None of these
7. The area of the figure bounded by y = sin x, y = cos x is the first quadrant is
(b)V3+1
(a) 2 (72^1)
(c)2(V3-1)
(d) None of these
8. The area bounded by the curves y = xex, y = xe~-Xx and the line x = 1 is
•
(a)e
(b)1——
e
.
(c)e
(d)l-e
9. The areas of the figure into which the curve y2 = 6x divides the circle x2 + y2 = 16 are in the ratio
(b)
(a) -
3
8n + 43
(c) --
(d) None of these
8n - 43
10. The area bounded by the Y-axis, y =cosx andy =sinx, 0 2:x < %/2 is
(a)2(V2-1)
(b)V2-1
(d)V2
(c)(V2+1)
3
11. The area bounded by the curve y = — and y +12 - x| = 2 is
|x|
(a)
4-log 27
3
(b)2-log3
(c) 2+ Iog3
(d) None of these
12. The area bounded by the curves y = x2 + 2and y=2\x |-cos + x is
(a) 2/3
(b) 8/3
(d) 1/3
(c) 4/3 ■
13. The are bounded by the curve y2 = 4x and the circle x2 + y2 -2x - 3 = 0 is
• (a) 2n + |
(b) 4n + |
(c) n + |
/
4 \
14. A point P moves inside a triangle formed by A (0,0), 8
1
,C (2,0) such that min {PA,PB,PC} =1,then the
area bounded by the curve traced by P, is
(b)V3+-^
(a)3V3-y
(d)3V3+y
(c)V3--|
15. The graph of y2 + 2xy + 401 x | = 400 divides the plane into regions. The area of the bounded region is
(b)800
(a) 400
(c) None of these
(c)600
16. The area of the region defined by 11 x | -1y 11 < 1 and x2 + y2 < 1 in the xy plane is
(a)ic
(b)2n
(c)3rc
(d) 1
17. The area of the region defined by 1 < |x -2| + |y + 1| <2 is
(a) 2
-(b) 4
(d) None of these
(c)6
18. The area of the region enclosed by the curve | y | = - (1 -1 x | )2 + 5, is
(a)-(7+5>/5)sq units
3
(b)-(7+5>/5)sq units
3
(c) - (545 - 7) sq units
3
(d) None of these
19. The area bounded by the curve f(x) = 11 tan x + cotx| -|tanx - cot x|| between the lines x =0,x =
X-axis is
(a) log 4
(b) logV2
2
and the
(d)V2log2
(c) 2log2
20. Iff(x) = max|sin x,cos x,^|, then the area of the region bounded by the curves y =f(x), X-axis, Y-axis and
5n .
x = — is
3
(a) ^V2 - 43 +
sq units
(b) ^42 + 43 + 5k sq units
2
(c) J2 + 43 +
sq units
(d) None of these
JEE Type Solved Examples:
Single Option Correct Type Questions
»
i '
If A denotes the area bounded by
sin x + cos x
, X-axis, x = K and x = 3k, then
/(*) =
• Ex. 1
.
Required area = 4 j
(
2
= 10 sin -7=-sin
X
j*
1 + sin x
...(1)
y=
2V2
I
3x | sin x + cos x |
, 242
dx<----
...(ii)
x = 0 and x = — is
4
x
x
kJ
2k
[IITJEE200S]
t
= dt
(i+t2)Vi^r2
5J2
4 3J2
----- < A<-----3n
k
V2-1
0.75 < A < 1.3
J . (i+t2)Vi^72
4f
V2+1
If f(x) > 0, Vx e (0,2) and y = /(x) makes posi­
• Ex. 2
and bounded by the lines
COS X
V2-1
Hence, (b) is the correct answer.
I
COS X
7t .
On adding Eqs. (i) and (ii), we get
=>
1 - sin x
and y =
n
111
vk<x<2k=> — < — < —
2k
----- <
3k
-4
• Ex. 4 The area of the region between the curves
2k
x
J5
Hence, (b) is the correct answer.
(d) None of these
,
2V2
So/.^<f 2K|sinx + cosx| dx
<-----
2k
1
V5
\
(b)0< A< 2
(a) 1 < A < 2
(c)2< A<3
<1:
(7s-x■:2-l)dx
tive intercepts of 2 and 1 unit on X and Y-axes respectively
and encloses an area of 3/4 unit with axes, then
J xf'(x)dxis
— dt
4t
= dt
(i+t2)Vw;2
V2+1
wj;J
t
(i+t2)Vw2
Sol. Required area = J
K/4
dt
1 + sin x
1 - sin x
COS X
COS X
o
(b) 1
dx
1 -sin x . A
>----------- >0
cos x
cos x
— —\
x
2 tan —
1-------------2x
1 + tan22 2 dx
x
1 - tan2
2.
1 + sin x
4
Sol. I = xf(x)| 2
0- f J f(x) dx = 0-’ =
4
3
/
4
2 tan —
1 +---------- 2
Hence, (d) is the correct answer.
• Ex. 3 The area^ofthe region included between the
"**'^~regionssatisfyingmin (Jx/, /y/) > 1 andx2 +y2 <5 is
1 + tan2 —
2
x
1-tan2
2.
r K/4
J0
, 2
1.2
1
, x 51
5
(a)- sin-1-7=—sin 1—j= -4 (b) 10 sin~ —j=-sin“ —-4
2
'5
2V
75
45) 5
I
75
45)
2(
,2
, 1 1
(
,22
2
(c)-rsin —7=-sin~—^-4 (d)15 sin-1-p=-sin
5
5l
V5
y/5j
I
755
Sol. Shaded region depicts min (| x |, |y |) £ 1
1 + tan2 —
2
TsJ
1 + tan—
2
C k/4
J0
1 - tan —
2
x
1+tan—
2 J
1 -tan2
Y
X
(1.2)
C K/4
3
1
X
1 + tan—1 + tan—
2
2
J0
1-tan2—
2
(2.1)
1
J2-1
„
X
dx=f
/4 —=X=dx
Jo
1-tan2—
2
4t
= dt
(1+t2)VT^7.2
Hence, (b) is the correct answer.
L
dx
2 tan —
x
Put tan — = t
2
Area = J o
2 J
—n
/
’-4
1 + tan2 —
Chap 03 Area of Bounded Regions
189
JEE Type Solved Examples:
More than One Correct Option Type Questions
• Ex. 5 Let T be the triangle with vertices (0,0), (0, c2) and
(c,c2) and let R be the region between y= ex andy -x2,
=> (b) is true; range is [—1,1) =$ into => (a) is false; minimum
value occurs at x = 0 and /(0) = -1 => (c) is false.
whereof), then
c3
c3
(b) Area of R = —
(a) Area (R) = —
6
(d) lim
c—»o+ Area (/?)
(C) |im
c-»o+ Area (/?)
Sol. Area(T) = —
2
A=2[”
Jo
xz + l
= [4- tan"1 x]“ = 4-y = 27t => (d) is false.
Hence, (a), (c) and (d) are the correct answers.
2
cos X, 0<x< —
2
• Ex. 7 Consider f(x\= -
2
,y=x2
dx
Jo x2 + l
dx = 4f
it
------x
2
,y=cx
\2
2
such that f
£x<n
is periodic with period it, then
(O.c2)
(c,c2)
(a) the range off is 0,—
. 4J
(b) /is continuous for all real x, but not differentiable for
some real x
/ (0,-0)
,3
Area (/?) = --[C x2dx=2 Jo
2
.
(c) / is continuous for all real x
c3
3
6
..
Area(T)
c3 6 „
Inn ------ — = hm----- - = 3
c-»o* Area (2?) c-»o* 2 c3
(d) the area bounded by y = /(x) and the X-axis from
(
\
x =-mt to x = mt is 2n 1 + — foragivenneN
Hence, (a) and (c) are the correct answers.
• Ex. 6 Suppose/is definedfrom R
[-1,1] as
x2 -1
where R is the set of real number. Then, the
x2 +1
statement which does not hold is
(a) /is many-one onto
(b) /increases for x > 0 and decreases for x < 0
(c) minimum value is not attained even though/is
bounded
71
0£x<—
cos x,
So/. Given, /(x) =
2
and / is periodic with
71
£ x <7t
---- x I , —
'
2
.2
period it. Let us draw the graph of y = /(x)
r n2y
From the graph, the range of the function is 0, — .
2
it
It is discontinuous at x = mt, n e /. It is not differentiable
mt
at x = —, n e I.
2
-y
3--.
(d) the area included by the curve y = /(x) and the line
<>x?/4
2/-
y = 1 is 7t sq units
So/.y = /(x) = 4^!.
1
x2 + l
4x
x > 0, f is increasing and x < 0 f is decreasing.
/'W =
(x2 +1)2
IX'-k -x/2
0
-=—►
x/2
4x X
Area bounded by y = /(x) and the X-axis from -mt to mt for
neN
f Xl2
= 2n J /(x)dx = 2n i
Jo
cos x dx +
fX
= 2n [1 + —1
(0.-1)
I 24 J
Hence, (a) and (d) are the correct answers.
2
n
---- x I dx
190
Textbook of Integral Calculus
• Ex. 8 Consider the functions fix) and g(x), both
• Ex. 9 Area of the region bounded by the curve y = e
definedfrom R —> R and are defined as fix) = 2x - x2 and
and lines x = 0 andy =e is
gix) = x" where neN. If the area between fix) and gix) is
1/2, then n is a divisor of
(c) 20
(b) 15
(a) 12
(d) 30
Sol. Solving, fix) = 2x-x2 and gix) = x" we have
[IIT JEE2009]
(a)e-1
(b) Jie|n(e + 1-y)dy
(C)e-J°e■* dx
(d) J ‘inydy
Sol. Shaded area = e -
2x - x2 = x" => x = 0 and x = 1
X
1 ex dx"] =
1
o
J
gto = x"
y=e
e
/(x) = Zx-x2
■>
O
A = j’(2x-x2-x")dx = x2
=1-1
3
Since,
2
3
1
n+1
1
2
1
n+ 1
2
3
2
3
1
2
0
11
x^
x"*1
3
n + 1Jo
1
n+1
1
n+ 1
Also, j ln(e + 1 -y) dy
Put
e+l-y=t
-dy =dt
4-3
+1 ==6
—-—
nn+1
6
n+ 1
=>
n —5
Thus, n is a divisor of 15, 20, 30.
Hence, (b), (c) and (d) are the correct answers.
1
= jQlnt(-dt) = j°lntdt
6
= J'lnydy = 1
Hence, (b), (c) and (d) are the correct answers.
JEE Type Solved Examples:
Passage Based Questions
Passage I
(Q. Nos. 10 to 12)
Consider the function fix) = x3 -8x2 + 20x -13.
• Ex. 10 Number of positive integers x for which fix) is a
prime number, is
(a) 1
(c)3
(b) 2
(d)4
Sol. f(x) =(x-l)(x2 -7x+13) for fix) to be prime atleast one of
the factors must be prime.
x-1 =1
Therefore,
x=2
=>
or
x2—7x+13 = l
=>
x2—7x + 12 = 0
x = 3 or 4
x = 2»3,4
Hence, (c) is the correct answer.
• Ex. 11 The function fix) definedfor R —> R
(a) is one-one onto
(b) is many-one onto
(c) has 3 real roots
(d) is such that /(x1)/(x2)< 0 where x, and x2 are the
roots of/'(x) = 0
Sol. f(x) is many-one as it increases and decreases, also range of
fix) € R => many-one onto.
Hence, (b) is the correct answer.
• Ex. 12 Area enclosed byy = /(x) and the coordinate
axes is
(a) 65/12
(c) 71/12
(b) 13/12
(d) None of these
65
Sol. A = jo/(x)dx =-jo(x3-8x2 + 2Ox-13)dx = ^~
12
Hence, (a) is the correct answer.
191
Chap 03 Area of Bounded Regions
On differentiating, we get 3y 2y' - 3y' -1 = 0
Passage II
(Q. Nos. 13 to 15)
Letb^x) = /(x) - g(x), where /(x) = sin4 Ttx and
g(x) = lnx. Lef x0,x1,x2,...,xn+l be the roots off(x) = g(x)
in increasing order.
y ("~10^)_3{l-(272)2}
=>
3{1-(2V2)2}
n
*r+1
fa) X Pr+’(“1)r/,(x)dx (b) y J *r+1(-1)r+,/’(x) dx
r=0
Sol.
2r=0
Xr
21
Xi-y2)
Xr
6-2^2 ■
J’”Vir/>(x)rfx(d) A jx’"'(-ir'«x)rfx
r=0
3(1-8)
Again differentiating Eq. (i), we get
6yf2
Y=gW is given by
X'
1
y'(—10V2) =----- = —P—-
• Ex. 13 The absolute area enclosed byy = f{x) and
r=0
...(i)
Xr
y'(-lO^) =
ffl-
4+J2
3(1-8)
Hence, (b) is the correct answer.
Y.
73-32
y = f(x)
• Ex. 17 The area of the region bounded by the curve
y = /(x), the X-axis and the line x = a and x = b, where
-00 < a < b < -2 is
0
1
/y = gW
3/2 Xt 2 x2 5/2 x3 3
(a)P-- ~2--
Jfl3[{/(x)}2-1]
b--- --------- dx - bf{b) + af(a)
(b)-J °3[{/(x)}
2-1]
Hence, (a) is the correct answer.
• Ex. 14 In the above question, the value ofn is
(a)1
(b) 2
(c)3
*b
(d)4
X
a 3[{/(x)}2
Sol. xn+l = x3 =>n=2.
• Ex. 15 The whole area bounded byy- f(x), y = g(x)
Sol. Required area = j /(x) dx = [xf(x)£ - J xf'(x) dx
= bf(b) - af(a) + f6-------------dx
andx=0 is
(a)I
—dx-bf(b)+af(a)
dx -bf(b)+af(a)
(d) -P
----J‘3[{/(x)}2-1]
Hence, (b) is the correct answer.
. J1
dx + bf(b)-af(a)
ia3[{f(x)}2-l]
(b)i
Hence, (a) is the correct answer.
(c)2
1,
11
0
8
Sol. Required area = | sin4rcxdx-J In xdx = —
Hence, (a) is the correct answer.
• Ex. 18 j1 g'(x)dx is equal to
(a)2g(-1)
Passage HI
(b)0
(c)-2g(1)
(d) 2/*(1)
Sol. I = p/(*) dx = [g(x)ti = g(l) - g(-l)
(Q. Nos. 16 to 18)
Consider the function defined implicitly by the equation
y3 -3y + x = 0 on various intervals in the real line. If
xg (-<»,-2) u(2, oo), the equation implicitly defines a
unique real-valued differentiable function y = f(x). If
x g (-2,2), the equation implicitly defines a unique
real-valued differentiable function y = g(x) satisfying
g(0)=0.
[IITJEE2008]
y3 —3y + x = 0
Since,
and
—(i)
Since,
y = g&
(g(x)}3-3g(x) + x = 0
At x = 1,
U(l)}3-3g(l)+l = 0
-(ii)
Atx = -1, {g(-l)}3-3^-l)-l = 0
...(iii)
On adding Eqs. (i) and (ii), we get
W))’ + W-l))’ -3(S<1) + «(-!» = 0
• Ex. 16 ///(-10V2) = 2^/2, then f"(-wj2) is equal to
/ \4^2
(a)-yy
7332
Sol..-.
4-72
7 33 2
(c)-7-
733
y3-3y+ x = 0
4V2
(d)—-y733
«1)+ s(-l)>Wl)2 +
=0
4(1) + 8(-l) = 0,8(1) = -«(-»
I =8(l)-8(-l) = 4(l)-(-S<l)) =2«(1)
Hence, (d) is the correct answer.
[byEq.(i)]
192
Textbook of Integral Calculus
Subjective Type Questions
log y = ax + log c
• Ex. 19 Find the total area bounded by the curves
y =cos x -cos2 x and y = x2
=>
x2
c =e
4 >
/
X2
Sol. Here, y = cos x - cos2 x and y = x2
\
which passes through P (1,1)
y ~ ce“,
y-e— a eax
=> Curve is
—4 JI could be
=> y=efl(l-1)
Y
drawn as in figure.
1
aY
a
\a
e''
1-----
y=cosx
- cos2x
Thus, the area = 2
f X/2
_2\1
n
(cos x- cos2 x) - x2 x2
Jo
—+X
/.Required area = — J a(log y + a) dy +
4>J
{ x/2
= 2]Jo
(
2
I
4
ir2
H
cos x - cos X - X +--- x2
4
= -{(-1 + a) - e~a (- a-1) - ae-0}
a
• Ex. 20 A curve y = f (x) passes through the point
P (1,1), the normal to the curve at P is a(y -1) + (x -1) = 0. If
the slope of the tangent at any point on the curve is propor­
tional to the ordinate of that point, determine the equation
of the curve. Also obtain the area bounded by the Y-axis, the
curve and the normal to the curve at P.
Sol. Here, slope of the normal at P (x, y).
=> Slope of the line a (y -1) + (x -1) = 0 is - a
:. Slope of the tangent at P = a,
dx
p
It is given that the slope of the tangent at any point on the
curve y = f(x) is proportional to the ordinate of the point.
dy
—
dx
dy_
dx
(
-a
2
y=----- 2r • Find the area.
1+x
2
1 + x2
Le.
x4 + x2 - 2 = 0
Le.
(x2 + 2) (x2 -1) = 0
Le.
x = ±l
ay
y=^
= X => a = X
2
y 1 +x2
(i.i)
dy
1
a
2+—
a. a
2
j
---- = a dx
y
.
i
• Ex. 21 Sketch the region bounded by the curves y-x2 and
dy
y =>
=* —
— == Xy
^y
dx
dy
— = ay
dx
1 + - I -I’
a,
1 1= 1 + ---------- I sq units
2a
a 2a J
Sol. For intersection point, x2 =
(0
=a
1+a
a
1
1+
= -(-l
+ a + e‘n0)+|a
a
2
a
2
+ (1 + a) 11 + i -1
\ a
it
n5
2—+—
2 120
(dJL
+ J (l + a)y-^y2 ■ a
I
2
Jd
= - {y (fog y -1) + ay)2
a
e
dx
Xs 1'?!“
x sin2x ----4-------= 2 sin x-----2
4
5
12 Jo„
'
{(1 + a)-ay|dy
dx
o
—-»X
Chap 03 Area of Bounded Regions
:. The equation of the tangent to the curve at the point
2
-x2 dx
1 + x2
>
Hence, required area = 2 J
is
f-.l
k 4
1
K
x3
= 2 2 tan 1 x----- = 2
2
3 o
1
1A
y -1 = 2 | x- — |
I
4J
~ x =------K 1 = OT
y = O,
0. = 4 2.
Now, the required area=Area of curvilinear AOPN - Area of APTN
t K,< 1
(tan x) dx----- NT • PN
Jo
2
when
3
• Ex. 22 Find the area enclosed between the curves
y = log(x + e);x = loge
and X-axis.
\.y)
1 n
2 4
= [log (sec x)];/4
Sol. The given curves are y = log (x + e) and
X
1
=> — = e =>y=e
y
x = loge
193
n -f. —1
1
• ,1 = -1 hlog o2 - -1
2I
4----- 2.
22,
• Ex. 24 Find all the possible values ofb > 0, so that the
-X
area of the bounded region enclosed between the parabolas
x2
Using graph transformation we can sketch the curves,
x = -e y
y = x-bx2 andy = — is maximum.
b
.2
>X
(1-e.O) 0
Sol. Eliminating y from y = — and y = x - bx2, we get
'
b
x2=bx-b2x2
b
x = 0,
1 + b2
fy
f 0
Hence, required area = J
r ••
log (x + e) dx + J
e~ x dx
-Jj log(t)dt + jo e~x dx
(putting x + e = t)
= [tlog t -1]/ ~[e"x] “ =1 + 1=2
• Ex. 23 Find the area of the region bounded by the curve
c:y = tan x, tangent drawn toe atx = n/4 and the X-axis.
Sol. The given curve is y - tan x
••
dyr
•—■ = sec 2 x
dx
Thus, the area enclosed between the parabolas,
x2^
6/(i + 62)
x - bx2----- dx
o
*>)
.. ■
bn + b2
Jo
= sec2 — = 2
x = k/4
.2
• X-X
y+
1 + b2
3
b
x bn + b2
0
>dx
1
Also, at x = —; y = 1
4
N
>X
1 b(l-b2)
6
(1 + b )4
3 (1 + b2)3
dA
Hence, — = 0 gives b = -1, 0,1 since b > 0
db
Therefore, we consider only b = 1
dA
Sign scheme for — around b = 1 is as below
db
+
co
0*
1
db
T
b2
6 (1 + b2)2
'
*
-^9
For maximum value of A, — = 0
v db
n dA 1 (l + b2)2-2b-b2-2(l + b2)-2b
R
1
b
\
4
x^
2
(1 + b2}
From sign scheme it is clear that A is maximum.
194
Textbook of Integral Calculus
• Ex. 25 LetC} andC2 be the graphs of the function
y - x2 andy = 2x,0 < x <1, respectively. LetC3 be the graph
of a function y = /(x); 0 < x < 1, /(0) = 0. For a point P on C3,
let the lines through P parallel to the axes, meets C2 and C3
at Q and R respectively. Iffor every position ofP on (CO, the
areas of the shaded region OPQ and ORP are equal, deter­
mine the function /(x).
[HTJEE1998]
Sol. On the curve Cb i.e. y = x2.
= lim
X—» 0 +
e log x
1/x
---- form
oo
= iim ZllZA = 0
(using L’Hospital’s rule)
x->o+ (-1/X2)
Thus, the first curve starts from (0, 0) but does not
include (0,0). Now, the given curves intersect therefore
log
x
ex ilog x = —
5—
ex
(e2x2 -1) log x = 0
Le.
Let P be (a, a2). So, ordinate of point Q on C2 is also a2.
y
Now, C2 (y = 2x) the abscissae of Q is given by x = —
2
.’.Qis
i.e.
(v x>0)
x = l,-
2
ex2 I
'
—, a and R on C3 is {a, f(x)}.
2
J
,a2
Now, area of &OPQ = j
(x1-x2)dy=j“
2 3
= -a
3
Again, area of AORP = J
a
dy
(i)
4
(y! - y2)dx = j
a2
{x2 - f(x)} dx ...(ii)
Therefore, using the above results figure could be drawn as
Required area = ( log x
- ex log x dx
ex
Y+
(1/2,1)
(0,1)
Ci
(1.1)
i
AX2
’ e2-5
= 1 (log x\.2
-e ^(21ogx-l) =—2
e
4
l/e
• Ex. 27 Let An be the area bounded by the curve
JC
y = (tan x)n and the lines x = 0, y = 0 and x = — • Prove that
4
(1.0)
0
C3 fl
forn>2\An + An.2 = ------ and deduce that
n-1
Thus, from Eqs. (i) and (ii), we get
—-— <
An < —— ■
<A„
2n - 2
2n + 2
2a3 a 4 =[“ {x2-/(x)}dx
J 0
3
4
Differentiating both the sides w.r.t. a, we get
Sol. First part We have, A = j
2a2 -a3=az - /(a)
/(a) = a3 - a2 => /(x) = x3 - x.2
Hence,
A.-2=J0
• Ex. 26 Find the area of the region bounded by the
curves y = ex log x and y - --------ex
[IITJEE1990]
Sol. Both the curves are defined for x > 0. Both are positive when
x > 1 and negative when 0 < x < 1.
We know that, lim log x -> - «>
X-> 0+
Therefore,
lim feS
x->o+
— oo
ex
Thus, Y-axis is asymptote of second curve.
and lim ex log x [(0) (- «>) form]
x-> 0*
(tan x) ■ dx
f ”/4
(tan x)'
loSx
j
(tan x)" dx
(tanx)-’ (tan2 x + 1) dx
Al + A-2 =
i
L 4e
.4
Q
,
• sec2 x dx
Let tan x = t, so that sec2 xdx = dt
-i A1
A + A,_2 = J> -2dt = <n-i j
Second part
O^x^ft/4
Since,
•(i)
"-1
o
1
tan" + z x<tann x<tan"-2 x
=>
(■ lt/4
=>
Q<, tan x
i
J0
f lt/4
tann+2 x dx < J
f
tan x dx < j
k/4
o
tan"-2 xdx
A + 2<A <A-2 => A +A + 2<2At <A +A-j
Chap 03 Area of Bounded Regions
1
<2An < —
n+1
n -1
Hence, the intersection points in the first quadrant is
(V2 - 1, V2 — 1).
1
__ 1
<A. <
2 (n + 1)
2(n-l)
Required area = 8 [Area of curvilinear AOLM]
[using Eq. (i)]
• Ex. 28 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 nearer to the origin than
to any edge. Sketch the region S and find its area.
[I1T JEE1995J
Sol. For the points lying in the AOAB the edge AB, i.e. x = 1 is
the closest edge. Therefore, if the distance of a point P (x, y)
(lying in the AOAB) from origin is less than that of its
distance from the edge x = 1 it will fall in the region S.
OP < PQ
y=x
D
195
=8
(V2 -1) (>/2 -1) + Jjl/2 p/1 -2x dx
=8
f 2(1—2x)3/2 X 1/2
-(3 —2^2)
+ [ 3(-2)
2
XV2-1
= 3(4^li}
• Ex. 29 Sketch the region included between the curves
x2+y2=a2andJ\T\+J\y | = 4a (a >0) andfind its.
_
area.
Sol. The grapKs'f x | + I y I = a and | x |2 + | y |2 = a 2 are as shown
in figure.
—
.Y
z..j
Q
fry) i
(O.a)
>X
0
c
jfi
(a.0)
(-a.0)
y = -x
X
^x2 + y2 51 - x
(O.-o) •'
=>
x2 + y2 <xz -2x+ 1
=>
y2 £1 -2x
Similarly, for points lying in the AOAD the side y = 1 is the
closest side and therefore the region S is determined by
From the figure itcaohejcorfcluded that when powers of | x |
and | y | both is reduced to half the straight lines get stretched
inside taking the shape as above.
Thus, required area = 4 [shaded area in the first quadrant]
x2 < 1 - 2y
Since, the edges are symmetric about the origin. Hence, by the
above inequality and by symmetry, the required area will be the
shaded portion in the figure given below
kY
/y=x
■
-/
=4
J0
(since in 1st quadrant x, y > 0), hence
• •
JiTi + J|yT=^
y[x + fy=4a
=>
y =(fa - Vx)2
W/'
L
4
XI
(O.a)
*X
(1/2,0)/!'
|x|2+|y[2 = a2
(a.0)
(-a.0)
''
z
x
0
*
'<y=-x
Now, when the curves y2 = 1 — 2x and y = x intersect each
other, then
|x| + |y|=a
(0- -a)
x2=l-2x => x2 + 2x-1 = 0
=>
X = V2 - 1, - V2 -1
Hence, required the area =
2
it --
3
a2
196
Textbook of Integral Calculus
Hence, required area
• Ex. 30 Show that the area included between the parabo8
lasy2 = 4a (x + a) andy2 = 4b(b- x) is- (a + b) Jab.
3
Sol. Given parabolas are
•••(*)
y2=4a(x + a)
and
*
y2 = 4b (b - x)
= [ {(x+ x Vx)-(x-x Vx)} dx
Jo
= [1 (2x Vx) dx
Jo
= 2 Jo’ xV2
3/2 dx
dx = - sq
sq unit
unit
-(ii)
Solving Eqs. (i) and (ii), we get
x = (b-a) and y = ±2-Jab
• Ex. 32 Prove that the areasSQ,S^,S2i... bounded by the
P and Q are (b - a, 2 Jab) and (b - a, - 2 -fab) respectively,
and the points A and A' are (- a, 0) and (b, 0), respectively.
X-axis and half-waves of the curve y -e-0^ sin px, x 10.
form a geometric progression with the common ratio
.. _ „-n a/p
g =e
+Y
p
Sol. The curve y = e~a x sin px intersects the positive X-axis at the
points where y = 0.
a x sin px = 0
|Y
AT
(-a.O)
(t>.o)
0
0
Now, required area = Area APA' QA = 2 Area APA' A
>x
n/p
= 2 [Area APMA + Area MPA' A4]
=2
2 fb (b - x) dx
* -a
yja + x dx + 4jb
b-a r
= 4^ -(a + x)M
.3
?
2
+ 4 Jb —-(b-x) 3/2
3
-0
b
. The function y = e~ a x sin Px is positive in the interval
(x2K, *2 k +1) and negative in (x2K + b x2K + 2), i.e. the sign of the
function in the interval (x„, x„ +1), therefore
g
= - fab (a + b) sq units
3
sinpx = 0 => x„ =^—, n = 0,1,2,...
=>
fb- x dx
_
3
p (n + 1) n/p
sin Px dx
■ (n + l)n/p
(-l)n + 1e -ax
=
• Ex. 31 Determine the area of the figure bounded by two
branches of the curve(y - x)2 = x3 and the straight line x = 1.
x
Jnn/p
"
2
[
“
a 2 + p:
(a sin Px + P cos Px)
>
Jn n/p
_pe-nna/p
So/. Given curves are (y - x)2 = x3
y - x = ± xjx
"(a2 + P2)
y = x + x Jx
pe-(n + l)n a/p
-.(i)
y = x - x Jx
+1
Hence, g =-----u
■Si
• ...(iii)
and
x=1
Which could be drawn as; shown in figure.
+ e-n a/Pj
(a2 + P2)
n a/p
Pe“ nna/p {1 -F e~ff o/p}
a 2 + p2
which completes the proof.
• Ex. 33 Letb^Q andfor j = 0,1,2,..., n. Let Sj be the area
y=x-xjx~
of the region bounded by Y-axis and the curve xeay = sin by,
+X
— <y <----------- .ShowthatS0,Si,S2,...,Sn are mgeometb
b
ric progression. Also, find their sum fora = -1 and b = n.
2
1/2
in
+
,
So/. Here, St= [
I
i
X=1
J
Jjn/b
xdy = f
J
• ( j + 1) nib e-ayMby
<jnlb
Chap 03 Area of Bounded Regions
■ ( J+ 1) K/&
a2 + b2
,2
=2
x(-b)(-l)/+1g-ajKlb
+
l2
a +b
2
J0
ajK lb
,-a(J + i)nlb
=
T>
- »i
1
2
= IN Wb2 {e~ mlb + l}-, {j=0,l,2>...,n}
e
~—
a2-~
+ bk2
dx ,
y — dt
dt
-If"
('* + e"2t -2)dt
2 J0
777
g-ajK/b
Now,
t fi
M-']
-J"
e~a)Klb
p-a (7+1) K/b
=IN
-
■ J*/b
e-a(J+l)K/t
a2 + b2
1 f e'1 + e~‘>
=2
(- a sin by - b cos by) ■
197
{e~ m,b+ 1) | b|
ea‘
1
2
777 k“
—
2
= e~m/b for all j = 0,1,2,.... n
Hence, So,
S2,....
are in GP with common ratio e~ax,b.
For a = -1 and b = n, we have
2
parabola y = x + x + 1,the curve y =
einfeil)
n
7t2 + 1
n
n
J=0
J=0
• Ex. 35 Find the area enclosed by circle x2 + y 2 = 4,
x
and
X-axis (where, [.] is the greatest integerfunction).
(e + 1)
nz + 1
(e + 1) 7C
it + 1
Sol.
e-1
. 2 x
x
y = sin — + cos —
4
4 .
.
- 2 x
x
1 < sin — + cos — < 2
4 4
• Ex. 34 For any real
et +e t
7 x
sin — + cos —
4
4
e* ~e t .
f,x = 2 + ------ ------,y = 2+----- - ------ is a point on the hyper­
. 2 x
x
y = sin — + cos —
4
4
=1
..y
bola x2 - y2 - 4x + 4y -1 = 0. Find the area bounded by the
(0. 2)
hyperbola and the lines joining the centre to the points
corresponding to t, and - f,.’
e' + e"
e'-e~ t
- is on the curve
,y =2 +
2
2
(x-2)2-(y-2)2=l or x2 - y2 - 4x + 4y -1 = 0
for x 6 (- 2,2]
Bi®
So/. The points x = 2 +
(-2,0) 1-^-1 —1/2 O
lkl(2,0)
(0.-2)
Now, we have to find out the area enclosed by the circle
x2 + y2 = 4, parabola ( y -
C
X
1
2
2
, line y = 1 and
X-axis. Required area is shaded area in the figure.
Hence, required area
= 73 X1 + (V3-1) X1 + j j(x2+ x + l)dx+2j^_(74-x2)dx
Put(x — 2) = x, (y — 2) = y
x 2 -y 2 = 1 and x =
.y =
"
2
er-e2
We have to find the area of the region bounded by the curve
x2 - y2 = 1 and the lines joining the centre x = 0, y = 0 to the
= (273-1) + 0-
point (tj) and (-tj).
.‘.Required area = 2 Area of APCN - J
2
y dx
c
,]]+2j(0+n)_^+^
o
x3 x2
= (2^-l) + — + — + x + 2 — V4-x2 + 2 sin
3
2
2
-1
L
44-
,n /r
5 2 It,
= (273 -1) + - + —
6
3
'3 =
2K
3
3
6
sq units
198
Textbook of Integral Calculus
• Ex. 36 Let f(x) = max {x2,(1 - x)2,2x(1 - x)}, where
ya
0 x < 1. Determine the area of the region bounded by the
curves y = /(x), X-axis, x = 0 and x = 1.
[HTJEE1997]
Sol. We have, f(x) = max {x2, (1 - x2), 2x(l - x)}
(0,73)
Graphically it could be shown as;
O’
(-2.0)
aY
(0.-^)
y=x2
\y=
: (1-X)2
max \
$
//
■
J
12-3x'.2
---- 1 dx
4
f2^'3
Required area = J
>y = 2x(1-x)
max
r 246/3
Tit
0,
(2.0)
>X
1
3 2 3
= 2sin"1 j| + ""
\%
x S>
For figure it is clear that maximum graph (Le. above max graph
is considered and others are neglected).
t 2^/3
I-------- 2 .
- x2 dx - • | - 246/3 Idx
2 J -246/3 di
v
• Ex. 38 Let
f(x) =
2
4
3
3
'6
-2,-3<x<0
, where
’ . 1 *
For x e
L
0, - , x2 < 2x (1 - x) < (1 - x)2
3 J
For xe . 13’11
2 J’ x2 £ (1 - x)2 £ 2x (1 - x)
For x e 11 , (1 - x)2 £ x2 <2x(l - x)
g(x) =W" {/(I x I) +1 /(x) I, /(I x 1) -1 /(x) I}. Find the
area bounded by the curve g (x) and the X-axis between the
ordinates at x = 3 arid x = - 3.
Sol. Here,
/(I* I)
-x-2,-3£x£0
x - 2, 0 < x <, 3
. 2*3 .
2,
’ 2
For x e
, (1 - x)2 £2x(l — x) £ x2
-.1
.3
l/(x)| = - x + 2, 0 < x £ 2
x - 2,
Hence, f(x) can be written as
2x(l-x),
for 1/3 <x <2/3
x2,
for2/3£x<l
•• /(|x|)-|/(x)| = 2x —4,
0,
r 1/3 ,
-
f 2/3
0 < x £2
2<x£3
+y
Hence, the area bounded by the curve y = f(x)-, X-axis and the
lines x = 0 and x = 1 is given by
= J,
2 < x £3
-x-4,-3<x<0
(1-x)2, for0<x£l/3
/(X)
-3£x£0
ri,
2x(l-x)dx+JM(?)&
17
= — sq unit
27
o
-3
2
13,1)
>x
3
------- <-2
• Ex. 37 Find the ratio in which the curve,
y = [- 0D1 x4 - 0.02 x2 ] [where, [■] denotes the greatest inte­
Graph of f (x)
+y
ger function) divides the ellipse 3x2 + 4y2 =12.
Sol. Here,
y = [- 0.01 x* - 0.02 x2]
Le.
, y = -1, when - 2 < x < 2
•
-2
0
y = -1 cut the ellipse 3x2 + 4y2 = 12
Atx2=?
3
- x = ± —j=or
'3
^(0.-2)
Graph of f (|x|)
y = (|x|)
2
>x
199
Chap 03 Area of Bounded Regions
■ fy
• Ex. 40 Let O (0,0), A (2,0) and B J1, -4= | be the vertices
I V3 J
(3,1
o
2
of a triangle. Let R be the region consisting of all those
points P inside A OAB which satisfy
d (P, OA) < min {d (P, OB), d (P, AB)}, when ‘d’ denotes the
distance from the point to the corresponding line. Sketch the
region R and find its area.
[IIT JEE 1997]
------- H
3
Sol. Let the coordinate of P be (x, y).
Equation of line OA = y = 0
Equation of line OB = fly = x
Graph of |f (x)|
+y
Equation of line AB = fly = 2 - x
1-3
I
--------- 1-
2
0
(-3-1)k^
3
>X
d(P, OA) = Distance of P from line OA = y
d(P, OB) = Distance of P from line OB =
I
d(P, AB) = Distance of P from line AB
I (0.-4)
——
_2
V3y + x — 21
2
Graph of g(x)
Since, | f(x) | is always positive.
^) = /(|x|)-|/(x)|
where the graphs could be drawn as shown in above figures.
From the graph, required area
1
fl1
A
23
23
= -(1 + 4) X3+ - x2 x 4 + 0 = — sq units
2
\22
J
22
• Ex. 39 Let ABC be a triangle with vertices
A=(6,2 (V3 +1)), B a (4,2) andC = (S, 2). Let R be the region
Given, d(P, OA) < min {d(P, OB), d(P, AB)}
y £ min ■
consisting of all those points P inside A ABC which satisfy
d (P, BC) > max {d (P, AB), d (P, AC)}, where d (P, L) denotes
the distance of the point P from the line L Sketch the
region R and find its area.
So/. It is easy to see that ABC is an equilateral triangle with side
of length 4. BD and CE are angle bisectors of angle B and C,
respectively. Any point inside the A AEC is nearer to AC
than BC and any point inside the A BDA is nearer to AB than
BC. So any point inside the quadrilateral AEGC will satisfy
the given condition. Hence, shaded region is the required
region, whose area is to be found, shown as in figure
A(6,2(J3+1))
| V3y - x| \ f3y + x-2\
2’
2
\^y -x|
y*
2.
...(i)
\f3y +x-2|
and
y£
f
. Case I If
y<
2
...(ii)
\Jiy -x|
2
i
(••• fly - x < 0)
(2 + f3)y£x
=>
y£(2-f3)x
E.
P
y £ x tan 15°
G
(4. 2)
X
A (2,0)
0 (0,0)
C(8, 2)
Thus, required area = 2 Area of AEAG = 2 x | AE x EG
= - ABx-CE=-x4x J4z-22
2
3
6
v
4^3
= —— sq units
...(iii)
(v y = x tan 15° is an acute angle bisector of Z.AOB)
\yfiy + x-21
Case n Ify <
2
2y <2-x~fiy
(Le. V3y + x-2<0)
=>
=>
(2 + ^) y < 2 - x
y<.-(2-fl) (x-2)
y
—(tan 15°) (x-2)
...(iv)
[v y = (x - 2) tan 15° is an acute angle bisector of CA]
200
Textbook of Integral Calculus
From Eqs. (iii) and (iv), P moves inside the triangle as shown’in
figure.
Sol. We have, x2 + y2 - 4x + 2y + 1 = 0
or
(x-2)2 + (y + l)2 = 4
•(i)
Yf
Ux-2)2 + (y+1)2 = 4
O (0,0)
-S X X X X XX,
A
(2.0)
c
(1.0)
\ (4.0)
§\(2+/3,0)
o Z‘(2-jS?0)
.""•.7
As
=>
'
unit
Area of shaded region = area of AOQA = - (base) x (height)
2
= | (2) (1 tan 15°) = tan 15°
= (2-<73) sq units
• Ex. 41 A curve y = f(x) passes through the origin and
lies entirely in the first quadrant. Through any point P(x, y)
on the curve, lines are drawn parallel to the coordinate axes.
If the curve divides the area formed by these lines and coor­
dinate axes in m: n,find f(x).
X \
.
for 05x54
Now, for 0 5 x < 4,
x3
x
x3
x
----- 4- —
=0
05 — + —<1
100 35
100 35
So, we have to find out the ratio in which X-axis divides the
circle (i).
Now, at X-axis,
y=0
(x-2)2=3
So,
So, it cuts the X-axis at (2 -
0) and (2 + -J3, 0).
Therefore, required area, -4 = J 2
(^4-(x-2)2 -1) dx
47C -35/3
f(t) dt
Sol. Area of (OAPB) = xy, Area of (OAPO) = j
/X3
f /M=(ioo+35TO
ZQOB = ZOBQ = 15°, AOQB is an isosceles triangle
OC = AC =
>x
3
/y=f^
H
B
P = fry)
_ 471-3^3
4rt - A
8tc + 3^3
• Ex. 43 Area bounded by the line y = x, curve
>X
A
Required ratio =
A
y = /(x), ( /(x) > x, V x > 1) and the lines x = l,x = t is
(t + ^1 + t2) -(1 + V2) for all t > 1. Find f(x).
Sol. The area bounded by y = /(x) and y = x between the lines
Therefore, area of (OBPO) = xy - J * f(t) dt
x = 1 and x = t is
According to the given condition,
(t + -Jl + t2) -(1 + -J2).
xy “J/
V/(0A
n xy = (m + n) J
m
n
j (f(x) - x) dx. But it is equal to
So, . J ‘ (fix) - x) dx = (t + yjl + t2)-(1 + 5/2)
Differentiating both the sides w.r.t. t, we get
f(t) dt
=> f(t) = l + t +
yjl + t2
Differentiating w.r.t. x, we get
n
x^~ + y ] =(m + n) f(x) =(m + n) y,asy = f(x)
dx
)
— — = — => — • (log x) = log y - log c, where c is a constant
n x
y
n
=>
y=cxm,n
• Ex. 42 Find the ratio of the areas in which the curve
y=
x3
x
----- 4---100 35
divides the circlex2 + y2 -4x + 2y +1=0
(where, [.] denotes the greatest integer function).
or
f(x) = 1 + X .+
t
■jl + t2
X
JT+ x.2‘
• Ex. 44 The area bounded by the curve y = /(x), X-axis
and ordinates x = 1 and x = bis(b-1) sin (3b + 4) ,find f(x).
Sol. We know that the area bounded by the curve y = f(x), X-axis
and the ordinates x = 1 and x = b is j /(x) dx.
From the question; |
f(x) dx = (b -1) sin (3b + 4)
Differentiating w.r.t. b, we get
f(b) • 1 =3 (b -1) cos (3b + 4) + sin (3b + 4)
f[x) = 3 (x -1) cos (3x + 4) + sin (3x + 4)
Chap 03 Area of Bounded Regions
201
* Ex. 45 Find the area of region enclosed by the curve
• Ex. 46 Let f(x) be a function which satisfy the equation
(x-y )2 (x+y
(x + y)) 2
-—2— ++ —^2
b2— = 2 (o > b), the line y = x and the posi­
f(xy) = f(x) + f(y)for allx>0,y>0 such that f '(1) = 2.
Find the area of the region bounded by the curves
y - f{*)> y -1 x3 - 6x2 +11x - 61 and x = 0.
tive X-axis.
(x — y )2 (x + y )2
Sol. The given curve ----- ----- + - ---- ----- = 2 is an eUipse major
a*
b*
and minor axes are x - y = 0 and x + y = 0, respectively.
The required area is shown with shaded region.
y
Sol. Take x = y = 1 => /(I) = 0
__
1
Now, y = —
x
1
0=/ X- — = jf(x) + /
-/(x)
X
M+f
yy=x
■(i)
/(x+h)-f(x)
h
x+h A
f x )
= Um
h-t 0
h
Now, /'(x) = Um
h—> 0
X
/O
Instead of directly solving the problem we can solve equivalent
problem with equivalent eUipse whose axes are x = 0 and y = 0.
The equivalent region is shown as (OA' S' O) where the
x2 y 2
equation of eUipse is — + = 1.
a
b
Required area = Area (AOA' A" + A' A!' &)
x
ab
ab
where,
A' =
+ b2 >
\
AOA' A”=- x
ai>
x-_^
ab—=i^ a„2.b 2
Area
(0
_2
2
f 1 + - -/(I)
x )
_HD
h
h —> 0
X
= Um
a
/'(x) = X
/(x) = 2 log x + c
/(x)=21ogx
• a
ab
a
= ba
b
a
y=|x3-6x2+11x-6|
1
0
*X
dx
■yja2 - x2 dx
C
2
“
2
. -1 X
- x + — sin
2
2
a2 K
0 +------2 2
Hence, required area
a
—
a
= f1 (x3— 6x2 + llx-6)dx + [ ° eyl2 dy
ib/yja2 + b*
ab
2 ^a2 + b2
2
it a
4
2
a2 . -i
------sin
2
rc ab ab . _j
------ sin
4
2
b
J°2 + bZ >
2(a2 + b2)
Hence, required area = Sum of Eqs. (i) and (U)
it ab ab .
=------------- sin
4
2
r
=
b
J
,
3
..
2
X 1
6x3 llx2 ,
--- ------- +--------- 6x
+ 2(e>,'2)’_
3
2
/ 0
I4 -
= (1-2 + 11-6
b
^Tb\ ■
I4
,2
-(0) + 2(e° - e"”)
1
11 o „
1
= — +----- 8 + 2 = — sq unit
4 2
4 *
a3b .
2(a2 + b2)
a2b2
b
J0
a2b2
a2 + b2^
a2
a2 . -i
—-sin
b
a
[since, /(I) = 0 => c = 0]
Thus, /(x) = 2 log xandy = | x3 - 6x2 + 11 x — 61 could be
plotted as
y = 2logx
1
b
[v/(l) = 0]
— • X
2 >/7+7 Ja2+b2 2 \a +b J
Area of A'B* A* = J’ ab
ab
[using Eq. (i)]
...(ii)
• Ex. 47 Find the area of the region which contains all the
points satisfying condition [ x - 2y | +1 x + 2y | < 8 andxy 2.
x
Sol. The line y = ± — divide the xy plane in four parts
Region I 2y - x < 0 and 2y + x
So that, | x - 2y | + | x + 2y | < 8
=>
0
(x-2y) + (x + 2y) <8 => 0£x<4
202
Textbook of Integral Calculus
Region II 2y - x I 0 and 2y + x a 0
Region HI 2y + x < 0,2y - x £ 0
So that, | x-2y | + | x + 2y | <8
-(x-2y)-(x + 2y)^8 => -4£x<0
Thus,
g(x) = max {x2, /(x), | x |}
which could be graphically expressed as
3
Region II
2y-x£0, 2y+xi0
y
|x} - -, if x <5 I
2
0, if xel
=
So that,
| x - 2y | +1 x + 2y | S 8
=>
- (x - 2y) + (x + 2y) < 8 => 0 S y < 2
y=x2
y=^
f
/
y=|x
>rY=(xl
max
-^=2
Region I
y
2y-x < 0,2y+x 2 0
Region III
2y-x £ 0, 2y+x £ 0
-2
x2, - 2 <, x < - 1
-x, - 1 < x < - 1 / 4
Region IV
2y-x <, 0, 2y+x £ 0
Region IV
So that,
Clearly,
+ -,-l/4<x<0
2
x, 0 < x < 1
glx)
2y + x < 0,2y - x < 0
|x-2y| + |x + 2y|<8=>-2<y<0
x2, l£x<2
Here, all the points lie in the rectangle.
t 2
Hence, required area = j
Ya
r -i ,
g(x) dx
o
r -i/4
.
(-
1
-1/4
(0. 2) \D
<
2
dx
+i«x,ix+r
Jo
— A(1,0)
B (4,0)
=
<3
units
Also, the hyperbola xy = 2 meets the sides of the rectangle at
the points (1,2) and (4,1 / 2) in the 1st quadrant graphically.
Hence, required area
= 2 (Area of rectangle ABCD - Area of ABEDA)
4 2
'l
= 2 3X2-/, — dx | = 2 (6 - 2 log 4) sq units
x
)
'xf
-1/4
2
x2 1
+ --+- X
2
<2>
2
\°
+
/-1/4
x2 dx
Y I x^ X2 275
= — sq
+
<2 Jo
X 3 Ji
48
• Ex. 49 Find the area of the region bounded by
y = fix), y = | glx) | and the lines x = 0, x = 2, where f, g
are continuous functions satisfying
fix + y) = fix) + fly) -8xy,Wx,yeR
and g(x+y) = g(x) + g(y) + 3xy(x+y),Yx,yeR
Also, /'(0) = 8 and g'(0) = -4.
Sol. Here, f(x + y) = f(x) + fly) - 8xy
• Ex. 48 Consider the function
/(*) =
x — [ xl---- if x C I
2’
, where [.] denotes the great-
0, if xel
est integer function and I is the set of integers. If
g(x) = max {x2, f(x), |x|},-2<x<2, then find the area
bounded by glx) when - 2 < x < 2.
Sol. Here,
fix)
x - [x] - -, if x e I
2
0, if xel
Replacing x, y -» 0, we get /(0) = 0
Now,
flx+h)-f(x)
fix) = lim ------------------------- = lim
h—> 0
h------------ y-> o
fix) + fly)-^xy-f(x)
= lim
y—> 0
y
fly)
8xy
= lim
0
. y
y
= lim
y-> 0
(fly)
I i
y
- 8x (using (/Hospital's rule)
= /'(0)-8x=8-8x
[given, /'(0)=8)
Chap 03 Area of Bounded Regions
=>
/'(x)=8-8x
Integrating both the sides, we get
_f(x) = 8x - 4x2 + c
As
f(0) = 0=>c = 0
=3
/(x)=8x-4x2
=>
y = x2 and y = it
...(0
y-> 0
= lim
y
g(x) + g(y) + 3x2 y + 3xy2 - g(x)
y—> 0
y
g(y) | y(3x2 4- 3xy)
= lim
y-> 0
1
.
i.e. when x2 = it or (x = - -fit to x = fit)
.
y
• Ex. 51 Sketch the graph o/cos-1 (4x3 -3x) and find
the area enclosed between y=0,y = f(x) and x > -1 / 2.
Sol. Here,
.
y
Let
+ 3x2 = - 4 + 3x.2:
=>
f(x) = cos-1 (4x3 - 3x)
and
x = cos 0
0 £ 0 < it
f(x) = cos-1 (4 cos3 0 - 3 cos 0)
= cos-1 (cos 30); 0
Points where f(x) and g(x) meets, we have
/(x) = g(x) or 8x - 4x2 = x3 - 4x
=>
3k
30
O<30^k
30,
[as g(0) = 0] ...(u)
g(x) = x3 - 4x
=>
(K - X2) dx = I KX - —
= it (fit + fit) - - (it fit + it fit) = — fit
33
g/(x) = —4 + 3x2
=>
r3
(
Required area = J
+
g-(x)=lim
g'(x) = lim
=
sec-1 [- sin2 x] = sec-1 (- 1) = K
Thus, to find the area bounded between
Also,
g(x + y) = g(x) + g(y) + 3xy (x + y)
Replacing x, y => 0, we get g(0) = 0
Now,
203
/(x) = 2k - 30, It < 30 £ 2K
30 - 2k, 2k < 30 < 3k
x = 0,2,-6
If
3 cos-1 x,
1/2SXJS1
2k -3 cos-1 x,
-l/2£x<l
3 cos-1 x - 2n, -1 S x < -1 /2
,y=M
-3/fl-x2,
.....
o7
:_____!\/
/\(2,0)
f(x) =
>X
1/2<x<1
3/yjl-x2, -1/2<x<1/2
It _ w2
_Q /
'
' '-
-3/Jl-x2, -l<x<-l/2
Now,
I g(x) | =
-3x
(l-x2)3/2’
x3 - 4x, x e [- 2, 0] U (2, oo)
4x - x3, x e [- oo, - 2] u (0,2)
and
3x
— 1/2<x<1/2
(l-x2)3/2’
f"W
Area bounded by y =/(x)andy = | g(x) | between x = 0 to x = 2
-3x
-1 <x<-l/2
(1-x2)3'2’
= f 2 {(8x - 4x2) - (4x - x3)} dx
Jo
4
_ 2
= Jo (x3 - 4x2 + 4x)dx = - squnits
1/2<x<1
Thus, the graph for f(x) = cos-1 (4x3 - 3x) is
:
IX
:
• Ex. 50 Find the area of the region bounded by the curve
y-x2 andy = sec”1 [- sin2 x], where [.] denotes the great­
:
n
est integer function.
Sol. As we know, [- sin2 x] = 0 or -1. But sec-1 (0) is not defined.
k/2
+y
-1/2
0
■1/2
1
*X
-►y = it
Thus, required area = J
0
X
=
f(x) dx
r1/2 ,
(2k -3cos xjdr + j^^(3 cos-1 x)dx
J -1/2
204
Textbook of Integral Calculus
= 2k-3
. _!
r1/2 ( n
I ---- sin x | dx + 3 fi1 2 (cos-1 x)dx
/
J11
2
(iii) For 1 = - a, we have y2 = 4a (x + a)
J -1/2
1/2
J’
as j -1/2 sin
cos"1 x dx
= - +3
J 1/2
2
= —+ 3 ■ (xcos-1 x)}/2 + ji/2
2
X
i
x dx = 0
= dx
-x.2:
x2 = 4a (y + a)
The point of intersection in the 1st quadrant
P = ((2 + 2>/2) a, (2 + 2V2) a)
Required .area = 2 (area of AOPQ - area APQA )
ofAOPQ = ~ • (2 + 2 -J2)2 a2 =2 (1 + J2)2 a2
3'^3
>
On solving, we get =-----sq units
2
(2+ 2^2) a
Area of APQA = J
2a
• ,Ex. 52 Consider two curves y2 - 4a (x - X) and
(2 + 2V2) a ( X 2
\
I 4a
J
ydx = j 2a
----- a
dx
-] (2+ 2^2)a
x2 =. 4a (y - X), where a > 0 and X is a parameter. Show that
1
4a
(i) there is a single positive value ofX for which the two
curves have exactly one point of intersection in the 1st
quadrant find it.
(ii) there are infinitely many negative values of X for which
the two curves have exactly one point of intersection in
the 1st quadrant.
(Hi) ifX=-a, then find the area of the bounded by the
two curves and the axes in the 1st quadrant.
(2+ Zjtya
3
~a(Xk
2a
=-^- [(2 + 2^2)3 - a3 -Sa3]-2^2 a,2
12a
( 4^2 + 12
a2
3
Required area = 2 2 (1 + V2)2 +
4^2 + 12
3
a2
4
= — (15 + 8 V2)a2:
Sol. The two curves are inverse of each other. Hence, the two
curves always meet along the line y = x.
Consider, •
y2 = 4a (x -1) and put x = y
• Ex. 53 Let f(x) be continuousfunction given by
y2 - 4ay + 4al = 0
4a ± 4 Ja2 - aX
I —-------------------*--------------------
y
= 2a ± y/a2^- aX
/(*) = ’
x2 + ax + b, |x[>1
2
Since, y is real => a22 - al I 0 or X
a
(i) If 0 < X < a, then there are two distinct values of y and
|x|<1
2x,
Find the area of the region in the third quadrant bounded
by the curves x = -2y2 and y = f(x) lying on the left of the
both 2 (a + yja2 - aX) and 2 (a - -Ja2 -al) are positive,
line 8x +1 = 0.
i.e. both points lie in the first quadrant.
Sol. Given, a continuous function f (x), given by
[iitjee 1999]
If X = a, then y = 2a only, i.e. only one point of
intersection (2a, 2a).
+y
Hence, there is exactly one point of intersection in 1st
quadrant for X = a. It is infact the points of tangency of
the two curves.
I
(ii) If X < 0, then y = 2 (a + ^a2 - aX) > 0 and
*x
y = 2 (a - fa2 - al) < 0. i.e. the only point of intersection
f
4
is in the first quadrant, the other in the 3rd quadrant.
Hence, there are infinitely many such values.
x2 = 4a(y + a)
-2
y=x
xY
x = -2 x=-1 x=-
y2 = 4a(x + a)
P
B
(0,2a)
f& =
Q
/ A
(2a. 0)
2x,
8
1X|<1
x2 + ax + b, | x | > 1
x2 + ax + b, — 00 < X < ~ 1
+X
i.e.
2x.
x2 + ax + b,
f is continuous.
“1^X<1
1 < X < oo
205
Chap 03 Area of Bounded Regions
It is continuous at x = -1 and x = 1.
• Ex. 54 £et[x] denotes the greatest integerfunction.
(-l)2+a(-l)+b = 2(-l)
and
Draw a rough sketch of the portions of the curves
x2 =4 [Vx] y andy2 = 4 [jy ] x that lie within the square
(I)2 + a (1) + b =2 (1)
{(x, y) 11 < x < 4,1 < y < 4}. Find the area of the part of the
square that is enclosed by the two curves and the line
x + y = 3.
1-a + b = -2
and
1 + a + b =2
and
-a + b = -3
a+b=1
Sol. We have,
b = -1 and a = 2
/(*)
Now,
1 £y £4
1 £ -fx £ 2 and 1 £ fy £ 2
=>
x2 + 2x -1, when - °° < x < -1
2x,
when -1^x51
1 £ x £ 4 and
=> [Vx] = 1 and [fy ] = 1 for all (x, y) lying with in the square.
y = x2 + 2x -1 = (x + I)2 - 2
or
(y + 2) = (x + l)2
We need the area of the region in third quadrant bounded by
the curves a = - 2y2, y = f(x) lying on the left of the line
8x+l = 0.
2
Cuts the T-axis at (0, -1) and the X-axis at
(-1-V2, 0) and (-1 + 41, 0).
1
When x = 1 and y = 2
-+
2
0
Solving x = - 2y2 and (y + 2) = (x + I)2, we get
(y + 2) = (-2y2 + I)2
For
y = -1, 4y4 - 4y 2-y-l=0
Thus, required area
At
y = -i,x = -2
= // (2V^-3 +
-J.'1 {-^-f-(x+1>!+21*
+J
— 1/8
-1
{“iR’2*}*
1-1/8
-j
(-x/2)’/2
3
;(i/2)
■iK-i)”
.
(x + 1)- + 2X’
3
*
(-x/2)a/2
(»!)’
3
|(l/2)
4
3
3-2V2
3
1
3
4
1
257
= - +----- = — sq units
3 3(64) 192
4
3-2>/2 + + 3 (64)
4 ^3/2
2 A I3
(¥-■)-(
12
' 2
4 o 1 1
4
—
-3 + 2 )
3
12 )
3
8
12
19
= — sq units
-1/8
-X2
-i
• Ex. 55 Find all the values of the parameter a (a >1) for
which the area of the figure bounded by pair of straight lines
y2 -3y + 2 =0 and the curvesy =[a] x2,y =^[a] x2 is
|
64 ‘
= ._• -2-1
13
r22 A2
«-3x+2
4
I
8>/2
.3/2
3
(a
4 [ 2-Vx—— dx
I 3
J-i
•■{W
2;
2x2
2
3
J-2
Ct
X
4
=>
x2 = 4y and y2 = 4x when 1 £ x, y £ 4 which
could be plotted as;
4y4 -4y2 -y -1 = 0
Required area
3
x2 = 4[Vx]y and y22 = 4[fy]x
Thus,
4y4 -4y2 + 1-y-2 = 0
=3
(4.4)
—♦—
3
(y + 2)=(x + l)2
=>
(1.4)
x+y=34
64 J
greatest, where [.] denotes the greatest integer function.
1
Sol. The curves y = [a] x■2‘and y = - [a]- x2 represent parabolas
which are symmetric about T-axis.
The equation y2 -3y + 2 = 0 gives a pair of straight lines
y = 1, y = 2 which are parallel to X-axis.
206
Textbook of Integral Calculus
Thus, the area bounded is shown as
kY
y = [a] x2 y =J [a] x2
-Lt--------tL------= 2
>y = 1
X
• Ex. 56 Find the area in the first quadrant bounded by .
[x] + [y ] = n, wheren&N andy = / (where, ieN Vi <n +1),
[ • ] denotes the greatest integer less than or equal to x.
Sol. As we know, [x] + [y] = n => [x] = n - [y ]
Wheny = 0, [x] = n=>n^x<n + l
Wheny = 1, [x] = n- l=>n-lSx<n-2
When y = 2, [x] = n - 2
=>
n - 2 < x < n - 3.... and so on.
Wheny = n, [x] = 0 => 0 < x<l
which could be shown as
From the above figure,
r y°2
Required area = 2 I
*
{Xi-x^dy
n= 1—
n—
I
J.i
■rit
L—!.
i
___ i____ t_____ i,
I
I
L.___ I.
I
I
d >I
llllHIIIL?
c ’'
2 (72 -1)2
_ 4 (72 - 1)
•(23/2 -1)
”3 [a]
_ 4 (5 - 72 - 272) = 4 (5-3^2)
“3
[a]
3
[a]
Area is the greatest when [a] is least, i.e. 1.
Area is the greatest when [a] = 1
Hence,
a e [1,2)
3/2. 2
4
3
2
1
I
I
I
I
I
J
J
L
I
1
”!—
f—!•
1
2
r
4
7"t
I
I
'!—i
i
T
I
I
3
4
■>
n n+1
From the above figure,
= (n + 1) area of square ABCD = (n + 1) • 1
Required area = (n + 1) sq units
g Area of Bounded Regions Exercise 1:
Single Option Correct Type Questions
1. A Point P(x, y) moves such that [x + y +1] = [x]. (where
[•] denotes greatest integer function) and xe (0,2), then
the area represented by all the possible positions of P, is
(a)V2
(b)2^
(c)4^
(d)2
2. If/: [-1,1]
1 1
.. .
./(*) = -------- . The area bounded
2 2J
1 + x2
byy = f-1(x), X-axis, x = -,x = --is
2
2
(a)-Ioge
(b) log f|)
(c);iog;
<-'*
(djl'ogfl)
o
\"Z
3. If the length of latusrectum of ellipse
7. Let ‘a’ be a positive constant number. Consider two
curves C1:y = ex,C2:y = ea-x. Let S be the area of the
5
part surrounding by
C2 and the Y-axis, then lim —
<*-*0 a2
equals
(a) 4
(b) 1/2
(c) 0
(d) 1/4
8. 3 points 0(0,0), P (a, a2), Q(-b,b2)(a>0,b>0)aie on the
parabola y = x2. Let Sj be the area bounded by the line
PQ and the parabola and let S2 be the area of the AOPQ,
the minimum value of S1/S2 is
(a) 4/3
(b) 5/3
(c) 2
(d) 7/3
9. Area enclosed by the graph of the function y = In2 x -1
Ex :4(x + y-l)2 + 2(x-y + 3)2 =8
lying in the 4th quadrant is
x2
v2
and E2 :----- 1= 1, (0 < p < 1) are equal, then area of
P
p
ellipse E2, is
(a)e
(b)e
(c) 2 [ e + -
(d)4 e —
I
e.
1
e.
3
<*>f
(c)^
<d>74 ,
4. The area of bounded by the curve
2017
4|x-2O17zo
171| + 51 y - 20172017
Z017|< 20, is
(b) 50
(d) 30
(a) 60
(c) 40
5. If the area bounded by the corve y = x22 +1, y = x and
the pair of lines x2 + yz +2xy - 4x - 4y + 3 = 0is K
units, then the area of the region bounded by the curve
y = x2 + l,y = t/x -1 and the pair of lines
(x + y -1) (x + y - 3) = 0, is
(a) K
(b) 2 K
%
(c) —
(d) None of these
2
6. 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 f(x) < g(x) for
2<x<4.If £[y(x)-g(x)]dx = 10
10. The area bounded byy = 2-|2-x| and y = — is
|x|
. . 4 + 31n3
„ . 19 „ . n
(a —
(b) — -3 1n2
2
8
3
1
(c)- + ln3
(d)- + In3
2
2
11. Supposeg(x) = 2x + landh(x) = 4x2 + 4x+5and
h(x) = (fog)(x). The area enclosed by the graph of the
function y = /(x) and the pair of tangents drawn to it
from the origin, is
(a) 8/3
(b) 16/3
(c) 32/3
(d) None of these
12. The area bounded by the curves y = -V-x and
x = -■j-y where x,y < 0
(a) cannot be determined
(b) is 1/3
(c) is 2/3
(d) is same as that of the figure bounded by the curves
y = 7-x; x < 0 and x = J-y ;y < 0
13. y = f(x) is a function which satisfies
(i)f(0) = 0 (n)r(x) = f(x)and (iii)/'(0) = l
and
(g(x)-f(x)] dx = 5, then the area between two
curves for 0 < x < 2, is
(a) 5
(c)15
(b)10
(d) 20
Then, the area bounded by the graph of y = f(x), the
lines x = 0,x-l = 0andy + l = 0,is
(a) e
(b)e-2
(c) e -1
(d) e +1
208
Textbook of Integral Calculus
14. Area of the region enclosed between the curves
x = y2-1 and x =|y|/l-y2 is
(b) 4/3
(d) 2
(a) 1
(c) 2/3
•4-
WT-*
(c)^-4
»T
15. The area bounded by the curve y = xe“x;xy = 0and
x = c where c is the x-coordinate of the curve’s
inflection point, is
(a)l-3e~2
(b) 1 — 2e~2
(c)l-e-2
(d)l
21. If/(x) = x-land g(x) = |/(|x|)-2|,then the area
(a) 1(4^2-5)
(b)j(4^2-3)
16. If (a,0);a> 0 is the point where the curve
y = sin 2x — -73 sin x cuts the X-axis first, A is the area
bounded by this part of the curve, the origin and the
positive X-axis, then
(b) 4A + 8 sin a = 7
(a) 4 A + 8 cos a = 7
(d) 4A-8 cos a = 7
(c) 4A-8 sin a = 7
(c)?(W2-3)
(d)|(4^-5)
2
17. The curve y = ax + bx + c passes through the point (1,2)
and its tangent at origin is the line y-x. The area
bounded by the curve, the ordinate of the curve at
minima and the tangent line is
(a)l
(b)l
(c)|
(d)l
24
12
8
6
18. A function y = f(x) satisfies the differential equation
dv
— - y = cos x - sin x, with initial condition that y is
dx
bounded when x -> «>. The area enclosed by
y = f(x),y = cos x and the K-axis in the 1st
quadrant
(a) -Ji-1
(b)/2
(c)l
(d)l//2
19. If the area bounded between X-axis and the graph of
y = 6x-3x2 between the ordinates x = 1 and x = a is 19
sq 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 the above
20. Area bounded by y = /-1(x) and tangent and normal
drawn to it at the points with abscissae 7t and 2tc, where
/(x) = sin x-xis
bounded by y = g(x) and the curve x2 - 4y+8 = 0 is
equal to
Q
22. Let S = < (x,y): ^3x
< oL
x(3x—2)
$' = {(x,y)€ AxB:-l£A£l,-l£B£l},
then the area of the region enclosed by all points in
SnS'is
(a) 1
(c)3
(b) 2
(d)4
23. The area of the region bounded between the curves
y = e||x|ln|x||,x2 +y2 -2(|x|+|y|)+l£0andX-axis
where | x | £ 1, if a is the x-coordinate of the point of
intersection of curves in 1st quadrant, is
(a) 4^Jq exlnxdx + J(1 - Jl-(x-l)2)dx
(b) 4^Jo exlnxdx+ £ (1 --^l~(x~l)2)dx
exlnxdx + J (1 -71-(x-l)2)dxj
(c)
(d) 2^jo exlnxdx+
(1 —Jl-(x-l)2)dxj
24. A point Plying inside the curve y = \2ax - x2 is moving
such that its shortest distance from the curve at any
position is greater than its distance from X-axis. The
point P enclose a region whose area is equal to
_„2
„2
(a)—
(b) —
(C)T
(d)
3tc-4
6
a2
Chap 03 Area of Bounded Regions
209
§ Area of Bounded Regions Exercise 2:
More than One Option Correct Type Questions
25. The triangle formed by the normal to the curve
“ /(*) = x - ax + 2a at the point (2, 4) and the coordinate
axes lies in second quadrant, if its area is 2 sq units, then
a can be
(a) 2
(b) 17/4
(c) 5
•
(d) 19/4
26. Let/and g be continuous function on a< x < b and set
p(x) = max {/(x), g(x)} and q(x) = min{/(x), g(x)},
then the area bounded by the curves y = p (x), y = q (x)
and the ordinates x = a and x = b is given by
(a) JaV(x)“5(x)|dx
(b)
(c) jj/W-g(x)}dx
(d)j
. . Ja‘(p(x)-9(x))dx
.
27. The area .bounded by the parabola y = x 2 -7x + 10and
X-axis equals
(a) area bounded by y = -x■‘2 + 7x -10 and X-axis
(b) 1/6 sq units
(c) 5/6 sq units
(d) 9/2 sq units
,
X
2
V
2
28. Area bounded by the ellipse — + ~ = 1 is equal to
(a) 6k sq units
(b) 3k sq units
(c) 12k sq units
x2 y 2
(d) area bounded by the ellipse — + = 1
29. There is a curve in which the length of the perpendicular
from the origin to tangent at any point is equal to abscissa
of that point Then,
(a) x2 + y2 = 2 is one such curve
(b) y2 = 4x is one such curve
(c) x2 + y2 = 2cx (c parameters) are such curves
(d) there are no such curves
□ Area of Bounded Regions Exercise 3:
Statement I and II Type Questions
■ Direction (Q. No. 30-34) For the following questions,
choose the correct answers from the codes (a), (b), (c) and
(d) defined as follows :
32. Statement I The area of region bounded parabola
32
y2 = 4x and x2 = 4y is — sq units.
(a) Statement I is true, Statement II is also true; Statement II
is the correct explanation of Statement I
(b) Statement I is true, Statement II is also true; Statement II
is not the correct explanation of Statement I
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
Statement II The area of region bounded by parabola
y2 = 4ax and x2 = 4by ts^ab.
30. Statement I The area of the curve y = sin2 x from 0 to
n will be more than that of the curve y = sin x from 0 to
n.
Statement II x2 > x, if x > 1.
31. Statement I The area bounded by the curves y = x2 -3
and y = kx+2 is least if k - 0.
Statement II The area bounded by the curves
y-x2-3 and y = kx+2isyk2 +20.
33. Statement I The area by region [x + y| + |x - y | s 2 is 8
sq units.
Statement II Area enclosed by region
|x+y| + |x -y | 2 is symmetric about X-axis.
34. Statement I Area bounded by y = x (x -1) and
y = x(l-x)is
Statement II Area bounded by y - f( x) and y = g( x) is
?b
(f(x)-g(x))dx is true when /(x) and g(x) lies
above X-axis. (Where a and b are intersection of
y = /(x)andy = g(x)).
210
Textbook of Integral Calculus
H Area of Bounded Regions Exercise 4:
Passage Based Questions
Passage I
(Q. Nos. 35 to 37)
Let f(x) =
ax +bx+c
x2 +1
such that y = -2 is an asymptote of the
curve y = f(x} The curve y = f(x)is symmetric about Y-axis
and its maximum values is 4. Let h (x) = /(x) - g(x) where
/(x) = sin4 Ttxandg(x)= logc x Letx0,x{ ,x2,....xn + 1 be the
roots of f(x) = g(x) in increasing order.
35. Then, the absolute area enclosed by y = /(x) and
(c) f(x) is increasing on (-1,1) but has neither a local
maximum nor a local minimum at x = 1
(d) f(x) is decreasing on (-1,1) but has neither a local
maximum nor a local minimum at x = 1
40. Letg(x) = J* - — dt. Which of the following is true?
J 1 + t2
(a) g'(x) is positive on (-°°,0) and negative on (0,“°)
(b) g'(x) is negative on(-<»,0) and positive on (0,«>)
(c) gT*) change sign on both(-<»,0) and (0,<»)
(d) g'(x) does not change sign on(-«>,oo)
y = g(x)is given by
'
n
(a)^ j **(-l)r-/i(x)dx
(Q. Nos. 41 to 43)
r=0
Computing areas -with parametrically represented boundaries:
If the boundary of a figure is represented by parametric
equations i.e. x = x (/), y = y (7 ), then the area qf the figure is
evaluated by one of the three formulas
r=0
(c)2£ f*-'*,(-l)r-h(x)dx
r= 0
“ij.C
Passage HI
S
(-1)'
h(x)dx
Ja
<5 = J^x(/)-y'(/)tZ/
a
36. In above inquestion the value of n, is
(a) 1
(b) 2
(c) 3
(d) 4
37. The whole area bounded by y = /(x),y = g(x) x = 0 is
(a) 11/8
(c) 2
(b) 8/3
(d)13/3
Passage II
where a and £ are the values of the parameter ‘ f
corresponding, respectively to the beginning and the end ofthe
traversal of the curve corresponding to increasing1t\
z
41. The area enclosed by the asteroid
313. Which of the following is true?
(a) (2 + o)2r(l) + (2-a)7'(-l) = 0
(b) (2-a)!r(l)-(2 + a)!r(-l) = 0
(c) /W'(-l)=(2-a)2
(d) /'(D/'(-l) = -(2 + a)2
39. Which of the following is true?
(a) f(x) is decreasing on (-1,1) and has a local minimum at
x=1
(b) /(x) is increasing on (-1,1) and has a local maximum at
x=l
(-1
2/3
/ \2/3
(a)-a2Jt
4
3
(c)-na2
8
’
=
= liS
\a)
(Q. Nos. 38 to 40)
Consider the function /:(-<», «>)-»(- co, o©) defined by
< x2-ax+l n
_
/(x), = —---------- , 0< a < 2.
x2 +ox+l
x 2/3
(b)— Tta ,2
18
3
(d)-an
4
42. The area of the region bounded by an arc of the cycloid
x = a (t - sin t), y = a (1 - cos t) and the X-axis is
(a)6na2
(b)3na2
(c) 4na2
(d) None of these
43. Area of the loop described as x = •*t
3
, '27
. (b)^
(a)?
/ \27
CT
(d)^
t2
8
Chap 03 Area of Bounded Regions
211
§ Area of Bounded Regions Exercise 5:
Matching Type Questions
44. Match the statements of Column I with values of
Column n.
45. Match the following :
Column I
Column II
Column I
(A) The area bounded by the curve
y = x + sinx and its inverse function
between the ordinates x = 0 to x = 2n is
4s. Then, the value of 5 is
(B) The area bounded by y = x e**1 and lies
(A) Area enclosed by y=|x|, |x|= 1 and (p)
y = Ois
(B) Area enclosed by the curve y - sinx,
(q)
x = 0, x = n and y = 0 is
(C) If the area of the region bounded by
(r)
k
x2 < y and y < x + 2 is —, then k is
2
Column II
(P)
(q)
2
1
Jx| = 1» y= Ois
(C) The area bounded by the curves
(r)
and |y|= 2xis
(D) The smaller are included between the
(s)
curves y/\x\ + ^\y\ = 1 and |x| + |y| = 1 is
16“
5
f
3
equal to
(D) Area of the quadrilateral formed by
tangents at the ends of latusrectum of
x2 >
ellipse of ellipse — + -y = 1 is
(s)
4
27
18
g Area of Bounded Regions Exercise 6:
Single Integer Answer Type Questions
46. Consider f(x)= x22 - 3x + 2 The area bounded by
ly| = iy (lxl)l -
1 is A ^en find the value of3A + 2.
47. The value of c + 2 for which the area of the figure
x = 1 and x = c and X-axis is equal to ~, is
1*1
i
(a) /(0) = 0
(b) has a maximum value of- at x = L
bounded by the curve y = 8x2 - xs; the straight lines
48. The area bounded by y = 2 -12 - x |; y = — is
polynomial of 2nd degree, satisfy the following
condition:
/c-31n3
2
then k is equal to
49. The area of the AABC, coordinates of whose vertices are
A(2, 0), B(4, 5) and 0(6, 3) is
50. A point P moves in XY-plane in such a way that
[| x | ] + [| y | ] = 1, where [ • ] denotes the greatest integer
function. Area of the region representing all possible of
the point P is equal to
51. Let f: [0,1] -» 0, - be a function such that f (x) is a
2
If A is the area bounded by y = f(x);y = /-1(x)and the
line 2x + 2y - 3 = 0 in 1st quadrant, then the value of 24A
is equal to
x, — ►, xe [0,1]. If area
6
bounded by y = /(x) and X-axis, between the lines x - 0
52. Let f(x) = min • sin
and x = 1 is
5(^ + 1)
x, cos
. Then, (a - b) is
53. Let f be a real valued function satisfying
/(- =/(*)-/(?) and lim
kJ/
1x-»0
'’
x
3. Find the area
bounded by the curve y - f(x), the T-axis and the line
y = 3, where x, y e R+.
212
Textbook of Integral Calculus
Area of Bounded Regions Exercise 7:
Subjective Type Questions
+y
54. Find a continuous function ‘ f’,
-4x
)< f(x)<(2x
(x4 -4
x2)</(
x)<(2x 2 -x3)such that the area
bounded by y = /(x), y = x 4 - 4x2, the Y-axis and the
line x = t, (0 < t < 2) is k times the area bounded by
y = f(x),y = 2x2 - x3,Y-axis and line x = t,(0< t<2).
55. Let/(t) = | t — 11 — 11 l + l t + 11, V t G Rand
g(x) = max {f(t):x + l<t<x + 2};V xg RFind g(x)
and the area bounded by the curve y = g(x), the X-axis
and the lines x = - 3 / 2 and x = 5.
I !'•
a2
A
Ay
O
a
y = sinx
P
2p
A3
y = f(x)
between x = a and x = K; i = 2 between x = it and x = 2tc; i =1
56. Let y(x) = minimum {ex,3/2, l+-e“x},0< x<l. Find the
area bounded by y = /(x), X-axis, Y-axis and the line
x = 1.
If Aj = 1 - sin a +• (a -1) cos a, determine the function
f( x). Hence, determine a and A j. Also, calculate A 2 and A 3.
62. Find the area of the region bounded by curve y = 25x +16
57. Find the area bounded by y = /(x) and the curve
2
y =--------, where f is a continuous function satisfying
1 + x2
and the curve y = b.5x + 4, whose tangent at the point
the conditions /(x) • f(y) = f(xy), V x, y g R
andy/(l) = 2,/(l) = L
58. Find out the area bounded by the curve
sin2 x.
"(sin-’VFjdt +J,r^° X (cos-1 Vt) dt (0< x< tu/2)
and the curve satisfying the differential equation
y (x + y 3) dx = x (y3 - x) dy passing through (4, - 2).
59. Let T be an acute triangle. Inscribe a pair R, S of
rectangles in T as shown :
x = 1 make an angle tan “1 (40 In 5) with the X-axis.
63. If the circles of the maximum area inscribed in the •
region bounded by the curves y = x2 - x - 3 and
y = 3 + 2x-x2, then the area of region
y-x2 + 2x + 3<0,y + x2 -2x-3<0ands<0.
64. Find limit of the ratio of the area of the triangle formed
by the origin and intersection points of the parabola
y = 4x2 and the line y = a2, to the area between the
parabola and the line as a approaches to zero.
65. Find the area of curve enclosed by :
|x + y| + |x-y|^4, |x|<l,y> Jx2 -2x+ 1.
66. Calculate the area enclosed by the curve
4<Sxz + y2<2(|x| + |y|).
s
R
Let A (x) denote the area of polygon X find the
maximum value (or show that no maximum exists), of
A (R) + A (S) ,
_
....
,
j n c
— ^ (7.)' »w«ere T ranges over all triangles and R, S
over all rectangles as above.
60. Find the maximum area of the ellipse that can be
inscribed in an isosceles triangles of area A and having
one axis lying along the perpendicular from the vertex
of the triangles to its base.
61. In the adjacent figure the graphs of two function
y ~ f(x) and y = sin x are given, y = sin x intersects,
y = /(*)at -A (a, /(a)); B (7t, 0) and C (2n, 0).
Aj (i = 1,2,3) is the area bounded by the curves y - f\x)
and y = sin x. between x = 0 and x = a, i = 1
67. Find the area enclosed by the curve [x] + [y] = 4 in 1st
quadrant (where [.] denotes greatest integer function).
68. Sketch the region and find the area bounded by the
curves|y + x|^l»|y-x|<l and 2x2 +2y2 =1.
69. Find the area of the region bounded by the curve,
2>| y | + 2*~1 < 1, with in the square formed by the
lines ] x | < 1 / 2, | y | < 1 / 2.
70. Find all the values of the parameter a (a < 1) for which the
area of the figure bounded by the pair of straight lines
y2 -3y+ 2 = 0and the curvesy = [a]x2,y = -[a]x2 is
the greatest, where [.] denotes greatest integer function.
71. If /(x) is positive for all positive values of X and
f'(x)<0,f"(x)>0,\f xg R+,prove that
n
/(l) + Jf fM dx<r^ f(r) < fW dx + /(I).
Chap 03 Area of Bounded Regions
213
§ Area of Bounded Regions Exercise 8:
Questions Asked in Previous 10 Years' Exams
[i] JEE Advanced & IIT-JEE
(b) |111
(0 1
(d) 1
3
(a) -
72. Area of the region {(x, y)} e R2 : y £ ^/l x + 3|,
5y<(x + 9)< 15} is equal to
[Single Correct Option 2016]
(a)l
(b)l
(c)|
(d)|
t>
3
2
3
73. Let F(x) = J
x2 +-
6 2 cos2t dtfor all xe Rand
f: 0, - —> [0, «>) be a continuous function. For
2J
a e 0, - , if F'(a) + 2 is the area of the region bounded*
2
byx = 0,y = 0,y = f(x)andx = a, then/(0)is
[Integer Answer Type 2015]
79. Area of the region bounded by the curve y = ex and
lines x = 0 and y = e is [More than One Option Correct 2009]
(a) e-1
(b) J ln(e +1 -y)dy
(c) e-Joexdx
(d)
80. The area of the region between the curves
ll + sin x
,
1 - sin x
-------andy =
and bounded by the
y=
cos x
cos x
lines x = 0 and x = — is
4
t
parabola y2 = 8x touch the circle at the points P,Q and
4t
the parabola at the points R, S. Then, the area
(in sq units) of the quadrilateral PQRS is
[Single Correct Option 2014]
(b) 6
(c) 9
(d) 15
75. The area enclosed by the curves y = sin x + cos x and
7C
y = I cos x - sin x | over the interval 0, — is
2
[Single Correct Option 2014]
(a) 4(72-1)
(b) 2V2(V2 -1)
(c) 2(72 + 1)
(d) 272(72 + 1)
76. If S be the area of the region enclosed by
[Single Correct Option 2008]
=dt
t2
74. The common tangents to the circle x2 +y2 = 2 and the
(a) 3
In ydy
i
==dt
-t2
= dt
.2
w/7’1
=dt
,2
■ Directions (Q. Nos. 81 to (83) Consider the functions
defined implicity by the equation y3-3y + x=0 on
various intervals in the real line. If x
-2)u(2, °°),
the equation implicitly defines a unique real-valued
differentiable function y =f (x). If x e (-2,2\ the equation
implicitly defines a unique real-valued differentiable
function y = g (x), satisfying g (0) = 0.
passage Based Questions 2008]
81. If /(- 1072) = 2 42, then f" (- 10 72) is equal to
472
442
73
73
y = e~x ,y = O,x = 0andx = LThen,
[More than One Option Correct 2012]
(a)Sii
(b) S>1-e
e
1
(c)S$lfi + -L
4
Je
1
(d)s^4=+4=f
i--7=i
72 Tel 42'2 J
77. Let/: [-1,2] —»[0, oo) be a continuous function such that
M = /(I - x), Vx € [-1,2 J If Rj = £ x/(x) dx and R2
are the area of the region bounded by y = f(x\
x = -1, x = 2 and the X-axis. Then,
[Single Correct Option 2011]
(a) Rx=2R2 (b)R!=3R2
(c)2Rx=R2
(d)3Rx=R2
78. If the straight line x = b divide the area enclosed by
y = (1 - x)2, y = 0 and x = 0 into two parts Rv(0x < b)
and R2(b^ x < 1) such that Rt - R2 = —. Then, b equals
4
to
[Single Correct Option 2011]
82. The area of the region bounded by the curve y = f (x\
the X-axis and the lines x = a and x'= b, where
-oo<a<b<-2,is
(a) fanrz U2 '71
+
~
Ja3[{/(x)}2 -1]
(b) -
ft
X
------ —— dx + bf(b) - af(a)
°3[{/(x)}2 -1]
(c) f6------- -------- dx - bf(b) + af(a)
Jfl3[{/(x)}2-l]
(d) - fb------- ----- — dx - bf(b) + af(a)
MW-l]
83. J1 g'(x)dx is equal to
(a)2g(-l).
(b) 0
(c)~2g(l)
(Wl)
214
Textbook of Integral Calculus
(ii) JEE Main & AIEEE
[2017 JEE Main]
84. The area (in sq. units) of the region
{(x,y):x£Q,x + y<3,x2 < 4y} and y < 1 + Vx} is
(b)^
(a) |
(0-
(a) 9
(c) 18
x2 + y2 <4x>x>0,y>0}is
(a)*-|
(b)n-|
/ X
4^2
(c) tc------3
n2V2
(d) - - —
2
3
[2016 JEE Main]
90. The area bounded between the parabolas x2 = - and
4
n
x = 9y and the straight line y = 2 is
[2012 AIEEE]
(a) 20J2
(b) 2^
(b)
(c)^
(d)10j2
91. The area of the region enclosed by the curves y = x,
86. The area (in sq units) of the region described by
{(^>y):y2 <2xandy£4x-l}is
[2015 JEE Main]
x = e, y = — and the positive X-axis is
x
(a) 1 sq unit
5
(c) - sq units
2
“S
87. The area (in sq units) of the quadrilateral formed by the
tangents at the end points of the latusrectum to the
x2
y2
ellipse — + — = 1 is
[2015 JEE Main]
9
5
7
, .27
T
[2011 AIEEE]
3
(b) - sq units
2
1
(d) - sq unit
2
92. The area bounded by the curves y = cos x and y = sin x
3?t
between the ordinates x - 0 and x = — is
[2010 AIEEE]
2
(a) (4^2-2) sq units
(b) (4-^2 + 2) sq units
(c) (4V2 -1) sq units
(b) 18
(d) (4-^2 +1) sq units
93. The area of the region bounded by the parabola
(y — 2)2 = x -1, the tangent to the parabola at the point
(d) 27
88. The area of the region described by
A = {(x,y):x2 + y2 < 1 andy2 < 1 - x}is [2014 JEE Main]
(a)- + 2
3
. . n 2
c)--2
3
4
(d)-
85. The area (in sq units) of the region {(x, y): y2 > 2x and
, a27
(b) 36
(b)--2 3
.k
2
(d- + 7
2 3
(2,3) and the X-axis is
(a) 6 sq units
(c) 12 sq units
[2009 AIEEE] •
(b) 9 sq units
(d) 3 sq units
94. The area of the plane region bounded by the curves
x+2y2 = 0andx+3y2 =1 is equal to
[2008 AIEEE]
89. The area (in sq units) bounded by the curves y = Jx,
2y - x + 3 = 0, X-axis and lying in the first quadrant is
[2013 JEE Main]
(a) - sq units
(b) -squnit
2
(c) - sq unit
4
(d) — sq units
3
3
Answers
Exercise for Session 1
1. 1 sq unit
. 4
4. - sq units
_ 436
2.---- sq units
15
_5. —
28 sq units
4
3. - sq units
8. 2z sq units
9. _ sq units
3
, 8.2
7. -a sq units
3
10. 3+16 log2sq units
II. (d)
16. (a)
2. (a)
7. (a)
12. (b)
17. (c)
3. (c)
8. (a)
13. (a)
18. (a)
4.(d)
9. (c)
14. (c)
19. (a)
55. g (x) =
6. — a2 sq units
3
Exercise for Session 2
I. (d)
6.(b)
-x-1,
x^-5/2
4+x,-5/2<xS-2 2,
-2<x5-l
1 - x, - l<x£ - 1/2
1 + x,
x>-l/2
5. (a)
10. (b)
15. (b)
20. (b)
1°1 sq units
and. area = —
56.
2 + log
4 A
3V3 J
57.
K-----
2
3
sq units
e
58.8
3n 4 sq units
16
2
59. Required maximum ratio = 3
V3 kA sq units
60. (X)^
9
61. At = 1 - sin 1, A2 ~ K - 1 - sin 1, A3 = 3k - 2
Chapter Exercises
5. (b)
4.(c)
2.(b)
3. (b)
1. (d)
10.
(b)
9.(b)
6. (c)
7.(d)
8. (a)
15. (a)
14. (d)
13. (c)
11- (b) 12. (b)
20.
(b)
19. (c)
16. (a) 17. (a)
18. (a)
25. (b, c)
24. (c)
23. (d)
21. (a) 22. (b)
27.
(a,
d)
28.
(a,
d)
29. (a, c)
26. (a, b, d)
34.
(c)
33. (b)
32. (d)
30. (d) 31. (c)
39. (a)
38. (a)
37. (a)
35. (a) 36. (b)
43.
(a)
42. (b)
40. (b) 41. (c)
44. (A) -> (p); (B) -> (p); (C) -> (r); (D) -> (s)
45. (A) -> (p); (B) -> (p); (C) -> (r); (D) -> (r)
49. (7)
48.(4)
46. (7) 47.(1)
52. (3)
50. (8) 51. (5)
54./(x) = ——[x4 - Ax3 + (2* - 4JX2]
53. 3 esq units
k+ 1
2 sq units
sq units
62. 4 log5
63. 4
— - 4n sq units
3
J
65. 2 sq units
64. 2
66. 8 sq units
(
it A
68. 2---- I sq units
I
69.
67. 5 sq units
2 J
) - I sq units
log 2
70. oe[l, 2)
76. (b,d) 77. (c)
82. (a) 83. (d)
88. (a) 89. (a)
94. (d)
72. (c)
78. (b)
84. (a)
90. (c)
73. (3)
74. (d)
79. (b,c,d) 80. (b)
86. (d)
85. (b)
92. (a)
91. (b)
75. (b)
81. (b)
87. (d)
93. (b)
x2
Solutions
Area of ellipse E2, is
r~
K-VP'P = n P
(i)
=>
4x-x4x5 = 40
••■(ii)
*X
2
1
0
n
=—
2
x=2
x= 1
y
3/2
4. Area of bounded region by
41 x - 20172017| + 5 | y - 20172017| £ 20, is same as area of the
region bounded by 4| x| + 5 |y | 20.
1. Here, [x + y] = [x] — 1
when x e [0, l)=>[x + y] = -l
-Ux + y<0
xe[l,2) => [x + y] = 0
when
OSx + y <1.
which can be shown, as
y2
: 2-W3 + 4~l/3 - 1
5. Here, y = x2 + 1 and y = ^x — 1 are inverse of each other.
\
\x+y=1
The shaded area is given K units
=> Area of the region bounded by y = x2 + 1, y = ^x -1 and
ix+y=0
(x + y - 1) (x + y - 3) = 0, is 2K units.
y=x2+ 1
x+y=-1
Required area = 2
2. Required area = 2J
Let,
y=x
1/2
f~l (x) dx
/-’(x) = t^x = /(t)
K
1
dx = /'(t)dt
A =2 J* (r-noft
1
y-)>•/■«)*
=2
i f (0
.
3\"X
\
x+y=3
x+y= 1
0
=2 /(d-j;
t
*2
dt
6. Given,
z
xl
= 2 /(l)-i1 log(1 + t2)
2
f(x)dx-j* g(x) dx = 10
(Aj + A3 + A4) ~ (A2 + A3 + A4) = 10
A!-A2 =10
^4
= 2 f(l)-llog2
y = ^(x)
(0,4)
=2
1 -1 log 2 = 1 - log 2 = log
.2
x + y-1
3. Here, Ej:
12
2
2
x-y + 3
V*2
__________
=1
+
"
(V2)z
'1
.a2
A3
iA4
O
2
Again, J * g(x) dx - j^f(x) dx = 5
2p2
Now,-7= =
(A2 + At) — At =5
A2 =5
Adding Eqs. (i) and (ii),
A, =15
-k/,3/2 = 2~1"2
=>
P = 2"1/S
y = g(x)
(2.2)
2-
Length of latusrectum = 2 — = —j= = V2
VP
■W
------kX
(4.0)
■(ii)
Chap 03 Area of Bounded Regions
a
7. Solving, ex = ea~x, we get e2x = ea => x = —
2
y
y = e*
x > 1, y increasing and 0 < x < 1, y is decreasing
A = jjln2x-l)dx
= C ln2xdx
Jl/e
M0,ea)
(0,1)
Jut
» fin
“• x
-1,(1
----- |xax-| e—
e,
1/e\ x
= HM-'
>x
s
-r
= [x ln2*K/e-2
y = e^x
01
217
- ex) dx = [-(ea • e~x + ex)] a0'2
1
e-e.
= -2 I [x In xtf/e-
= (efl + 1) - (ea/2 + efl/2) = ea - 2efl/2 + 1 = (efl/2 -1)2
e°/2-l 2 = lfea/2-l)
a
4 a/2 J
S
a2
2
10.
Iim4 = £
»-»0a2
y=
4
a2-b2‘‘
Also, y =
8. mpQ=----- - = a - b equation of PQ
a+b
Q(-b.b2)
x, if x£2
4-x, if x£2
—, if x>0
x
----3 , if x < 0
. x
y+
3-
I
0
2-(2 —x), if x£2 = x
2-(x-2), if x 2*2 = 4 —x
P(a,a2)
2-3/2
1-
>2-b2:
y-a2
-(x-a) or y-a2 = (a-b)(x-a)
*a + b
013/2 2
y =a2 -h.x(a-b)~a2 + ab
31J
Sj = f ((a - b)x + ab - x2) dx
x)
|3
x2
—-31nx
+ 4x-------- 31nx
2
2
3/2
L
2
(fl+b)3
a
a2
-b
b2
0
0
1
1
1
1 = -[ab2 + a2b] = -ab(a + b) ...(H)
2
2
1
_ (a + b)2 _ 1 ja b
2
-+-+2
lab ~3Lb a
ab (a + b)
6
xj
,
2
6
S2
3 1
■3
2 i
x— dx +
A=[Jr3/2'
(4-x)-- dx
2
J —b
Also,
•y = 4-x
y = X‘
y =(a-b')x+ab
which simplifies to
>X
3
5,
min
9.y = ln!x-l =» y =
=^>
9
+ 12---31n3-(8-2-ln2)
h(x) = /[g(x)]
(2x+l)2 + 4 = /(2x+l)
X=1
X
Ya
\^4)
^0=2
o
e
>X
•
7
3
19
= --3 In 2 + 3 ln3-31n2 + --3 In3 + 31n2 =----- 3 In2
8
2
8
11. Given, g(x) = 2x+1; h(x) = (2x+1)2 + 4
Now,
4
3
=0
/9
3
= 2-31n2- --31n\8
2.
0
*X
218
Textbook of Integral Calculus
2x+l=t
Let
T(0 = t2 + 4
=>
-xe ] = e~x [-1 -1 + x] = (x -2)e
/(x) = x2 + 4
(i)
Solving,
y = mx and y = x2 + 4
Put D = 0;
x2 - mx + 4 = 0
m2 = 16 => m = ±4
Tangents are y = 4x and y = -4x
A = 2 f 2[(x2 + 4) - 4x] dx = 2 f (x - 2)2dx
For point of inflection y” = 0 =>
x=2
2
2
Jo
Jo
■ (-2e-2)-(e’x)?
A = Joxe-Xdx = [-xe"
[-xe x] o+ J e~x =
16
= ^(x-2)3 2 = —
sq units
3
Jo
12. y =
-x
= -2e~2 ~(e~2 -1) = 1 -e~2-2e "2 = 1 -3e"2
=> y2 = -x, where x and y both negative
=> x2 = -y, where x and y both negative
16. (a, 0) lies on the given curve
0 = sin2a - ^3 sin a => sina = 0 or cosa = V3 / 2
K
Y|
a = — (as a > 0 and the first point of intersection
6
with positive X-axis)
0
and A =
x= -V=y
x/4
/6(sin2x-^3 sin x) dx= f- cos 2x
/-------- + V3 cosx
2
o
•(-H
r-
7
„
'3 = —2 cos a
4
2=a+b+c
x = 0, y - 0 => c = 0 =>a + b =2
Y
A=1
3
13.
7
4
=> 4 A + 8 cos o = 7
17. x = l;y=2
, 16a& ,
, 1
A =------ where, a = b = —
3
4
Since,
X
x=2
-(i)
f'M
Integrating, In f'(x) = x + C, f'(0) = 1 => C = 0
Y.
0
>x
7
y+1=0
/'(x) = e7(x) = ex + fc,/(0) = 0 => fc = -l
Now,—
= 2a x + b = 1
dx (o.o)
b = l,a=l
Hence, the curve is y = x2 + x
f0
/(x) = ex-l
Area = |°(ex -1 + 1) dx = [ex]o = e -1
14. A =2^[y^yz-(y2-i)]dy = 2
2
1
2
- x) dx = | j (x ) dx = — sq units
2
24
18. IF = e
ye
= J e-x(cos x-sinx) dx. Put -x = t
= -j e‘(cos t+ sint) dt= -e‘ sin t + C
4
i <
x =|y|ViV
ye * =e~x sinx+ C
Ya
(-1.0)
>X
o
X = /2-1
0
15. y = xe
y'=e~> -xe
= (1 - x)e x increasing for x < 1
>x
Since, y is bounded when x —> «> => c = o
Chap 03 Area of Bounded Regions
y=sinx
f-2+2-72
rn/4
Area = Jo
Required area = 2 JQ
(-x + 3)-
x2 +8
(cosx -sin x) dx = v2 -1
2
r
3
19. /=f(6x-3x2)dx = —- —= 3x2-x3 = x2(3-x)
J
2
3
A = J(2)-I(l) = 4-2 = 2 units
A2 = 1(2) -1(3) = 4-0 = 4 units
A3 = 1(3) -1(4) = 0 -(-16) = 16 units
_ 2 r-2+2V2
4 Jo
219
dx
4
J-1-2+2V2
(4- 4x - x2) dx = - 4x-2x2- —
22 3 Jo»
=1(i2x-6?-xj];!*2^
6
= -[12(— 2 4- 2^2) - 6(- 2 + 2^2)2 - (-2 + 2V2)3]
6
Ai
+
3
2
= - (-24 + 24^/2 -6(4 + 8-8>/2)-(-8 + 16^) + 24^2 - 48]
6
= | [- 24 + 24^2 - 72 + 48-72 + 56 - 40^2]
4
l2
^3
= - [32>/2 - 40] = - (472 - 5) =-(4^ - 5)
6
6
3
22. Shaded region represents S O S' clearly area enclosed is 2 sq
units.
One value of a will lie in (3, 4).
(0.1)
Using symmetry, other will lie in (-2, -1).
20. Required area, A = f ^((sin x-x) +2rt) dx
Jn
0
(-1.0)
nZ
= — -2sq units
1/3 2/3 (1.0)
21- g(x) = |/(|x|)-2| = ||x|-l-2|=||x|-3l
x<-3
-x-3,
(|x|-3),
-(-x-3), -3<x<0
-(|x|-3), -3<x<3 =
0<x<3
-(x-3),
|x|-3
x£3
x£3
x-3,
-x-3,
x+3,
x < -3
-3£x<0
- x + 3,
0^x<3
x-3,
x£3
23. Required area is 2
ex In x dx+
x - 4y + 8 = 0
x2 = 4(y-2)
Y
x2=4(y-2)
y=-x-3
\ y=x+3
y=-x+3
——i------- f
-3
-2
y=x-3
3
2-'
■+
1
-1
-t-
2
3
>X
24. y=\2ox-x.22 => (x-a)2 + y
Let P(h,k) be a point, then BP > PN
For the bounded condition BP = PN ~k
Now,
AP = a-k = ^(h-a)2 + k2
L
L
k — /I
(-2+272, 0)
~~~
2a
-r
For point of intersection,
x2 - 4(- x + 3) + 8 = 0
=>
(1 -Jl -(x-1)2) dx
„2
Now,
X'-*
(0.-1)
a
x2 + 4x - 4 = 0
x=2ii7*ElL_2±2j2
2
■*. Point of intersection is at x = - 2 + 2>{2
s\"
4-
P
N C(a,0)
4
220
Textbook of Integral Calculus
2
x2
Boundary of the region is y = x----2a
•a
Required area = 2
0
dx
x 2\
x----- dx = —
3
dy _
dx
. 2aJ
x2-y2
2xy
25. f(x) = 2x - a. At (2,4), f'(x) = 4 - a
1
Equation of normal at (2, 4) is (y - 4) = —
(x-2)
(4-a)
dy _l-x_x-x2
dx
y
xy
Let point of intersection with X and Y-axes be A and B
respectively, then
A =(- 4a + 18,0) B = I 0, 4u-18
a -4
Since, a > -9 as
2
Area of triangle = ±(4a-18)—^—~ = 2
=>
(4a-17) (a-5) = 0
17
4
a = 5 or
Min {/W. gto) = | (l/to + g(z)| -1 Z(z) - g(z)| ]
:. Area = f* [max {/(x), g(x)} - min {/(x), g(x)}] dx
Ja
27. Area bounded by parabola y = xz - 7x + 10 and X-axis is given
,
2 -xz
xy
2xy
Since, option (a) is true (equation of circle).
If option (a) is true (b) can’t be (It is parabola).
30.
9
JJx -7x+10| dx = JJ-x +7x-10| dx = -squnits
x2 y 2
28. Area bounded by the ellipse — +
-1 will be the same as
a
v
x2 y 2
the area bounded by the ellipse — + ~ = 1 and 7t ab
b
a
A Required area = n (2)(3) = 6n sq units
Therefore, area of y = sin2x will be lesser from area of y = sinx.
Statement II is obviously true.
Hence, (d) is the correct answer.
31. Let the line y = kx + 2 cuts y = x2 -3 at x = a and a = p, area
B
P
bounded by the curves = f (yj -y2) = f {(fcx + 2) -(x2 -3)} dx
Ja
=>
/(x) =
(k2 + 20)3/2
6
which clearly, shows the Statement II is false but f(k) is least
when k = 0.
Hence, (c) is the correct answer.
32. As of region bounded by parabola y2 = 4x and x2 - 4y is
4
J4 2^-- dz = iz3'2
JoI
4
3
x^
12 0
32
3
16 16
= — sq units
3
Hence, Statement I is false.
33. As the area enclosed by |x| + |y | < a is the area of square
(i. e., 2a2).
/.Area enclosed by | x + y | + | x - y | < 2 is area of square shown as
y“
,yy = x
29. OP = x
If slope is
then equation of tangent is
Y-t = ^(X-x)
dx
xy = -x
P(*.y)
Area = 4 Q X2 X2j = 8 sq units
p
o
Length of perpendicular from origin to this tangent is
v
dy
Y-x—
dx
>
If option (a) is true, option (c) is also true where c = 1.
sin2 x <sinx, V x e(0,n)
,2
by
fs
,
2
x 2 +y
-z2
2
Ja
26. Max{f(x),g(x)}
Max (/(z), g(z)) =
= ^[|/(x)
1 [|/(z) +
+ g(x)|+|/(x)-g(x)|]
g(z)|+|/(z)-g(z)|]
26.
f5 ,
[homogeneous form]
Also, the area enclosed by | x + y | + | x - y | £ 2 is symmetric
about X-axis, Y-axis, y = x and y = -x.
.■.Both the Statements are true but Statement II is not the
correct explanation of Statement I.
is true f°r quadrants.
34. J 6{/(x) “ #(*)}
Statement II is false.
The area bounded by y - x(x -1) and y = x(l - x).
221
Chap 03 Area of Bounded Regions
y = x(x-i)
40. v
It is seen from the above that f'(ex) and so g'(x) is positive on
(0,00) and negative on (-<», 0).
41. Clearly, x = a sin31,y = a cos3 t, (0 £ t £ 2n)
« y =x (1-x)
Area enclosed = 2jQx(l-x)dx = 2
x^
2
2a-(e2l-l)
(e^ + ae' + l)2
Now,
>X
x 3V
3 0
r a sin3 t-a-3 sin2t(-sin t) dt
S=f
Jo
3-2 = —1
6
3
= 2j--<2 3
2
f
= -3a2 x | *4
! t cos2 tdt
sin
Jo
35. Since, absolute area
= j ’ /i(x) dx + j 1 ~h(x) dx + J ’ h(x) dx
3 1 /— 1 1—
-X-V7C x-vn
3^2 2 2
2
2x3x2xl
= -3a2x4 + 24-2
2
2
-£ /’"‘(-ir-Mx)*
- * *r
r=0
3,3
= —na =-na 2
8 8
36. Also, x„ +, = x3 => n = 2
. 4
37. Required area= f sin
Tlx axJo
42.
.
Iogexdx.11
=—
[absolute value]
(2k
S = ~J a(l-cost)a(l-cost)dt
= -a2Jo (l-2cost+coszt)dt
Jo (
X
0
x2-ax+l
x2 + ax +1
For differentiation, better write /(x) as
r. x ,
2ax
/(x) = l------------- x +ax + l
2a(x2-l)
Now, on differentiation f(x) =
(x2 + ax+l)2
M /(x) =
/'(i) = o = H-i)
Then, the options (b) and (d) are eliminated.
Again, for the differentiating Eq. (i) gives
(x2 + ax+l)2-2x-(x2-l)
/*(x) = 2a-
2(x2 + ax+1) (2x +a)
(x2 + ax + l)4
1 -2 cos t +
1 + cos2t
2
a2 r 2*
=---- [3-4cos t+cos2t]o
2 Jo
=-3na2 = 3na2
dt
•>,
[absolute value]
43. x = |(6-t) = |(6t-t2), y=y(6-t) = |(6f2-t3)
3
3
8
8
Area = f - (6 -1) • -(12t - 13t2)dt
J03 8
= - P (t4 + 24t2 - 10t3) dt = 8Jo
8
s=
(x+sinx) dx— j QX dx
_2
=----- cosn + cos 0------ =2squnits
2
2
y=x
>y=f(x)=x+sinx
s
>r1(x)
Combining both /*(1) (2 + a)2 + /*(-1 )(2 - a)2 = 0
x
(x-l)(x+l)
(x2 + ax+l)2
If is easily seen that /(x) decreases on (-1,1) and has a local
minimum at x = 1, because the derivatives changes its sign
from negative to positive.
1
44. (A) Required area = 4s
y*(-l) =---- f*(i) =
-----------J
(2-a)2 /'(!)
J U = - (2 + a)2
(x2-1) _
39. /'(x) = 2a
(x2 + ax+l)2
L5
1 7776
. ,,J
216
27
------- 1-1728 — 3240 =----- = — sq units
8
5
5X8 5
222
Textbook of Integral Calculus
(B) Required area = 2 f'xex dx = 2[xex -ex]’o = 2
(C) The line y = x + 2 intersects y = x2 at x = -1 and x = 2
Jo
The given region is shaded region area
Ya
Jr
O
---- ->X
x =1
1
2 .
2
2
9
x dx = =-4
2 J
2
(C) y2 = x3 and |y |= 2x both the curve are symmetric about
Y-axis
(D) Here, a2 =9, ft2 = 5, fe,2 = a2(l-e2)
2
4
2
e' =- => = —
9
3
y2 = x3
. 2x y
is — + - = 1
Equation of tangent at
9
3
9
x-intercept = -, y-intercept = 3
4x2 = x3 => X = 0, 4
16
Required area = 2j & (2x - x3/2) dx = y
9
1
Area = 4x-x3x-=27sq units
2
2
5
(D)Vx +Jiy|=l
Above curve is symmetric about X-axis
46.
Y.
(0.1)
1
0
2
>x
d.O)'
Area is given by A = 2 [ \x2 - 3x + 2) dx = Jo
3
y/\y \ =l-VxandVx sl-^/iyf
=> forx>0,y >0, 7y =1-Vx
1 dy _
1
2yfy dx
2Vx
yy
dy A
— <0
dx
Function is decreasing required area = J °(2Vx - 2x) =
45. (A) The area = 2 unit
fK
sin x dx = 2
Jo
(B) Area enclosed = I
=>
3A + 2 = 3- + 2=7
3
47. For c < 1; J (8lx.22-x5)
=“
c
8
3
c3
1
8? ? = 16
6
3
8
3
c_ 3
6
6
6
16 8 1
--- +
3
3 6
17
6
c = -1
Again, for c > 1 none of the values of c satisfy the required
condition that
[(8x2 - x5) dx = — => c + 2 = 1
3
Chap 03 Area of Bounded Regions
x;
X <2 t
3/ r, x > 0
and y =
4-x; • x>2
-3 / x x < 2
48. y =
a
(0.3/2)
fl;
2
^3
A
Required area = PQRSP = Area PQRP + Area PRSP
r2_ r x---- dx + j^(4_x)_2px
M
N (3/2.0) (2.0)
(3/4.0) (1.0)
Since, 2x + 2y = 3; passes through A
4-31n3
sq units
2
49. Equation of AB y = - (x - 2)
2
3-5
Equation of BCy - 5 =------ (x -4) =>y = - x + 9
6-4
0-3
3
Equation of CA y - 3 =----- (x - 6) =>y = - (x -2)
2—6
4
Required area
e6
3 r6
= -J2(x-2)dx + J4-(x-9)dx--J2(x-2)dx
_5 (x - 2),2
2
2
4
6
(x-9)2
2
,
3 (x - 2),2
2
4
6
2
= ;[2!-01-|[(-3)2-(-5)!]-;[42-0]
o
5
1
3
= 2x4-i[9-25]-qi6-0]
4
2
8
lr
,3
= 5 —[-16] — X 16 = 5 + 8 - 6 = 7 sq units
2 8
50. If[|x|] = 1 and[|y|] = 0, thenl <|x| <2,0 <|y| < 1
=>
If
o
7|(0,0)
x)
5 f4
x e(-2,-1] u [1, 2), y g(—1,1)
M = 0,[|y|] = l
.
2x —x2
Y-.
(0.2) >
P
=
f(x) =
51. Clearly,
Yk
24A=5
52. f(x) = min < sin x,cos ,x —kxe[0,1]
6J
(J3
Area = f»« sm-i x, wax + 1Jo
2 I^2
ri
'3-1
2
9 - 7t
a-n
127 ~ 6(V3 + 1) ~ b(>/3 + 1)
J3-1
= ^(i-^)+ -----2
711 =(^-i)f|\4
a = 9, 5=6 => a - b =3
53. Given, f ( - | = f (x) - f (y)
...(i)
Putting, x = y = !,/(!) = 0
f(x+h)-f(x)
Now, /*(x) = lim ----------------- — inn
h-t 0
h
*->o
3
Z(x) = -
X,
\
h
[from Eq. (i)]
\
x)
h
—•x
X
since, lim f(l + x) = 3
x
X
(e,3)„.
D
Zic
B
Area of required region = 4 (2 -1) (1 - (-1)) = 8 sq units
I']
cos lxdx
2) J^/2
V3 -1
= (xsin-1 x + -^1 - x2)q2 +-------4
____
+(xcos-1x-71-x2)^/2
h-t 0,
xe(—1, l),y e (-2,-1] o(l, 2]
and B
so bounded area A
= Area OAB = 2 [Area OCM 4- Area CMNA - Area ONA]
’13 3 1 3 1
1
5
=2-X-X-+- - + - x- -- fl(2x-xz)dx = —
2 4 4 2 4 2,
4 2Jo
24
= lim
Then,
223
------ 7 5
0 =/(1.0)
A
(e.0)
y=3
224
Textbook of Integral Calculus
=>
Putting
=>
.’.
r
Now, required area =1
f (x) = 3 log x + c
x = 1 => c = 0
/(x)=3 1ogx = y
Required area = J
g (x) dx
l_’„(2 + x)dx
[say]
x dy = |
ey/3 dy = 3 [ey/3]_3m
1
1
5 + -|+ — | 25 2
2
= 3 (e - 0) = 3e sq units
101
=---- sq units
54. According to given conditions,
jo'[/(x)-(x’-4x■:2)]dx = lIJo'[(2x2-x))-y(x)]dr
55. It is easy to see that,
Differentiable both the sides w.r.t. t, we get
y(t)-(t*-4t2)=K(2t!-f5)-/(t))
or
ex,
/(x) = | 3/2,
0£x<log(3/2)
log (3 / 2) £ x < log (2)
(1 + fe)y(r) = 2**2--4t2.
log (2) £ x £ 1
f(') = jA-(t<-tt2 + (2fc-4)l2)
Let A be the required area. Then,
A=
Hence, required f is given by;
/(x) = — [x4 — fcx3 + (2fc - 4) x2]
J
k +1
-1<t£0
0<t<l
t.
57.
/'(x)= lim
1
1
- dx +
fl
_
Jlog2
(1 + e x) dx
1>fe
2
f(x+h)-/(x)
h
/(x(i + h/x))-y(x)
= lim
h-> 0
h
= /W lim /(l + h/x)-f(l) _ /(x)
•/'(D
X
fc-»0
hl x
x
,,2/(x)
f'M = 2
or
X
/(x) x
2
1/2
3
Jlog3/2 2
h— > 0
Y
-1/2 0
flog 2
3(
3
1 - - - log 2 + |
+ - log 2 - log e
2
u \
V2
= ^2+ log ( 4 A 1
sq units
3^3 J e
1 <-l
2-t,
e dx +
-(«')!* M +;(x)|:H„+(x-e-«)'te!,
55. y(t) = |t-l|=|t| + |t + l|
’ -t,
<■ (log 3/2)
Jo
>X
Ya
Case I
/(x)=x2
1
.5
x + 2<— => x£—
2
2
g (x) = max {f (t): x + 1 £ t £ x + 2}
= f (x +.1) = - x -1, x^-5/2
>TO
Case fl --<x + 2^0=>--<x^-2
2
2
0
(-1.0)
>X
(1.0)
g(x) = /(x + 2) = 4+x,-|<x^-2
Casein
0<x + 2£l =>-2<x<-l
g(x)=2
CoseIV l<x + 2£3/2=>-l<x^-l/2
g(x) = /(x+l) = l-x
CoseV x + 2>3/2=>x> —1/2
g(x) = f (x + 2) = 2 + x
-x-1,
x^-5/2
Integrating both the sides, we get f (x) = Cx2, since
f (1) = 1 =>C = 1.
.2
So,
f (x) = x
Now,
2
= x2 => x4 + x2 - 2 = 0
1 + x2
x2 = l
Required area = 2
=> x = ± 1
2
J
1 + x.2‘
-5/2<x£-2
Hence,
S(x) = «
2,
1-x,
-2<x£-l
= 2 2 tan'
-1 <x£-l/2
x>-l/2
=
x3
-x2
- 1
X------
3 -0
j sq units
=2
dx
n_l
.2 3.
I
i
Chap 03 Area of Bounded Regions
58. y (x + y3) dx = x (y3 - x) dy
=> xy dx + y2 dx = xy3 dy - x2 dy
>y = 3n/16
*
=>
xd (xy) = x2y3
— dy -dx
x
x
^y
^■d
x
(xy)2
1
2
1
2
xy
+C
At, x = 4, y = - 2
1
8
So,
1
2
1
2
Hence,
x , y
z
By similar triangles — = —— = —
h b+c c
a(b + c) x , ex
+b—
A(R) + A(S)_
h
h
So,
A(T)
hx/2
2
= — (ab + ac + be)
h2
We need to maximize (ab + be + ca) subject to a + b + c = h.
One way to do this is first to fix a, so b + c = h - a.
Then,
(ab + be + ac) = a (h~ a) + be
and be is maximized when b = c. We now wish to maximize
2 ab + b2 subject to a + 2 b = h. This is a straight forward
calculus problem giving a = b= c = l/3. Hence, the maximum
ratio is 2/3 (independent of T).
60. Consider a coordinate system with vertex P of the isosceles
APQR at (a, 0) and Q and R at (0, b) and (0, - b) respectively.
A = - a-2b =ab
„.(i)
2
Let the centre of ellipse be (a, 0) and the axes be of lengths, 2a
and2p.
(x-a)2 , (y)2
So, the equation of ellipse is
2
+ c => c = o
y3 + 2x = 0
/(x)=(-2x),1/3
’
So,
225
a2
p2
The second equation given is
• cos’ X
rasin’1 Stdt+
[
y = *1/8
Jl 1/8
cos 1
dt
Q
(0,b)
=> y' = x • 2 sin x cos x + x • 2 cos x (- sin x) = 0
So, y is constant.
Put sm x = cos x =
U
f 1/2
Hence, y = J
V2
R
(sin-1 4t + cos-1 Jt) dt
=r
dt
J1/8
K 3
2 8
3n
3n
andg(x) = —
3tc
So, we must find the area between y = / (x), y = —
.. y = —
3n ; x = - -11 3% I
At
16 2 \
16
Hence,
3
= P (say)
f0 (
area = j^ I — + (2x),1/3
16
=
>----- -X
P
(a.0)
(a. 0)
( Qir
37t
-1/3
,3/3
X
— x + 211/J
'3-----16
4/3
Now, the line PQ is tangent to the ellipse. To apply condition
of tangency, let us take a new system / y' whose origin is at
(a, 0).
Then, x = x' + a and y = y'.
x'2 y'2
So, the ellipse becomes —- + —— -1
a
dx
o
,40squni,s
A (R) + A (S) _ ay + bz
59. As in the figure
A(T)
" hx/2
where h = a + b + c, the altitude of T.
(0,-b)
p*
and the line PQ becomes
a
b
which can be written as y/ = -- x< + b
a
2
So,
b2
a
2
l-“ I = a2
a
I +P2
-- I
aJ
p2 = z>2 l-“l
(ii)
a )
Now, area of ellipse = n a p => A2 = n2a^2
Using Eq. (ii), A2 = n2a2b2f 1 - — | = f (x)
z
y
X
(say)
\
a J
( .
6a2 )
f'(x) = n2b2 2a-------
I
a J
2
a
f'(a) = 0 => 2a = — => a = ^(a*o)
3
Textbook of Integral Calculus
226
n 12a
2------ a
(at a = a / 3)
f"(a) = it2b2
Since,
„2
a
l2
= it — • b
= -2n2b2 = (- ve)
7t2A 2
- [using Eq. (i)]
27
__2
So,
9
3
Hence,
_ 7t A _ -J3lt A
sq units
~3 V3 "
9
61. A] = f
{sin x - f (x)} dx = - [cos x]£ - f
Jo
Jo
= - (cos a — 1) — J
=> - cos a -
ra
Jo
a
3 + 2X-X2
f (x) dx = 1 - sin a + (a - 1) cos a (given)
So, the area of shaded region = 4 X
f (x) dx = - sin a + (a - 1) cos a
area of circle = 4
f 2n
Jx
A3 = I
,y = 4x2
'fi
[as sin a = 0]
I x sin x - sin x | dx = 3tc - 2
:C
a/2
-a/2
■>
Area= Area OCBD
62. For x = 1, y = • 5X + 4 = 5fr + 4
_
. dy
and — = b • 5X log 5 => 5b log 5 = 40 log 5 => b = 8
dx
The two curves intersects at points where
8-51 + 4=25x + 16
=> 52x
2x-8-5 + 12 = 0 => x = log5
x = log5 6
Hence, the area of the given region;
= f1Og’6{8-5x + 4-(25x + 16)}dx
J)og,2
2
3
6
Area of triangle
*->o Area between the line and parabola
= lim
x-t o
q3/2 3
a313 ~ 2
1_____ = -2 sq units
...
65. Required area = -x2x2
H.2)
(1.2)
(8-5x — 25x —12) dx
8-5x
25x
--------- 12x-----------log, 5
log, 25
logs 6
(-1.0)
Jk>gS 2
(-1.-2)
---------- kX
(1.0)
(1.-2)
logs 6
_ jlogs 36
8-5
- 12 (log 5 6 - log 5 2)
+---log, 5
log, 25
5log’4
s-s108*2
+----------log, 5
log, 25
4
16
-36
48
---------- 4---------- - 12 [log5 3] +
2
(log,
5)
(log,
5)
2 log,5 log,5
66. Required area = 4 x
it (V2)I:2
---- (it-2) =8 sq units
2
(0, 2)
- 12 log5 3 = 4 logs e4 - 4 log5 25
log, 5
= 4 log5
a>_
2
1 a
2
-x - xa
2
(xsin x - sin x) dx = it -1 - sin 1
A2 = Jj
16
£3 (3 + 2x - x2) dx
16 A A
— - 4rc I sq units
64. Area of &AOB = 2 x
Differentiating w.r.t. a.
f (a) = cos a - (cos a - a sin a) = a sin a
:. f (x) = x sin x
y = sin x and y = f (x) intersects at
Now,
a sin a = sin a => (a -1) sin a = 0
=>
a=1
Aj = 1 - sin 1
Hence,
•Ilogs 2
=> Radius of circle = - x AB “ - x 8 = 4 units
2
2
f(x)dx
-I f (x) dx = - sin a + a cos a
Jo
ca
Jo f (x) dx = sin a - a cos a
_ flogs 6
63. By the symmetry of the figure circle(s) of maximum area will
have the end point of diameter at the vertex of the two
parabola.
(2.0)
e4
27
sq units
Chap 03 Area of Bounded Regions
^2
2
67. 5 sq units
Now, curves are y = bx and y = — x
±Y
5
y = b4
Ya
4y = t>x2/2
4
->y=2
3
>y = 1
2
>x
Vl
1
>x
0
1
3
2
5
4
f 2
Bounded area = 2
68. Area of the square ABCD = 2 sq units
|
J, (xz-x^dy I
C
2
|_U 3/2
D
4b 3/2
e
Area will be maximum when b = [a] is least.
Asa £2 => [a]leMt = 1 => l<a <2
71. Since, f' (x) < 0 => f (x) is a decreasing function and also
f " (x) > 0 => f (x) is concave upwards.
Hence, the graph of the function y = f (x) is as follows :
A
Area of the circle = n x - = ^ sq units.
Required area =
2 - — I sq units
Ya
69.
...(i)
2|x|.|y| + 2|x|“1 SI
Clearly, this region is symmetrical about X and Y-axes.
Let x < 0, Eq. (i) gives,
2X-y + 2X-1 <1
1-2 x-l
---- =2 - X
2X
1
2
1
1/2
B
0
*x
n-1 n
(0
x-1/2
C
2 3 4 5
Y
y=1/2
>X
1/2
-1/2
Clearly, bounded region in the first quadrant is OABC. The
required area is 4 times the area of the region OABC.
Required area = 4
r 1/2
Jo
2~x
f 2~x
1
dx= 4
< ln2
2
x
2 /o
1
n-1 n
>X
(ii)
Let SL denotes the shaded area in figure (i).
Sl=/(2)+/(3) + - + /(")= £
= -L(l-2-''2 )-l squnits
ln2
70. Pair of lines y2 - 3y + 2 = 0
=> Lines are y = 2, y = 1..
Let
[a] = b
2 3 4 5
r=1
From the figure (i) it is clear that, SL < J
=>
f (x) dx
i f(r)-/(I)<f f(x)dx
r=l
227
228
Textbook of Integral Calculus
y
...(i)
r■1
(1.2)
c
Let Su denotes the area of the shaded region in figure (ii).
1
5U = lf(l) + /(2)j + l(/(2)+/(3))+...
Ct
J
Ct
2
X*
.,(1.0)
£(-3,0)0 0
(-4,0) a
-9
i
-X
6
+ |(/ («-!) + /(«))
= [(/(I) + /(2) 4-/(3)) 4-... 4- /(n — 1) 4- f („)]
Y'
-1 Cf (l) + /(n))
.
C
£
r=1
Required area = Area of trapezium ABCD
- Area of ABE under parabola
- Area of CDE under parabola
= | (1 + 2)(5) - J_"437- (x + 3) dx - J* 37(7^ dx
From figure (ii) it is clear that;
n
s« >[" f(x)dx
15
2
=> Zrw-|cr(i)+/■("»>/”f&j*
>
, 1 .. .
f« _. , ,
(-3-x)3/2
f(x)dx+ 4--/(I)
-(H)
From Eqs. (i) and (ii), we get
3 ~
_3
2
Using Newton-Leibnitz formula,
/(a) = Ha) and /(0) = r(0)
72. Here, {(x, y) e R2: y 2 ^|x + 3|, 5y S (x + 9) < 15}
y > 7|x + 3|
On differentiating,
-3, when x < - 3
y2>
9
- 3 - x, when x < - 3
Shown as
+ {4 cos x sinx}
I
¥
-3
2l
2
7C
xz + —
6,
?+?)l
.2
■X
0
+ 2sin2x
F*(0) = 4 |cos2(y) j- = 3
r
Also,
9
6
y2=x+3
= 4 < cos
X'*
7C I
F'(x) = 2cos2 x + — -2x -2 cos,2 x l
6J
Again differentiating,
x,22 4.. —
sin^. x2+ “J2-* ’
7C -2xcos(xz +
F"(x)=4 cos2
x + 3, when x > - 3
,y
•(>)
7 2cosztdt
F(x)=[
Given,
y]x+ 3, when x > - 3
or
(fl) + 2
=
r=1
J_:
16 _ 15 _ 18 =3
= —+ -[0-l]--[8-0] = —--3 ~2
2 3
3
2 3 3 ” 2
73. Since, F'(a) + 2 is the area bounded by x = 0, y = 0, y = /(x)
and x = a.
1/W + J," /(x)dx<£ /«</_" f(x)ix + /(l).
•
(3/2
2
=> i/(r)>J,"/(X)dx+!(/(!)+yw)
r=1
j ^(y + 3):
3
5y <(x + 9)<15
(x +9) 2 5y and x < 6
/(0)=3
2
74. Let equation of tangent to parabola be y = mx + —
m
It also touches the circle x2 + y2 = 2.
Shown as
Y
2
(0. 9/5)
= V2
m y]l + m
(-9,0)
X
0
x=6
=> m4 + m2 = 2 => m4 + m2 - 2 = 0
=>
=>
••• {(*, y) e Rz: y 2:
+ 3l, 5y <(x + 9)
15}
(m2 -1) (m2 + 2) = 0
m = ±l,m2=-2
So, tangents are y = x + 2,y = -x -2.
They intersect at (-2, 0).
[rejected m2=-2]
Chap 03 Area of Bounded Regions
R
y
pj7(-2,0)
X'<
Since, x2 £x when x e [0,1]
2
-x £-x or e
-
£e'
J’ X'dx > Jof'e^dx
X
o
Q-
(i)
r
Also,
Equation of chord PQ is -2x = 2 => x = -l
f e~x dx < Area of two rectangles
Jo
1
x“r
Equation of chord RS is 0 = 4(x - 2) => x = 2
Ve
Coordinates of P, Q, R, S are
P(—1,1), Q(—1, — 1), R(2, 4), 5(2,-4)
Area of quadrilateral = (2 + x 3. = 15 Sq units
1
'2
a
Using
\baf(x) dx = j7(a + b -x)dx
a
Ri=\2_fi-x)f(x)dx
b
■x
= Jj/(x)-«(x)|dx
cosx - sinx.
could be shown as
r 2
0£x£ —
..y
y = sinx + cosx
= V2 sin (x+
f(x)
Lf(x) = f(l - xX given]
Given, R2 is area bounded by /(x), x = -1 and x = 2.
R^^fWdx
(iii)
2Ri =J_i/(x)dx
4
n
K
sinx - cosx, — 5 x< —
4
2
g(x) -y =1 cosx-sinx| =
1
W)
n/4
------ I—
■X
n/2
Area bounded = f ‘ {(sinx + cosx) -(cosx - sinx)]dx
Jo
+ [
Jx/4
fx/4
n/4
2Rj = R2
78. Here, area between 0 to b is Rt and b to 1 is R2.
<£x-£* (J-x)2 dx = l1
•••
4
(1-X)1
b
1
tK/2
Jlt/4
0
=>-|((l-(’)’-l] + j[0-(l-i’)J]=|
4
1
--(l-i>)
’
=-i
+
l
=
-=>
3
3 4: 12
d-^4
=» »4
2cosxax
= -2[cosx]”/4 + 2[sinx]“/4
= 4 - 2J2 = 2y/2{y/2 -1) sq units
£
(1-x)3 •
-3
4
b
{(sinx+ cosx)-(sinx-cosx)} dx
2sinxdx+
...(iv)
From Eqs. (iii) and (iv), we get
-3
9(x)
79. Shaded area = e-
■‘e’dxKl
0
J
Also, j ln(e + l-y)dy [pute + l-y = t=> -dy=dt]
76. Graph for y=e~xi
,,y
1
y =e
7
M-
O
1
1
...(ii)
On adding Eqs. (i) and (ii), we get
Here, /(x) = y = sinx + cosx, when 0 £ x < y
JO
...(i)
Rl ~ J,/1 ~
Area bounded = Jjgfx) - /(x)]dx + jV(x) - g(x)]dx
=
[from Eqs. (i) and (ii)]
77. Rj = J x f(x)dx
gW
c
0
1
e
r2
fto\
<2-
...(ii)
■fi)
e
J2
gW
and
'i
1
1
... > + >
y/2 4e
75. To find the bounded area between y = /(x) and y - g(x)
between x = a to x = b.
y
/(x)
o
229
0
>x
1
■X
In tdt= J Inydy = 1
230
Textbook of Integral Calculus
p lt/4
/
80. Required area = J °
1 + sin x
1 -sinx
cos x
cosx
83. Let/ =dx=[«(*)]-1 = g (1) - s (-1)
dx
/
p It 14
J0
x
2 tan—
14---------- 2—
x
14-tan22 —
2.
2x
1 1 —tan2 —
_______ 2_
|
U
/
p it /4
J0
14- tan2 —
2
>0
At x = 1,
x
x
14- tan---- 14- tan—
p Jt/4
2
2
J0
2x
1-tan22
o
p V2-1
J0
=>
=>
g(l) + g(~ 1) = 0
g(l) = - g(-1)
I = g(l) - g(- 1) = g(l) -
X
1 - tan2
\x+y=3
(0, 0) TO)(2,0) (3,0)
X+ —
3/2
[v tan - = 42 -1]
8
[dx'J
\dxj
.(ii)
dx2
At x = - 10 42, y = 2^2
On substituting in Eq. (i) we get
dy _
3(2V2)2 —-3-—4-1=0 =>
dx ~
dx
dx
=>
1242
dx2
(21)2
6-2-34)-£)
3
2
On substituting y2 = 2x in Eq. (ii), we get
x2 4- 2x = 4x => x2 = 2x
=>
=>
d2y
-3-4 = 0
dx2
x = 0 or x = 2
y = 0ory=±2
82. Required area =
[using Eq. (i)]
Now, the required area is the area of shaded region, i.e.
Y
A(2,2)
x2+y2=4x
.3 -32
73
(21)3
dx = j /(x) dx = [/(x) • x]£ - J f(x)x dx
= bf(b) - af(a) 4- Jf‘b---—----3[{/(x)}2-1]
dx
3(y2-l)
X'
(0.0)
B (2,01
:^k/y2=2x
= bf(b) - af(a) - ^/(xjx dx
v /(x) = ^ = --1
-(ii)
which is a circle with centre (2, 0) and radius = 2
1
21
d2y = - 12/2 _ - 442
dx2
-(i)
which is a parabola with vertex (0, 0) and axis parallel to
X-axis.
And
x2 4- y2 = 4x
Again, substituting in Eq. (ii), we get
d2y =
o
85. Given equations of curves are y2 = 2x,
d 2y
-3-4 = 0
3(2-72)2 --^4-6(2>/2) • f- —
dx2
< 21
0
2
5 3 2 , 3 5
= -+------- = 1 + - = -sq units
3 2 3
2 2
—(i)
dx
J‘
2
.* J, ii2J
4-
= 1+?
I
■>X
, *T IZI
1
x3/2
dy + 1=0
3/^-3^
dx
/4y=x2
42.1)
(0,1
X'*
t2
4
y=1+^X
2
y3 -3y 4- x = 0
81. Given,
o
Y+
x
= 2g(l)
(3 - x)dx - J(
(0,3)s P.2)/
t
x = dx
4t dt
.(iii)
W))3 + W-1))3 - 3 Ig(1) + g(-1)) = 0
=> WD+M- 1)][«1)P + W-«)’ -Ml-1) -3] = 0
«
x
1
2 x .
.
Put tan — = t => - sec — dx = dt
2
2
2
_ r taiZ
4f dt
Jo
n X f2
1 ~t2
As
■■(ii)
On adding Eqs. (i) and (ii), we get
2 tan —
nl4
[from Eq. (i)]
{g(l)}3-3g(l) + l = 0
84. Required area = J (1 + Vx)dx 4„
*=1
y = g(x)
{&(x)}3-3g(x) 4-x = 0
Atx = -1, {g(—I)}3 — 3g(—1) — 1 = 0
-----------x
1-tan—
______ 2. dx
x
14- tan—
2)
x
14- tan —
2.
x
1-tan—
2
and
cosx
cosx
-----------------„
x \
2 tan—
1----------- —
2
2x
14- tan 2 dx
2x
1-tan2—
2.
2x
14-tan2 —
2
(i)
y3 - 3y 4- x = 0
Since,
.. 1 + sinx >---------1 -sinx
Y'
=----- Zl----3[{/(x)}2 -1]
„
. ,
Area of a circle
Required area =--------------------
Chap 03 Area of Bounded Regions
x(2) | y(5/3)^i
9
5
|2
_ *(2)2
4
'2 I x,/: IX = 7C 0
2->^ rn fn
(
4£I
0
8^I
= 7C---- — [2V2 - 0] = I n - - I sq units
3
Al
(-2. 5/3)
L{2, 5/3)
y < 2x represents a region inside the parabola
(-2.0)
—(i)
and y 2 4x -1 represents a region to the left of the line
y = 4x -1
The point of intersection of the curve (i) and (ii) is
(4x-l)2 = 2x
=>
16x2-10x + 1 = 0
1 1
x = -, 2 8
0
(2.0)
y^o/s, o)
(2. -5/3)
(-2. -5/3)
-(ii)
—) 3>
y'
2x + 3y = 9
(ii)
Q, 0^ and (0,3), respectively.
Eq.(ii) intersects X and /-axes at
or
Area of quadrilateral = 4 x Area of &POQ
fl
9
A
= 4 x I - x - x 3 I = 27 sq units
and
The points where these curves intersect, are
1S
fl
YSP v
X'+
y2=2x
16xz + 1 - 8x = 2x
(0.3)
p
86. Given region is {(x, y): y2 < 2x and y S 4x -1}
=>
231
88. Given, A = {(x, y) :x2 + y2 Slandy2 £l-x}
\8* 2/
y = 4x -1
H
Yi
1--
y2=2x
X
1_
X'+-t
_1
2
/J,
F
ol f-—2i—
X
(-1.0)
2"
—F
1
>x
■+
'1 1>
.8' 2>
1
2
Y
1
4
2
Required area = - nr +2
Y
Hence, required area = j
4l2
xl
and
-^(y3)-i/2
6
Z-l/2
_ 1 If 1+1
4)HH
1
8
4 [I2
15 3
1 J9 1>=-1 x--------=
1 [31 31
= -(- + 4 |2 8
6 [8 I
11 .2z + 2(y-^-3Y
1 — y22)dy =-Jt(l)
o
2
■i
n 4
= —+ 2 3
89. Given curves are y = Vx
'y + i Z
I 4 ~
2) dy
1 (y2
~ T- + >'
(0.1)
4
8
16
•(i)
2y - x + 3 = 0
Y
4
(ii)
3
1
1
y = 'lx
X
32
9
■X
87. Given equation of ellipse is
2
2
^+^=1
9 5
Now,
...(i)
Y
a2=9,82=5
=> a=3,fe = V5
e=
u
9
l2
2
3
(Vx)2 — 2-Jx - 3 = 0
=>
e
Foci = (±ae, 0) = (± 2, 0) and — = a 3
Extremities of one of latusrectum are
Equation of tangent at ^2,
On solving Eqs. (i) and (ii), we get 2Vx - (Vx)2 + 3 = 0
is, '
(Vx — 3)(Vx + 1) = 0
Vx=3 [v x = -1 is not possible]
=>
y =3
Required area = JQ (line - curve) dy = £ {(2y + 3) - y2} dy
-3
= y2 + 3y-y =9+9-9=9
232
Textbook of Integral Calculus
90. Given Two parabolas x2 = — and x2 = 9y
4
To find The area bounded between the parabolas and the
straight line y = 2.
The required area is equal to the shaded region in the drawn
figure.
Y
1
it
-d----- 1
It
.
T
y = 4x.2
y=|x2
I
3n/2
it
U
Z—y = 2
n/2
0
fK/4
Required area = j °
X
f5rt/4
(cos x - sin x) dx + J
(sinx-cosx)dr
+ f 3*/2 (cosx- sin x)dx
J5KH
The area of the shaded region (which can be very easily found
by using integration) is twice the area shaded in first quadrant.
= [sinx + cosx]JM + [-cosx-sinx]’’f +[sinx+ cosx]^2
= (412-2) sq units
Required area = 2
dy = 2
3^/y -
93. The equation of tangent at (2,3) to the given parabola is
x = 2y - 4
y3/2 -iy= 2
=5
= l10“(2J«-0) = 2Qy/2
3/2 Jy = „0 3
3
91. Given, y = x, x = e and y = —, x 2i 0
x
Since, y = x and x > 0 =>
y Si 0
Area to be calculated in I quadrant shown as
Y
(2.3)
t
i
i
i
i
i
11
(
I
I
\
I
\ I
X i
x
i
—
X'
Y' (y-2)2 = (x-1)
Required area =
y=x
{(y - 2)z + 1 - 2y + 4}dy
= (y~2)I3
3
B
X'* 0
Y'
D(1,0)C(e, 0)
x=e
- —y2 + 5y
-o
1
8
= -- 9 + 15 + - = 9sq units
y = 1/x
X
94. Given, equations of curves are x+3y2 = 1
•••(*)
x + 2y2 = 0
...(ii)
and
Area = Area of AODA + Area of DABCD
_1
- dx
~2
i x
On solving Eqs. (i) and (ii), we get
y = ±1 and x = - 2
=|+0og|x|);
:. Required area = J ^Xj-XzJdy
= j+{log|e|-logl}
[v log|e| = l)
♦Y
(-2, !)•
1.3
= - + 1 = - sq units
k(1, 0)
X'*----------
Cl
1
\ 8’ /
92. Graph of y = sin x is
y
(-2, -1
i’z
X'
—i—
-2n
I3
3x/2
o<
♦- —j
*/2
-i—
3it
*x
lY'
= jjl-3y2 + 2y2)dy = j’^l-y2)^
■I
and graph of y = cosx is
I
11
44
= 2f’(l-/)dy = 2y-y- = 21
— I = - sq units
Jo
1
3 Jo
I 3) 33
CHAPTER
Differential
Equations
Learning Part
Session 1
• Solution of a Differential Equation
Session 2
• Solving of Varibale Seperable Form
• Homogeneous Differential Equation ■
Session 3
• Solving of Linear Differential Equations
• Bernoulli’s Equation
• Orthogonal Trajectory
Session 4
• Exact Differential Equations
Session 5
• Solving of First Order and Higher Degrees
• Application of Differential Equations
• Application of First Order Differential Equations
Practice Part
JEE Type Examples
• Chapter Exercises
Arihant on Your Mobile!
Exercises with the @ symbol can be practised on your mobile. See inside cover page to activate for free.
234
Textbook of Integral Calculus
A differential equation can simply be said to be an
equation involving derivatives of an unknown function.
For example, consider the equation
dy
2
-- + xy = x
dx
This is a differential equation since it involves the
derivative of the fimtion y(x) which we may wish to
determine. We must first understand why and how
differnetial equations arise and why we need them at all. In
general, we can say that a differential equation describes
the behaviour of some continuously varying quantity.
Scenario 1 : A Freely Falling Body
A body is release at rest from a heigh h. How do we
described the motion of this body?
The height x of the body is a function of time. Since the
=g
dt
This is the differential equation describing the motion of
the body. Along with the initial condition x(0) = h, it
completely describes the motion of the body at all instants
after the body starts falling.
acceleration of the body is g, we have
Scenario 2 : Radioactive disintegration
Experimental evidence shows that the rate of decay of any
ratioactive substance is proportional to the amount of the
substance present,
dm
y
— = - km
dtwhere m is the mass of the radioactive substance and a
function of t. If we know m(0), the initial mass, we can use
this differential equation to determine the mass of the
substance remaining at any later time instant.
i.e.
Scenario 3 : Population Growth
The growth of population (of say, a biological culture) in a
closed environment is dependent on the birth and death
rates. The birth rate will contribute to increaseing the
population while the death rate will contribute to its
decrease. It has been found that for low populations, the
birth rate is the dominant influence in population growth
and the growth rate is linearly dependent on the current
population. For high populations, there is a competition
among the population for the limited resources available,
and thus death rate becomes dominant. Also, the death rate
shows a quadratic dependence on the current population.
Thus, if N(t) represents the population at time t, the
different equation describing the population variation is of
the form
—=x,n-x2n2
dt
where Xj and X2 are constants.
Along with the initial population N(0), this equation can
tell us the population at any later time instant.
Session 1
Solution of a Differential Equation
These three examples should be sufficient for you to realise
why and how differential equations arise and why they are
important.
In all the three equations mentioned above, there is only
independent variable (the time t in all the three cases).
Such equations are termed ordinary differential
equations. We might have equations involving more than
one independent variable:
V
2
dx
Sy
d
where the notation — stands for the partial derivative, i.e.
df
the term — would imply that we differentiate the function
dx
f with respect to the independent variable x as the
variable (while treating the other independent variable y as
a constant). A similar interpretation can be attached to —.
Such equations are termed partial differential
equations but we shall not be concerned with them in
this chapter.
dy
Consider the ordinary differnetial equation
d2y
dy
dx2
dx
2=c
The order of the highest derivative present in this
equation is two; thus we shall call it a second order
differential equation (DE, for convenience).
The order of a DE is the order of the highest derivative
that occurs in the equation
Chap 04 Differential Equations
Again, consider the DE
235
is a solution to the second-order DE
^y+^y=x‘2,y,2
dx3
dx2
dx
The degree of the highest order derivative in this DE is
two, so this is a DE of degree two (and order three).
The degree of a DE is the degree of the highest order
derivative that occurs in the equation, when all the
derivatives in the equation are made of free of
fractional powers.
2
[fdy^ -1 4-x
k
l\dXy
is not of degree two. When we make this equation free of
fractional powers, by the following rearrangement,
dy
dx
2
— 1 + k-x
f ,2 A2
dy
2
tdx2J
we see that the degree of the highest order derivative will
become four. Thus, this is a DE of degree four (and order
two).
Finally, an n111 linear DE (degree one) is an equation of
the form
dn y
dn~}
dy
,
a0-^- + al—----—+...
+ ...+a
+ a n_l^- + any = b
n-1
dx
dx 1
dx
where the a-s and b are functions of x.
Solving an n* order DE to evaluate the unknown function
will essentially consists of doing n integrations on the DE.
Each integration step will introduce an arbitrary constant.
Thus, you can expect in general that the solution of an
n111 order DE will contain n independnet arbitrary
constants.
By n independent constants, we mean to say that the most
general solution of the DE cannot be expressed in fewer
that n constants. As an example, the second order DE
but even through it (seems to) contain two arbitrary
constants, it is not the general solution to this DE. This is
because it can be reduced to a relation involving only one
arbitrary constant:
y = aex+b =aex.eb =ceXx (where c = a • eb)
Let us summarise what we have seen till now : the most
general solution of an n111 order DE will consist of n
orbitrary constants; conversely, from a functional relation
involving n arbitrary constants, an nth order DE can be
generated (we shall soon see how to do this). We are
generally interested in solutions of the DE satisfying some
particular constraints (say, some initial values). Since the
most general solution of the DE involves n arbitrary
constant, we see that the maximum member of
independent conditions which can be imposed on a
solution of the DE is n. As a first example, consider the
functional relation
y = x22 +Cie2x + c2e3x
...(i)
This curve’s equation contains two arbitrary constants; as
we vary Cj and c2, we obtain different curves; those curves
constitute a family of curves. All members of this family
will satisfy the DE that we can generate from this general
relation; this DE will be second order since the relation
contains two arbitrary constants.
We now see how to generate the DE. Differentiate the
given relation twice to obtain
y' = 2x +2cleZx + 3c2e3x
...(ii)
/, = 2 + 4c1e2x+9c2e3x
e2x
e3x
x2 -y
2e2x
3e3x
2x-y'
4e2x
9e3x
2-y
1
x2-y
2 3
2x-y'
4 9
2-/z
1
...(1)
(verify that this is a solution by explicit substitution).
Thus, two arbitrary and independent constants must be
included in the general solution. We cannot reduce (1) to a
relation containing only one arbitrary constant. On the
other hand, it can be verified that the function
y = aex+b
...(iii)
From Eqs. (i), (ii) and (iii), q and c2 can be eliminated to
obtain
dy
----- +y =0
dx2
has its most general solution of the form
y = A cos x + Bsin x.
y
=> 6-3"-18x+9y'+8x-4y'-4+2y" + 7x2 -6y =0
=>
y"-5y'+6y=6x2-10x+2
...(iv)
This is the required DE; it corresponds to the family of
curves given by Eq. (i). Differently put, the most general
solution of this DE is given by Eq. (i).
236
Textbook of Integral Calculus
As an exercise for the reader, show that the DE
corrersponding to the general equation
y = Ae2x + Bex +C
Linear and Non-linear
Differential Equation
where A, B, C are arbitrary constants, is
/"-3y" + 2/=0
A differential equation is a linear differential equation if it
is expressible in the form
By expected, the three arbitrary constants cause the DE to
the third order.
I Example 1 Find the order and degree (if defined) of
the following differential equations :
3
(i) y=1
+ |^
-4 2 +ifdy'| + ...
(0y
= 1+
y +±
■
1 ( dy
( dy')
3! ^dx?
2!^dx
A2/3
f| = —+ 2
(ii)
dx
d2y
(dyA
(m) —- = x In —
dx12
IdxJ
dx
(i) The given differential equation can be rewritten as
y = edy,dx
Sol.
dy
(
d y A2
3
f dy
= —+ 2
I dx
<dx\
I Example 2 Find the order and degree (if defined) of
the following differential equations :
dx2
" dx
Thus, if a differential equation when expressed in the form
of a polynomial involves the derivatives and dependent
variable in the first power and there ae no product of
these, and also the coefficient of the various terms are
either constants or functions of the independent variable,
then it is said to be linear differential equation, otherwise,
it i a non-linear differential equation.
< ,3
The differentiable equation
d y
\2
f ,2 A
d y — 4y = 0, is
-6
<dx2,
a non-linear differential equation, because its degree is 2,
more than one.
( ,3 A
e.g. The differential equation,
dy
d y
+2
dx
<dx\
2
+ 9y = x,
is non-linear differential equation, because differential
Hence, its order is 3 and degree 2.
(iii) Its order is obviously 2.
Since, the given differential equation cannot be
written as a polynomial in all the differential
coefficients, the degree of the equation is not defined.
(i)
where
a2,...,an and Q are either constants or
functions of independent variable x.
<dx3
i
— = In y.
dx
Hence, its order is 1 and degree 1.
(ii) The given differential equation can be rewritten as
d2y
dny
dn-1yy
dn~2y
dy
n
a0-^- + a1-—^+a
+ a22—^-+... + an-i~ + any^Q
dx dx 1
dx 2
dx
(«' d2y-^(dy \
(ii) —4= sin
' dx2
coefficient ~ has exponent 2.
e.g. The differential equation (x2 + y2) dx - 2xydy = 0 is a
non-linear differential equation, because the exponent of
dependent variable y is 2 and it involves the product ofy
and —. e.g. Consider the differential equation
dx
( ,2 >
( dy\
d y
- 5 — + 6y = sin x
{dx)
dx^
This is a linear differential equation of order 2 and degree.
md
/ = ^5
dx
Sol.
(i) The given differential equation can be rewritten as
( dy
= —+ 3
2
I dx
Hence, order is 2 and degree is 3.
(ii) The given differential equation has the order 2. Since,
the given differential equation cannot be written as a
polynomial in the differential coefficients, the degree
of the equation is not defined.
(iii) Its order is obviously 1 and degree 1.
Formation of Differential Equations
If an equation in independent and dependent variables
involving some arbitrary constants is given, then a
differential equation is obtained as follows :
(i) Differentiate the given equation w.r.t. the
independent variable (say x) as many times as the
number of arbitrary constants in it.
(ii) Eliminate the arbitrary constants.
(iii) The eliminant is the required differential equation,
i.e. If we have an equation f (x, y, Cj, c2,..cn) = 0
Chap 04 Differential Equations
Using Eqs. (i) and (iii), we get
Containing n arbitrary constants clyc2,c3,...,cn, then by
differentiating this n times, we shall get n-equations.
Now, among these n-equations and the given equation, in
all (n +1) equations, if the n arbitrary constants
ci>c2>c3>-««»cn are eliminated, we shall evidently get a
differential equation of the nth order. For there being n
differentiation, the resulting equation must contain a
derivative of the nth order.
237
r d2y , 9 dy _
x —— + z — = xy
dx2
dx
Which is the required differential equation.
I Example 5 Find the differential equation whose
solution represents the family c (y + c)2 = x3.
Sol. Differentiating c (y + c)2 = x3
...(i)
We get c [2 (y + c)] — = 3x2 but from Eq. (i), we have
dx
Algorithm for Formation
of Differential Equations
2x3 (y + c)^- = 3x2
dx
(y + c)2
2x3 dy _ 3x2
=>
y + c dx
2x
3
Substituting c in Eq. (i), we get
Step I Write the given equation involving independent
variable x (say), dependent variable y (say) and the
arbitrary constant.
Step in Differentiate the relation in step I, n times with
respect to x.
The equation so obtained is the desired differential
equation. The following examples will illustrate the above
procedure.
y + c dx
2x
c=—
[3— -
Step II Obtain the number of arbitrary constants in Step
I. Let there be n arbitrary constants.
Step IV Eliminate arbitrary consstants with the help of n
equations involving differential coefficients obtained in
step DI and an equation in step I.
i.e.
3
[8]-’
2
[t(3-’][t8]
= x3
Which is the required differential equation.
I Example 6 Find the differential equation whose
solution represents the family y = ae5x +bex.
y = ae3x + bex
Sol.
Differentiating the given equation twice, we get
I Example 3 Form the differential equation, if
y2 = 4a(x+b), where a,b are arbitrary constants.
— = 3ae3x+bex and
= 9ae3x + bex
dx
dx2
From the three equations by eliminating a and b, we obtain
d2y 4dy _
—v----- - + 3y = 0
dx2
dx
Remark
Sol. Differentiating y2 = 4a(x + b) w.r.t. x,
2y • — = 4a i.e. y — = 2a
dx
dx
Again, differentiating w.r.t. x, we get
The order of the differential equation will be equal to number of
independent parameters and is not equal to the number of all the
parameters in the family of curves.
2
dx
I =0
I Example 7 Find the order of the family of curves
which is the required differential equation. Thus, the
elimination of arbitrary leads to the formation of a
differential equation.
y = (q + c2)ex +c3 ex+e*.
■(i)
Sol. Here, the number of arbitrary parameters is 4 but the
order of the corresponding differential equation will not
be 4 as it can be rewritten as, y = (c^ + c2 + c3 ec<) ex,
which is of the form y = Aex. Hence, the corresponding
differential equation will be of order 1.
-(ii)
I Example 8 The differential equation of all
non-horizontal lines in a plane is given by
I Example 4 Find the differential equation whose
solution represents the family xy = aex + be~x.
xy = aex + be~ x
Sol.
Differentiating Eq. (i) w.r.t. x, we get
y
x
> —x
x — + y = ae -be
dx
Differentiating Eq. (ii) w.r.t. x, we get
x d2y , dy .l + ^ = ae* + te-x
dx
dx2 dx
'
(a)^ = o
(b)£ = 0
dx
(iii)
(c)
dx2
= 0 and
dy
dy2
=0
(d) All of these
Textbook of Integral Calculus
238
1
or sin a =
tan a =-----
Sol. The equation of the family of all non-horizontal lines in a
plane is given by,
ax + by = 1
(where a * 0) ...(i)
Differentiating w.r.t. y, we get
7i
and
(as a * 0and be R)
cos a = -
(as a
0and be R)
Hence, (c) is the correct answer.
I Example 11 The differential equation of all circles of
radius r, is given by
(b)(1 + (y,)2)3=r2y23
(a)(1 + (y,)2)2=r2yi
(d) None of these
(c)(1 + (y,)2)3=r2y22
I Example 9 The differential equation of all
non-vertical lines in a plane is given by
(a)^ = 0
(b)^ = 0
dx
dy
dy2
dx2
=0
(d) All of these
Sol. The equation of the family of all non-vertical lines in a
plane is given by ax + by = 1, where b # 0 and a e R.
Differentiating both the sides w.r.t. x, we get
dv
a + b—=0
(as b^Q and a e R)
dx
Again, differentiating both the sides w.r.t. x, we get
, d2y
(as b * 0 and a e R)
b-^- = 0
dx2
d2y
(as b * 0)
dx2
Differential equation of all non-vertical lines in a plane.
=P
=> (7 - xyt)2 = p2(l + y2) is required differential equations.
(as a * 0)
.-. Differential equation of all non-horizontal lines in a
d2x
plane is---- = 0. Hence, (b) is the correct answer.
dy2
= 0 and
7
- xy.
^=0
(c)
...(ii)
From Eqs. (i) and (ii), we get
dy
Again, differentiating w.r.t. y, we get
d 2x A
a dy2 ~
i
So/. Equation of circle of radius r,
(x - a)2+(y - b)2 = r 2
(Here, a, b are two arbitrary constants)
Differentiating Eq. (i), we get
2 (x - a) + 2 (y - b) y1 = 0
•
—(i)
•••(ii)
Again, differentiating Eq. (ii), we get
1 + (7 - b) y2 + 7? = 0
,2\
(7"^) = -
...(iii)
I 72
Putting (y - b) in Eq. (ii), we get
(x-a)
__ (1 + 7i2)7i
.(iv)
72
From Eqs. (i), (iii) and (iv), we get
(1+7i)2,7i2 , (l + 7i2)2
-4=o
yl
72
=>
= r2
(i + (yi)2)3 = rM
Hence, (c) is the correct answer.
dx2
Hence, (a) is the correct answer.
I Example 10 The differential equation of all straight
lines which are at a constant distance p from the
origin, is
(a)(y 4-xyJ2 -p2 (1 + yf) (b) (y - xyf) = p2 (1 + yJ2
(c) (y - xy!)2 = p2 (1 + y2) (d) None of these
Sol. As, we know
x cos a + y sin a = p
..•(i)
Represents the family of straight lines which are at a
constant distance p from origin. Differentiating Eq. (i) w.r.t.
x, we get
dy
cos a + sin a • — = 0
dx
I
I Example 12 The differential equations of all circles
touching the x-axis at origin is
(a) (y2 -x2) = 2xyf~>l
<dx J
(b) (x2 -y2)^ = 2xy
dx
(c) (x2 -y2) = 2xy|^
VdxJ
(d) None of the above
Sol. The equation of circle touches x-axis at origin.
=>
(x - 0)2 + (y - a)2 = a2
or
x2 + y2 - 2ay = 0
■(i)
Chap 04 Differential Equations
Differentiating w.r.t. x, we get
I Example 14 The differential equation satisfying the
x2
y2
curve ----- + —L— = i, where X being arbitrary
o +X b2 + X
unknown, is
2x + 2y — - 2a — = 0
dx
dx
=>
y-(a-y) = x
dx
(dy
dy
x+y —
(,dx
dx
a =-----7—\—
dy
-(ii)
dx
From Eqs. (i) and (ii), we get
(^y\
x+y —
<dxj
x2 + y2 - 2y
(a) (x + yy,) (xy, -y) = (o2 -b2)y,
(b) (x + yy,)(x-yy,) = y,
(c) (x-yy,)(xy, +y) = (o2 -b2)y,
(d) None of these
Sol. Here,
x2
y2
a2 + X
b2 + X
=1
-.(i)
=0
Differentiating both the sides, we get
dy _
2x , 2y
a",2 + X
(x2 -y2)^ = 2xy
dx
or
239
0
b2 + X dx
x (b2 + X) + y (a2 + X) yx = 0
Hence, (b) is the correct answer.
( ,2
I Example 13 The differential equation of all circles in
the first quadrant which touch the coordinate axes is
<
Also,
Sol. Equation of circles touching coordinate axes is
(x - a)2 +(y - a)2 = a2
A
X + W'i
>
,2 _ xb2 + a2yyl
a2 + X = a
x + yyi
(a) (x-y)2(1 + (y')2) = (x + yy')2
(b) (x + y)2(1 + (y')2) = (x + yz)2
(c) (x-y)2 0 + y') = (x + yy')2
(d) None of these
2
xb + a yy,
X=-
2
(a2-b2)x
a,2 + A =-------------x + yy,
(a2-b2)yyl
b2 + X =
x + yyi
-(ii)
-(iii)
From Eqs. (i), (ii) and (iii), we get
Differentiating, we get
x2(x + yyt) t y2(x4-yy,) _
2(x - a) + 2 (y - a) y' = 0
dy
; where y = —
i + y'
(a2-b2)x
(ii)
=>
(a2-b2)yyl
1
(x + yyi)(xyj-y) = (a2 - b2)y}
Hence, (a) is the correct answer.
I Example 15 The differential equation of all conics
whose centre lies at origin, is given by
(a) (3xy2 +x2y3)(y-xyi) = 3xy2 (y-xy, -x2y2)
(0,a) .... "......*(a\a)
n i)
(b) (3xy! + x2y2) (y, -xy5) = 3xy! (y-xy2 -x2y3)
(c) (3xy2 + x2y3) (y1 - xy) = l>xy} (y - xy, - x2y2)
>x
(a.0)
0
(d) None of the above
Sol. Equation of all conics whose centre lies at origin, is
From Eqs. (i) and (ii),
2
2
z
(
.
A2
[x-^- I + y- x + yy' JI xi ++ yy
y' >
l + y'
/
x2
xy -yy
l + y' J
=>
+
\2
(
•
(x-y)2 (y')2 +(y-x)2 = (x + yy')2
(x-y)2 (i + (y')2) = (x + yy')2
-(i)
Differentiating Eq. (i) w.r.t. x, we get
2ax + 2hxy} + 2hy + 2byy} = 0
'\2
= x+w
U+yJ I i+y' >
Hence, (a) is the correct answer.
ax2 + 2hxy + by2 = 1
=>
ax + h (y + xyj + byyx =0
Multiplying by x equation becomes,
ax2 + h (xy + x2y,) + bxyyr = 0
Subtracting Eqs. (i) and (ii), we get
h (xy - x2y,) + b (y2 - xyy,) = 1
•••(ii)
240
Textbook of Integral Calculus
(hx + by)y - xyi (hx + by) = 1
(hx + by) (y-xyi) = l
Hy-xyt) = y
(y “ *yi)
1 =--------• . hx + by
...(iii)
y-xyi
Again, differentiating w.r.t. x, we get
.
h 4 by, = -
t
h + by} =
xyz
(y-xyx)2
y - *?■ (y-xyi)3
...(iv)
=> (3xy2 + x2y3)(y - xyj = 3xy2 (y - xyx - x2y2)
Hence, (a) is the correct answer.
From Eqs. (iii) and (iv), we get
Exercise for Session 1
•M*> ■*:-■—ii.i V*—
i ■ ,^n—■ w-m^X^y,T»r».ir ■* <■
iZ1.'
1. The differential equation of all parabolas whose axis of symmetry is along X-axis is of order.
(a) 2
(b) 3
(c) 1
(d) None of these
2. The order and degree of the differential equation of all tangent lines to the parabola x2 =4y is
(a) 1,2
(d)4,1
(c)3,1
(b)2,2
yY
d2y
fcfyV
21 (d2y
3. The degree of the differential equation —y + 3
=x2log —y is
\dxj
r1vi 1
dx
^dx
(c)3
(a)1
(b)2
(d) Not defined
4. The degree of the defferential equation satisfying the relation V1+ x 2 + 71+y2 -
(a)1
/ ,2 \2
—
d y
O. The degree of the differential equation —y +
<dx2J
(a)1
(b)2
(®2-xsln
y2 -y-J1+*2)is
(d)4
(c)3
(b)2
<Cfx2?
is
(c)3
(d) Not defined
6. The differential equation of all circles touching the y-axis at origin, is
(a) y2 - x2 = 2xy
dx
(b) y2 - x2 = 2xy —
dy
(c) x2 - y2 = 2xy dy
dx
(d) x2 - y2 = 2xy —
dy
7. The differential equation of all parabolas having their axes of symmetry coincident with the axes of x, is
(a)yy2 + y2 = y +
(b)yy2 + y2 = o
(v)
Again, differentiating both the sides w.r.t. x, we get
n _ yi - yi ~ 3xy2 - xzy3 ! 3(y-xyi-x2y2)xyz
(y-*yi)3
(y-*yi)*
- **
(y-W
or
x*yz
(c)yy2 + y2 = y,
(d) None of these
8. The differential equation of all conics whose axes coincide with the coordinate axes, is
(a) xyy2 + xyf - yy1 = 0
(b) yy2 + yj2 - yy, = 0
(c) xyy2 + (x - y) K = 0
(d) None of these
9. The differential equation having y = (sin-1 x)2 + A (cos-1 x) + B, where A and B are arbitrary constant, is
(a) (1- x2) y2 - xy, = 2
(b) (1- x2) y2 + yyy = 0
(c) (1 - x) y2 + xy, = 0
(d) None of these
10. The differential equation of circles passing through the points of intersection of unit circle with centre at the
origin and the line bisecting the first quadrant, is
(a) yi (x2 + y2 - 1) + (x + yyO = °
(b) (yi - 1) (x2 + y2 -1) + (* + w) 2 (x - y) = o
(c) (x2 + y2 - 1) + yy2 = 0
(d) None of these
Session 2
Solving of Variable Seperable Form,
Homogeneous Differential Equation
Solving of Variable
Seperable Form
Solution of a Differential Equation
The solution of the differential equation is a relation
between the variables of the equation not containing the
derivatives, but satisfying the given differential equation
(i.e from which the given differential equation can be
derived).
Thus, the solution of — = e x could be obtained by simply
dx
integrating both the sides, i.e. y = e x+ C and that of,
dy
x2
—=px + q is y = p — + qx +C, where C is arbitrary
dx
2
constant.
(i) A general solution or an integral of a differential
equation is a relation between the variables (not
involving the derivatives) which contains the same
number of the arbitrary constants as the order of the
differential equation. For example, a general solution
d2x
of the differential equation —— = - 4x is
A differential equation of first order and first degree is of
the form — = f{x, y).
dx
Remark
All the differential equations, even of first order and first degree,
cannot be solved. However, if they belong to any of the standard
forms which we are going to discuss, in the subsequent articles
they can be solved.
Equations in Which the
Variables are Separable
dy
The equation — =/(x, y) is said to be in variables separable
dx
form, if we can express it in the form f(x)dx = g(y) dy.
By integrating this, solution of the equation is obtained
which is, j* f(x) dx = j g(y) dy + C
I Example 16 Solve
sec2 x tan y dx + sec2 y tan x dy = 0.
Sol. Dividing the given equation by tan x tan y, we get
sec 2 x ,
tan x
x = A cos 2t + B sin 2t, where A and B are the
arbitrary constants.
(ii) Particular solution or particular integral is that
solution of the differential equation obtained from the
general solution by assigning particular values to the
arbitrary constant in the general solution.
For example, x = 10 cot 2t + 5 sin 2t is a particular
d2x
solution of differential equation —— = - 4x.
Differential Equations of the
First Order and First Degree
In this section we shall discuss the differential equations
which are of first order and first degree only.
sec 2 y
dy = 0
tan y
This is variable-separable type
sec2 x ,
t sec2 y j
Integrating, | -------dx + ----- — dy = C
tan x
J tan y
In | tan x J + In | tan y | = In C; where C' = In C
In | tan x • tan y | = In C; (C > 0)
| tan x • tan y | = C
This is the general solution.
or
dy
I Example 17 Solve —= ex"
dy
ex
x2
2e-y
v .
,2
xx j
— = — + — => ey dy = (x + e )dx
dx ey ey
This is variable-separable form.
Sol. Here,
Textbook of Integral Calculus'
242
Integrating both the sides,
dy~ = In (x +1)
dx
fdy = |ln (x + 1) dx
3
J ey dy = | (x2 + ex) dx => ey = — + ex + C
Which is the general solution of the given differential
equation, where C is an arbitrary constant.
I Example 18 Solve -Jl+ x2 + y2 + x2y2 + xy
Sol. The given differential equation can be written as
7(l + y2)(l + x2) = - xy
dx
+x
x
_
2
ydy
Ji + y2
This is the variable-separable form.
/. Integrating both the sides, we get
-J
- Jl + x2 + log
7i + x2
y
, ..2
X
dy
’1 + x2 - 1
’1 + X2 + 1J
This is the general solution to the given differential
equation.
dy (
dy
»/
w
I Example 19 Solve y-x-~
„dx = a y2 + -^~ •
I CvTmnln IO
C---------- r---------------------- 7
dx
dy
- dx
y(l-ay) (x + a)
This is the variable-separable form.
Integrating both the sides, we get
JJ —
£-=f dx
y (i - ay) J (x + a)
fl- +
=>
ly
=>
=>
or
Differential Equations Reducible
to the Separable Variable Type
Sometimes differential equation of the first order cannot
be solved directly by variable separation but by some
substitution we can reduce it to a differential equation
with separable variable. “A differential equation of the
form— = f(ax + by + c) is solved by writing
dx
ax + by + c = t.”
I Example 21 Solve ^ = sin2 (x+3y)+5.
Sol. Rewriting the given equation as
=>
i.e.
y = xln(x + l)-x + ln(x + l) + C
This is the general solution.
To find the particular solution, put x = 0, y = 3 in the
general equation.
3=0-0+0+C
C =3
.'. The required particular solution is,
y = (x + 1) In (x + 1) - x + 3
uX j
OX
y - ay2 = (x + a)
y = x In (x + 1) - | -^—dx
x+1
i.e.
= 0.
(integration by parts)
=—
dx dx
The given differential equation becomes,
1 dt
— -1 | = sin2(t) + 5 => — = 3 sin2t +16
3 \dx J
dx
dt
= |dx
3 sin2 t + 16
Sol. Let x + 3y = t, so that 1 +
. f
r
sec2 t dt
=x+C
J 3 tan2 t + 16 sec2 t
a
dx
dy = J x + a
1-ay^
sec2 t dt „
-------- = x + C
f ------19 tan2 t + 16
In y - - In (1 - ay) + In C = In (a + x)
a
(a + x)(l - ay)^ _
In
ln(C)
y
du
|~---- = x + C; where tan t = u
J 19u2 +16
sec2 t dt = du
Cy = (a + x) (1 - ay) is the general solution.
I Example 20 Solve edy/dx = x+1, given that when
du
=>
x = 0,y = 3.
~2
16
U +---
T9
— tan
19 4
4 7
tan
V19
—— tan (3y + x)
+C
„
1
19
Sol. This is an Example of particular solution.
e(dy/dx) = x + l
=>
1
+c
Chap 04 Differential Equations
I Example 22 Solve (x + y)2 y^- = a22.
Sol.
§ Example 24 Solve
2
(* + y)2
-.(i)
n x
dy dt
Put x 4 y = t => 1 4 — = —
dx dx
dx
dx
x = t - a tan
a7
4-C
x+y
x = x + y - a tan-1
i.e.y = a tan
(i)
tan 0 = —
x
—(ii)
and
dx
• *2 dt
i.e. r — = ai2 4-12, separating the variable and integrating.
dx
y
t2
a2
1dt
a2 + t
a2 + t2
i.e.
x2 4- y2 = r2
= a2
Eq. (i) reduces to t2
x 4- y
a
+C
a
a2 — x2 — y2
x2 + y2
x dx 4- y dy
xdy - y dx
Sol. Let x = r cos 0, y = r sin 6
So that
=
or
243
From Eq. (i), we have d (x2 + y2) = d (r2)
x dx + y dy = r dr
i.e. •
From Eq. (ii), we have
d
...(iii)
d (tan 0)
i.e.
x dy - y dx
2 n jn
——
= sec 0 d0
x2
i.e.
x dy - y dx = x2i sec-12 0 dQ
d0 = r*
r2 cos2 0 sec2 0 d0...(iv)
Using Eqs. (iii) and (iv) in the given equation, we get
- C is the required general solution.
r dr
a2 - r2 .
dr
r2 dQ
= do
-r2
I Example 23 Solve
(2x4- 3y -1) dx 4- (4x 4- 6y - 5) dy = 0.
Sol. (2x + 3y - 1) dx 4- (4x + 6y - 5) dy = 0
—(i)
u = 2x + 3y - 1
Substitute
3dy
*=2 + ^
dx
dx
or
or
■Jx2 + y2 = a sin {C + tan 1 (y / x)}
0+C or r = asin(0+C)
It is advised to remember the results (iii) and (iv).
u 4- (2u - 3) - I — - 2 I = 0
3 Ux
i.e.
sin
dy _ 1 (du
-I —-2
dx 3 I dx
:. Eq. (i) reduces to
i.e.
Le.
J
u - - (2u - 3) 4- - (2u - 3) — = 0
3
3
dx
- u 4- 6 1
du
——— 4 ~(2u-3) —= 0
3
3
dx
Writing this in the variable-separable form
Homogeneous
Differential Equation
By definition, a homogeneouos function f(x,y) of degree
n satisfies the property
/(Xx, Xy)=Xn/(^y)
For example, the functions
fI 2uu “~63 > du = dx
2u -3 ,
------- du
u-6
2 (u —6)4 9
fi(x,y) = x3 + y3
/2(x,y) = x2 + xy+y2
f3(x,y)^x3ex,y +xy2
du
are all homogeneous functions, of degrees three, two and
three respectively (verify this assertion).
x4C=2u491n|u-6|
Observe that any homogeneous function f(x, y) of degree
n can be equivalnetly written as follows :
/ \
y^
/(x,y) = x" L =ynf xj
\y)
x+C~J
(u -6)
x 4 C = 2 (2x 4 3y - 1) 4 9 In | 2x 4 3y - 7 |
3x 4 6y - 2 4 9 In 12x 4 3y - 7 | = C is the general solution.
Remarks
Sometimes transformation to the polar coordinates facilitates
separation of varibales. It is convenient to remember the
following differentials.
2. x dy - y dx = r2 dQ
1. x dx + y dy = r dr
3. dx2 + dy2 = dr2 + r2d >2
Forexample,
f(x, y)-x3 +y3
/
= x3 1 +
3>
=/
x, 7
14-
V
x
\yj
3\
7
244
Textbook of Integral Calculus
Having seen homogeneous functions we define
homogeneous DEs as follows :
Step III Shift v on R.H.S and seperate the variables in v
and x.
Any DE of the form M(x, y) dx + N (x, y)dy = 0
Step IV Integrate both sides to obtain the solution in
terms of v and x.
or — = -
y). is called homogeneous if M (x, y) and
N(x,y) are homogeneous functions of the same degree.
Step V Replace v by — in the solution obtained in step IV
x
to obtain the solution in terms of x and y.
What is so special about homogeneous DEs? Well, it turns
out that they areextremely simple to solve. To see how,
Following examples illustrate the procedure.
we express both M(x, y) and N(x, y) as, say xnM — and
I Example 25 Solve y dx + (2^xy -x)dy = 0.
y dx + (2-Jxy - x)dy = 0
Sol.
xnN — . This can be done sicne M(x,y) and N(x, y) are
x)
both homogeneous function of degree n. Doing this
reduces our DE to
dy _ _ M(x,y) _
dx
N(x,y)
xnM p
M X)
I XJ
N
~M(t)
N(t)
Now, the simple substitution y = vx reduces this DE to a
VS form
=>
y = vx
dy
, dy
dx
dx
dv
.
v + x — = P(v)
dx
dv
dx
x
This can now be integraed directly since it is in VS form.
Let us see some examples of solving homogeneous DEs.
Alogorithm for Solving
Homogeneous Differential Equation
Step I Put the differential equation in the form
dy <t>(x,y)
dx
i.e. x • {udx + (2 Vu - 1) u dx + x du (2 Vu - 1)} = 0
i.e.
dx (2u312 - u 4- u) + x du (2 Vu - 1) = 0
dx
Separating the variables, —
x
( 2 Ju- 1 'I
du = 0
Integrating both the sides In | x | + In | u | + -4= = C
dr
=C
In | xu | +
or ln|y| + ^| = C ^/u = y
X
■du
Which is the general solution.
I Example 26 Solve (x2 + y2) dx - 2xy dy = 0.
Thus, — = p — transforms to
dx
P(v) - v
This is homogeneous type. Substitute y = ux
dy
+ x du
dx
dx
Equation ux dx + (2 -Jx22 u - x)(u dx + x du) = 0
X,
k X,
(The function P(t) stands for
7
■(•)
V(x,y)
Step II Put y = vx and — = v + x — in the equation in
dxdx
step I and can out x from the right hand side. The
dv
equation reduces to the form v + x — - /(v).
dx
.
dy x2 + y2 1
Sol. Here, — =------ — =.dx . 2xy
2 . y
x.
With y = ux,~ = u + x — ,so that the differential
dx
equation becomes
du 1 1
A
u+x—=- -+u
dx 2 u
J
x du
dx
1 + u2
----------u
2u
x du
dx
1-u2
2u
2u
1-u2
J
X
- log 11 - u2 | = log | x | - log | C |
.
x(l-uz) = C
x 2 - y2
=c
x
Hence, x2 - y2 = xC, is the required solution.
/
'
X
Chap 04 Differential Equations
2dy_y_
I Example 27 Solve
dx
x
V
x2 ■
Sol. The above equation is homogeneous so that we put y - ux.
r
du
2 u + x — = u + u 2 => 2u + 2x— = u + u 2
dx
dx
Sol. Since, xy = v, y = — and d (x y) = d v
x
x dy + y dx = dv
i.e.
x dv -v dx
dy = d
and
72
x dy = dv - — dx
x
i.e.
du =--dx
u2 -u 2x
„ du
2x — -u 2
dx
V
— f(y)dx + g(v) dv- — dx
x
X
t du _ 1 rdx
Ju(u-l)~2J x
*u~l
du-(Ldu=l(^.
1
Which is in variables separable form.
=> . log I u - 11 - log I u I = - log I x I + log ICI
“■
=>
=>
'
”
2
log ~—1 =log|cVx|
u
U~1
Reducible to Homogeneous Form
^ = cV7
u
y
y - x = C 4x -y
Which is the required solution.
I Example 28 Solve
(1+2 ex/y)dx + 2 ex/y (1-x/y)dy = 0.
Sol. The appearance of x / y in the equation suggests the
substitution x = vy or dx = v dy + y dv.
The given equation is
(1 + 2 ev)(v dy + y d v) + 2ev (1 - v)dy = 0
i.e.
y (1 + 2 ev) d v +(v + 2ev)dy = 0
i.e.
l + 2ev ,
dy n
--------- d v + — = 0
v +2ev
y
Many a times, the DE specified may not be homogeneous
but some suibtale manipulation might reduce it to a
homogeneous form. Generally, such equations involve a
function of a rational expression whose numerator and
denominator are linear functions of the variable, i.e., of
the form
dy = f ' ax + by +7
(i)
<dx+cy+f;
dx
Note that the presence of the constant c and f causes this
DE to be non-homogeneous.
and select h and k so that
ah + bk + c = 0
f £-0
,(ii)
dh + ek + f = 0
log | v + 2 ev | + log | y | = log | C |
x
v=—
.
y)
— + 2ex,y y = C
y
Type I
To make it homogeneous, we use the substitutions
x
X+h
l + 2ev ,
Integrating, J --------dv +
v + 2ev
J y
=>
=0
v{/(v)-g(v)} dx + g (v) dv = 0
x
dx
_Q
— + _ S (v) dv
X
V {/(v)-g(v)}
Le.
2J x
Ju
245
This can always be done ^if
(i) now reduces to = f
J
=/
(x + 2yex,y )=C, is required solution.
I Example 29 Show that any equation of the form
y f(xy) dx + x g(xy) dy = 0
can be converted to variable separable form by substi­
tuting xy = v.
j. The RHS of the DE in
' a(X + h) + b(Y + k) + c'
<d(X + h) + e(Y + k + f)J
'aX + bY'
^dX + eY;
(Using Eq. (ii))
This expression is clearly homogeneous! The LHS of Eq. (i)
. dy , . ,
dy dY dX
dy dx
.
is — which equals —------------- . Smce —------- -- 1, the
dx
dY dX dx
dY dX
LHS
equals
. Thus, our equation becomes
246
Textbook of Integral Calculus
"=f aX + bY'
dX
Thus,
ax + by + c _ X(dx + ey) 4- c
dX + eY;
We have thus succeeded in transforming the
non-homogeneous DE in Eq. (i) to the homogeneous DE in
Eq. (iii). This can now be solved as described earlier.
I Example 30 Solve the DE ^- = ———-•
dx y-3x+3
Sol. We substitute x —> X + h andy —>Y + k where h,k need
to be determined
dy _dY _ (2Y - X) + (2fc - h - 4)
dx ~ dX ~ (Y -3X)+ (fc-3ft+3)
h and k must be chosen so that
2k - h - 4 = 0
k - 3h + 3 = 0
This gives h = 2 and k = 3. Thus,
x = X +2
y = Y +3
Our DE now reduces to
dY _2Y -X
dX~ Y - 3X
Using the substitution Y = vX, and simplifying, we have
(verify),
v-3
-dX
dv =----X
v2 5v +1
We now integrate this DE which is VS; the left-hand side
can be integrated by the techniques described in the unit of
Indefinite Integration.
Y
Finally, we substitute v = — and
X = x-2
Y = y-3
to obtain the general solution.
dx +ey + f
This suggests the substitution dx + ey = v, which will give
d +e — =—
dx dx
dy 1 (dv
— = —------ a
dx e dx
J
=>
Thus, our DE reduces to
1 ( dv
, i Xv + c
-- d —
e \dx
7 v+J
dv _ Xev + ec .
---------- + d
~di~
_(Xe+d)v + (ec + d)
v +y
=>
——.dv=dx
(Xe + d) v + ec + df
which is in VS form and hence can be solved.
I Example 31 Solve the DE — = X + 2^ 1 •
dx x + 2y + l
Sol. Note that h, k do not exist in this case which can reduce
this DE to homogeneous form. Thus, we use the substitu­
tion
x + 2y = v
dy =—
dv
=>
1. + 2„—
dx dx
Thus, our DE becomes
lfdv_A_v-l
v -1
2 Ux J v + 1
dv 2v - 2 , 3v -1
---- =------- + 1 =------dx v + 1
v+1
Type II
Suppose our DE is of the form
f ax + by + c
T
dx =f dx +ey + f ?
We try to find h, k so that
ah + bk + c = 0
dh + ek + f = 0
What if this system does not yield a solution? Recall that
this will happen if - = - . How do we reduce the DE to a
b e
homogeneous one in such a case?
Let ~ = - = X (say).
de
dx +ey + f
. v + 1 ,----- ,
-------dv = dx
3v — 1
=>
1
3
l + -±- dv - dx
3v-l
Integrating, we have
4
A
1
v + - In (3v - 1) I = x +
3
Substituting v = x + 2y, we have
4
x + 2y + — In (3x + f>y - 1) = 3x + C2
?
y - x + - In (3x + 6y - 1) = C
247
Chap 04 Differential Equations
I Example 32 The solution of the differential equation
dy
sin y + x
dx
sin 2y - x cos y
+ 2x2
(a) V2
X
is
,±V2 _y+ -Jy2 +V2x2
x
(b) V2x
(a) sin2 y = xsiny + —+ C
± ’
(c) V2 y
x2
(b) sin2 y = xsiny- — + C
X
(d) None of the above
(c) sin2 y = x + siny + — + C
2
Sol. The given differential equation can be written as
X2
,
y2
(d) sin2 y = x-siny + — + C
Sol. Here,
=>
=>
dy
sin y + x
dx
sin 2y - x cos y
dy
sin y + x
cos y -j- =----- ------- , put sin y = t
2 sin y - x
dt
dx
or
=(y2 + 2x2) 1 +
dx
y
„.(i)
Let y = vx
dv
dy
dx
dx
Eq. (i) becomes
V+1
dv
v+x—=v±
dx
2v - 1
2v - 1
+ 4x2 + 4xy
v+x—=—
dv v + 1
v + 1 - 2v2 + v
x — =---- v =--------dx
O'
^y = y±
dx x
2^xJ
1 —- , put t = vx
2t-x
xdv
vx + x
----- + v =---------dx
2vx - x
2
=
2v - 1
2v2 - v
,
dx
--- ------- dv = —
x
- 2v2 + 2v + 1
- v2 +1
2
or
J X
V2 log | v + yv z+2| = Iog|xC|
On solving, we get
=> Ji log
2
sin y = x sin y + — + C
y + yy2 +2x2
= log | xC |,
X
putting x = 1 and y = 0
Hence, (a) is the correct answer.
I Example 33 The equation of curve passing through
(1, 0) and satisfying
2
9
•) f fdy} 2\
( dy
, is given by
y^+2x = (y2 + 2x2) 1+ ~
dx
< [dx)
7
c = (4iyr2
+ 2XZ
r~
±-7z
------ = V2 x 72
Curves are given by
X
Hence, (a) is the correct answer.
248
Textbook of Integral Calculus
Exercise for Session 2
.
(x + y)2
1. The solution of — =
is given by
dx (x + 2)(y~2)'
2(y-2)
(b)(x + 2)'[n-2(y~2) = k e(* + 2)
I (X + 2) J
(a)(x+2]
1+ — ^ = ke2y/*
x J
(c)(x+2)3
1+2<1^1 = ke * + 2
2(y-2)
(d) None of these
x+ 2 ?
2. If (y3 -2x2y)dx + (2xy2 -x3 ) dy = 0, then the value of xy Jy2 -x2, is
(a)y2 + x
(b)xy2
(c) any constant
(d) None of these
3. The solution ofdy /dx =cos (x + y) + sin (x + y), is given by
(a) log 1 + tan
x+y
2
=x+C
(b) log | 1+ tan (x + y) | = x + C
(c) log 11-tan (x + y) | = x + C
(d) None of these
4. The solution of — = (x + y-1) + —, is given by
dx
log(x + y)
(a){1+ log (x + y)) - log{1+ log (x + y)) = x + C
(c){1+ log (x + y))2 - log{1+ log (x + y)| = x + C
(b) {1 - log (x + y)) - log {1- log (x + y)) = x + C
(d) None of these
5. The solution of (2x2 + 3y2 - 7) xdx - (3x2 + 2y2 -8) ydy = 0, is given by
(a) (x2 + y2 - 1) = (x2 + y2 - 3)5 C
(b) (x2 + y2 - 1)2 = (x2 + y2 - 3)5 C
(c) (x2 + y2 - 3)= (x2 + y2 - 1)5 C
(d) None of these
'-’M
X — 1J
(x -1)2 + (y - 2)2 tan
6. The solution of — =
dx
(xy - 2x - y + 2) tan
(a){(x- 1)2 + (y- 1)2} tan"1
y-2
x- 1
is equal to
I x-1J
- 2 (x - 1) (y - 2)= 2 (x - 1)2 logC (x - 1)
(b){(x - 1)2 + (y - 1)2) - 2 (x - 1) (y - 2) tan"1
y-2'
= 2(x- 1)2 logC
X- 1 ?
(c) f(x - 1)2 + (y - 1)2) tan"1
——^x-1
- 2 (x - 1) (y - 2)= logC (x - 1)
(d) None of the above
7 tu
■
* dy ( x + 2y - 3 \2
/. The solution of—= ------- -----.is
dx {2x + y + 3 )
(a) (x + 3)3 - (y - 3)3 =C (x - y + 6J4
(b) (x + 3)3 - (y - 3)3 =C
(c) (x + 3J4 + (y - 3f = C (d) None of these
a _.
...
,dy -cos x (3 cos y-7sin x-3) .
o. The solution of — =---------- , is
dx
sin y (3 sin x - 7 cos y + 7)
(a)(cosy-sinx- 1)2 (sinx + cosy- 1)5 =C
(b) (cos y - sin x - 1 )2 (sin x+cosy-1)3=C
(c) (cos y - sin x - 1 )2 (sin x + cos y - 1)7 = C
(d) None of these
249
Chap 04 Differential Equations
9. A curve C has the property that if the tangent drawn at any point P on C meets. The coordinate axes at A and B,
then P is the mid point of AB. The curve passes through the point (1,1). Then the equation of curve is
(a) xy = 1
(b) — = 1
y
(d) None of these
(c) 2x = xy - 1
10. The family of curves whose tangent form an angle — with the hyperbola xy = 1, is
4
(a) y = x - 2 tan-1 (x) + K
(b) y = x + 2 tan"1 (x) + X
(c)y = 2x- tan"1 (x) + K
(d)y = 2x + tan"1(x) + K
11. A and B are two separate reservoires of mater capacity of reservoir are filled completely with water their inlets
are closed and then the water is released simultaneously from both the reservoirs. The rate of flow of water out
of each reservoir at any instant of time is proportional to the quantity of water in the reservoir at that time. One
1
hour after the water is released, the quantity of water in the reservoir A is 1 - times the quantity of the water in
reservoir B. The time after which do both the reservoirs have the same quantity of water, is
(a) log3/4 (2)
(b) log3/4(—1 ]
2
(c) logi/2^j
(d) None of these
12. A curve passes through (2,1)and is such that the square of the ordinate is twice the rectangle contained by the
abscissa and the intercept of the normal. Then the equation of curve is
(a) x2 + y2 = 9x
(c) 4x2 + 2y2 = 9x
. • ,
(b) 4x2 + y2 = 9x
(d) None of these
2
13. A normal at P (x, y) on a curve neets the X-axis at Q and N is the foot of the ordinate at P. If NQ =
.
(1+x2)
Then the equation of curve passing through (3,1) is
(a) 5(1+ y2)= (1+ x2)
(c) (1+ x2) = (1+ y2).x
(b) (1+ y2) = 5(1+ x2)
(d) None of these
14. The curve' for which the ratio of the length of the segment intercepted by any tangent on the Y-axis to the length
of the radius vector is constant (k), is
(a) (y + 7*2-y2)x*“1 = c
(b) (y + 7*2 + y2 )x*-1 = c
(c) (y - 7*2 -y2)x*-1 = c
(d) (y - Jx2 + y2 )x*"1 = c
15. A point P(x,y) notes on the curve x2J3 + y2/3
>Ofor each position (x,y) of p, perpendiculars are drawn
from origin upon the tangent and normal at P, the length (absolute valve) of them being P5(x) and P2(x)
brespectively, then
(a) ^1.^2 <0
<0
•dx dx
dx dx
(c)^l.^>0
(d)^Pt ^P2 >o
dx dx
dx dx
Session 3
.
W—i
—.-Trzri
...
.-. • ••w.sarx.-
4-rTjjtv<. ? ’*** •’' i inTi^gt;
.
^- rr- .-r-.Tv^.^>vsrew^»T«w?-4ST^r»aw«!aMMu
Solving of Linear Differential Equations,
Bernoulli's Equation, Orthogonal Trajectory
Solving of Linear
Differential Equations
First Order Linear
Differential Equations
A differential equation is said to be linear if an unknown
variable and its derivative occur only in the first degree.
An equation of the form
^ + P(x)-y=Q(x).
dx
Where P(x) and Q(x) are functions of x only or constant
is called a linear equation of the first order.
To get the general solution of the above equation we
proceeds as follows. By multiplying both the sides of the
above equation by e $Pdx
, we get
^=Qe j Pdx
l^.^ye+yp.e
dx
!^.^+y±(e
^)=ec/^
d
dx
dx '
Jpdx
d ,
i.e.
—(ye
dx
T
•
(Wx
\Pdxdx+C
Integrating, we get ye1
= jQe
i.e.
e
)=Q-S*
Here, the term e
which converts the left hand
expression of the equation into a perfect differential is
called an Integrating factor. In short it is written as IF.
Thus, we remember the solution of the above equation as
y (IF) = jQ (IF) <tr + C.
Algoritm for Solving A Linear
Differential Equation
Step I Write the differential equation in the form
dy I dx + Py = Q and obtain P and Q.
Step II Find integrating factor (I.F.) given by I.F. = e^Pdx.
Step III Multiply both sides of equation in Step I by I.F.
Step IV Integrate both sides of the equation obtained in
step HI. w.r.t x to obtain y (I.F.) = J Q.(I.F.) dx + C
This gives the required solution following examples
illustrate the procedure.
dy
S Example 34 Solve -~-+2y = cos x.
Sol. It is a linear equation of the form
^ + Py = 2(x)
dx
where P = 2 and Q = cos x
Then
IF = JPdx = e^2dx = e 2x
Hence, the general solution is y (IF) = J Q (IF) dx
i.e.
y • e2x = J e2x cos x dx + C
y-e 2x =
e2x
[2 cos x + sin x] + C
dy y
I Example 35 Solve — + — = log x.
ax x
Sol. It is a linear differential equation of the form
^- + Py = QM
dx
Here,
Then
P = —, Q = log x
x
efl/xdx = elogx
=x
IF =
Hence, the general solution is
y (IF) = j Q (IF) dx + C
i.e.
y* = J(log x) x dx + C
i.e.
z,
, x2 rl
yx=(logx) —- I-- — + C
2 Jx 2
i.e.
x2 ,,
. x2
yx = — (log x) - —- + C
2
4
251
Chap 04 Differential Equations
I Example 36 Solve ~
— =
= —-—- ------dx 2y\ny+y-x
3^3
Sol. The equation can be written as
8
dx_2ylny + y- x
= (21ny + l)- —
dy
y
y
i.e.
— + —-x = (21ny + l)
dy y
3^3
n
Now, when
1-1
-.-. — + c => c=o
2
3
2
sin 2x
y=
2(1 - tan2 x)
t
dy
I Example 38 Solve y- + y <|)' (x) = 0(x) • 0'(x), where
In this equation it is clear that P = — and Q = (2 In y + 1).
y
Which are function of y only because equation contains
derivatives of x with
respect to y.
= efpdy = efl'ydy = ein y =
4>(x) is a given function.
P=0'(x) and Q = 0(x) 0' (x)
So/. Here,
IF = e^(x)<ix =eo(x)
=e
The solution is
y
= |(])(x)-(|)'(x)-e0(x,dx = | t ■ e‘ dt,
y
The solution is; x (IF) = j(2 In y + 1) (IF) dy
where ^(x) = t
i.e.
xy = |(2 In y + 1) • y dy = y2 In y + C
i.e.
x = y In y + —
y
Note In some cases a linear differential equation may be of the
cfx
form — + x = Q, where % and Q are function of y alone. In
dy
such a case the integrating factor is
dy.
I Example 37 Solve cos2 x
- y tan 2x = cos4 x,
6
Sometimes a differential equation is not linear but it can
be converted into a linear differential equation by a
suitable substitution. An equation of the form
^ + Py=Qy".
8
Where P and Q are functions of x only, is known as
Bernoulli’s equation (for n =0 the equation is linear.)
2
2
It is easy to reduce the above equation into linear form as
below:
— - sec x • tan 2x • y = cos‘ x
dx
2 tan x
IF = e]
(n*0,1)
dx
Sol. The given equation can be written as
f- tan 2x • sec2 x dx
+C
Bernoulli's Equation
Tt j = 3V3
where |x|<ll and y
dy
ye0(x) = {(Xx) - 1}
i.e.
c
i
ye^^e^t-^ + C
i
,
sec x dx
b ----------tan2x-l
=e
Dividing both the sides by y", we get
J2
= e f, where t = tan x - 1
-’^ + Py'
y
= elnlt| = |r | = | tan2 x -11
It is given that | x | < — and for this region tan2 x < 1.
4
IF = (1 - tan2 x)
The solution is
y (1 - tan2 x) = jcos 2 x (1 - tan2 x) dx
Putting y
1 -n = z
and hence, (l-n)y n^ = — the
dx
= j(cos 2x) dx = ^ + C
2
dx
equation becomes — + (1 - n) Pz = (1 - n) Q which is linear
dx
inz.
™
f(l-n)Pdx
IF = e}
Here,
cos2 x - sin2 x)dx
-n=Q
dx
The solution is,
f(l -n)Pdx
z eJ
= j {(1-n)-Q-e^1~n)Pdx }dx
252
Textbook of Integral Calculus
,
dy _y£W-y 2
-, where <|)(x) is
I Example 41 Solve — =
dx
(|)(x)
i a given function.
I Example 39 Solve (y log x -1) y dx = x dy.
Sol. The given differential equation can be written as
x
dx
...(i)
4- y = y2 log x
Sol. The equation can be written as
dy <K(*)
y_
= - y2
dx
<Xx)
<Xx)
Dividing by xy2, hence
1 i
1
y
— = - log x
xy x
dx
1
1 dy dv
— = v =>---- — = —
y
yz dx dx
Let
,-2 dy
—y
dx
i.e.
dv 1
1 .
V=
logx
dx x---------x
Which is the standard linear differential equations, with
_ 1„ 1 ,
P =---- ,Q =----- logx
Let l = z.Sothat,-±^=*
dx
y
y2 dx
So that
J- 1/X dx
IF =
1
dz + £(x)
■z =------ dx <Kx)
<X*)
f^dx
. 0(x)
IF = e'^
X
X
-In X
= e - In x —
— c
X
4>'(x) 1 _ 1
<Xx) y <Xx)
/. The solution is z • (Xx) = [-1— • (J)(x) dx - x + C
• J <&x)
The solution is given by
logx
1
v • — = [- f- - log xl dx = - [
dx
J X < X
J
J x2
X
X
X
J X
= 1285 + l + c
X
X
v = 1 + log x + Cx = log (ex) + Cx
— = log (ex) + Cx or y {log (ex) + Cx} = 1
y
or
So that
Remark
Another type of equation which is reducible to the linear form is
/'(y)^+P(x)./(y) = Q(x)
dx
An equation of this type can be easily reduced the linear form by
taking z = f(y).
dy 1
-2 —
4---- x = x
dx y
i
—=z
y
1 dy _ dz
y2 dx dx
The solution is, z • e
The given equation reduces to
dz
----- xz = - X
dx
IF = eJ[" X<lX - --x’/2
i.e.
tan y • e
*2 = Jx3 • ex dx
= ljx
2 e x1 •(2x)dx = -ft-er dt,
2J
tan y ■ ex = -(t-e‘-ef) + C
2
.-. The solution is
2
_ „-xi/2 _
i.e.
—J
x•e
- x*/2
1 = 1 + Cex2,z
• y
dx-e
. • I
where t = x2
V
ze
+ 2x tan y = x3.
Sol. Let tan y = z so that sec2 y • — = —
dx dx
Thus, the given equation reduces to
dz n
3
— 4-2x-z = x
dx
IF = J2xdx = e x1
Sol. Dividing by y2, we get
Let
^1 = xX +
+C
C ie,
ie,
=y
y
x+C
i.e.
I Example 42 Solve sec2 y
I Example 40 Solve ^- + xy = xy2.
y
=eln0(x)=<|)(x)
-x2/2+c
i.e.
• — X*
g
tan y = Ce~ x 4-------- e
2
tan y = Ce~ x
2
4(x!-1)
(x2-l)
Chap 04 Differential Equations
253
I Example 43 Solve ^-+x(x+y) = x3 (x+y)3 -1.
Orthogonal Trajectory
So/. The given equation can be written as
Any curve, which cuts every member of a given family of
curves at right angles, is called an orthogonal trajectory of
the family. For example, each straight line passing through
the origin, ie, y = kx is an orthogonal trajectory of the
| — +1 | + x (x + y) = x3 (x + y)'■3
< dx
J
d(x + y)
+ x(x + y) = x3 (x + y)3
dx
d(x + y)
+ x(x + y) 2 = x3
i.e. (x + y)-3 •
dx
d (x + y) dz
Let (x + y) 2 = z so that -2(x + y)-3
dx
~~
dx
The given equation reduces to
1 dz
3
------- + xz = x
2 dx
i.e.
dz
i.e.
----- 2xz = - 2x
dx
TT?
The solution is
2x3-e-x’
= Cex’
I Example 45 Find the orthogonal trajectories of the
hyperbola xy = C.
=cos y(1 - xcos y).
sin y — = cos y (1 - x cos y)
dx
dy - cos y = - x cos 2 y
sin y —
dx
dy - sec y = - x
tan y • sec y • —
dx
dy dv
sec y = v => sec y tan y • — = —
Let
dx dx
dv
So that
------- V = - X
dx
Which is linear differential equation with P = -1, Q = - x
f-ldc
= eJ
= e~x
- x e-x dx = xe~ x
= e-x(x + l) + C
or
v = (1 + x) + Cex
or
sec y = (1 + x) + Cex
■(ii)
x dx
.(Hi)
------- + y = 0
dy
This is the differential equation for the orthogonal
trajectory of given family of hyperbola. Eq. (iii) can be
rewritten as x dx = y dy, which on integration gives
x - y = C.
This is the family of required orthogonal trajectories.
I Example 46 Find the orthogonal trajectories of the
curves y = ex2.
Sol. Here,
The solution is given by
v-e~x =
(0 ’
Substitute - — for — in Eq. (u), we get
dy
dx
Dividing by cos2 y, we get
IF = eJ
So/. The equation of the given family of curves is xy = c
Differentiating Eq. (i) w.r.t. x, we get
x dy
—- + y = 0
dx
So/. The given differential equation is
or
(ii) Differentiate f =0; w.r.t x’ and eliminate ‘c’, i.e. form
a differential equation.
(iv) By solving this differential equation, we get the
required orthogonal trajectories.
+C
(x + y)2
I Example 44 Solve sin y-
(i) Let /(x, y, c) = 0 be the equation of the given family of
curves, where c is an arbitrary parameter.
(iii) Substitute - — for — in the above differential
dy
dx
equation. This will give the differential equation of
the orthogonal trajectories.
f-2xdx
1
Procedure for Finding the
Orthogonal Trajectory
i
IF = ei
z e
family of the circles x2 + y2 = a2.
+C
y = cx2
Differentiating w.r.t. x, we get
^=2cr
dx
Eliminating c from Eqs. (i) and (u),
1 dy
x2
y=
2x dx
(ii)
254
Textbook of Integral Calculus
_ =x—
Jy
2y
...(iii)
dx
This is the differential equation of the family of curves
given in Eq. (i).
dx
dy by - —
Now, to obtain orthogonal trajectory replace —
in
dx
dy
Eq. (iii).
=>
2y = -x —
dy
=»
-y = ~ + C =* x2+y2 + Cy = 0
y
Represents family of orthogonal trajectory.
So/. Here,
y2
I Example 47 Find the equation of all possible curves
that will cut each member of the family of circles
x2 + y2 - 2cx = 0 at right angle.
-.(i)
■(ii)
Ti
Substituting ‘a’ in Eq. (i), we get
x
x2 +
+ y2
y' _ 2(* + yyi) y = 0
yi
x2 + y2 - 2cx = 0
(x2-y2)yi-2xy =0
Differentiating w.r.t x, we get
2x + 2yy} - 2c = 0
=>
c = x + yyx
From Eqs. (i) and (ii), we eliminate c
x2 +y2 -2(x + yy1)x = 0
- x2 + y2 - 2xy
x2 + y2 - ay = 0
Differentiating, we get
2x + 2yyx - ayx=Q
a = 2(x + yyt)
=> x2 + 2y2 = Cb is the required family of orthogonal
trajectory.
or
Integrating both the sides, we get
x2 + y2 - ay = 0.
y2 = “v + C1
=>
I? J
I Example 48 Find the orthogonal trajectory of the
circles
or
2y dy = - x dx
Integrating both the sides, we get
Sol. Here,
f x 2\
- dy = d —
=>
=0
This is the differential equation representing the given
family of circles. To find differential equation of the
.,
i. . . .
,
dy
dx
orthogonal trajectories, we replace---- by-------.
dx
dy
„
dx
y 2 - x 2 = -2xy—
dy
This is the differential equation of the family of circles
given in Eq. (i).
-(ii)
The differential representing the orthogonal trajectory is
obtained by replacing — by - —.
dx J dx
i.e.
-(x2 -y2)^-2xy = 0
dy
2xy dy - y2 dx = - x2 dx
xd(y2)-y2dx_
x2
( 2\
d y_ = - dx
y2 dy = x2 dy-2xydx
dy
yd (x2) - x2 dy
y2
Integrating both the sides, we get
y2 + x2 = Cx, is required family of orthogonal trajectories.
Chap 04 Differential Equations
255
Exercise for Session 3
dy
t
1. The solution of (1 + x2)-^+ y =e tan"1 x , is given by
■ dx
.tan"1 x
7
,2 tan"1 x
(a)2yetan x =e
= e 2tan''x +C
(b)yetan y =e2tan x +C
=e
e2tan
2 tan- ’y +C
(c)2yetan’yX =
(d) None of these
2. The solution of — + ——y y = x Vy. is given by
dx 1-x
(a)37y + (1-x2) =C (1-x2)1'4
(b) j Vy + (1-x2) =C (1-x2)3'2
(c)3 7y-(1-x2)=C(1-x2)3/2
(d) None of these
3. The solution of — + x sin 2y = x3 cos2 y, is
dx
(a)ex’ = (x2 - 1)e*z tany + C
(b) e*’ tan y =
(c)ex2 tan y = (x2 - 1) tan y + C
(d) None of these
(x2 - 1) ex’ + C
4. The solution of3x (1-x2)y2 dy Idx + (2x2 — 1) y3 =ax3 is
(a) y3 = ax + C -^1- x.22
(b) y3 = ax + Cx ^1- x.2
(c) y2 = ax + C ^1- x2
(d) None of these
5. The solution of — + — log y =
(log y )2, is
dx x
x
(a) x = — log y + C
2x
(b) x2 + log y = C
(C)
—!_= 1 +c
x log y
(d) None of these
2xz
6. The solution of — + yf'(x)-f(x)-f'(x)=0,y *f(x)is
dx
(a)y = f(x) + 1+ce‘r(x)
(b)y-ce’r(x)
(c) y = f (x) -1+ ce~f( * 1
(d) None of these
7. The solution of(x2-1)dy/dx-siny-2x cosy =2x -2x3, is
(a) (x2 -1) cos y =
- x2 + C
(c)(x2-1)cosy = 2--y + C
(b) (x2 -1) sin y =
- x2 + C
(d) (x2-1) sin y = ^--y+C
8. The Curve possessing the property text the intercept made by the tangent at any point of the curve on the
y-axis is equal to square of the abscissa of the point of tangency, is given by
(a) y2 = x + C
(b) y = 2x2 + ex
(c) y = -x2 + ex
(d) None of these
9. The tangent at a point P of a curve meets the y-axis at A and the line parallel to y-axis at A, and the line
parallel to y-axis through P meets the x-axis at B. If area of A0A8 is constant (0 being the origin), Then the
curve is
(a) ex2 -xy + k = 0
(b) x2 + y2 = ex
(c) 3x2 + 4y2 = k
(d) xy - x2y2 + kx = 0
10. The value of k such that the family of parabolas y = ex2 + k is the orthogonal trajectory of the family of ellipse
x2 +2y2-y =C, is
(a)y2
(b)y3
(c)y4
(d)y5
Session 4
Exact Differential Equations
Exact Differential Equations
A differential equation of the form
M (x, y) dx + N (x, y) dy = 0 is said to be exact (or total) if
its left hand expression is the exact differential of some
function u (x,y).
i.e.
du = M- dx + N • dy
Hence, its solution is u (x, y) = c (where c is an arbitrary
constant). But then there is a question that how do we
confirm whether the above mentioned equation is exact.
The answer to this question is the following theorem.
Theorem The necessary and sufficient condition for the
differential equation M dx + N dy = 0 to be exact is
dM _dN
dy dx
The solution of M dx + N dy = 0 is ,
[
Mdx + [(terms of N not containing x) dy - C
Jy-constant
J
., , dM dN
provided---- = —.
dy
dx
I Example 49 Solve (x2 -ay)dx + (y2 -ax)dy = 0.
The solution is,
Jy-constant (2x log y) dx + J 3y2 dy = C
x2 log y + y3 = C
Equations Reducible
to the Exact Form
Sometimes a differential equation of the form
M dx + N dy = 0 which is not exact can be reduced to an
exact form by multiplying by a suitable function /(x, y)
which is not identically zero. This function /(x, y) which
then multiplied to a non-exact differential equation makes
it exact is known as integrating factor.
One can find integrating factors by inspection but for that
some experience and practice is required.
For finding the integrating factors by inspection, the
following identities must be remembered.
1. x dy + y dx = d (xy)
2. x dx+y dy =- d (x2 +y2)
2
x dy - y dx ,
3.
7
Sol. Here, we have M = x2 - ay and N = y2 - ax
4.
dM
dN
---- = — a~ —
dy--------- dx
Thus, the equation is exact.
The solution is, J
3
3
3
3
(x2 - ay) dx + J y2 dy = C
y dx - x dy
xy
d log
d log -
L AJJ
x2
So/. Here, we have M = 2x log y and N = — + 3y
dM = 2x
y
= — and hence the equation is exact,
dx y
72
7 xdy-ydx
. AsL■=d tan’1 /
x2 +y2
>
iy
and
x
x dy -y dx
(x2
I Example 50 Solve(2xlogy)dx+ —+ 3y2 dy = O.
dy
y dx -x dy
y2
xdy-ydx _dy _dx
5.
xy
y
x
6.
^-<uy+r=c
=>
V
1+^—
X2
2
8.
dx + dy
x +y
1+ y
x
d(ln(x+y)]
9. d(ln(xy)) =
x dy+y dx
1
10. d -ln(x2 + y2)
2
xy
_x dx +y dy
x2 + y2
\2
lx.
Chap 04 Differential Equations
11. d
12. d
13. d
1 ] _ x dy + y dx
< *yj
e'
2
x y
I Example 54 Solve
f
y + sin xcos2 (xy)
X
dx +
+ siny dy = 0.
cos2(xy)
.cos2 (xy)
7
2
_ x ey dy - ey dx
7
Sol. The given differential equation can be written as
^i^ + Sinzdx+sinyl/y = 0
cos (xy)
ex _yexdx -exdy
<y)
y2
=>
14. d(xm yn) = x m -1 yn 1 (my dx + nx dy)
15.
d (tan (xy)) + d (- cos x) + d (- cos y) = 0
tan (xy) - cos x - cos y = C
d[f(x,y)]1~n _ /z(x,y)
1-n
(/(x,y))n
Sol. The given differential equation is
x2dx + y2dy - a(y dx + x dy) = 0
d
( x 3A
3 J
x+y-7I Example 55 Solve------- dx
7T~X 2+2y2 + 4dy
X
y-x^dx
Sol. The given equation can be written as
xdx + ydy_ydx-xdy y2
r3>
+ d| — -ad(xy) = 0
(x2+y2)2
7
y'
f d(x2+y2)_
J (x2+y2)2 ~
i
< 3>
x2 y
v3
Integrating, we get — + - ----a xy = k
=>
x3 + y3 - 3axy = 3k = C
fx2
"l
I Example 52 Solve (2xlogy)dx + —+3y2 dy = 0.
Sol. The given differential equation is
dy
„,
I Example 51 Solve (x2 ~ay)dx+(y2 -ax)dy = 0.
=>
sec2 (xy) d (xy) + sin x dx + sin y dy = 0
1y
X
(logy)d(x2) +x2 d(logy) + d(y3) = 0
=>
d(x2 logy) + d(y3) = 0
=*
x2 log y 4- y3 = C
I Example 56 The solution of
x (y2 -D
e y {xy2 dy +y5dx}+{ydx-xdy] = 0, is
(aje-y + c»/> +c=0
(b)exy-ex/y+C = 0
(c)ex>'+e^x+C = 0
(integrating both the sides)
y
__ 1_
=C
x2 +y2
y_
x2
=>
d
Integrating both the sides, we get
1
—— + c
"(x2+y2)
(x/y)
>
(log y)2x dx + — dy + 3y2 dy = 0
y
x2/y2
7
(d)exy-ey/x+C = 0
„ (y1 -1)
I Example 53 Solve x dx+y dy = x dy - y dx.
Sol. Here,
Sol. The given equation can be written as
■
p(x2+y2) = x2dW
or
2
' =>
.
,
2x2 d y
d(x2 +y2)_____ X
, 2
x2+y2
x 2 +y
=>
d(x2 +y2) _ 2d(y/x)
x2+y2
l+(y/x)2
log(x2 + y2) = 2 tan-1
y
e
or
■ y2 {xdy + ydx} + {ydx ~ xdy} = Q
y2 • {xdy + ydx} + ex,y {ydx - xdy} = 0
-{xdy +ydx}+ ex,y
{ydx-xdyl
y2
ee ”-d(xy) + ex,y -d
X
=0
y>
or
d (exy) + d (ex/y) = 0
Integrating both the sides, we get
+C
exy + ex,y + C=0
Hence, (a) is the correct answer.
257
258
Textbook of Integral Calculus
Sol. The given equation is
I Example 57 The solution of
x2dy
< - y 2dx + xy2 (x - y) dy = 0, is
2
(a) log
(c) log
xy'
x- y
xy■
=ii+c
2 2
x y
dy
dx
2
2
Iy
y
[2 £
Iy X
+ x2y3
x )
2_
-d
d
i )
+y
y2exy (xdy + ydx) = ex,y (ydx - xdy)
=>
(d(xy)) = ex/y ■
X
X J
d
7 _2
ly__ £ + ydy = 0 or — <y
X
1-1"!
xj
Integrating both the sides, we get
exy = ex/y + x
PM
mJ
xJ
y
y2
11
log------- = o +C
x y
22
Hence, (c) is the correct answer.
I Example 60 The solution of the differential equation
2 tan-1’ £ = C
(a) ^i ++ 2tan-
2
Hence, (a) is the correct answer.
py
2
I Example 58 The solution of the differential equation
(b)
x2 +y
- + 2tan-1
2~~
2
ydx - xdy + xy2 dx = 0, is
/lx x
(a) — + x2 =X
y
. . x
x2
(c) —+ —= X
xy = log (ex,y + X)
(y + Xy[xy (x + y))dx+(y Jxy (x + y)-x)dy = 0,is
Integrating both the sides, we get
i
yJ
d(exy) = d(ex,y)
=>
=>
=d
x |
dy = 0
dy = 0
y
ydx - xdy
y2
e3^ (d (xy)) = ex,y ■ d
I
—
1_1
y
=>
(d) log x~y = x + C
xy
2
Sol. Here,
x2
xy = —
+C
2
x-y
(b) log
2
(xy2exy) dy + (xex/y) dy = (yex,y) dx - (y3exy) dx
x2
.
x +y
- + 2tan"r
(c)
V2
=C
(b)- + — = X
y
2
(d) None of these
(d) None of these
Sol. The given equation can be written as
(ydx - xdy) + x Jxy (x + y) dx + y -/xy (x + y) dy = 0
Sol. Given equation is, ydx - xdy + xy2 dx = 0
(ydx - xdy) + (x + y) Jxy (xdx + ydy) = 0
2y2
4
Which could be converted into exact form
ydx - xdy
+ xdx = 0
i.e.
ydx - xdy *
y2
~7~
+d
d
V =0
y
X
d
+d
)M
x2 +y2
x2 +y2
MP
2
y
k2
y>
±+l
X
/
d
2
x +y
\
y
2
XX2-
-+—=X
y
2
r
■
Hence, (b) is the correct answer.
I Example 59 The solution of differential equation
xdy (y2exy+ex/y) = ydx(ex/y -y2exy), is
or
(b)x2 /y = log(ex/y +X)
(c) xy = log (ex/y +
+ X)
X)
(d) xy2 = log (ex/y + X)
x2+y,2\ + 2d tan-1
d
\
~2
J
Integrating both the sides, we get
=>
(a) xy = log (ex + X)
0
y
X
Integrating both the sides, we get
x x2
— + — = constant
y
2
or
=o
2
x 2 + y2
-i
------- — + 2 tan 1
2
Hence, (b) is the correct answer.
=0
Chap 04 Differential Equations
259
Exercise for Session 4
1. The solution of xdy + ydx + 2x 3dx = 0, is
(b)xy+-x4=C
2
(a)xy+x4=C
(c) — + —=C
y
4
(d) None of these
2. The solution of, ydx - xdy + (1 + x2) dx + x2 sin y dy = 0, is given by
(a)x+ 1-y2 + cosy + C = 0
(b)y + 1-x2 + x cosy + C = 0
(c)- + -- y+cosy + C = 0
y y
(d)^ + -- x + cosy + C = 0
xx
3. The solution of (1 + x y]x2 + y2)dx + (-1 + -]x2 + y2) ydy = 0, is
(a) 2x - y2 + - (x2 + y2)3/2 =C
3
(b) 2x - y + - (x2 + y2)3/2 =C
3
(c)2y - x2 + - (x2 + y-2):3/2=C
3
(d) None of these
4. The solution of,
xdy
~
x +y
(a) tan’1
x
+ x =C
y
2 ~
{x22 + y22
(b) tan’1
y
-1 dx, is given by
/
(?)-=
(c) tan-1
(d) tan’1
-
+ x2 =C
5. The solution of yex,y dx = (xexly + y2 sin y)dy, is given by
(a)ex/y =-cosy+ C
(b)ex/y + 2
2 cosy =C
j dy =
6. The solution of x sin
(a)logx-cos
\X
-UH
=logC (b)logx-sin
=C
x)
7. The solution of — +
=
xdy - ydx
yjx2 + y2
(d)ex/y = 2cosyex/y +C
dx,is given by
(c)log - |-cos|- = logC (d) None of these
x)
y
2
I 2
a -x -y2
is given by
x2 + y2
(a) sin’1 (jx2 + y2) = a tan’ 1^ + C
(c) sin"1
(c)exly = x cosy + C
(b) sin’1 (Jx2 + y2) = - tan’1 ( Z
a
lx
= tan 1
+C
(d) None of the above
a
8. The solution of (1 + exly) dx + exly ( 1 - — | dy = 0, is given by
I
(a)x - yex,y =C
y
(b)x + yexly = C
(c)y--exly =C
y
(d) None of these
9. The solution of -x.^d)',dx = xsln8(x^yx)
y -x dy/dx
/
(a) - cot (x2 + y2) =
x
y
\2
+C
(b) tan (x2 + y2) = x2y2 + C
y
(c)cot(x2 + yz)-- + C
y
(d) None of these
1
10. The solution of — + —:
is given by
dx
x (1 + log x + log y)2 ’
(a)xy (1+ log (xy)) = C
(b)xy2 (1+ log (xy)) = C
(c)xy(1+ log (xy))2 = C
(d)xy (1+ (logxy)2) = C
Session 5
Solving of First Order and Higher Degrees, Application of
Differential Equations, Application of First Order
Differential Equations
Solving of First Order
and Higher Degrees
Differential Equation of First Order
and Higher Degrees
A differential equation of first order is of the form
f(x, y, P) where P = dy I dx. If in the equation degree of P
is greater than one, then the equation is of first order and
higher degree. •
The differential equation of first order and higher degree
can be written in the form
P" + F^x.y) P"-1 + ... + Fn_1(x, y) P + F„(x,y) =0
The differential equations of these category can be solved
by one or more of the following methods :
(i) Equations solvable for P.
Hence, the general solution is given by
(x, y. c), <t>2 (x, y, c)
(ft, (x, y, c) = o
Here, the arbitrary constant c1,c2,...,c„ are replaced by a
single arbitrary constant c because every first order
equation has only one arbitrary constant in its solution.
I Example 61 Solve (p - x) (p — ex) (p — 1/y) = 0;
.
dy
where p = -^~dx
1
Sol. The component linear equations are p = x, p = ex, p = y
dy
.
,
,
x2
If — = x, then dy = x dx => y = — + Q
dx
2
dy
_■
,
,
, ,
If
— = e , then dy = e dx => y-e x +C2
dx
If
dy 1 ,
,
y2
— = —, then y dy = dx => — = x + C3
dx
(ii) Equations solvable for y.
y
2
The required solution is
fy~ —
x2 + C (y - ex + C) ( —2 -x + C 'I =0
r.
. 22 )
I2
J
(iii) Equations solvable for x.
(iv) Clairaut’s equations.
Now, we shall discuss these cases.
I Example 62 Solve x2p2 + xyp - 6y2 = 0.
(i) Equations Solvable ForP
If the equation
n'1 + ... + F„_
F, 1(x,y)P + F„(x,y)=0,
Pn+F,(x,y)P"-'
is solvable for P, then LHS expression can be resolved
into n linear factors and hence can be put in the form
(P " j\(x, y)) (P - f2(x,y))... (P - fn(x, y)) = 0.
Equating each of these factors to zero, we get n differential
equations of the first order and first degree.
Sol. The given equation is
x2p2 + xyp - 6y2 = 0
Solving as a quadratic in p, we get
(- xy ± yjx2y2 + 24x2y2)
'■
2?
P
2y a. dy 2y idy
‘
If p = —, then — = — => —
x
dx x
y
2y
2dx
Tr
X
y = t=>y = C, x2
log -4
X"
~ = fi(x,y),^- = f2(x,y\...^ = fn(<x,y)
dx
dx
dx
Let the solutions of these obtained equations are
Ifp =
(})i(x,y,c1)=0, 02(x,y, c2)=0,...,^(x,y,cn)=0
respectively.
=>
-^Zjhen —
x
dx
x
dy
3dx
y
x
x3y = C2
The required solution is (y - Cx2) (x3y - C) = 0.
3y
,
Chap 04 Differential Equations
(The equation p2 - a = 0 gives us singular solution in
which we are not interested).
The substitute p in Eq. (i), we get the required solution
I Example 63 Solve xy2 (p2 + 2) = 2py3 + X3.
Sol. The given equation can be written as
(xy2p2 - x3) + 2(xy2 - py3) = 0
x (y2p2 - x2) + 2y2 (x - py) = 0
2y = Cx + —
C
(py ~ x) {x (py + x) - 2y2} = 0
I Example 65 Solve y = 2px - p2.
=>
If py - x = 0, then y dy - x dx = 0 => y2 - x2 =
Sol. Differentiating the given equation w.r.t. x, we get
If xyp + x2 - 2y2 = 0, then 2y — = - 2x
dx
x
dt 4
=>
-------- t = - 2x,
dx x
-Ji*
1
4 In x
IF = e J x
where t = y2
or
or
X*
Its solution is t
1
xX
1
X
)
*
261
dy „
« dp n dp
— = 2p+2x — — 2p —
dx
dx
dx
2(x-p)^ + p=Q
dx
dx 2
—+ —x = 2
dp p
It is a linear equation in x and p.
/-*
IF = e''P =e2logp=p2
± = _1 + C2 i.e. y2=x2+C2x4
X
x2
Hence, the required solution is
(y2-x2-C)(y2-x2-Cx4) = 0
i.e.
A
The solution is xp2 = J p2 • 2 dp = - p3 + C
Thus, the solution of the given equation is
2
x = - p + Cp
-2,
, where p is parameter.
(ii) Equations Solvable For y
Equation that comes under this category, can be expressed
in the form
This type of equation can be put in the form
y=g(x,p)
(i.e. an explicit function y in term of x and p) ...(i)
Differentiating Eq. (i) w.r.t. x, we get
^y =
dx
dx
Which is a differential equation of the first order
containing x and p. Let us suppose that its solution is
4>(x,p,c)=0
...(ii)
Then, the solution is obtained by eliminating p between
y = g(x, p) and <|>(x, p, c) = 0. However, if eliminating of p is
difficult express x and y as a function of the parameter p.
I Example 64 Solve xp2 - 2yp + ax = 0.
Sol. The given equation can be written as,
xp ax
y = — +—
2 2p
dy^_P + x dp
dx 2 2 dx
=>
2p
p(p2 -a)=x(p2 -a)-^
dx
dp p
/ = ^-=>p = Cx
dx x
(0
x = g(y.p)
Differentiating w.r.t y, we get
P = F\ X'P'¥~
l
(iii) Equations Solvable For x
•(i)
1 dx —=—=G
p dy
dp
dx
which is a differential equation of 1st order containing y
and p and its solution is
(Xy,p,c)=0
•,
Then, the solution is obtained by eliminating p between
x = g(y, p) and (Ky, p, c) = 0. However, if eliminating of p is
difficult express x and y as a function of the parameter p.
I Example 66 Solve y = 2px+y2p5.
Sol. Solving for x, we get
ax dp
2p2 dx
,22
2p
2
Differentiating Eq. (i) wx.t y, we get
dx -------1
y
dp
----i— • dp - yp22i - y,22ip . -L.
dy 2p 2p dy
dy
(i)
262
Textbook of Integral Calculus
1
1
1
2
or
- - — + yp = -y
p 2p
or
(1 + 2yp3) p = - y (1 + 2p3y) •
I 2p
J
dp
dy
dy
C
dp_ + ^y = 0=>py = C=$p = —
or
y
p
y
Substituting this in the Eq. (i), we get
v2
C2
x = ------------=> y2= 2Cx + C 2
2C
2
(iv) Clairaut's Equation
The differential equation y = px + f(p) is known as
Clairaut’s equation. The solution of equation of this type
is given by y = ex + f(c).
Which is obtained by replacing p by c in the given
equation.
Remark
Some equations can be reduced to Clairaut's form by suitable
substitution.
p
I Example 67 Solve y =px+
Sol. Its solution is, y = ex +
Application of Differential
Equations
Differential Equation of First Order
But not of First Degree
1. The most general form of a first order and higher
degree differential equation is
pn +Pj p"'1 + P2pn~2 + ...... + Pn =OwherePi,P2,
...... Pn are function of x, y and p - dy/dx. If a 1st
order any degree equation can be resolved into
differential equation (involving p) of first degree and
1st order, in such case we say that the equation is
solvable for p.
Let their solution be
gi(x,y,Ci) xg2(x,y,c2) x... xgn(x,y,cn) =0,
(where C], c2,......... cn, are arbitrary constant) we
take Ci = c2 =..... = cn = c because the differential
equation of 1st order 1st degree contain only one
arbitrary constant. So solution is
gi (x, y, c) x g2 (x, y, c) X... x gn (x, y, c) = 0
2. The most general form of a first order and higher
degree differential equation is
p" + p"'1 + P2 pn’2 +... + P„ =0, where P15 P2,
c
y/l + c2
I Example 68 Solve -^1 + p2 = tan (px - y).
Sol. The given equation is
-Jl + p2 = tan (px - y)
or
px-y = tan'1 (-J1 + p2)
or
y = px - tan'1
1+
Its solution is,
y = ex - tan'1
1 + c2
..... , Pn are function of x, y and p = dyldx. If
differential equation is expressible in the form
y = f(x,p), then
d
z
x
dp
Step 1 Differentiate w.r.t. x, we get p —=f\ x, p, — •
'dx
dx \
dx)
Step 2 Solving this we obtain 0(x, p, c) = 0.
Step 3 The solution of differential equation is
obtained by eliminating p.
2
I Example 69 Solvey2 logy = pxy + p2.
Application of First Order
Differential Equations
Sol. Let log y = t. Then — — = —
y dx dx
Growth and Decay Problems
dt
P
So, if— - p, then — = p
dx
y
Substituting these in the given equation, we have
y2t = y- p xy + p2y2 or t = px + p2
Which is in Clairaut’s form.
Thus, the required solution is
t = cx + c2 or log y = ex + c
(c being an arbitrary constant.)
2
Let N(t) denotes the amount of substance (or population)
that is either growing or decaying. If we assume that
dN/dt, the time rate of change of this amount of
substance, is proportional to the amount of substance
present, then
dN ,„r
dN
A
---- = kN
, ‘ T or ------ kN=0
.(i)
dt
dt
Where k is the constant of proportionality. We are
assuming that N(t) is a differentiable, hence continuous,
function of time.
Chap 04 Differential Equations
I Example 70 The population of a certain country is
known to increase at a rate proportional to the
number of people presently living in the country. If
after two years the population has doubled and after
three years the population is 20000, estimate the
number of people initially living in the country.
45 = 50e2k
263
or it = - In — = - 0.053
2
50
Substituting this value into Eq. (ii), we obtain the
amount of mass present at any time t as
N = 5Oe-0053'
...(iii)
Where t is measured in hours.
Sol. Let N denotes the number of people living in the country
at any time t, and let No denote the number of people
initially living in the country. Then, from Eq. (i)
(b) We require N at t = 4. Substituting t = 4 into Eq. (iii)
— -kN = 0
dt
(c) We require t when N = 50/2 = 25. Substituting N = 25
into Eq. (iii) and solving for t, we find 25 = 50e”° 053<
Which has the solution N = Cekt
(i)
At t = 0,N = No; hence, it follows from Eq. (i) that
No = Cek(0) or thatC = N0.
Thus,
N = Noekt
...(ii)
At t = 2, N = 2N0. Substituting these values into Eq. (ii), we
have
2N0 = Noe2k from which k = | In 2 = 0.347
Substituting this value into Eq. (i) gives
N = Noe(O‘347)t
...(iii)
At t = 3, N = 20000. Substituting these values into Eq. (iii),
we obtain
20000 = N0e(0’347)(3)
I Example 71 A certain radioactive material is known
to decay at a rate proportional to the amount present.
If initially there is 50 mg of the material present and
after two hours it is observed that the material has
lost 10% of its original mass, find (a) and expression
for the mass of the material remaining at any time t,
(b) the mass of the material after four hours, and (c)
the time at which the material has decayed to one half
of its initial mass.
Sol.
(a) Let N denotes the amount of material present at time
t. Then, from Eq. (i)
dN
A
------ kN =0
dt
This differential equation is separable and linear, its
solution is
...(i)
N = Cekt
At t = 0, we are given that N = 50. Therefore, from
Eq. (i), 50 = Cek(0} or C = 50.
Thus,
N = 50 ekt
and then solving for N, we find N = 50e(-0053) (4^
or
-0.053 t = In - or t = 13 h
2
I Example 72 Five mice in a stable population of 500
are intentionally infected with a contagious disease to
test a theory of epidemic spread that postulates the
rate of change in the infected population is
proportional to the product of the number of mice who
have the disease with the number that are disease
free. Assuming the theory is correct, how long will it
take half the population to contract the disease?
Sol. Let N(t) denotes the number of mice with the disease at
time t. We are given that N(0) = 5, and it follows that
500 - N(t) is the number of mice without the disease at
time t. The theory predicts that
dN
—— = kN (500 - N)
•(i)
dt
Where k is a constant of proportionality. This equation is
different from Eq. (i) because the rate of change is no longer
proportional to just the number of mice who have the
disease. Eq. (i) has the differential form
______kdt=0
N(500 - N)
Which is separable. Using partial fraction decomposition,
we have
1
_ 1/500 t 1/500
N(500 - N) ~ N
500 -N
Hence, Eq. (ii) may be rewritten as
1
1
'
_1
— +------------ idN -kdt=O
500
500 - N J
\
1
_ , .
1
1
dN - jkdt = C
It solution is — — +
500
500-N )
or
(In | N| - ln| 500 - N|) - kt = C
...(ii)
Which may be rewritten as
At t = 2,10% of the original mass of 50 mg or 5 mg has
decayed. Hence, at t = 2, N = 50 - 5 = 45. Substituting
these values into Eq. (ii) and solving for k, we have
In
N
500-N
= 500(C + kt)
(ii)
264
Textbook of Integral Calculus
N
500 - N ~
500(C + ir)
(iii)
y-intercept of the tangent = y j — x1
Bute500^ + fc,) = e 500 eJet . Setting Ct = e 500C, we can write Eq.
(ii) The equation of normal at P is,
(iii) as
N
= C1e500tr
500- N
y-y\ =- ----- ------ (x - Xi) x and y-intercepts of
(dy / dx)
(iv)
normal are; Xi +yi — and yi +Xj —
dx
dy
At t = 0, N = 5. Substituting these values into Eq. (iv), we
find
(iii) Length of tangent = PT -1 yt | Jl+(dx /dy)2* yi)
_L = c. e500^ = C,
495
(iv) Length of normal = PN = [ y i | ^Jl+(dy / dx)2^ y )
So, Ci = 1/99 and Eq. (iv) becomes
_2L_ - 1 eS00H
500-N
99
(v) Length of subtangent = ST - yr
(vi) Length of subnormal = SN =» y x
Geometrical Applications
dy
dx
1 _ 1 „500kt.
1=—e
;
In 99 = 500kt
99
or t = 0.0091 Ik time units. Without additional
information, we cannot obtain a numerical value for the
constant of proportionality k or be more definitive about t.
rdx^
< dy
We could solve Eq. (v) for N, but this is not necessary. We
seek a value of t when N = 250, one half the population.
Substituting N = 250 into Eq. (v) and solving for t, we
obtain
.
I Example 73 Find the curve for which the area of the
triangle formed by the x-axis tangent drawn at any
point on the curve and radius vector of the point of
tangency is constant equal to o2.
Sol. Tangent drawn at any point (x, y) is
slope of the tangent at P (= tan y) = ( —
Y-y^(X-x)
dx
and
I dx
y*
hence we find the following facts.
(*i.yi)
y = /(x)
PM
N'
0
P(x1iyi)
0
T
S
N
When
+-x
M
dx
Y=0, X = x- y —
dy
Area of A = 2a2 (given)
r
i.e.
-•X-Y = 2a.2‘
2
i.e.
2 dx
xy-y — = 2a2
dy
i.e.
2 dx , n 2
xy-y — = ±2a
dy
Fig. 4.2
(i) The equation of the tangent at P is,
y-yi = — (x - Xj) when it cuts x-axis,y = 0.
dx
dx'
x-intercept of the tangent = x, -jyl
Jy,
(xi.yi)
(vii) Length of radius vector - -Jx^2 + y2
Let P(Xi, yi) be any point on the curve y = f(x), then
xy
dy
dx
i.e.
dx
dy
x
,2a2
y
y
2
>x
Chap 04 Differential Equations
j--dy
IF = e y
1.
dy _( dx
Sol. We know,
dx
y
,2a2 1
The solution is x • — = f ± — -~dy
y J y y
-2
-2
d^ f dx .dy
dy I dy
(dxY2 d2x dy
y
I dy,
I Example 74 Find the curve for which the intercept
cut off by any tangent on y-axis is proportional to the
• square of the ordinate of the point of tangency.
or
dx
dy2 dx
d 2y _
d 2x
dy_
dr2
dy2
dx
4+
dx
<ty2
dx2
Sol. The equation of tangent at any point (x, y) is
■
, differentiating both the sides
dy >
a2
*=±.2a2-y'2 + C, i.e. x = Cy + —
y
-1
d2y _ _ < dx
dS~ \dy J
or
3
Hence, (d) is the correct answer.
y-y = ^(X-x)
dx
It is given T «= y2. i.e. Y = ky2
I Example 76 The solution of y = x
.
dx dx
is
(a)y = (x-1)2
(b)4y = (x + 1)2
(k being constant of proportionality)
(c) (y — I)2 = 4x
dy
When X = 0; Y = y - x — = y-intercept
dx
i.e.
1
Let — = z sothat,-4.^=*
y2 dx dx
y
IF = eJJ x
(d) None of these
y = xp + p-pi-,
-dy
-------1 y =
dx x
x
-2 dy 11 k
-y
dx x y x
i.e.
I <'dx J '
Sol. The given equation can be written as
dy L 2
y - x — = ky
dx
i.e.
265
,
dy
where p = —
(i)
Differentiating both the sides w.r.t. x, we get
xdp dp „ dp
p = p + — + — ~ZP-T
dx
dx
dx
^(x + l-2p) = 0
dz z k
—+—=
dx x X
Either
=eiiogx = x
r kx
The solution is zx = — dx = kx + C
J x
or
— = 0 i.e. p = C
dx
x + 1 - 2p = 0
-(ii)
ie, p = -(x + l)...(iii)
2
Eliminating p between Eqs. (i) and (ii), we get
- = kx + C
y
y = Cx + C - C2
x = kxy + Cy
As the complete solution and eliminating p between Eqs. (i)
and (iii)
=>
=>
-C 1
1
x - Cy = kxy => — +---------= 1
ky
k x
-^+^=1
x
y
1
C
where — = C2 and---- -- C.
k
k
y = - (x + 1) x + - (x + 1) - — (x + l)2
2
2
4
ie,
4y = (x + l)2 as the singular solution.
Hence, (b) is the correct answer.
I Example 75 For any differential function y = /(x),
d2y ( dy y d\.
the value of—< + — —- is equal to
dx2 k dx J dy2
I Example 77' The solution of
( dy V
dy
-7- + (2x + y)-~
y)~ + 2xy = 0, is
ydx J
dx
(a) 2y 5^
dx
(b)y^
(a) (y + x2 -CJfr + logy + y2 + C2) = 0
f ,2 \2
, \ dy
(c) y -7- +
dx Yy2>
(d) None of these
dx
(b) (y + x2 - CJ (x - log y - C2) = 0
(c) (y + x2 -C1)(x + logy-C2) = 0
(d) None of the above
266
Textbook of Integral Calculus
Sol. The given equation can be written as
p2 + (2x + y) p + 2xy = 0, where p
i.e.
= dy
dx
(p + 2x)(p + y) = 0
p + 2x = 0, otherwise p + y = 0
dy
—+ 2x=0 or
—+y=0
dx
dx
Jdy + 2 |x dx = Ci or
j^ + px = C2
y
y + x2 = c,
or
log y + x = C2
(y + x2 -Ct) = 0
or
g Example 79 The equation of the curve passing
through the points (3o, o) (o > 0) in the form x = f(y)
which satisfy the differential equation;
o22 dx
xy
.
---- — = —+----- 2, IS
xy dy
yx
f W1
f 1 + ey'-k >
(a) x = y + a
(b)
x
=
y
+o
k 1-2ey -k 7
J
f 1 + ey~k
(c)y = x + o
(d) None of these
l-ey~k
a,2“
Sol. Here,
(x + log y - C2) = 0 '
is the required solution.
y
xy dy
y
x
dy _ dv
2 dy
2
(x - y) • — = a , put x - y = v .’. 1 dx dx
dx
Hence, (c) is the correct answer.
I Example 78 A curve y = f(x) passes through the
origin. Through any point (x,y) on the curve, lines are
drawn parallel to the coordinate axes. If the curve
divides the area formed by these lines and coordinate
axes in the ratio m :n. Then the equation of curve is
(a) y = Cxm/n
(b) my2 = Cxm/n
(c) y3 = Cx m,n
(d) None of these
v2
=>
1-^
dx
= a2
v-a
a i
v + - log
=>
B
~2
v -a
2
= dx
,
x
=x+C
v+a
n
y+
v2dv
dv
v2 - a 2 = V 2 —
dx
\
a2
1+ 2
dv
dx
2
I vV 2-a
-a : J
/
Integrating both the sides, we get
Area of OBPO _ m
Area of OPAO
x
a2 ^(x2 +y2-2xy)
dx
(y + x2 -CJfx + logy-C2) = 0
Sol.
dx
x-y-a
a 1
(x " y + 2 °8 <x-y+a, = x + C
' x-y-a
„ a,
y + C = - log
<x-y + a )
/’(x.y)
(i)
It passes through (3a, a)
1 1
a + C = - log
2
==> nxy = (m + n) £ ydx
£ ydx
3J
1
C = - a + - log
2
3
2 + log 3
C = -a
2
A
0
„
"
y=
a i
(2 + log 3) + log
Differentiating w.r.t. x, we get
dy .
n x — + y=(m + n)y
dx
J
dy
nx — = my
dx
m dx
= —, integrating both the sides
n x
y
y = Cxm,n
Hence, (a) is the correct answer.
=>
x - _y =- ea
----x-y+a
x-y
a
y-k
, where k =
x-y~a
x-y+a
(2 + log 3)
l + ey~k
l-ey~k
Z V-k A
a
x=y+a
, where k = - (2 + log 3)
l-eyTk
2
\ * v
7
Which is required equation of curve.
Hence, (b) is the correct answer.
267
Chap 04 Differential Equations
I Example 80 The family of curves, the subtangent at
any point of which is the arithmetic mean of the
coordinates of the point of tangency, is given by
(a)(x-y)2=Cy
(b)(y-x)2=Cx
(c) (x - y)2 = Cxy
(d) None of these
So/. Let the family of curves be y = f(x)
tan6 = ^
Z(TP')
dy
2y
dx
x+y
It is a homogeneous differential equation.
Put x - vy
Differentiating w.r.t y, we get
dx
dv
—=v+y—
dy
dy
...(ii)
In Eq. (i) replacing — by Eq. (ii), we get
dy
y=W
p
. dx _x + y
2xy
dy
WW)
dy
2y
dv
y dy
1+v
2
1 4- v
2
1 + v - 2v
2
1-v
2
2 j
*y
------- dv = —
1-v
y
e
■> X
0
tanO
I (subtangent)
2 log 11 - v |
-1
log | y | + log Cl (Ci > 0)
- 2 log | y - x | + 2 log | y | = log | y | + log Cj
_l(PP')
~ I (TP')
=>
log | y - x |2 = log | y | - log Cj
f(x)
=>
log | y - x |2 = log I y | + log C,
f '(x)
where log C = - log Ci
(given)
y
Integrating,
2
2y
/=
x+y
=>
log | y - x |2 = log | yC |
=> (x - y)2 = Cy, is the required equation of family of
curves.
Hence, (a) is the correct answer.
268
Textbook of Integral Calculus
•
■
Exercise for Session 5
I 1
1. The equation of curve for which the normal at every point passes through a fixed point, is
(a) a circle
(b) an ellipse
(c) a hyperbola
• : (d) None of these
2. If the tangent at any point P of a curve meets the axis of x in T. Then the curve for which OP = PT, O being the
origin is
(a)x=Cy2
(b)x =Cy2 orx =C/y2
(c)x =Cy or x =C/y
(d) None of these
s<.;:
3. According to Newton’s law, the rate of cooling is proportional to the difference between the temperature of the
body and the temperature of the air. If the temperature of the air is 20°C and body cools for 20 min from 100°C
to 60°C, then the time it will take for it temperature to drop to 30°C, is
(a) 30 min
(b) 40 min
(c) 60 min
(d) 80 min
,
4. Let f(x, y) be a curve in the x-y plane having the property that distance from the origin of any tangent to the
curve is equal to distance of point of contact from the y-axis. If f (1,2) = 0, then all such possible curves are
(a)x2 + y2=5x
(b)x2-y2=5x
(c) x2y2 = 5x
(d) All of these
.
( 1A
f(t) passing through the point 0, - • The
V 2 J
tangents drawn to both the curves at the points with .equal abscissae intersect on the x-axis. Then the curve
. y =f(x\is
5. Given the curves y = f(x) passing through the point (0,1) and y = [
x
(a)f(x) = x2 + x + 1
x2
(b)f(x) = ^
e
(c)f(x) = e2)t
(d)f(x) = x -ex
6. A curve passing through (1, 0) is such that the ratio of the square of the intercept cut by any tangent on the 1 't
y-axis to the Sub-normal is equal to the ratio of the product of the Coordinates of the point of tangency to the
product of square of the slope of the tangent and the subtangent at the same point, is given by
(a) x = e*2^
(b) x = e±'^Tx
(c)y.e^-1
-1
(d)xy+ex,x-1= 0
7. Consider a curve y = f(x) in xy-plane. The curve passes through (0, 0) and has the property that a segment of
tangent drawn at any point P(x,f(x)) and the line y =3 gets bisected by the line x + y = 1 then the equation of
curve, is
(a)y2 = 9(x-y) •
(b)(y-3)2 = 9(1-x-y)
(c)(y+3)2 = 9(1-x-y)
(d)(y-3)2-9(1+x + y) •
8. Consider the curved mirror y = f(x) passing through (0, 6) having the property that all light rays emerging from
origin, after getting reflected from the mirror becomes parallel to x-axis, then the equation of curve, is
(a)y2 = 4(x-y)ory2 = 36(9+x)
(b)y2 = 4(1-x)ory2 = 36(9-x)
(c)y2 =4(1+x)ory2 = 36(9-x)
(d) None of these
Chap 04 Differential Equations
269
JEE Type Solved Examples:
Single Option Correct Type Questions
• Ex. 1 The order of the differential equation offamily of
curvesy- C, sin“1x+ C2 cos-1 x+C3 tan-1 x + C4 cot-1 x
Sol. Here, (xy3 - x2)dy = (xy + y*)dx
y3(xdy - ydx) - x(xdy + ydx - 0)
=>
C2, C3 andC4 are arbitrary constants) is
(where
■W9
(b) 3
(d) None of these
(a) 2
(c) 4
dividing by x3y2, we get
Sol. Here, y = Q sin”1 x + C2 cos-1 x + C3 tan-1 x + C4 cot-1 x
(it
i
i
i
(it
n
=$y = Qsin x+C2—sin x +C3tan x+C4---- tan
x2
J
\2
•
- x d(xy) = 0
t
1
xy
x yl
yd
x \xj
x
2 2
\y J
2
n
= (Ci - C2)sin-1 x + (C3 - C4) tan" x + (C3 - C4 )—
Now integrating, we get
There are only two independent arbitrary constant order of the
differential equation is 2.
Hence, (c) is the correct answer.
’+i-‘
It passes through (4, -2)
• Ex. 2 The solution of the differential equation
dy
1
— =----------------------- is
dx xy(x2 siny2 +1)
(a) x2(cosy2 - siny2 - 2ce~y ) = 2
(b) y2(sinx2 - cosy2 - 2ce~y ) = 2
(c) x2(cosy2 - siny2 - e~y ) = 4c
- — -=c=>C = 0
8 8
y3 = -2x is required curve.
=4
• Ex. 4 Spherical rain drop evaporates at a rate propor­
tional to its surface area. The differential equation corre­
sponding to the rate of change of the radius of the rain drop,
if the constant of proportionality is K >0, is
(d) None of the above
Sol. Here, — = xy(x2siny2 + 1)
dy
1 dx
1
2
—2y =ysmy
x3 dy x
1 _
2 dx _ dt
Let,
x2
x3 dy dy
But
=>
t*ey = fay siny2 xey dy--ey (siny2 - cosy2) + C
(d) None of these
L,
dV
Sol.
— + 2t-y = y • siny2, Lf. = e/+2j'J}'=e+’'2
dy
So, required solution is
(b)y-K = 0 ■
dt
(a) —+ K = 0
dt
^-Kr
2
— = - k4nr
dt
..
4 .3'
V =-nr
3
dV
, ,2■ dr
— = 4nr
dt
7t
Therefore,
——= — K
dt
Hence, (a) is the correct answer.
2t = (siny2 - cosy2) + 2C e~y
=>
2 = - x2(siny2 - cosy2 + 2ce~y )
• Ex. 5 A function y = /(x) satisfies the differential equa­
x2(cosy2 - siny2 - 2ce~y2) = 2
tionf(x) • sin 2x - cos x + (1 + sin2 x) /' (x) = 0 with initial
condition y (0) =0. The value of f(n/6) is equal to
• Ex. 3 The curve satisfying the differential equation
— _ y(x+X ) ancj p^^g through (4,-2) is
dx
(b) 3/5
(d) 2/5
(a) 1/5
(c) 4/5
2
x(y3-x)
(a)y2=-2x
(b)y = -2x
(c)y3 =-2x
(d) None of these
dy
Sol. y sin2x - cosx + (1 + sin x) — = 0 where y = f(x)
dx
sin2x
cosx
y
=
7
—
~
dx
1 + sin2 x
1 + sin x
270
Textbook of Integral Calculus
sin 2x
IF = e
.
.dt
— dx
J -------1 + sin* x = A=C,ln (1 + sin2 x)
• Ex. 8
Number of straight lines which satisfy the differ-
,
t.
dy
ential equation — + x
dx
^y
pnri/il omintinn _ z
= 1 + sin2 x (by putting 1 + sin2 x = t)
y (1 + sin2 x) = Jcosx dx
y (1 + sin2x) =sinx + C\y (0) = 0=> C = 0
sinx
Therefore,
y = --------- *y
1 + sin x
(b)2
(a) 1
it
6
2
5
-y =0 is
dx y
(d) 4
(c)3
,
, dy ,
y =kx + b-, — = k
dx
Sol.
=>
=s k = k2 and b = k
kx+ b = k+ xk2
k = 0 or k = 1
Hence, (d) is the correct answer.
• Ex. 6
Hence, (b) is the correct answer.
The general solution of the differential equation
• Ex. 9
Consider the two statements:
~ = -—- is a family of curves which looks most like which
dx
y
Statement I y = sin kt satisfy the differential equation
y" + 9y=Q.
of the following?
Statement 11 y = ekt satisfy the differential equation
y" + y' -6y =0.
The value ofk for which both the statements are correct is
(a)
(a)-3 (b)0
2
- k2sinkt + 9 sinfct = 0
2
— = x- —+ C
2
2
x2 + y2 ~2x = C
sin kt [9 - k2] = 0 => k = Q, k = 3, k = -3
Statement II y = ekt, y' = kekt;y" = k2ekt
(x-l)2 + y2 = C+l=C
k2ekt + kekt - 6ekt = 0
Hence, (b) is the correct answer.
ek,[k2 + k-6] = 0
Remark
(k + 3) (k - 2) = 0
Family of concentric circles with (1, 0) as the centre and
variable radius.
• Ex. 7
(d) 3
Sol. Statement I y = sinfcr, y' = k cos kt-, y" -- k2sinkt
J ydy = J (i ~ x) dx
Sol.
(c)2
k = - 3 or 2
Common value is k = - 3.
Hence, (a) is the correct answer.
Water is drained from a vertical cylindrical tank
by opening a valve at the base of the tank. It is known that
the rate at which the water level drops is proportional to the
square root of water depth y, where the constant of propor­
tionality k > 0 depends on the acceleration due to gravity and
the geometry of the hole. Ift is measured in minutes and
k-~, then the time to drain the tank, if the water is 4 m
(b) 45 min
(c) 60 min
Sol. - = - kyfy-, when t - 0; y = 4
dt
J4 Jy
JO
[2y[y]°=~kt = -^
t
0-4 =----15
=>
t = 60 min
Hence, (c) is the correct answer.
(d) 80 min
—
(where c is an arbitrary constant) is
In |cx|
the general solution of the differential equation
/
X
X
x
— = — + (J) — , then the function (|)| — I is
dx
x
y
x2
(a)^-
deep to start with is
(a) 30 min
• Ex. 10
y
x2
(c)4
(b)-~
y
X
(d)-4
X
Sol. In c + In | x| = —
y
Differentiating w.r.t. x, — sy-*yt
y2
x
y2
„^y
—=y -x—
x
dx
J'
y2:
dy = y _ y_
dx
x
Hence, (d) is the correct answer.
.(x
— => 0 -x
x
y
x2
Chap 04 Differential Equations
?
• Ex. 11 if^ *y(0 dt = x 2 + y(x), then y as a function of
al
-a2=2 + Ce2;Ce2 =-(2 +a2)
Hence,
x is
x2-a2
(a)y = 2-(2 + <z2)e
(b)y = 1-(2 + a2)e
2
22
2
2
• Ex. 12 The differential equation — = ———— deterdx
y
mines a family of circles with
dx
V=J~xdl
-x2
= e“
_
ye 2 = j-■2xe
2xe 2 dx
i
dy = 71 ~y2
Sol.
dx
=> -xe 2 dx-dt
e 2 =t
[HT JEE 2007]
(a) variable radii and a fixed centre at (0, 1)
(b) variable radii and fixed centre at (0, -1)
(c) fixed radius 1 and variable centres along the x-axis
(d) fixed radius 1 and variable centres along they-axis
-x2
y
I = ^2dt
x2
C = -(2 + a2)e 2;y=2-(2+az)e
(d) None of these
Sol. Differentiating both the sides, we get
xy (x) = 2x-y'(x)
dy
Hence,
— - xy =-2x
dx
Let
x2 -a2
Hence, (a) is the correct answer.
x2 - a2
(c)y = 2-(1 + a2)e
q2
x2-a2
2
271
y
= dy=jdx => ---2^1~^ y2 = x + C
2
V1 ~y
v
-^l-y2=x+C => l-y2=(x + C)2
X2
ye 2 = 2e 2 + C
(x+C)2 + y2 = l
=>
y = 2 + Ce 2
If x = a => a2 + y = 0 => y = —a 2 (from the given equation)
Therefore, the differential equation represents a circle of fixed
radius 1 and variable centres along the x-axis. Hence, (c) is the
correct answer.
JEE Type Solved Examples:
More than One Correct Option Type Questions
• Ex. 13 A curve y = f(x) 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 the point P
from the x-axis. Then the differential equation of the curve
(a) is homogeneous
(b) can be converted into linear differential equation with
some suitable substitution
(c) is the family of circles touching the x-axis at the origin
(d) the family of circles touching the y-axis at the origin
4y
dx
- x.2
- => homogeneous
2xy
y2
dx
„ dy
2y-/ =
dx dx
Also,
Puty2 = f,;
dx
— ~ — t = -x which is linear differential equation.
dx x
Hence, (a), (b) and (d) are the correct answers.
• Ex. 14 A differentiable function satisfies
Sol. Equation of normal
Y-y = - — (X-x) => -my + my = X -x
m
X + my - (x + my) = 0
P(x.y)
f(x) = j* {/(f) cos t - cos (t - x)} dt. Which is of the follow­
ing hold good?
(a) /(x) has a minimum value 1-e
(b) /(x) has a maximum value 1-e-1
( 71
0
x + my
Perpendicular from (0, 0) =
■Jl + m2
=e
(d)/'(0) = 1
Sol. f(x) = J {/(t) cos t - cos (t - x)} dx
y => x2 + 2xym = y2
= £ /(t) cos t dt - £ cos(- t) dt j^/(x)dx= £/(a - x) dx
272
Textbook of Integral Calculus
1 = lim (e-2 + Ce~x)
lim
(t) cos t dt - sin x
f(x) =
eX
x -» 2
Differentiating both the sides, we get
forx>2
y =e
dy
— - y cosx = - cosx
dx z
(L.D.E.)
rinx
IF = e
Therefore, y-e-sinx _ _ j^-rinx cos x dr,
y.e-
-rin>
Therefore,
y(3) = 2e~3 + e
= e'2(2e~’ + 1)
y'(3) = -2e"3
Hence, (a), (b) and (d) are the correct answers.
;y = Ce”nx + l
y=0
C = -l
/(x) = l-erinx
Ifx = 0;
Hence,
+ 2e
y' = - 2e~x
-jcojxdx _
rinx _ p ,
2
3e~2 =e~
= e 2+Ce~2 => C = 2
f'(x) = f(x) cosx ~ cosx
Z(x) = y;/'(x) = ^
dx
Let
X
• Ex. 16 A curve y = /(x) passes through (1, 1) and
(from the given relation)
Now, minimum value = 1 - e
Maximum value = 1 - e-1
(when x = n/2)
(when x = - 71/2)
tangent at P(x, y) cuts the x-axis andy-axis at A and B
respectively such that BP: AP = 3:1, then
fllTJEE 2006]
(a) equation of curve is xy' - 3y = 0
(b) normal at (1,1) is x + 3y = 4
(c) curve passes through ^2,
. f'(x) = -etinx COSX
Therefore,
/'(0) = -l
f"(x) = - [cos2 x • e“” x _e»inx •sin x]
(d) equation of curve is xy' + 3y = 0
Sol. Equation of the tangent to the curve y = /(x) at (x, y) is
Z-y=^(X-x)
dx
yi
Hence, (a), (b) and (c) are the correct answers.
B
dy
• Ex. 15 Let— +y = /(x) wherey is a continuous funcdx.
e~xt if0<x£2
tion ofx with y(0) = 1 and f(x) = . Which
e-2, ifx>2
Sol.
P(x,y)
1\
is of the following hold(s) good?
(a)y(1) = 2e-1
(b) y'(1) = - e-1
(c)y(3) = -2e-3
y=M
3
0
Thus, cuts the x-axis at A and y-axis at B.
r dy
j..
\
x — -y
(
dv
dx
A —
, n0 and B 0, - x — + y
k
dx
< dx
>
(d)y'(3) = -2e
— + y = f(x) => IF = ex
dx
yex=Jexf(x)dx+C
BP: PA =3:1
f dy
L dx
Now, if 0 < x < 2, then yex = je1 e~ dx + C
=>
yex = x + C
x = 0,y(0) = l,C = l
yex = x + 1
3 x— -y
(i)
Ifx>2,
ye
e-2e
e2
x=fex~2 dx
+ 1x0
dy
x -^ + 3y = 0
dx
2
ex — (x + l)ex
;y(l) = - => y' =
e
e 2x
/(!) =
(dy/dx)
4
x=
J y
1
e
=>
J
x
logy = -31ogx + logC
C
y=—
X
ye x = ex~2 + C •
y = e~2 + Ce
As y is continuous.
Curve passes through (1,1) C = 1
Curve is x3y = 1 which also passes through ^2,
Hence, (c) and (d) are the correct answers.
.
Chap 04 Differential Equations
273
JEE Type Solved Examples:
Statement I and II Type Questions
• Ex. 17 Let a solution y =y(x) of the differential equa­
Integration yields, J
tion x/x2 -1 dy - y -Jy2 -1 dx = 0 satisfy y (2) = —=.
I
73
Statement I y (x) = sec sec
6J
1
2^3
Statement II y (x) is given by—=------
*
y
sec ' x = sec-1 y + C
HIT JEE 2008]
it A
x— .
sec-1(2) = sec-1 (2/-T3) + C
- = - + C => C = 3
6
6
=>
-xL.2
dy
dx
-/
x^x2 -1
yjy2 -1
Thus,
(a) Statement I is true, Statement II is also true; Statement
II is the correct explanation of Statement I.
sec
cos cos
(d) Statement I is false, Statement II is true.
Sol. v x-Jx2 -1 dy - y yjy2 -1 dx = 0
dy
y^y2-^
6,
__ 1___
(c) Statement I is true, Statement II is false.
dx
x^x2 -1
It
x----
y = sec sec
(b) Statement I is true, Statement II is also true; Statement
II is not the correct explanation of Statement I.
Which can be rewritten as
7t
x = sec
2x
y
1
-i 1
n
X
6
—
3
3-1
2
Hence, (c) is the correct answer.
JEE Type Solved Examples:
Passage Based Questions
Passage
(Q. Nos. 18 to 20)
A curve y- f(x) satisfies the differential equation
2 dy
2
(1 + x ) — + 2yx = 4x and passes through the origin,
dx
Sol. (Q. Nos. 18 to 20)
4x2
2x
y=
1 + x2
1 + x 2,
dx
IF = e 1 + 3^
= eta(1+xI)=(i + x2)
18 The function y = /(x)
(a) is strictly increasing, V x e R
(b) is such that it has a minima but no maxima
(c) is such that it has a maxima but no minima
(d) has no inflection point
y (1 + x2) = px2 dx = -y- + C
Passing through (0,0) => C = 0
4x3
y=
19 The area enclosed by y = f~\x)t the x-axis and the
ordinate atx = 2/3 is
(a) 2 In 2
3(1 + x2)
dy_4 (1 + x2)3x2 -x.332x
(b)-ln2
(c) —- In
In 22
(c)
(d)-ln2
3
3
3
dx
3
(1 + x2)2
_ 4 3x2 + x4 _ 4x2(3 + x2)
~ 3 L(1 + x2)2
3(1 + x2)2
20 For the function y = /(x) which one of the following
does not hold good?
(a) /(x) is a rational function
(b) /(x) has the same domain and same rage
(c) /(*) is a transcendental function
(d) y = /(x) is a bijective mapping
— > 0, V x * 0;— = 0at x = 0
dx
dx
and it does not change sign => x = 0 is the point of inflection
y = /(x) is increasing for all x c R.
Hence,
274
Textbook of Integral Calculus
A = ---f’-^dx
Q Jo
3 3
J°l + x2
yx
Put
1 + x2 = t => 2x dx = dt
2/3
3 3
t
2 2 f2 1 - -1 dt
=—— I
3 3
tj
0
3
3
2
Area enclosed by y = f~ (x), x-axis and ordinate at x = 3
3
3
= ---[l-ln2] = -ln2
3 3
3
JEE Type Solved Examples:
Single Integer Answer Type Questions
• Ex. 21 Lety = /(x) be a curve passing through (4, 3)
y=|5-lxl|
such that slope of normal at any point lying in the first
quadrant is negative and the normal and tangent at any
point P cuts the Y-axis at A and B respectively such that the
mid-point of AB is origin, then the number of solutions of
y = f(x) andy = |5-1 x||,/s
Sol. Equation of tangent at any point (xb yj of curve y = _f(x) is
(y - yi) = Z'(xi)(x - Xi), so B (0, yj - Xi/Txi))
number of solutions for y = f(x) and y = 15 -1 x||
number of solutions are 2.
Equation of normal at (xb yj) is (y, y,) =----- -— (x, Xj)so,
/'(xi)
A = 0, yi +
*i
/'(x.),
mid point of AB is origin, so
1
2?i - xJ /'(xi) -
Hxi)
=0
Thus differential equation of curve y = /(x), is
x
I 71 1
Sol. Here,
(x fz(x) +/(x))dx
l+(x/(x))2
*_ > n
In first quadrant, x > 0, y > 0, _—
0,
dx
dy _y + 7x2 + y 2
—, put y = vx
dx
x
So
dv
v + x— =
dx
dv
,2
on solving we get
xf'(x) = 1 + x2f2(x) - /(x)
Integrating both sides
x
dx
4
and f — = —, then lim f(x), is
\ 4 J 7C
*-»0
x f'(x) 4- y(x) _
1
1 + x2f2(x)
dy _y± ^x2 + y2
Thus,
f TtA
• Ex. 22 A real valued function, f(x), f: 0, — —»/?+
\ 2J
-1
satisfies the differential equation x/'(x)=1 +/(x){x2/(x)~ )
dx
{dx)
0
-2
=>
x+C
tan-1(xf(x) = x + C, as
-
71
i
n
tan 1 = — + C
4
C=0
VX + Xy]l + v2
X
x/(x) = tanx
J
x!
X
_4
.. .
=>
,
and
tanx
/(x) =-----x
Ihn /(x) = lim
x—>0
x-»0
x
=1
Chap 04 Differential Equations
1
-v3
'3
• Ex. 23 If the area bounded byy = /(x), x=-, x = -— and
„
2n
2
the X-axis is A sq units where f(x) = x + -x +------ x 5
2 4 6
3
3 5
+--------- x7 +... oo, | x| < 1 ,Then the value q/[4A] is (where[-J
x
1
,2 y ~ x2' I.F. = e l~x
1-x
dy
dx
y-^l-x2 =J
isl.l^F)1
Sol. Here,
1
275
,2
=e2
_.7T7 + c
1 -x
yfl-x2 =sin-1 x + C, as /(0) = 0 => C = 0
f'(x) — l + 2x2 + --4x4 + ----6x6 + ...«
3
-W
=i+
y=
3 5
•V3/2 sin
=> A = Ji[
dx
J
/'(x) = l + x|xf(x) + /(x)|
=> (1 - x2)/'(x) = 1 + x/(x)
sin-11 x
■JT-x7‘
it
=dx= f 1 1tdt= -
,,/2 TT-x-
Jk/6
.2
2
I2 Ji
1 7?
2
4
36
6
(4XJ = 1
Subjective Type Questions
• Ex. 24 For a certain curve y = /(x) satisfying
• Ex. 25 If^x) is a differentiable real-valued function
d_\ =
satisfying 0'(x) + 2 0(x) < 1, prove that 0(x) - - is a
2
non-increasing function ofx.
6x - 4; /(x) has a local minimum value 5 when x = 1.
dx2
Find the equation of the curve and also the global maximum
and global minimum values of f(x) given thatQ < x < 2.
Sol. Integrating, —4 = 6x - 4, we get — = 3x2 - 4x + C
dx2
dx
dy
when x = 1, — = 0. So that C = 1
dx
— = 3x2 - 4x + 1
dx
Integrating, we get
y = x3 - 2x2 + x + Cb when x = 1, y = 5,
so that Ct = 5
Thus, we have
0'(x) + 2 <D(x) < 1
e
j
»2
Hence,
Sol.
y = x3 - 2x2 + x + 5
Form Eq. (i), we get the critical points x = |, x = 1
=>
2x
0'fx) + 2 <p(x) e2x ^e2*
d (
— e< 2z 0 (x) — — e2x J <0
dx k
2
J
=> e,2x <Xx) - J is a non-increasing function of x.
—(i)
=> 0(x) - - is a non-increasing function of x.
2
• Ex. 26 Determine all curve for which the ratios of the
length of the segment intercepted by any tangent on the
y-axis to the length of the radius vector is a constant.
Sol. Let y = f(x) be the equation of the required curve.
dy
1 d2y. . .
At the critical point x = -, —r is (-ve).
3 dx2
= k (a constant)
Given that
Therefore, at x = , y has a local maximum.
d2y
At x = 1, —y is (+ ve).
dx
Therefore, at x = 1, y has a local minimum.
Also,
=>
/(I) = 5
2
log j V + -J1 + V'2 =±fclnx+C
=>
xy + -jx2 + y2
/(0)=5,/(2)=7
Hence, the global maximum value = 7, the global minimum
value = 5.
y
x
Let y = vx, then v + x — = v±fc -JFT 2
dx
dv
, , dx .
= ± k — , integrating we get
x
■Jl + v',2
139
27
x
dx
1+
log
x
±klnx+C
Which are the equations of the required curves.
276
Textbook of Integral Calculus
2
• Ex. 27 Letu(x) andv(x) satisfy the differential equa­
tions — + p(x) u = /(x) and — + p(x) v = g(x), where
dx
■
dx
p(x), /(x) and g(x) are continuous functions.
Ifu(xf)>v(xf)forsomex} and /(x) > g(x),for all x> xv
prove that any point (x, y), where x > xb does not satisfy the
equation y = u(x) andy = v(x).
[JITJEE1997]
dy
y2 + y2 dy I =k2 or y---7 dx
< dx J
■■
d(u-v)
+ (U-v)-Ax)JpW'fc
dx
^(/(xJ-jfx)}
^((u-v)J’d’) = (/(x)-4(x))e»<'w‘b
i.e.
dx
Since, exponential function takes only positive values and
J(x) > g(x) for all x > xb RHS is + ve; x > Xj
dx
is increasing function.
(u - v) ■
ie,
-,/Fv=±x
’•a?
kor2-y2 = x2
=>
x2 + y2 = k2, is required equation of the curve.
• Ex. 29 A curve passing through the point (1, 1) has the
property that the perpendicular distance of the normal at
any point P on the curve from the origin is equal to the
distance of P from x-axis. Determine the equation of the
curve.
IIITJEE19991
Sol. Let P (x, y) be any point on the curve y = /(x). Then, the
equation of the normal at P is,
Y-y =------------ (X-x)
(dy/dxy
or
i.e.
u(x) - v(x) >
<Xx)
dy
y—+x
dx
X + Y-dx
1+
2
dy
dx
2
=>
dy
y~+x
dx
=>
? + 2xy^ = y2
meets the x-axis atQ. IfPQ is of constant length k, then
show that the differential equation describing such curves is,
y — = ± yjk2 -y2 and the equation of such a curve passdx
ing through (0, k).
fUTJEE 1994]
Sol. Let y = /(x) be the curve such that the normal at P(x, y) to this
curve meets x-axis at Q. Then,
PQ = length of the normal at P
~2
=y
But
1+
&
dx
PQ = k
y
dy
dx
=y
dy
dx
1+
2
• Ex. 28 A normal is drawn at a point P(x, y) of a curve. It
(i)
dy
y-f+x
dx
Le.
> 0 [V l^Xj) > vfo)]
Thus,
u(x) > v(x), V X > X]
i.e.
u(x) * v(x), V X > Xj
Hence, no point (x, y) such that x > Xi can satisfy the equations
y = u (x) and y = v(x).
=0
It is given that distance of Eq. (i) from origin = Distance from
x-axis (i. e. y)
0-
{u(x) - v(x)} 0 (x) > {u(xj) - v(x,)} 0 (Xj)
2
•i,'
Integrating both the sides, we get
-4k2-y 2 = ± x + C, since it passes through (0, k) -> C = 0.
Hence, if>W<b = 4>(x), then for x> xx
We have,
=±7*^77 <,
■r
= = ± dx
Jk2-y
Sol. Given that
du
, .
, dv
, .
, .
— + p(x) • u = f(x) and — + p(x) ■ v = g(x)
dx
dx
Subtracting, we get
i<^ + f(x).(u-v) = /(x)-s(x)
dx
Multiplying by e^p dx, we get
y dy
=>
••
.2
dy
~x
dx
2xy
which is homogeneous differential equation and we can solve
by homogeneous or by total differential.
Here, using total differential,
2xy dy - y2 dx = - x2 dx
or
xd(y2)-y2dx
---------- 5-------- = - dx
x1
2 \
=>
d
y_
V x >
= -dx
Integrating both the sides, we get
y2
—=-x+C
x
It passes through (1,1) => C = 2
2
y2
=k
=>
(U)
22
— = - x + 2 or y = - x + 2x
x
x2 + y2 - 2x = 0, is required equation of curve.
Chap 04 Differential Equations
• Ex. 30 A country has a food deficit of 10%. Its population
grows continuously at a rate of 3% per year. Its annual food
production every year is 4% more than that of the last year.
Assuming that the average food requirement per person
remains constant, prove that the country will become
self-sufficient in food after n years, where n is the smallest
I02 10 - Ioe 9
integer bigger than or equal to ——--------- -- — {IITJEE 2000]
(log e 1.04)- 0.03
• Ex. 31 A right circular cone with radius R and height H
contains a liquid which evaporates at a rate proportional to
its surface area in contact with air (proportionality constant
= k > 0), find the time after which the cone is empty.
[UTJEE2003]
Sol. Let the semi-vertical angle of the cone be 0 and let the height
of the liquid at time Y be 'h' from the vertex V and radius of
the liquid cone be r. Let V be the volume at time t. Then,
■
-
____
■
(
4 ¥
1+—
k
100 J
• Q = 0.9 kP0 (1.04)'
...(h)
This gives quantity of food available in year t. The population
in year t is,
P = P0e003'
[from Eq. (i)J
Consumption in year, t = kPgC 003 ‘
...(in)
The country will be self sufficient, if
=>
=>
o'
H
h
V = - itr2h
3
=>
V = - nr3 cot 0
3
Q^P
0.9k Po (1.04)' ^kPoe0 03'
— (1.04)' £e0 03'
10
(1.04)' e-003' >12
9
t log (1.04) - 0.03 t £ log ^12^
t {log (1.04) - 0.03} £ log
t> log 10-log9
“ log (1.04)-0.03
Thus, the least number of year in which country becomes self
sufficient.
log 10-log 9
" log (1.04)-0.03
tan 0 = h
Let S be the surface area of the liquid in contact with air at time t.
Then,
S = nr2
dV
dt
e
---- ocS
dV
=> — = - kS, k is constant of proportionality.
dt
I
d 1 ,
- itr cot 0 = - lair.2’
dt 3
J
=>
nr2 — cot 0 = - fair2 => cot 0 dr = - k dt
dt
On integrating, we get cot 0 J dr = - k J
dt
=> R cot 0 = + kT, where T is required time.
=>
T = H/k
(as tan 0 = R/H)
• Ex. 32 Solve the equation
x Jo* y(t) dt = (x +1) Jox* t y(t) dt, x > 0.
Sol. Differentiating the equation wr.t. x, we get
xy(x) + 1 • f1 y(t) dt =(x + 1) xy(x) + 1 • [ * t y (t) dt
J0
Le.
=>
B
tK-i-S KSK#
=*
Let the average consumption per person be k units.
90
= 0.9 kP0
Qo - kP0
100
=>
R
- ■
It is given that the annual food production every year is 4%
more than that of last year.
Q = Qo
O
A
Sol. Let Po be the initial population, Qo be its initial food produc­
tion.
Let P be the population of the country in year t and Q be its
food production in year t.
dP
dP 3P
or —
dt 100
P 100
Integrating, we get
3
log P = — t + C
100
At t = 0, we have P = Po
=>
C = log Po
=>
P = PoeOO3f
...(i)
277
JO
joXy(t)dt = x2y(x) +JqX ty(t)dt
Again, differentiating wr.t x, we get
y(x) = x2 y\x) + 2xy(x) + xy(x)
Le.
(l-3x)y(x) =
x2 dy(x)
dx
(1 - 3x) dx dy(x)
Lc.
x2
y(x)
C
Integrating, we get y = — e~ Vx
x
278
Textbook of Integral Calculus •
=>
• Ex. 33 lf(y1t y 2) are tw0 solutions of the differential
equation
^ + P(xfy = Q(x)
dx
Then prove that y = y, 4- C (y ] - y 2) is the general solution
of the equation where C is any constant. For what relation
between the constant a, $ will the linear combination
a/i + Py 2 a^° be a solution.
.-x/'1(x) + /i(x) = -x/‘'2(x) + /2(x)
=>
fM - fM -x(f\ (x) - f '2(x))
Integrating both the sides, we get
=>
ln|/[(x)-/2(x)| = ln|x| + C
=*
/i(x) - f2(x) = ± q x
Now, equations of normal with equal abscissa x, are
y “ /i(x) = - —(X - x)
f i(x)
1
(X-x) .
(y-/2(x)) = /'2(x)
Sol. As yt, y2 are the solutions of the differential equation;
and
7-+ P(x) • y = Q(x)
dx
^ + P(x)-yi=Q(x)
dx
...(ii)
~- + PM-y2=Q(x)
dx
^-^'] + P(x)(y-y,) = 0
From Eqs. (i) and (ii),. —
\dx ax )
...(iii)
y-(y-yi) +P(x)-(y-yi) = o
dx
...(iv)
and
From Eqs. (ii) and (iii),
y-(yi ~y2) + P(x)(yt -y2) = o
dx
d .
-riy-yf!
v_ v
From Eqs. (iv) and (v),
-----------= ———
As these normal intersect on the x-axis,
x + fM • f \(x) = x + /2(x) • f z2(x)
fM • fi '(x) = /2(x) • f2 \x) Integrating
/',2(x)-/22(x) = C2
C2
= ±^- =
x '4
C,x
fM - fM
[using Eq. (i)] ...(ii)
From Eqs. (i) and (ii), we get
_fi(x) + /2(x) =
2/,(x) = ±f ^ + qx . 2/2(x) = ±
(v)
x
We have,
dx
d ,
d ,
— (y-yi) — (yi -y2)
dx
_ dx
=>
y-yi
yi -y2
Integrating both the sides, we get
log(y -yi) = log (yi -y2)
y = yi + C (y, - y2)
Now, y = ay] + 0y2 will be a solutions, if
or
Hence,
7- (ayi + py2) + ^(x) (ay, + Py2) = Q(x)
dx
( dyt + P (x) y, | + P [
af
+ P(x) y2] = 0x)
\dx
<dx
)
V dx
)
aQ(x) + pQ(x) = Q(x)
(a + P) Q(x) = Q(x)
a+P=l
. [using Eqs. (ii) and (iii)]
X
• Ex. 35 Given two curves y = /(x) passing through (0,1)
andy = I
/(t) dt passing through (0,
I n). The tangents
drawn to both the curves at the points with equal abscissae
intersect on the x-axis find the curve y = /(x).
Sol. Equation of the tangent to the curve; y - f(x) is
(Y-y)=fW-x)
Equation of tangent to the curve g(x) = yi = J| — oo f(t) dt is
(T - yi) = g'(x) (X - x) = /(x) (X - x)
Given that tangent with equal abscissas intersect on the x-axis.
y _
yi
=>
X‘/'W~X”/(x)
• Ex. 34 Find a pair of curves such that
Sol. Let the curve be y = fM and y = /2(x) equation of tangents
with equal abscissa, x are
(y~fM) = f'M (X -x)
and
Y-fM = f'M(X-x)
These tangent intersect at y-axis,
2
y2(x) = — + x
x
fM = - - x and
=>
(a) the tangents drawn at points with equal abscissae
intersect on the y-axis.
(b) the normal drawn at points with equal abscissae
intersect on x-axis.
(c) one curve passes through (1,1) and other passes
through (2, 3).
X
/z(2) = 3
and
'
or
(i)
fM _ y,
/'(x) /(x)
[vy = /(x)]
Z(x)_r(x)
gM_fM
gW
/(x)
fM
yi
g'(x)
=>
g(x) = Cet)t
g(x)
=>
g/(x) = fcCe.far
kx => f(x) = kCe.far
‘
y = fM passes through (0,1) => kC = 1
y, = g(x) passes through
(0,1 / n) => C = — =>k = n
n
fM = enx
I
Chap 04 Differential Equations
• Ex. 36 A normal is drawn at a point P(x, y) of a curve. It
dy y + x
— = -—, put y = vx
dx y - x
or
meets the x-axis and the y-axis in point A and B, respec­
tively, such that-^ +
=1, where O is the origin, find the
dv 1 + 2v - v2
x—=
v-1
dx
(l-v)dv [ dx_Q
1 + 2v - v2
x
equation of such a curve passing through (5, 4).
Sol. The equation of the normal at (x, y) is
log (1 + 2v - v2) + log x = Q
(X-x) + (K-y)^ = 0
dx
x
x+
y
,
dy
dx
(x + ydy/dx)
dy ! dx
x2 + 2xy-y2=C
=1
where Q = log fc
Since, curve passes through (1,1) —>C = 2
dy
x+y—
_____ dx
OA = x + y — ,OB =
dy_
dx
dx
• Ex. 38 The tangent and a normal to a curve at any point
P meet the x andy axes at A, B, C and D respectively. Find
the equation of the curve passing through (1, 0) if the centre
of circle through 0, C, P and B lies on the line y -x (where 0
is origin).
=> (y-1) —+ (x-l) = 0
dx
Integrating, we get
(y -1)2 + (x -1)2 =C
Sol. Let P (x, y) be a point on the curve.
dy 'I
C = x + y— , 0
dx )
Since, the curve passes through (5, 4), C = 25.
Hence, the curve is (x -1)2 + (y -1)2 = 25.
Bs
• Ex. 37 A line is drawn from a point P (x, y) on curve
Sol. The equation of the line through P(x, y) making an angle
with the x-axis which is supplementary to the angle made by
the tangent at P(x, y) is
dy
y_y=_JL(X-x)
-(i)
dx
where it meets the x-axis.
-.(ii)
or
.
»z
, dx
(y-x)=(y + x) —
dy
0,y-x —
k
dx J
fi
'P
A
J
0
C
D
Let the centre of the circle be (a, p).
_ =x+y—
dy
2a
=>
7 dx
The line through P (x, y) and perpendicular to Eq. (i) is
y_y=^(X-x)
dy
where it meets the y-axis.
dx
X = 0,y=y-x — => OB = y - x —
dy
dy
Since, OA = OB
dx
dx
x + y — =y~x~r
dy
dy
dy
i
Circle passing through 0, C, P and B has its centre at
mid-point of BC.
Y ~ f(x), making an angle with the x-axis which is supple­
mentary to the one made by the tangent to the curve at
P(x, y). The line meets the x-axis at A. Another line perpen­
dicular to the first, is drawn from P(x, y) meeting the y-axis
atB. IfOA = OB, where O is origin, find all curve which
passes through (1, 1).
X = x + ——— => OA = x + y —
dy / dx dy
x2 - y2 + 2xy = 2
;. Required curve,
Also, — + — = !=> 1 + — = x+y —
y dx
OA OB
dx
y = 0,
279
and
4y
2p = y-x^
dx
dy
i.o
...(iii)
dy
and smce p = a, y - x — = x + y —
dx
dx
4y = y ~ x
dx y + x
Let
=>
dv
(1 + v2)
x — =----- dx
1 +v
1+v ,
dx
- -- dv vz + 1----- x
y = VX =>
280
Textbook of Integral Calculus
Integrating both the sides, we get
f±±2Liv = -f*
J V2 + 1
J X
1 r 2v
dv + J dv
v2 + 1
2 J v2 +
1
lim /(x) = 3™. Also, /'(x) £ f 3(x) +
, then
/(*)
(a) b - atnl 4
(c) b - a £ n 124
J x
So/. Since,
r(x)a/3(x)+-i-
- log | v2 + 11 + tan-11 v | = - log x + C
2
f(x)
f'(x) f(x) £ 1 + f 4(x)
log {(-^v2 + 1) x} + tan-1 v = C
fMf'lx)^
log Jx2 + y2 + tan-1 — = C
x
l + /4(x)
As x = 1 and y = 0,
On integrating w.r.t ‘x’ from x = a to x = b.
log 1 + tan-1 0 = C => C = 0
Required curve, (log -^x2 + y2) + tan-1 I —
(b) b - a £ n 14
(d) None of these
i(tan-‘(y2(x)))J
=0
• Ex. 39 If f(x) be a positive, continuous and differentia­
ble on the interval(a, b). If lim /(x) = 1 and
x —»a+
lim (tan-1(/2(x))) - lim (tan-1(/2(x)))
or
(b - a) £ -
or
(b - a) £ it 124
Hence, (c) is the correct answer.
?
g Differential Equations Exercise 1:
/ 4
Single Option Correct Type Questions
1. If the differential equation of the family of curve given
by y = Ax 4- Be2x, where A and B are arbitrary
constants, is of the form
7. The real value of m for which the substitution,
y = um will transform the differential equation,
4
dy 4- ly | = 0, then the ordered
(1 - 2x) — | — 4- ly 14- k [ —
dx \dx
Jdx
pair (fc, /) is
(a) (2,-2)
(b)(-2,2)
(c) (2, 2)
(d)(-2,-2)
4
dy
6
2xy — 4-y = 4x into a homogeneous equation is
dx
•
•.
(a) m = 0
(b) m = 1
(c) m = 3/2
(d) No value of m
8. The solution of the differential equation,
2dy
1
.1
.
x — cos — ysiji--= -1,
dx
x
'x~
2. A curve passes through the point 1, — and its slope at
V 4J
y
2i y i
any point is given by — - cos — . Then, the curve has
x
I xj
where y—>-lasx—>°°is
. .
.1
1
(a)y =sm— cos—
X
X
the equation
(a) y = xtan-1 In —
x)
z .
1
<]n(c) y = — tan
x
k x.
(b)y = xtan-1 (In 2)
(b)xe
(b) >'=e
y
(d)yex=e
4. A function y = /(x) satisfies the condition
f' (x)sin x 4- /(x) cos x = 1, /(x) being bounded when
rn/2
x —» 0. If I = £
/(x) dx, then
(a)-<1<
e
(b)-</< —
4
2
4
it
(c) 1 < I < —
2
f
(d) 0 < I < 1
5. A curve is such that the area of the region bounded by
the coordinate axes, the curve and the ordinate of any
point on it is equal to the cube of that ordinate. The
curve represents
(a) a pair of straight lines (b) a circle
(c) a parabola
(d) an ellipse
6. The value of the constant 'rri and c’ for which
y = mx 4- c is a solution of the differential equation
O2y - 3Dy - 4y = - 4x.
(a) is m = - l;c = 3/4
(c) no such real m, c
(b)y =
. 1
xsin—
x
' X.+ 1
(d)y =------- xcos —
(b) ism = l;c = -3/4
(d) is m = 1; c = 3/4
*. '
X
(d) None of these
x
(a)ye?=e
2
(c)xex=e
X
X
3. The x-intercept of the tangent to a curve is equal to the
ordinate of the point of contact. The equation of the
curve through the point (1,1) is
X
. . • .1
1
(c) y = sin- 4- cos—
. ur i. .
j
9. A wet porous substance in the open air loses its
moisture at a rate proportional to the moisture content.
If a sheet hung in the wind loses half its moisture during
the first hour, then the time when it would have lost
99.9% of its moisture is (weather conditions remaining
same)
(a) more than 100 h
(b) more than 10 h
(c) approximately 10 h
(d) approximately 9 h
10. A curve C passes through origin and has the
property that at each point (x, y) on it, the normal line at
that point passes through (1,0). The equation of a
common tangent to the curve C and the parabola
y2 = 4xis
(a) x = 0
(c) y = x 4-1
(b) y - 0
(d) x 4- y 4-1 = 0
11. A function y = _f(x) satisfies
(x + l)-f'(x)-2(x2 + x)/(x) =
e
X2
•,Vx>-L
(x + 1)
If/(0) = 5, then f(x) is
(a)
(3x4-5
[x + 1
•e
' 6x 4- 5 '
(c) <(X4-1)2?
x2
(b)
(d)
6x4-5 ■S
k x 4- 1
5-6x
282
Textbook of Differential Calculus
12. The curve, with the property that the projection of the
ordinate on the normal is constant and has a length
equal to ‘a’, is
(a) x - a In (-Jy2 - a2 + y) = C
(b) x+-Ja2-y 2=C
16. The substitution y~za transforms the differential
equation (x2y2 -1) dy + 2xy3dx = 0 into a homogeneous
differential equation for
(a) a = -1
(b) 0
(c) a = 1
(d) No value of a
17. A curve passing through (2, 3) and satisfying the
(c) (y - a)2 = Cx
differential equation j* ty(t) dt = x2y(x), (x > 0) is
(d) ay = tan-1 (x + C)
13. The differential equation corresponding to the family of
curvesy = ex (ax + b)is
(a) x2 + y2 = 13
. ,^
d2y +2
n^dy ’’=0
(a)
(c)^ + 2^ + y = 0
dx1
dx
dx
dx
/JX d2y
(d)
dx
ndy
r2
2
(c)£.-/ = C
2
8
18
(d)xy =6
18. Which one of the following curves represents the
solution of the initial value problem Dy = 100 - y, where
. y(0) = 50?
dx
y.,
14. The equation to the orthogonal trajectories of the
system of parabolas y = ax2 is
(a)y + y2=C
2
(b)y2=-x
2
v2,2
100
2
100
50
a
(b)x2 + ^- = C
y,k
(b)
+-x
0
(d)x2-^-=C
0
100
x>0,is
1-n
(a)/(x) = C • x~
(C)
100
(d)
50
(b)/(x) = C-x"-1
i
(c)/(x) = C*x"
>x
y..
15. A function /(x) satisfying £ /(tx) dt = n f(x\ where
n
50
>x
0
(d)/(x) = C-x(1"fl)
50
0
>x
§ Differential Equations Exercise 2:
More than One Option Correct Type Questions
19. The differential equation x
(a) is of order 1
(c) is linear
3
+ — = yz
dx
(b) is of degree 2
(d) is non-linear
20. The function f(x) satisfying the equation
/2(x) + 4/'(x)-/(x) + Lf'(x)]2=a
(a) /(x) = C-e(2^/5)x
(b) /(x) = C-e(2+75)x
(c) /(x) = C-e(75-2)x
(d) /(x) = C-e“(2+75)x
where C is an arbitrary constant
21. Which of the following pair(s) is/are orthogonal?
(a) 16x2 + y2 = C and y16 = kx
(b) y = x + Ce~x and x + 2 = y + ke
(c) y = Cx2 and x2 + 2y2 - k
(d) x2 - y2 = C and xy = k
22. Family of curves whose tangent at a point with its
Tt
intersection with the curve xy = c2 form an angle of — is
(a) y2 - 2xy - x2 =k
(b) y2 + 2xy -x2 -k
(c) y = x - 2c tan'
(d)y = cln
C+ X
c-X
-x + k
283
Chap 04 Differential Equations
23. The general solution of the differential equation,
x
]=
point (0, -1) and satisfying the differential equation
dy
+ y cos x = cos x is such that
dxj y-log lx— J is
(a) y = xe1-Cr
(a) it is a constant function
(b) it is periodic
(c) it is neither an even nor an odd function
(d) it is continuous and differentiable for all x
(b) y = xe1 + Of
(c) y = ex-eCx
(d) y = xeCx
where C is an arbitrary constant.
24. Which of the following equation(s) is/are linear?
(a)^ + Z = tox
• ' dx
(b)y
27. The graph of the function y - f(x) passing through the
(3--
28. A function y = /(x) satisfying the differential equation
dy .
sin2 x
— •sin x - y cos x + -------- = 0
dx
xz
is such that, y 0 as x —> «>, then the statement which is
correct?
(a) lim /(x) = 1
(c) dx + dy = 0
d2y
(d) —= cosx
dx
(c)
25. The equation of the curve passing through (3,4) and
satisfying the differential equation,
IT
fH.12
(b) £ /(x) dx is less than y
flt/2
/(x) dx is greater than unity
(d) f(x) is an odd function
29. Identify the statement(s) which is/are true?
f dy
y ly
dx
+ (x-y}^-x=Q
dx
can be
(a) x - y + 1 = 0
(b) x2 + y2 = 25
(c) x2+ y2-5x-10 = 0
(d) x + y-7 = 0
I
d*y
(a) The order of differential equation .1 + —y = x is 1.
V
dx
(b) Solution of the differential equation
xdy - ydx = -Jx2 + y2 dx is y + Jx2 + y2 = Cx2.
d 2y
(dy
Y —
.. -- .
(c) —~ = 2 — -y is differential equation of family of
dx
dx2
\dx
)
curves y = ex (Acosx + Bsinx).
26. Identify the statement(s) which is/are true?
(d) The solution of differential equation
(a) f(x, y) = ey,x + tan— is homogeneous of degree zero,
x
y
v2
i v
(b) x- log — dx + —sin" — dy = 0 is homogeneous
XXX
differential equation.
(c) f(x, y) = x2 + sinx* cosy is not homogeneous.
(d) (x2 + y2) dx - (xy2 - y3) dy = 0 is a homogeneous
differential equation.
(1 + y2) + (x - 2eton ‘ y) —=
= 0isxe'
n ;<
dx
> _ j 2 tin' y + fc
30. Let y = (A + Bx) e3x
3x is a solution of the differential
d 2y
dy
equation —— + m — + ny = 0,m,ne I, then
dx2
dx
(a)m = -6
(b)n = -6
(c) m = 9
(d) n - 9
284
Textbook of Differential Calculus
[51 Differential Equations Exercise 3 •
u Statement I and II Type Questions
Directions
(Q. Nos. 31 to 40)
For the following questions, choose the correct answers from
the codes (a), (b), (c) and (d) defined as follows :
(a) Statement I is true, Statement II is true and Statement II is
the correct explanation for Statement I.
(b) Statement I is true, Statement II is true and Statement H is
not the correct explanation for Statement I.
(c) Statement I is true, Statement II is false.
(d) Statement I is false, Statement II is true.
31. A curve C has the property that its initial ordinate of
any tangent drawn is less than the abscissa of the point
of tangency by unity.
Statement I Differential equation satisfying the curve
is linear.
Statement II Degree of differential equation is one.
32. Statement I Differential equation corresponding to all
lines, ax + by + c = 0 has the order 3.
35. Statement I The order of the differential equation
whose general solution is
y = q cos2x + c2sin2 x + c4e2x+ cseZx + Ci is 3.
Statement II Total number of arbitrary parameters in
the given general solution in the Statement I is 6.
36. Consider differential equation (x2 + !)• —-y = 2x-—
dx
Statement I For any member of this family y —> °° as
x—
Statement II Any solution of this differential equation
is a polynomial of odd degree with positive coefficient of
maximum power.
37. Statement I Order of differential equation of family of
parallel whose axis is parallel to Y-axis and latusrectum
is fixed is 2.
Statement II Order of first equation is same as actual
number of arbitrary constant present in differential
equation.
Statement II General solution of a differential
equation of nth order contains n independent arbitrary
constants.
38. Statement I The differential equation of all
33. Statement I Integral curves denoted by the first order
d2x
non-vertical lines in a plane is —— = 0.
linear differential equation — - — y = - x are family of
dx x
parabolas passing through the origin.
Statement II Every differential equation geometrically
represents a family of curve having some common
property.
34. Statement I The solution of (y dx - x dy) cot —
z s
W
x
nx
dxissin — =Ce
(yj
Statement II Such type of differential equations can
only be solved by the substitution x = vy.
Statement II The general equation of all non-vertical
lines in a plane is ax + by = 1, where b 0.
39. Statement I The order of differential equation of all
conics whose centre lies at origin is, 2.
Statement II The order of differential equation is same
as number of arbitrary unknowns in the given curve.
40. Statement I y = asin x + b cos x is general solution of
y" + y = 0.
Statement IIy = asinx + & cos xis a trigonometric
function.
Chap 04 Differential Equations
285
H Differential Equations Exercise 4:
Passage Based Questions
Passage I
Passage III
(Q. Nos. 41 to 43)
(Q. Nos. 47 to 49)
Let y = /(x) satisfies the equation
f(x)=(e~x +eJt)cosx-2x-|o (x-t) f (t)dt.
Ifany differential equation in the form
(X, y)) + «/2(x, J))rf(/2(x, >-)) + ...= o
f(.f (X,
then each term can be integrated separately.
41. y satisfies the differential equation
. . dy
(a) — + y = e (cosx - sinx) - e (cosx + sinx) .
dx
dy
(b) ----- y = ex (cosx - sinx) + e (cosx + sinx)
dx
dy
(c) — + y = ex (cosx + sinx) - e (cos x - sin x)
dx
(d) — -y = ex (cosx-sinx) + e (cos x - sin x)
dx
42. The value of f' (0) + f "(0) equals to
(a)-l
(c) 1
(b) 2
(d) 0
For example,
(x>
1 (x
d — = -cosxy + - —
\yj
Jsin xytf(xy) + J
xdy-ydx = Jx2 -y2 dxis
sin'
y
sin'
1
1 =C
sin'
(b)xe
(a) Cx = e
(c) x + e
, i x3
I
3
2
x
I =c
ex
2
(b) e *(cosx + sinx) + — (3cosx - sinx) — e
5
5
ex
2
(c) e-x(cos x - sinx) + — (3 cosx - sinx) + - e‘
x4 1 x
(c) —-4 2 y.
I =c
(a) 0
(c) 2
3
. x4 1 X
=C
(b) — + 7
4 3U
x3 11 x
(d)i-4 2(7
2
I =c
49. Solution of differential equation
(2x cosy + y2 cos x)dx + (2ysinx - x2 siny)dy = 0is
(a) x2cosy + y2sinx = C
Passage II
(b) xcosy - ysinx = C
(c) x2cos2 y + y2sin2x = C
(Q. Nos. 44 to 46)
(d) None of the above
(b)1
(d)3
*=c
48. The solution of the differential equation
(xy4 + y) dx - xdy = 0 is
(a) —+ 4 I*
44. Number of critical point for y- f{x) for x e [0,2]
y
(d) None of these
ex
2
(a) e *(cosx -sinx) + — (3cosx + sinx) + - e
d2y
For certain curves y = /(x) satisfying —— = 6x - 4, /(x) has
dx2
local minimum value 5 when x = 1
+c
47. The solution of the differential equation
43. f(x) as a function of x equals to
ex
2
(d) e-I(cos x + sin x) + — (3 cos x - sin x) - - e‘
5
5
2
Passage IV
(Q. Nos. 50 to 52)
dy
Differential equation — = f(x)g(x)can be solved by
dx
.
separating variable
g(y)
45. Global minimum value of y = f(x) for x 6 [0,2] is
(a) 5
(C) 8
(b)7
(d)9
46. Global maximum value of y = J(x) for x e [0,2] is
(a) 5
(c)8
(b)7
(d) 9
50. The equation of the curve to the point (1,0) which
satisfies the differential equation (l + y2)dx- xydy = Q
is
(a) x2 + y2 = 1
(b)x2-y2=l
(c) x2 + y2 = 2
(d)x2-y2=2
286
Textbook of Differential Calculus
dy
1 + y2
= 0 is
51. Solution of the differential equation — + ,
dx
l-x2
53. IfC},C2EC
Cf Curve is passing through (1,0)
(a) tan-1 y + sin-1 x = C
C 2: Curve is passing through (-1,0)
(b) tan-1 x + sin-1 y = C
The number of common tangents for C] and C2 is
(a) 1
(b) 2
(c) 3
(d) None of these
(c) tan"’ y - sin-1 x = C
(d) tan-1 y - sin-1 x = C
52. If^ = l + x + y + xy and y (-1) = 0, then y is equal to
54. IfC3 eC
C3: is passing through (2, 4). If — 4- - = 1. is tangent to
a b
C3, then
(a) 25a + 10b2-ah2 = 0
(b) 25a + 10b-13ab = 0
(d) 29a + b + 13ab = 0
(c) 13a+ 25b-16ab = 0
(1 + x)2
(l-x)2
(a)e
y
x
dx
(b)e 2
(d)l + x
2
(c) In (1 + x) -1
-1
Passage V
55. If common tangents of Cj and C2 form an equilateral
(Q. Nos. 53 to 55)
Let C be the set of curves having the property that the point of
intersection of tangent with y-axis is equidistant from the point
of tangency and origin (0, 0)
triangle, where CVC2 G C and Cf Curve passes through
(2, 0), then C2 may passes through
(a) (-1/3,1/3)
(b) (-1/3,1)
(c)(—2/3,4)
(d)(—2/3,2)
E Differential Equations Exercise 5:
u Matching Type Questions
56. Match the following :
Column I
(A)
xdx + ydy
xdy- ydx
a2-x2-f
Column II
C -1/x
(P)
x' + y2
(B)
Solution of cos2 x — - tan 2x • y = cos4x, where jx| < — and y [ — j = 3j3
dx
4
\ 6J
6
8
(q)
(C)
The equation of all possible curves that will cut each member of the family
of circles x2 + y2 - 2cx = 0 at right angle
(r)
x2+ y2 + Cy=0
(D)
Solution of the equation x
(s)
sin2x
y= 2 (1 - tan2x)
dt=(x + 1) j^ry (/) dt, x > 0 is
y2 = asin C + tan
0
57. Match the following:
Column I
Column II
(A) Circular plate is expanded by heat from radius 5 cm to 5.06 cm. Approximate increase in
area is
(P)
4
(B) Side of cube increasing by 1%, then percentage increase in volume is
(q)
0.6n
x2
(C) If the rate of decrease of — - 2x + 5 is twice the rate of decrease of x, then x is equal to
(r)
3
(D) Rate of increase in area of equilateral triangle of side 15 cm, when each side is increasing
at the rate of 0.1 cm/s; is
(s)
3V3
4
Chap 04 Differential Equations
287
58. Match the following:
Column II
Column I
(A) The differential equation of the family of curves y=er (A cosx + fisinx), where A, B
are arbitrary constants, has the degree n and order m. Then, the values of n and m are,
respectively
(P)
2,1
(B) The degree and order of the differential equation of the family of ail parabolas whose
axis is the x-axis, are respectively
(q)
1. 1
(C) The order and degree of the differential equations of the family of circles touching the
x-axis at the origin, are respectively
(r)
2,2
(D) The degree and order of the differential equation of the family of ellipse having the same
foci, are respectively.
(s)
1,2
g Differential Equations Exercise 6:
u Single Integer Answer Type Questions
59. Find the constant of integration by the general solution
of the differential equation (2x2y -2y4)dx
+ (2x3 + 3xy3) dy = 0 if curve passes through (1,1).
60. A tank initially contains 50 gallons of fresh water. Brine
contains 2 pounds per gallon of salt, flows into the tank
at the rate of 2 gallons per minutes and the mixture kept
uniform by stirring, runs out at the same rate. If it will
take for the quantity of salt in the tank to increase from
40 to 80 pounds (in seconds) is 206X, then find X.
61. If f: R - {-1} —> and f is differentiable function which
satisfies:
f{x + f(y) + xf(y)) = y + f(x)+yf (x) V x,yeR-{-l},
then find the value of 2010 [1 + /(2009)].
62. If <Xx) is a differential real-valued function satisfying
<J>'(x) + 20(x) < 1, then the value of 2p(x) is always less
than or equal to.........
63. The degree of the differential equation satisfied by the
curves ^/l + x - a^i + y = 1, is..........
64. Let f(x) be a twice differentiable bounded function
satisfayi 2f5 (x) • f' (x) + 2 (f' (x))3 • f5 (x) = f" (x). If
f(x) is bounded in between y = k, and y = k2, Then the
number of integers between kx and k2 is/are (where
/(0) = r(0) = 0).
65. Let y(x) be a function satisfying d 2y I dx2 - dy I dx
+e2x = 0,y(0) = 2 and y'(0) = L If maximum value of y(x)
is y(a), Then Integral part of (2a) is.......
Differential Equations Exercise 7:
Subjective Type Questions
66. Find the time required for a cylindrical tank of radius r
and height H to empty through a round hole of area 'a'
at the bottom. The flow through a hole is according to
the law U(t) = u y]2gh(t) where v(t) and h(t) are
67. The hemispherical tank of radius 2 m is initially full of
water and has an outlet of 12 cm2 cross-sectional area at
the bottom. The outlet is opened at some instant. The
flow through the outlet is according to the law
v(t) = 0-6 yj2gh(t), where v(t) and h(t) are respectively
respectively the velocity of flow through the hole and
the height of the water level above the hole at time t and
g is the acceleration due to gravity.
velocity of the flow through the outlet and the height of
water level above the outlet at time t and g is the
acceleration due to gravity. Find the time it takes to
empty the tank.
288
Textbook of Differential Calculus
68. Let f :R+ —> R satisfies the functional equation
y(xy) = exy“x”y (ey/(x)+ex/(y))V x,yG R+.If
f '(1) = e, determine /(x).
69. Let y = /(x) be curve passing through (1,73) such that
tangent at any point P on the curve lying in the first
quadrant has positive slope and the tangent and the
normal at the point P cut the x-axis at A and B ,
respectively so that the mid-point of AB is origin. Find
the differential equation of the curve and hence
determine f(x).
70. If yx and y2 are the solution of differential equation
dy I dx + Py = Q,
where P and Q are function of x alone and y 2 = yx z,
-f °dx
then prove that z = 1 + c • e yi
where c is an arbitrary constant.
71. Let y = f(x) be a differentiable function V x G R and
satisfies :
/(x) = x + Jo* x2 z /(z) dz + J1 xz2 f(z)dz.
Determine the function.
72. If f: R - {-1) —> R and f is differentiable function
satisfies:
f(to + /(y) + x /(y)) = y + f(x) + yf(x) V x,
yG R-{-l) Find/(x).
g Differential Equations Exercise 8:
Questions Asked in Previous 10 Years' Exams
(i) JEE Advanced & IIT-JEE
73. A solution curve of the differential equation
(x2 + xy + 4x +2y+ 4)—-y2 = 0, x > 0, passes
dx
through the point (1,3). Then, the solution curve
[More than One Correct 2016]
(a) intersects y = x + 2 exactly at one point '
(b) intersects y = x + 2 exactly at two points
(c) intersects y = (x + 2)2
(d) does not intersect y = (x + 3)2
74. Let f: (0,00) —> R be a differentiable function such that
for all x G (0, ~) and f(1) * 1. Then
f '(x) = 2 x
(a) lim
x-»(H
(b) lim x
[More than One Correct 2016]
76. Consider the family of all circles whose centres lie on the
straight line y = x. If this family of circles is represented by
the differential equation Py' '+ Qy' + 1 = 0, where P, Q are
»
2
the functions of x, y and y' (here, y' = —, y " = — -),
dx
dx2
then which of the following statement(s) is/are true?
[More than One correct 2015]
(a) P =y + x
(b) P = y-x
(c) P + Q = l- x + y + y'+(y')2
(d) P-Q = x + y-y'-(y')2
77, The function y = /(x) is the solution of the differential
dy
xy
x4+2x
equation — + —-— =
in (-1,1) satisfying
\xj
dx x2—l
=2
/(0) = 0. Then,
(c) lim x2/'(x) = 0
. \ n V3
(a)T~ —
3 2
(d)|/(x)|<2forallxe(0,2)
75. Let y(x) be a solution of the differential equation
(1 + ex )y' + yex = L Lf y(0) = 2, then which of the
following statement(s) is/are true?
[More than One Correct 2015]
(a) y (-4) = 0
(b) y (-2) = 0
(c) y(x) has a critical point in the interval (-1,0)
(d) y(x) has no critical point in the interval (-1,0)
V3/2
_v3
2
71-x2
/(x)dx is
J3
(b)J- 4
[Only One correct 2014]
<o”6 4-^ c:"
(d)--—
6 6
2
78. Let f: [1/2,1] —> R (the set of all real numbers) be a
positive, non-constant and differentiable function such
that f' (x) < 2f(x) and f (112) = 1. Then, the value of
X/2
dX ^eS
^n^erva^
[Only One Correct 2013]
(a) (2e -1,2e)
., fe-1
(c)
k 2
. e -1
(b) (e -1,2e -1)
■(d) i0^)
Chap 04 Differential Equations
79. A curve passes through the point 11, — j. Let the slope of
\ 6)
y
the curve at each point (x, y) be — + sec y x>0. Then,
X
X
the equation of the curve is
[Only One Correct 2013]
f y'l 1
i VI
(a) sin — = log x + (b) cosec: — = log x + 2
\x)
2
(c) sec
(a) Both I and II are true
(c) I is false and II is true
(b) I is true and II is false
(d) Both I and II are false
84. If y(x) satisfies the differential equation
y' -y tan x = 2 xsec x and y(0\ then
[More than One Correct 2016]
it2
8^2
(x)‘loex+2
289
(d) cos
(b) /[7U
V 4/ 18
4n
3
It2
(c) y
9
2n2
3n/3
■ Directions (Q. Nos. 80 to 83) Let f: [0,1] -» R (the set of
all real numbers) be a function. Suppose the function f is
twice differentiable, f(Q) = /(I) = 0 and satisfies
f"M-2f'M + f(x)>ex,xe[Q, 1]
[Passage Based Questions 2013]
80. If the function e"x/(x) assumes its minimum in interval
[0,1] at x = 1 / 4, then which of the following is true?
13
1
(a) f\x) < /(x), -<x<- (b) f\x) > /(x), 0 < x < —
4
4
4
1
’
3
(C) f\x) < f(x), 0 < X < 4 (d) f'(x) < f(x), - < X < 1
4
4
81. Which of the following is true?
(a)0<f(x)<~
(c)-l<f(x)<i
(b)-l</(x)<l
z
z
(d) - <» </(x) < 0
4
82. Which of the following is true?
(a) g is increasing on (1, «>)
(b) g is decreasing on (1, 00)
(c) g is increasing on (1,2) and decreasing on (2, <~)
(d) g is decreasing on (1,2) and increasing on (2, <»)
85. Let /(x) + y(x)g'(x) = g(x)g'(x), y(0) = 0, x G R, where
f' (x) denotes
- and g(x) is a given non-constant
dx
differentiable function on R with g(0) = g(2) = 0. Then,
the value of y(2) is.....
[Integer Type 2011]
86. Let ft R -4 R be a continuous function, which satisfies
f (x) = f f(t)dt. Then, the value of f (In 5) is....
Jo
[Integer Type 2009]
■ Direction For the following question, choose the correct
answer from the codes (a), (b), (c) and (d) defined as
follows
(a) Statement I is true, Statement II is also true; Statement II is
the correct explanation of Statement L
(b) Statement I is true, Statement II is also true; Statement II is
not the correct explanation of Statement L
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
87. Let a solution y = y(x) of the differential equation
x^x2 -1 dy - y^y2 -1 dx = 0satisfyy(2) =
'3
83. Consider the statements.
I. There exists some x G R such that,
/(x) + 2x = 2(1 + x2)
Statement I y(x) = sec sec
x - —1 and
6J
1 2^3
Statement II y(x) is given by - =----y
x
II. There exists some x G R such that,
2f(x) + l = 2x(l + x)
-7
[Statement Based Questions 2008]
(ii) JEE Main & AIEEE
88. If a curve y = f(x) passes through the point (1, -1) and
satisfies the differential equation, y(l + xy)dx = x dy,
then f^~
is equal to
90. Let the population of rabbits surviving at a time t be
governed by the differential equation
[2016 JEE Main]
~” 2001 ff p(0) = 100, then p(t) is equal to
.dt
(»)-£
(b)-l
(b)-T
(C)|
(d)l
2
[2014 JEE Main]
t
(a) 400-300e2
t
89. Let y(x) be the solution of the differential equation
dy
(x log x) — + y = 2x log x, (x £ 1). Then, y(e) is equal to
[2015 JEE Main]
(a)e
(b) 0
(c)
(c)22
(d) 2e
(b) 300-200e 2
t
(c) 600-500e2
_£
(d) 400-300e 2
290
Textbook of Differential Calculus
91. At present, a firm is manufacturing 2000 items. It is
estimated that the rate of change of production P with
respect to additional number of workers x is given by
dP
— = 100 - 12Vx. If the firm employees 25 more workers,
dx
then the new level of production of items is
the total life in years of the equipment. Then, the scrap
value V (T) of the equipment is
[2010 AIEEE]
2
(c)ef"*
‘ T
(b)/-
2
(d)T’-l
[2013 JEE Main]
(a) 2500
(c) 3500
95. Solution of the differential equation
(b) 3000
(d) 4500
7C
cosxdy = y(sinx-y)dx, 0< x < —, is
[2010 AIEEE]
92. The population p(t) at time t of a certain mouse species
satisfies the differential equation
= 0-5(0 ~ 450. If
p(0) = 850, then the time at which the population
becomes zero is
[2012 AIEEE]
(b) log 9
(a) 2 log 18
(c) | log 18
(d) log 18
93. If — = y + 3 > 0andy (0) = 2, theny (log2)is equal to
dx
[2011 JEE Main]
(a) 5
(b) 13
(c)-2
(d)7
94. Let I be the purchase value of an equipment and V (t) be
the value after it has been used for t years. The value
V(t) depreciates at a rate given by differential equation
<*V(0 = - k (T -1),
dt
where k > 0 is a constant and T is
(a) sec x = (tanx + C)y
(b) ysecx = tanx+C
(c) y tanx = sec x + C
(d) tan x = (sec x + C)y
96. The differential equation which represents the family of
curves y = c1eC2X, where Cj and c2 are arbitrary
constants, is
[2009 AIEEE]
(a)y'=y2
(b)y" = y'y
(c)yy// = y/
(d) yy" = (y')2
97. The differential equation of the family of circles with
fixed radius 5 units and centre on the line y = 2 is
[AIEEE 2008]
(a) (x-2)y'2=25-(y-2)2
(b) (y-2)y'2 =25-(y-2)2
(c) (y ~2)2y'2 =25-(y-2)2
(d) (x-2)zy'2=25-(y-2)2
Answers
Exercise for Session 1
1. (a)
6. (a)
2. (a)
7.
(b)
7.(b)
3. (d)
8. (a)
Exercise for Session 2
l.(b)
3. (a)
2.(0
6. (a)
11. (b)
7-(a)
12.(0
8. (a)
13. (a)
4. (a)
9. (a)
5.(d)
10. (b)
4. (a)
9.(0
14. (b)
10. (a)
15. (b)
4. (b)
9. (a)
10. (c)
5.(0
Exercise for Session 3
1. (a)
6. (0
2.
(a)
2.(0
7. (a)
3. (b)
8. (0
5.(0
28. (a,b,c)
27. (a,b,d)
26. (a,b,c)
31. (b)
30. (a,d)
29. (b,c,d)
36. (a)
35. (c)
34. (c)
32. (d) 33. (d)
41. (a)
40. (b)
39. (d)
37. (b) 38. (d)
46. (b)
45. (a)
44. (c)
42. (d) 43.(c)
50.
(b)
49. (a)
51. (a)
47. (a) 48. (b)
55. (a)
54. (a)
52. (b) 53.(0
56. (A) -» (q), (B) -> (s), (C) -> (r), (D) -> (p)
57. (A) -> (q), (B) -> (r), (C) -> (p), (D) -> (s)
58. (A) -> (s), (B) -> (s), (C) -» (q). (D) (p)
59. (1) 60.(8)
61.(1)
62.(1)
63.(1)
64. (3) 65. (1)
66./ = —
Exercise for Session 4
Pa V g
3. (a)
8.(b)
lit x 10s
1.(0
6. (a)
2.(d)
7.(0
4. (b)
9. (a)
5. (a)
10. (d)
4. (a)
5.(0
Exercise for Session 5
1. (a)
6. (a)
2.
(c)
2.(0
7.
7.(b)
(b)
3. (c)
8. (c)
Chapter Exercises
4. (a)
2. (a)
3.(0
5.(0
1.(0
6.(b)
10. (a)
8. (a)
9.(0
7.(0
11. (b) 12. (a)
15. (a)
14. (a)
13. (b)
16. (a) 17. (d)
18. (b)
19. (a,b,d)
20. (c,d) 21. (a,b,c,d)
22. (b,c,d)
25. (a,b)
23. (a,b,0
24. (a,c,d)
■
67. t =
135
68. /(x) = Z logx
69. x + /(x)/'(x) = Jx2 + /2(x) and /2(x) = 1 + 2x
20x
71. /(x) = —(4+9x)
119
72. /(x) = -^1+x
73. (a, d) 74. (a,d) 75. (a, c) 76. (b, c) 77. (b)
82. (b)
80. (c)
81. (d)
78. (d) 79. (a)
86. <0)
87. (c)
85. (0)
83. (c) 84. (d)
92. (a)
90. (a)
91. (c)
88. (d) 89.(0
97. (c)
96. (d)
95. (a)
93. (d) 94. (a)
Solutions
dv
dx
v
1—v
X----=------------ V =
J
v - V + V2
1 -v
J X
V
---In v = lnx + C
v
1. y ■ e~2x - Ax e 2x + B
e"2* • y, - 2ye~2x = A(e~2x - 2x e~2x)
Cancelling e~2x throughout
yi - 2y = A(1 - 2x)
Differentiating again y2 -2yl=- 2A
—(i)
- —-In —= lnx+C => ■ — = lny + C
y
x
y
x = l,y = 1 => C = -l
x ,
1 — = Iny => y = e-e~x,y
y
AAxixyi
=>
e~x,y = -
2
On substituting A in Eq. (i)
2(y2 - 2y) = (2yi - ?z) (1 -2x)
2yi - 4y = 2y,(l - 2x) - (1 - 2x)y2
dy - 2y^ = 0
(1 - 2x) — f — - 2yl + 2 (—
dx \ dx
J
k dx
J
•
Jy + y cosx = 1
4.. sinx
—
dx
n/2
Hence,
k = 2 and I = - 2
=>
Ordered pair (k, I) = (2, - 2)
o dy y
2y
2.
=
cos — y = vx
dx x
x
dv
2
v + x — = v - cos.2 v
dx
dv
[—= C => tanv + In x = C
J cos V J x
i
i
i
-t
i
i
i
i
i
i
1 6-
■>
o
jt/2
dy
~ + y cot x sb cosec x
J—~ +
!F = cbobc‘fc = eln(sinx)=sinx
tan — + In x = C
x
ye
If x = 1, y = — => C = 1 => tan — = l- lnx = ln —
4
x
x
y = x tan
=i> yex,y = e
y
In—1
y sin x = jcosec x-sinx dx
y sinx = x + C
If x = 0, y is finite /. C = 0
xj
y = x (cosec x) = ——
sinx
3. Y - y = m(X - x). For X-intercept Y = 0
x=x~y
n2
4
Now,
m
y
Therefore, x - — = y
m
Hence,
y±
J
I*
^n2
4
n
2
5- £/(*)<& = y3. Differentiating f(x) = 3y2
P(x,y)
£.y)
(x,y)
>x
O
y = 3y2
or
r
and I > —
2
dx
y
x-y
v
dv
Put y = vx v + x — =
dx 1 -v
or
=> y = 0
3y dy - dx
3v2
2
= x + C => Parabola
(rejected)
Chap 04 Differential Equations
dy _ 1 ~ x
dx " y
d2y .
6. y = mx + c, — = m;-7 = 0
dx
dx
^ = x—+ O
2
2
Eq. (ii) passes through (0, 0).
Thus, C = 0
>2 -2x = 0
- 3 — - 4y = - 4x
dx
dx
Substituting in
0 - 3m - 4 (mx + c) = - 4x
- 3m - 4c - 4mx = - 4x
-(3m+ 4c) = 4x(m-l)
■0)
7. y = u" => — = mu”"1 —
dx
dx
- du
Since, 2x4 • tf1 • mif*^ — + u,4m = 4x4
dx
1
y = mx + —
m
dx
y±
4x6 - u4w
2m x4 u2”-1
J „ 1
1 1
-^-tan— = -sec — .2
-2
X
X
XX'
X
X
r 2f l') 11 ..
1
= - sec — —
dx = tan— + C
.2
J \xjx
X
1 + m,22 = m2
2x(x + 1) f-/ \
e
11. f'(x)- x+1 f(x)
n =
~(x+l)2
IF =
dx
(X+1)’
=>
when
=1
71 + m2
Hence, tangent is y-axis le, x = 0.
9. — = -KM => M = Ce* when t = 0; M = Mq
dt
=>
O = Mo
M = M^^
=>
when t = 1, M = —
2
k = In 2
Therefore, M = Mje"*ta 2
m + (1 /m)
=>
tfy —> -1, then x —>00
1
1
C = -l => y=sin—cos—
x
x
(1.0)
2 -2x = 0
Then,
1
1
= sec — =$ y -sec—
+-X
0
3
2m -1=2 => m = 2
—f— tan— dx
...(iii)
If it touches the circle
3
4m =6 => m = 2
and
.(ii)
Now, tangent to y2 = 4x
Eq. (i) is true for all real x, if m = + 1 andc = — 3/4.
du
~dx
293
fw-'"' ”777 + c
At x = 0,/(0) = 5 => C = 6
f(x) =
M=
, then t = log2 1000 = 9.98
1000
= 10 h approximately
6x + 5
.s
12. Ordinate = PM. Let P ■ (x, y)
Projection of ordinate on normal = PN
10. Slope of the normal = —- -
P
P(x,y)
a/
(*.y)
N.
o
M
>x
294
Textbook of Integral Calculus
(given)
PN = PN cos 8 = a
y
=a
•^1 + tan2 8
y=a^l + (yi)2
dy Jyz - a 2
dx
18. J
a
f-Fi±_=tdx
aln|y + -Jy2 -a2| = x+ C
=>
y = ex(ax + b)
13.
...(i)
Differentiating w.r.t. x, we get
— = ex(ax+ b) + ex-a or — =J
dx ' '
'
dx
Again, differentiating both the sides
“K
- In (100 - y) = x + C
ln(100-y) = -x + C
Also
x = 0, y = 50
C = ln50
x = In 50 - In (100 - y)
50
In
100-y ~
- =50
e
100-y
=>
y = 100 - 50e"x
From Eq. (iii) - Eq. (ii)
2
d2y 2dy
— + y =0
dx2
dx
is required differential equation.
19. Here, x
has order 1, degree 2 and non-linear.
20. Here, (f '(x))2 + 4/'(x) • f(x) + (f(x))2 = 0
M.^=2ax = 2x.4;& = a;
dx
Now,
x
dx
x
x
i
m — = -l => m =----- =>
dx
2y
dy
2
X
dy
dx
x
2y
J
1
f'(x)
-4 ± 716 - 4
/(x)
2
+1=0
^^- = -2±j3
=>
fM
.
Put, x = y => dt = — dy
x
Integrating both the sides
log|/(x)|=(-2±73)x+q
f0V(y) dy = n/(x)
’
[M + 4 f'W
fW
=>
2
y =----- + C
2
15. ^/(tx)dt = n-/(x)
rv
x
100-y =50e
...(iii)
dx
dx
Iny + In x = ln C
xy=C
At point (2,3),
2 x3 = C =>C =6
xy =6
dy
100 -y
/(x) = e(-2±V3)x-C
jX
Qf(y)dy = x-n-f(x)
Differentiating,
/(x) = n [/(x) + xf '(x)]
f(x)(l-n) = nx f'(x)
f\x) _ 1 - n
/(x)
nx
1 -n
Integrating, In /(x) = ------ In Cx = In (Cx)I "
V n )
1 “n
21. (a) 32x + 2y —= 0
dx
dy
16x
— = mi =---------dx
y
and 16v‘s^=*
dx
k
dy
— - m2 16y
15
dx
/(x) = Cx~
16x
k
= -—
m[m2 ------16 k
*
y 16y15
y
16. (x2z2a -1) aza-1 dz + 2XZ30 dx = 0
or
x
y16
a(x2 z33a-l
*1"1 -za-1) dz + 2xz3a dx - 0
for homogeneous every term must be of the same degree,
3a + 1 = a -1 => a = -1
17. Differentiate, xy (x) = x2y'(x) + 2.xy(x)
or xy(x) + x2 y'(*) = 0
dy
x— + y = 0
dx
ss —-----
X
dy =1 -ce~x= 1 - (y - x) =-(y - x -1)
(b) —
dx
and
dx
dx
dx
[using ce
= y-x]
Chap 04 Differential Equations
[l-(x + 2-y)]^ = l
[using ke~y = x - y + 2]
dx
1
dy =
™2 =
y-x-1
dx
or
7n177Z2 = -l
(c) — =2cx = 2x~
—" 2
dx
X
Also,
Hence,
=>
dy
x
= 771,
dx----- 2y
22. Let m = ~~ be the slope of tangent (x, y) to the required curve.
dx
zrij = slope of the tangent at xy = c2
c2
= —--- =.y
X
•
x2
•
c2
m+—
m+. —
y
m
,
ym
m
11 —
x
---- xL = ± 1 or —*-=±1
c2
1------2m
X
or
y = x-e1-Cx
and
log — = log e + Cx
y
dy + Py = Q, which is
24. Clearly, (a) and (c) are of the form —
dx
linear in y.
d 2y
dy
Also, (d) is —— = cosx, on integrating — = sinx + C
dx2
dx
which is also linear in y.
— dy
25.—
dx
(y - x) ± 7(x - y)2 + 4xy
2y
^ = 1 or dy
dx
dx
(i)
=>
x-y + l = 0 and xz + y2=25
Homogeneous differential equation.
y2
y.dy^o
(b)logpf| dx+ ^-z-sin
x
\Xj
X
i(
dy 1
=> 2y — - 21 y + x — I -2x~0
- 2x = 0
dx
kk
dx)
dxj
x
— (y - x) = x + y
dx
dy x + y
---- — —'-------— TH
dx y - x
(say)
log
^=dx y2 . .
^ysin
x
0
log0
From Eq. (i)
x+y ! y
y-x x
LHS =
y( x + y
1X y-xj
=4±4=1RHS
.2 .
xz + y2
Similarly option, (b), (c) and (d) safisfy
23.x^=ylogf^^ = Zlog[>' , put y = vx
dx
=>
<x.' dx
J"
x
y
26. (a) /(x, tx)~e‘ + tan-1(t), independent of x.
y2 -2xy - x2 =k
dy i
log — = 1 + xC
x
y - ex-eCx
xy = k
~dy.
dy
y m
x — + y = 0 => -m2
dx
dx
x
= -1
Hence, (a), (b), (c) and (d) are all correct.
dy
rdx
J X
y = xe1 + Cx
= -1
x•2 -y 2 = c
Consider
r dt
-i-
let log v = t
=>
log (t -1) = log (x) + log C
log (t -1) = log (xC)
t = l + xC
---- =
dy x
2x-2y& =
0 => — = - = jm
dx
dx y
Hence,
r
dv
J v(log v -1)
X
dy
2x + 4y ~ = 0
(d)
=>
x--
1 XJ
-
-j
dv
dy
—=v+ x—
dx
dx
dv
dv
n
V + x — = V log V => x —= v(logv-l)
ax
dx
f(x,y} = —2—
y ein'
—T sin
x
log (t) . .
t2 sin"1 (t)
,
/(x, tx) = ———;—, independent of x
Homogeneous differential equation.
(c) f(x, y) = x2 + sinx ■ cosy
/(x, tx) = x2 + sinx -cos (tx), not independent of x.
=> Not homogeneous differential equation.
_r2 + y2
(d)/(x,y)
xy 2 -y 3
=> f(x tx) is not independent of x
=> Not homogeneous differential equation.
295
296
Textbook of Integral Calculus
cojxctx = e»nx
27. Integrating Factor =
y.e*inx = J
or
e+«inx
„
_+ »inx
cosx = — e
(d) (! + /)&+x=2e"'
dx
+C
ften--y
dx
1
=> — +
X =--------~2:
dy 1 + y2
1+y
At point (0, -1),
-le° = -e° + C=>0
f-L_
IF = e ,+ /
A-®1'1' —
..•
- e
ye
y=-l
no dy
sinx
28. --y cotx =----- —
dx
x
ten1 y
1+y
X
X -»0
fK/2 sinx ,
I=
----- dx
x
Jo
is decreasing, when x > 0
y+k
30. — = 3 (A + Bx) e3x = Be3x
dx
=> ?y
-y- + my = (3 4- m) (A + Bx) e,3x + Be3'
dx
y.-2-=l
+c
sinx x
As x-+«>,y0=>C = 0
sinx
,
y =----- => lim j(x) = 1
xe,ten"
““ Jy'=e2'“'
=»
1
sinx ■
1
r sinx 1 .
„
Required solution is y • -— = J---- ;---- :— dx + C
x
y
xe tan' y = 2^ y~2dy
Integrating Factor = ef-cotx<lx = e“ ,0«sin*
Since,
y
dy
,3*
_
=> dl
dx2 + m — + ny «(9 + 3m + «)(A + Bx) e
dx*
dx
♦ B (6 + m) e3x = 0
(i)
=> 3m + n + 9«0 and m + 6-0
=*
and n«9
31. Equation of tangent
y-y-m(X-x)
Y =y-mx
PutX = 0,
T(x)<f(o)
r^/2
n
it
f /(x)<-andx<Jo
2
2
=>
'M9
=>
/(x)dr>l
29. (a) Order of the differential equation is 2.
xdy - ydx
xdy - ydx
(b)
■Jx2 + y2
In — +
x
1 + 4 = ln|Cx|
x
y , -jx2 + y- = Cx
X
X
y + y]x2 + y2 = Cx2
Le.
(c) y = ex (Acosx + Bsinx)
dy =
ex (Acosx + Bsinx) + ex(-Asinx + Bcosx)
dx
= y + ex (-Asinx + Bcosx)
d2y
dx2
dy
dx
Asinx + Bcosx)
0
Since, initial ordinate is
y - mx - x -1
=>
mx-y=l-x
dy-------1 y------1-x
which is a linear differential equation.
dx x
x
Hence, Statement I is correct and its degree is 1.
=> Statement II is also correct. Since, every 1st degree
differential equation need not be linear, hence Statement II is
not the correct explanation of Statement I.
32. Statement I The order of differential equation is 2.
Statement I is false.
33. Integral curves are y = cx - x2
The differential equation does not represent all the parabolas
passing through origin but it represents all parabolas through
origin with axis of symmetry parallel to y-axis and coefficient
of x2 as -1, hence Statement I is false.
Statement H is universally true.
yd* ~ xdy
+ ex (-Acosx - Bsinx)
’.
= —+ ex (-Asinx + Bcosx) -y
dx
dx
dy
dy +JL-y-.y
dy
^L
=2 a~y
dx dx
dx
or
x
j
cot — = x dx
f cot — • d
*
y
Chap 04 Differential Equations
J cot t dx = nx + C
or
41. /(0) = 2
/(x) = (ex + e~x) cos x - 2x -
In (sint) = nx + C; sin — = Ce™
y
y =Ci cos2x + c2sin2 x + c3 cos2 x + c«e2x + c5
35.
e2x+Ci = c1 cos2x + c2 1 - cos 2x
2
= c -C
-l+C
-l cos2x +
I
2
2,
1 T
/(x)=(ex+e
cos 2x +1
2
+ c4e2x + c^-e*
2x
A
43. IF of Eq. (i) is ex
y-ex = je^cosx -sinx) dx - j(cosx + sinx) dx
y-ex = je2x(cosx-sinx) dx -(sinx - cosx) + C
<'•;
A
Obviouslyy
as x —> «>; as C > 0
37. (x - h)2 = 4b (y - k) here b is constant and h, k are parameters.
38. The general equation of all non-vertical lines in a plane is
ax + by = 1, where b * 0.
Now,
ax + by = 1
a + 1,^0 (differentiating w.r.t. x)
dx
d2^
y =Q
b—
=>
(differentiating w.r.t x)
dx2
d 2y
=>
(as b * 0)
-4 = 0
dx2
d2y
Hence, the differential equation is
= 0.
39. The equation ax2 + 2hxy + by2 = 1 represents the family of all
conics whose centre lies at the origin for different values of a,
h, b.
:. Order is 3.
Thus, Statement I is false and Statement II is true..
Hence, option (d) is the correct answer.
40. y = asinx + b cosx
— = a cosx - bsinx
dx
=>
(ii)
Htence, — + y = ex (cos x - sin x) - ex (cos x + sin x)
dx
y=C —+ x +C',C'eB
dx2
•(*)
f'(x) + f(x) = cosx (ex - e“x) -(ex + e-x)sinx
—---- dx
x2 + 1
/ 3
=>
- 2x - xf(x) + 2x +Tx /(x) - j'f(l) *1
42. f'(0) + /(0) = 0-2-0 = 0
In — | = In (x2 + 1) + In C, C > 0
.dx/
.
^ = C(x2 + l)
=>
/(x) = (ex + e-x) cosx
On differentiating Eq. (i)
=> Total number of independent parameters in the given
general solution is 3.
36. The given differential equation is
dy
dx
)cosx-2x-[x(/(x)-/(0))-(f-/'(< -£7W*)1
f(x) =(ex + e~x) cosx-
c2 + £3 + (c4 + c'5)e2x
.2
2.
= X] COS2x + X2«2x + X3
da.
297
-acosx-bsinx = -y
d 2y
But Statement II is not the correct explanation of the Statement I.
Let
I = Je2x(cosx-sinx) dx = e2x(A cosx + Bsinx)
A = 3/5 and B =-1/5 and C =2/5
Solving,
3
1 .
y=ex - cosx — sinx -(sinx - cosx) e
5
5
2 .
+-e
5
Solutions (Q. Nos. 44 to 46)
d2y
Integrating, —y = 6x - 4,
dx
we get — = 3x2 - 4x + A
dx
dy
When x = 1, — = 0 so that A = 1.
dx
dy =
Hence,
dx
Integrating, we get y = x3 - 2x2 + x + B.
•(i)
When x = 1, y = 5, so that B = 5
Thus, we have y = x3 - 2x2 + x + 5
From Eq. (i), we get the critical points x =
x=1
1 d 2y
At the critical point x = -, —~ is negative.
3 dx
Therefore, at x =
At x = 1,
d2y
y has a local maximum.
is positive. Therefore, at x = 1, y has a local
minimum.
Also, f(l)=5,
=
157
/(0) = 5,f(2) = 7
Hence, the global maximum value = 7.
and the global minimum value = 5.
298
Textbook of Integral Calculus
54. C3:(x-5)2 + y2=52
47. xdy - ydx = x.22-y2dx
= ^-(y/x)2
x
sin
48.
d(y/x)
rdx
dx => j ■j\-(y/x)2 J x
Tangent, bx + ay-ab = 0, length of perpendicular to tangent
from centre = radius.
|5t> - ab| = 5^a2 + b2
=>
.sin-1 ylx
— = In x = In C or Cx = e
x
(xy4 + y) dx - x dy = 0
xy4dx + ydx-xdy = 0
2
/ \
J x3 dx -
I d
X
a2b2-10ab2 =25a2
ab2-10b2=25a
55. Cf.x2 + y2-2x = 0
C2:x2 + y2 -Cx = 0
=0
p.
x4
1
—+_ x
4 -3 y
=>
\3
Q
=C
R
(1.0)
M
49. 2x cosydx- x2 siny dy + y2 cosxdx + 2y sin xdy = 0
jd(x2cosy) + Jd(y2sinx) = 0
x2cosy + y2sinx = C
ZQRM =30°
dx _ ydy
x 1 + y2
50.
sin30° = —-— =s> OR = 1
1 + OP
sin30° = -^- =}£M = l/3
1-QM
In x = | • In (1 + y2) + C
=>
From the given condition, C = 0
x2-y2 = l
dy , i + y2
51.
dx
2
radiusofC2 = 1/3 ^C = -2/3,C2:xz + y2 + -x = 0
Point (-1 / 3,1 / 3) will satisfy C2.
=0
56. (A)Letx = rcosfty = rsin0
dx
dy ,
==0
1 + y2 ylt-X'.2
x2 + y2 = r2 (sin2 0 + cos2 0) = r2
and
tan-1 y + sin' x = C
d
{1*x)2
52. ~~ =(1 + x)-(l + y)givesy =e 2 -1
53. Tangent at point P(x,y), is y - y = f'(x)(x - x)
From Eq. (i), x dx + y dy = rdr
From Eq. (ii),
d
=>
From Eqs. (iii) and (iv)
rdr
SdQ
or
X2
C,:x2 + y2-x = 0
=>
C2:x2 + y2 + x = 0
=>
Cj and C2 touch externally => number of common tangents = 3.
_dr
Ja2 -r2
sin
— = l + ^-=>x2 + y2-cx = 0
X
d (tan 6)
= r2 cos2 0 sec2 0 d0 = r2d0
dt dy
t + x— = —
dx dx
dt t2-l
t + x— =-----dx
2t
dt -(1 + t2)
dx
2t J
x— =---------- =* ----- ?dt = X
i+r
dx
2t
.(iii)
xdy -ydx
2
------ ----- = sec o du
x2
xdy -y dx = x2 sec20d0
=> x2 =y2-2xy-f'(x)
_y2-x.2‘
put y = tx
2xy
.(ii)
d(x2 + y2) = d(r2)
Q:(0,y-x/"(x))
Then,
PQ = OQ
=> x2 + x2(/'(x))2 = y2 + x2(/'(x))2 -2xy(/'(x))
or dy I dx
tan0 = —
x
'a2 -r,2
r2
= dQ
- =0+C
aJ
r = asin (0 + C)
^x2 + y2 = asin ■ C + tan
-.(iv)
Chap 04 Differential Equations
(B) — - sec2 x tan 2xy = cos2 x
dx
—
x2dy(x)
dx
c
(1 - 3x) dx _ dy(x)
or y = —
x
x2
y(x)
or
iun2xsec2 xdx _ e tun x-I
x nee2 xdx
(l-3x)y(x) =
or
IF = eJ
I= e ‘, where t = tan2 x -1
299
■l/x
57. (A) r = 5 cm, 8r = 0.06, A = nr2
8A = 2nr 8r = lOn x 0.06 = 0.6n
(B)v = x3
= ehltl =| t|= |tan2 x — 11
71
Given that, | x | < —
4
tan2 x < 1
8v = 3x2 8x
— x 100=3 — X 100 =3 XI =3
IF = l-tan2x
X
V
Solution is y (1 - tan2 x) = | cos2x (1 - tan2 x) dx
(C)(x-2) —=2— => x = 4
dt
dt
= f (cos2 x - sin2 x) dx
(D)
sin2x „
-------+ C
2
,
7t
3V3
when x = —, y =---6
3^3 , 1^ 1 73 „
1-- =-x — + c
8
3
2
2
sin2x
C = 0 => y =
2(1 — tan2 x)
x2 + y2-2cx = 0
(C)
2x + 2y
dy
;.
...(i)
...(ii)
or - x + y - 2xy — = 0 is the differential equation
dx
representing the given family of circles. To find the orthogonal
trajectories.
2
2
or
dy
or = yd^±dy
or
„
dx
y-x2=-2xy —
dy
y2 dy = x2 dy -2xy dx
373
dt
2
10
4
2
dt
dx
= — + ex(-Asin x + B cos x) +ex(-A cos x - Bsin x)
dx
=^y + (^y
dx
dx
dy
-y=2 —-y
dx
)
\dx
So, degree = 1 and order = 2
(B) The equation of the family is y2 = 4a (x - b)
where a, b are arbitrary constants.
d2y
dy = 4a or
2y ~
dx
dy
2
1
= y + ex (-Asinx + Bcosx)
ydy x = 0
From Eqs. (i) and (ii), x2 + y2 - 2 x + 2-2dx
2
Ji
58. (A) — = ex (Acosx + Bsinx) + ex(-Asinx + Bcosx)
dx
- 2c = 0
C = x+^
dx
or
A = — x2
4
dA 73 dx
So, degree = 1 and order = 2
(C) The equation of the family is
(x - c)2 + y2 = c2
or
x2 + y2
x2 + y2-2cx = 0 or ------ — =2c
x
f?” J- *>’■ —
y2
fx2>|
-dy=d —
\y)
x2
or
-y=~ + C
y
=>
x2 + y2 + Cy = 0
=> Orthogonal trajectory.
(D) Differentiating the equation w.r.t x, we get
"
So, degree = 1 and order = 1
(D) The equation of the family is
x*.2
y2
+
= 1 because
a2 + X b2 + X
they have the same foci (± -^a2 - b2, 0).
2x
2y
dy
On differentiating, -------- + —-------- - = o
d2 + X b2 + X dx
xy (x) + 1 Joy(t) dt = (x + 1) xy (x) + 1 j* ty(t) dt
Again, differentiating, w.r.t x, we get
y (x) = x2y'(x) + 2xy(x) + xy(x)
2x + 2y -y-1 x-(x2 + y2)-l
dx)
=0
----------------
or
x
, yp
=0
a2 + X b2 + X
or x(b2 + X) + yp(a2 + X) = 0
dy
300
Textbook of Integral Calculus
-b2x - a2yp
5i
I
The required time (t2 - ti) = 251 In 5 - In - I = 25 In 3
x + yp
= 25 X 1.0986 = 26 min 28 s
/. The differential equation is
= 1648 s - 206 X 8 = 206 X X
x2
y2
--- ±-----------++---------1------------ = i
----- b2x - a2yp b2 | -b2x-a2yp
b2 +
a2 +
x + yp
x + yp
x(x + yp) y(x + yp)
a2_b2 (b2-a2)p
or
Differentiating w.r.t. x and y is constant
f\x + f(y)+xf (y)) (1 + /(y)) = f'(x) + yf'(x)
...(ii)
From Eq. (i) again differentiating w.r.t y as x is constant
f\x + /(y) + xf(y)) (1 + x) /'(y) = 1 + /(x)
(iii)
From Eqs. (ii) and (iii)
i + /(y) _ (i + y) f\x)
l + /(x)
(l + x)/'(y)
,
59. 2x2y dx - 2y4 dx + 2x3dy + 3xy3 dy = 0
Dividing throughout by x3y, we get
„ dx -^dx
2y3 , + 2- dy 3y2 ,
2------- dx + 2-—+dy = 0
x3
xx
y x
=>
(i + y)/'(y)_ i + /(x) _x
i + /(y)
(i + x) /'(x)
„ dx . „ dy
dy+. 3y
3/2 x4 dy -2y3x3dx _ Q
2 —+ 2 —
,6
X
y
x
2_ _
dx + 2— + 3y2x2jy-2y3xdx = 0
x
y
•••
f'(y)
=
md r(x) = *l/W
X and f\x)
1+y
A(1 + x)
X = 1 =» X-±l
x4
2d(lnx) + 2d(ln y) + d
( 3>)
I* J
2 In |x| + 21n |y| +
(0
f(x + /(y) + xf(y)) = y + /(x) + y/(x)
order = 1 and degree = 2
So,
X=8
.
y3
x
=0
±—
1 + A«)
x
■
1+x
Integrating both the sides f(x) = C(1 + x)11 -1
From Eq. (i) put x = y = 0
=C
Z(/(0))=f(0)
when x = 1, y = 1. So, C = 1
60. Let the salt content at time ‘f be u lb, so that its rate of change
is du! dt.
= 2 galx 2 lb = 4 lb / min
If c be the concentration of the brine at time t, the rate at which
the salt content decreases due to the out flow
= 2 galx c lb / min
= 2c lb / min
du
— = 4-2c
...(i)
dt
Also, since there is no increase in the volume of the liquid, the
u
concentrations
c=—
50
du
2u
;. Eq. (i) becomes — = 4 - —
.
From Eq. (ii), /(0) = C -1
So,
/(C-1) = C-1
C = 0,1
(taking +ve sign)
So, f(x) = r-1 and f(x) = (1 + x) -1 = x and C = 1
/(x)=(l + x)-‘-l = -^1+X
l + /(x) =
1
1+x
(2010) (1 + _f(2009)) = 1
1 + /(2009) =
62.
4>'(x) + 20(x) <. 1
1
e2x 0'(x) + 2<|> (x) e2* £ e,2x
— I le2* tfx)-^ <0
dx I
2
Separating the variables and integrating, we have
e2x 0 (x) -1j is a non-increasing function of x.
frff=25r_^_
J
J 100 - u
or
t =-25 In (100-u) + K
Initially when t = 0, u - 0
0 = - 25 In 100 + K
Eliminating ‘K’ from Eqs. (ii) and (iii), we get
' 100
; = 25 In
(iii)
+
S3.
=0
+
dx
=>
JOO-uJ
Taking t = tj when u = 40 and t = t2 when u = 80, we have
. (100 )
, riooa
and t2 = 25 In —
^251nW
=> 0 (x) - - is a non-increasing function of x.
2
A 20 (x) is always less than or equal to 1.
...(ii)
uo;
1
_ a
dy
.*y
yjl + x ^1 + y dx
-yjl + y
1
^1 + x dy/dx
Putting this value in the given equation,
iLjm. *+y
dx
y/l + x
dx
Chap 04 Differential Equations
— dy
x—
dx
(l + x)^ = l + y + ^
dx
301
Now, when t = 0, h = H and when t = t, h = 0
f° h~U2dh
Thus,
J0
=> (1 + x - Jl + x) — = 1 + y
dx
ga yj2g
_2
:. Degree of the given equation is 1.
64. Heie,2f5(x)‘f'(x)-\l + (f'(x))2\ = f"(x)
i+(/'(*))
|/*(xW2(x))| = J</(tan-'(/'(x)))
=>
Ax) + C = tan'^^x)), as /(0) = f'(0) = 0 => c = 0
[2H
71 r-2
t =-----]ia N g
67. Let at time t the depth of water is /1; the radius of water surface
is r.
r2 = R2-(R-h)2
Then,
r2=2Rh-h2
3
tan ‘(f'(x))
=>
n
2
_f37lY/6
=>
/6(x)
(x) < ft
/
33
2
.. . (37cV1/6
>
«----- R
+
R-h
3
=dh
It] </(x)<(t]
=> Number of integers between kt and k2 are 3.
65. Put, dy/dx = t
Now, if in time dt the decrease in water level
is dh, then
-it r2 dh = 06 figh • a dt
dt/dx-t + e2x = O,I.F. = ef^=e‘
:.
Solutionis, t-e~x = j
;2x-e
dx+C
(0.6) a fii
=> te~x = -ex + Cy'(0) = l =>C=2
, . ,
2-ex
dy I dx -------- fdy = j(2ex-e2x)dx
e-x
e2x
=> y=2!“
‘ ------ + C1,y(0)=2=>C1 =1/2
2
.’.
.’.
e2x 1
y(x) = 2ex------ + - =>
2
2
(a is cross-sectional area of the outlet)
(2Rh-h2)^= = dt
-it
y(x) < - for x = log 2
2
[2x] = [2 log 2] = [log 4] = 1
66. Let in time dt the decrease in water level in the tank is dh, then
amount of water flown out in time dt = itr2 ■ dh
Now, through the hole the amount of water flown
= (Volume of cylinder of cross sectional area ‘a’
and length vdt).
it
(0.6) a fig
/X
it
-----------
?/ls'2 --Rh™
-2RhV2)dh = ^ dt
0
(0.6) a ^j2g [ 5
it
=>
~ (0.6) a 42g _
3
=t
R
lit X 105
________
2_ 4
0 - R5'2
5 3JJ- 135 7i
68. We have been given,
f (xy) =
-1 - > {e> f (x) + ex f (y)}, V x, y & R*
Replacing x = 1, y = 1 in Eq. (i), we get
/(l) = e’-,-,{e/(l) + e/(l)}=^/(l) = 0
f(x+h)-f(x)
Now, f' (x) = lim
fc-»0
J
r
h
-Ax)
= dh
= lim
h
h-»0
h
H
=
e
h->o
~l ~ xfe1^ A/x. f(x) + ex • /(I + hJx)) - f(x)
h
*-l-* + x
e7(x) + e
►
•0
X
\
= lim------------= a • v • dt = a p yj2gh dt
A-+0
.
Hence,
=>
« -2
-ft
r
■h~m dh
Hay/2g
xj
h
h
A-1--+X
/(x)(e‘-l) + e
pa ^2gh dt = -lt r2dh
dt =
j[l + -'|-/(x)
= lim
h-tO
h
» + -]-/(!)
xj
-.(i)
302
Textbook of Integral Calculus
=
■
h-1 - - + x
= lim
h
h-+ o
1 +-’1-/(1)
e
x
+ lim------------A-> 0
xy
h
—• X
X
ex
= /(x) + i—A^ = /(x) + X X
f'(x)-/(x) = -
dyt
dz
=>yi-~ + z —+ P?i = Q
dx
. dx
dz
yi— + z.Q = Q
dx
dz = Q(l-z)
y,^
=>
dx
dz
Q
=>
------= — dx
ex f'(x)-f(x)ex
e2x x ’
=> In | z -11 = - J — dx + 1 ‘A’ being constant of integration
•
d f /(x)^.!
dx
yi
z=1 + ce w
x
ex
Jo
Z(x)_ In| x| + C
f (x) = x + x2 Xj + x X2
ex
f (1) = 0 => C = 0
f (x) = ex • In (x)
Now,
69. Let P (x, /‘(x)) be a point lying on the curve in first quadrant
Equation of normal and tangent at P are
=>
Jo
Jo
1 + X2 |
•3
4
9 Aj - 4 X2 = 4
X2 = Jo z22/(z)dz
f (x)
= J0‘ ((1 + X2)Z3 + Z4 l!)dz
(T - f (x)) = f ' (x) (X - x) respectively.
r
/(x)
/'(X)
,0
,Bi(x+f(x)f '(x), 0)
Since, the mid-point of segment AB is origin.
2x--^- + /(x)-/'(x) = 0
=>
(1 +12) A]
=--------- + —
4
5
15X2-41! =5
From Eqs. (i) and (ii),
f (x)
1
_ -2x± ^4x2 + 4/2(x) _ - x + Jx2 + f2(x)
2/W
"
/(x)
Negative sign been neglected as f' (x) > 0
Thus, we have, x + f (x) • f ' (x) = ^/x2 + /2 (x)
=>
-^-(x2 + /2(x)) = 7x2 + /2(x)
2dx
v
d(x2+r2(x))_^
2 ^x2 + /2 (x)
Integrating both the sides, we get ^x2 + f2 (x) = x + 1
It passes through (1, ^3).
Hence, 1+1=2=>X = 1
Curve is x2 + f 2 (x) = (x + I)2 or y2 = 1 + 2x
70. We have been given,
f'(y)(i + y)
=>
=>
.
Keep y constant and differentiating the expression w.r.t. ‘x’, we .
get
f ' (x + f (y) + x (y)) (1 + f (y)) = f ' (x) (1 + y)
-(i)
Similarly, differentiate the expression w.r.t. *y’ and keep x
constant, we get
f' (x + f(y) + x/(y)) (/' (y) (1 + x)) = (1 + f (x))
•••(ii)
Dividing Eq. (i) by Eq. (ii), we get
z(x)(i + y)
i + /(y)
/'(y)(i + x)
(l + /(x))
y/(x)(i + x)
i + /(y)
=» /'(y) = 7
dz
dy,
dy!
7!=>'1Z=’ —= yl- + z-^dxdx
dx ’ dx
dz
dyt
— + * -j- + Py, z = Q
dx
dx
61
72. /(x+/(y) + x/(y)) =y +/(x) + y/(x)
^ + Py1=e^ + fy! = e
Now,
1
80
.(ii)
A, = — and X2 = —
119
119
.. .
80 2 61
20x .
.
f (x) = x +---- x2 +----- x =----- (4 + 9x)
119
119
119
f (x)(f'(x))2 + 2x f'(x) - f (x) = 0
f'M "
(say)
X] = f zf(z)dz = [ ((1 + X2)z + z2 Xi)z dz
Also,
(y-/(x)) = - —(X-x) and
=> A =
z2 f (z) dz
+ xf
Jo
71. We have, f(x) = x + x.2 f1 z f (z) dz + x f
Integrating both the sides, we get
Since,
Thus,
I
yj
1-z
X
iI
l+y
(i + T(x))
and f' (x) = C
C = 1=>C = ± 1
C
f'M r l
l + /(x)
=>
l + /(x)
(1 + x)
1
1+x
/(x) = X1(l + x)±,-l
!
Chap 04 Differential Equations
Replacing x, y —> 0 in the Eq. (i), we get
Thus, (a) is the correct answer.
To check option (c), we have
/(/(0)) = /(0)
f (x) = Xi (1 + x)*1 -1 => f (0) = X! -1
■ Now,
and
/Cf(O)) = /(X1-l) = X1X1±,-l
Since,
/(/(0)) = /(0),X1X1±1-l = X1-l
+ -^- = o
log
|X+ 2|2
3e
By taking +ve sign, we get Xj = 0,1
log
/(x) = -l or
=>
/(x) = x
By taking - ve sign, we get Xj = 1
-
1+X
73. Given,
(x 4- 2)'
3e
/(x) = -A--l=----x—
=>
y=(x+2)2
106 (£)•
and
X11±I=X1
=>
303
x+ 2
(x + 2)2
x+2
n
|x + 2|2
= -(x+2)
3e
^2} or(x+2)2ex+2=3e
1+X
ex*2 =
dy -y2 = 0
(x2 + xy + 4x + 2y + 4)—
dx
3c
(x+2)2
=>
[(x2 + 4x + 4) + y(x + 2)]— -y2 = 0
dx
[(x+2)2 + y(x + 2)]^-y2 = 0
dx
Put x + 2 = X and y = Y, then
e2
3e/4
0
(x2 + xy)—-y2 = o
dX
X2dY + XYdY -Y2dX = Q
=>
Clearly, they have no solution.
To check option (d), y = (x+ 3)2
X2dY + Y(XdY - YdX) = 0
Le.
dY ‘ XdY - YdX
Y ~
X2
On integrating both sides, we get
y
- log | Y| = — + C, where x + 2 = X and y = Y
X
-10g|y| = ^- + C
x+ 2
R
+^=o
k 3e J x + 2
Now, to check option (a), y = x + 2 intersects the curve.
=>
log
|x+2n x+2 „
3e J
log
+ <5±^
=o
(x + 2)
.
’
2 [(x + 3)(x+l)
x+3
(x+2)2
(i)
log|y| + -^--log(3e) = 0
x+ 2
=>
|x + 3|2
3e
f (x + 2)2(x + 3)-(x+3)2-1
-0
g'(x) = — + l
(x + 2)2
5 ’ x+3
Since, it passes through the point (1, 3).
-log 3 = 1 + C
=>
C = -l-log3 = - (loge + log 3)
= - log 3e
Eq. (i) becomes
log
log
To check the number of solutions.
(x + 3)2 .
g(x) = 21og (x + 3) + "iTTy’ 08
Let
-d(iog|y|)=</W
=>
3e/(x + 2)2
- --------->X
------ = 0
x+ 2
|x + 2H_ -1
3e J
|x + 2|
1
=e
3e
e.
=>
|x + 2|=3 or x + 2 = ±3
x = 1, -5 (rejected), as x > 0 [given]
A x= 1 only one solution.
Clearly, when x > 0, then, g'(x) > 0
A
g(x) is increasing, when x > 0.
Thus, when x > 0, then g(x) > g(0)
(3} 9
g(x) > log - + - > 0
key 4
Hence, there is no solution.
Thus, option (d) is true.
74. Here, f'(x) = 2-^±
x
dy + y =2n
[Le. linear differential equation in y]
—
or
dx x
Integrating Factor, IF = e» x
A
-
x—x
A Required solution is y - (IF) = J Q(IF)dx + C
=>
y(x) = J2(x) dx + C
=>
yx = x2 + C
.•.
C
y=x+ —
x
[•/ C * 0, as f(l) * 1]
304
Textbook of Integral Calculus
(a)
76. Since, centre lies on y = x.
lim (1 — Cx2) = 1
-»o*
lim f‘
x-4 0*
.’. Equation of circle is x2 + y2 - 2ax - 2ay + c = 0
Option (a) is correct.
On differentiating, we get
2x + 2yy'-2a -2ay' = 0
x + yy'- a - ay'= 0
= lim (1 + Cx2) = 1
(b) lim x
x -> 0+
X -> 0*
Option (b) is incorrect.
(c) lim x2/'(x) = lim (x2 - C) = - C * 0
x
x -» 0+
0*
Again differentiating, we get
0 = (i + /)[i + yyz,+(y/)2]-(x+yyz)-(y//)
(i + /)2
.’. Option (c) is incorrect
(d) /(x) = x+-,C*0
X
For C > 0,
=> (1 + /) [1 + (y')2 + yy"]-(x + yy') (y ~) = 0
lim /(x) = «>
x-»0+
=>
=>
On comparing with Py* + Qy' + 1 = 0, we get
P=y-x
Q = (y')2 + y'+l
and
77. (i) Solution of the differential equation — + Py = Q is
dx
y-(IF) = JQ-(IF) dx + c
dx
dx
dy + exdy + yex dx = dx
dy + d(exy) = dx
(ii) £ /(x) dx = 2 £ /(x) dx, if/(-x) = /(x)
y(0)=2
Given,
Given differential equation
dy
x
x* + 2x
y=
dx x2 -1
2 + e°-2 = 0 + C
C=4
y (1 + ex) = x + 4
This is a linear differential equation.
x+ 4
y = ------ 7
IF = e xx2 -11
-4+4
x = - 4, y = i
=0
For critical points,
ex (x + 3) -1 = 0
7T7
-
r5
f
2 = J(x4 + 2x)dx= — + x.22 + c
or
/(0) = 0 => c = 0
rs
e~x =(x + 3)
=> /(x)Ti^.2 = —5 + X 2
Yi
Now, f _ / x dx=f r
\y = e-
(-1. ek
(-1.2p<
x*<
—
y^-x2 =/ xfx3 4- 2) •yjl-x2 dx
...(i)
dy _(1 + ex)-l — (x + 4)ex _
0
dx
(1 + ex)2
or
fj3l2
x2
(41/2
■■ ■
dx [using property]
1-4312 Jl-x2
J-43!2J
y=x+3
fj3l2
= 2 Jo[
0
=7F-7
= e2
=> Solution is
y(-4) = 0
^=0
dx
i.e.
IF=7/><ix
where,
On integrating both sides, we get
y + exy = x + C
Now at
1 + / [(y')2 + / +1] + y"(y - x) = o
=>
Function is not bounded in (0, 2).
Option (d) is incorrect,
(1 + ex)y' + yex =1
75. Here,
>X
x2
| y,
flt/3
Jo
Q & (taking x = sin 0]
|sin2 0 d0 = J
sin20 1
= (_u------
k
— dx
♦77^7
«/3sin 0
cos 9
_
=2
Clearly, the intersection point lies between (- 1, 0).
:. y(x) has a critical point in the interval (-1,0).
• ,
2
(1 - cos 20) d0
*/3 n sin2n/3 n
o
_ —___ ___ — =-------3
2
3 4
Chap 04 Differential Equations
78. Whenever we have linear differential equation containing
inequality, we should always check for increasing or
decreasing,
dv Py < 0 => ^ + Py > 0
i.e.
i.e. for
for — +
+ Py < 0 =>
Multiply by integrating factor, i.e.
and convert into total
differential equation.
Here, f\x) < 2/(x), multiplying by e"(zdx
f\x)-e~u -2e~2xf(x)<0 => A(/(x) • e'21) < 0
dx
.*.
f(x)>o,xe
and
=*'~xf\x) - e‘7(x) < 0, x e (o, 1)
=* f\x) < f(x) ,0 < x <4
81. Here, /'(x) - 2f\x) + /(x) £ ex
dx
. A I 1
" ' 'U
j2
{e~x fM} > 1, Vx 6 [0,1]
dx
0 (x) = e”x/(x) is concave function.
=» e-!«/(x)<e-1-/ll
/(x) < e2x 1 •1, given
=>
1
21-1 dx
0<Ji'2/(x)dx<
=>
<e2x-1
<
2
>1/2
o<l>)^<£r
/(x)<0
82. Here, /(x) = (1 - x)2 • sin2 x + x2 £ 0, V x
g'(x) =
2(t — 1)
-logt \f(t)dt
t+1
2(x-l)
-logx./(x)
(x+1)
X
dv
v + x— = v + sec(v)
dx
dv
. .
f dv
rdx
x— =sec(v) => I----- = I —
dx
J secv J x
jcosvdv = J— => sinv = logx+logc
£
x
— (x~l)
x(x+l)2
0' (x) < 0, V x > 1
=>
0 (X) < 0 (1) => 0(x) < 0
From Eqs. (i) and (ii), g'(x) < 0, x e(l,«)
g(x) is decreasing on x e (1, <»).
83. Here, f(x) + 2x = (1 - x)2 • sin2 x + x2 + 2x
where, I: /(x) + 2x = 2(1 + x)2
2(1 + x2) = (1 - x)2 sin2x+ x2 + 2x
y I = log(cx)
sin i —
\x)
(1 - x)2sin2 x = x2 - 2x + 2
(1 - x)2sin2 x = (1 - x)2 + 1
As it passes through
logc
logc = |
.
1
“0- logx + 80. Let0(x) = e' ff(x)
Here,
4>z(x) < o, xe^o,
(i)
♦ VB
For g'(x) to be increasing or decreasing.
T
x/ » 2(x-l) .
Let
0(x) = -~^--logx
4
<$>'(*) = (x+l)2‘
Put
=>
=>
=>
\x)
x
y(0) = /(l) = 0
0(O) = O = /(1)
0(x) < 0
e~x /(x) < 0
and g(x) =Jt
79. To solve homogeneous differential equation,
y
i.e. substitute — = v
x
dy _ v + x dv
y = vx =>
dx
dx
Here, slope of the curve at (x, y) is
dx
dx
dx
Thus, when x > 2
X
+ /(x>*x £ 0
=> r(x)e-x-f'(x)e-x-f'(x)e
0(x) = _f(x)e“2x is decreasing for x e -, 1
2
A/
305
=>
(1 - x)2 cos2 x = -1
which is never possible.
I is false.
Again, let h(x) = 2f(x) + 1 - 2x(l + x)
where, h(0) = 2 /(0) + 1-0 = 1
/<1) = 2(1) + 1 - 4 = - 3 as [h(O)h(l> < 0]
=> A(x) must have a solution.
II is true.
...(ii)
...(i)
•••(ii)
306
Textbook of Integral Calculus
84. Linear differential equation under one variable.
~ + Py = Q, W = JFdx
dx
:. Solution is, y (IF) =
(IF)dx+C
Alter
/(x)=Jo7(oa
Given,
/(0) = 0 and /'(x) = /(x)
if y(x) # o
f'& = 1 => In /(x) = x + c
fW
y'-ytanx = 2xsecx and y(0)=0
dy
-y tan x = 2xsec x
=>.
IF=fe- Unxdx= eloR|co*^ = cosx
f(x) = e'-ex
/(0) = 0
=> ec = 0, a contradiction
Solutionis y-cos x=j2xsecx-cos xdx+C
y-cosx = x2+C
y(0) = 0
C=0
y = x2secx
As
=>
Now,
=>
87. Given,
„2
It
8^2
0
n
7tz
f(x) = 0,Vxe« =» f(ln5) = 0
dy _ yjy2 -1
<*x
f
Sy^
41+&42
y
9
2%2
3yfi
6.
t(43
= cos cos 1 — +
\2x
Solution is y (e r(x)) = jg(x) • g\x) • e
y eg(x) = (g(x) -1) eR(x) + C
xdy -y dx
—---- = xdx
y
(i)
Given,
y(0) = 0, g(0) = g(2) = 0
Eq. (i) becomes,
y(0)-eR(o)=(g(0)-l)-eR(o) + C
t
=>
'
= (g(2) -1) eR(2) + 1, where g(2) = 0
y(2)-l=(-l)-l + l
y(2) = o
86. From given integral equation, /(0) = 0.
Also, differentiating the given integral equation w.r.t x
f'W = f(x)
log/(x) = x + c
/(X)
,X
/(x) = eceJ
/(0) = 0 => ec = 0, a contradiction
y(x) = 0, VxgK => /(ln5) = 0
■
_b±^dy)=xdx=i_d x = xdx
y
On integrating both sides, we get
x x2
+c
y
2
<y.
•(i)
V It passes through (1, -1).
1=1 + C => C = 2
2
x x2 1
Now, from Eq. (i) ---- = — 4- —
y 2
2
=>
/(x)#0
=>
'-7
88. Given differential equation is
y(l + xy)dx = xdy
=>
y dx+ xy2dx = xdy
= t-e‘ -e‘ + C
0 = (-1)1 + C => C = 1
y(x)-eR(x)=(g(x)-l)eR(x) + l
X
43 1
y=— + 2x 2
dx + C
Put g(x) *t, g'(x) dx = dt
y(esM) = jt-et dt + C = t-e‘ - J1 • ef dt + C
/Z(x)_1
J3
2
Jt
-i -1 cos
Now, y - sec sec' x---- = cos cos
«(x)
If
dx
xjx2 -1
sec y =sec
2 n it
n
At x = 2, y = -=■; — = — + c => c = —
43 6 3
6
2ft2
85. ~^y-g'(x) = g(x)g'(x)
dx
y(2) •
- Jf
=>
4n
3
=>
xjx2 -1
2
«
2x
2x
x + 1 =----- =^y =x
2
+1
y
4
5
89. Given differential equation is
dv
(xlogxj-f^ + y = 2xlogx,
ax
dy
-1+ —iy— = 2
dx xlog x
(x^l)
Chap 04 Differential Equations
This is a linear differential equation.
/—!—dr
,Jxlogx
IF = e'xlogx
= elog(logx) = logx
Now, the solution of given differential equation is given by
307
To find The time at which the population of the mouse will
become zero, i.e. to find the value of‘f at which p(t) = 0.
Let’s solve the differential equation first
p'(0 = ^ = 05p(0~450
y-logx = jlogx-2dx
p(t) - 900
y-logx = 2pogxdx
=>
y • log x = 2[x log x - x] + c
At
x = l, c = 2
y • log x = 2[xlog x - x] + 2
At
x = e,
i
y = 2(e - e) + 2
y=2
=>
dp 1
90. Given differential equation — --p(t) = -200 is a linear
dt 2
differential equation.
Here,
p (t) = y ,Q(t) = -200
=»
J p(t) - 900
J
=> 2 log | p(t) - 9001 = t + C, where C is the constant of
integration.
To find the value of 'C, let’s substitute t = 0.
=> 2 log | p(0)-9001 = 0 + C
=>
C =2 log I 850-9001
=>
C = 2 log 50
Now, substituting the value of C back in the solution, we get
2 log | p(t) - 9001 = t + 2 log 50
Here, since we want to find the value of t at which p(t) = 0,
hence substituting p(t) = 0, we get
2 log | 0-9001 = t + 2 log 50
,
900
t = 2log ---50
Hence, solution is
p(t).IF=/Q(t)IFdt
t
t = 2 log 18
t
93. Here, — = y + 3 > 0 and y(0) = 2
dx
p(t)-e 2 =1-200-6 2dt
t
t
p(t)-e 2 = 400e 2 + K
Jy + 3
p (t) = 400 + ke~U2
=>
If p(0) = 100, then Ac = -300
t
p(t) = 400-300e2
dP
97. Given, —=(100-12>/x)
dx
=>
dP=(100-12Vx)dx
J
=>
Since,
log |y + 3| = x + C
y(0) = 2
=>
loge |2 + 3| = 0 + C
C = log, 5
=>
loge |y + 3| = X + log,5
x = Iog,2
When
On integrating both sides, we get
J dP = J (100 - 12Vx) dx
=>
=>
P = 100x-8x3/2 + C
loge |y + 3| = log,2 + log,5 = log, 10
y + 3 = 10
y=7
When x = 0, then P = 2000
=>
C = 2000
Now, when x = 25, then
94. Given,
P = 100 x25 -8 x(25)3'2 + 2000
~k(T-t)
dt
d{V(t)} = -k(T-t)dt
On integrating both sides, we get
= 2500-8 X125 + 2000
V(t) = -k
= 4500-1000 =3500
2(-l)
92. Given
(i) The population of mouse at time*t* satisfies the
differential equation p'(t) =
= 05p(t) - 450
(ii) Population of mouse at time t = 0 is
p(0) =850
V(r) = — (T—t)2 + C
2
At t = 0, V(f) = /
Z=-(T-0)2 + C
2
308
Textbook of Integral Calculus
96. Given,
C = I--T2
2
L.
I-
y' = Czqe*2*
„
V(t) = ^(T-02 + /~^T2
Now,
y =c1eC2X
y' = C2y
.(ii) ■
y" = w'
V(T) = /-|r2
from Eq. (i), c2
95. Since, cosxdy =ysinxdx-y2dx
y
=>
——tanx = -secx
y dx y
The equation of family of circles with centre on y = 2 and of
radius 5 is
(x-a)2 + (y-2)2=52
•(0
1 dy _dz
y2 dx dx
=>
dz ,
.
— + (tanx)z = -secx
dx
This is a linear differential equation.
Therefore,
=secx
Hence, the solution is
z • (sec x) = j-sec x • sec x dx + Q
x2.2+ a2-2ax + y2 + 4-4y =25
On differentiating w.r.t. x, we get
n n
n dy
dy
2x-2a + 2y—-4— = 0
dx
dx
dy.
=>
a = x+-y-(y-2)
dx
On putting the value of a in Eq. (i), we get
dy
T^(y-2)1
+ (y-2)2 =52
x-x--^~
dx
—sec x = -tan x + Q
y
=>
yy"=(y'?
97. Equation of circle having centre (h, fc) and radius a is
(x-h)2 + (y-k)2 =a2.2
Put -— = z
y
ZF=eJunxdx =el
4yj ■
(S (y"2)2=25"(y"2)Z
secx = y(tanx + C)
=>
y'2(y-2)2 = 25-(y-2)2
i
• ■
I
i
■
1 -
■t
IK
Skills in Mathematics for
Integral Calculus
This textbook on Integral Calculus is exactly based on the premise of Calculus problems asked in
JEE book covers the strategies for solving the problems by breaking them down into easily
manageable steps. The elaboration is simple and engaging with a clear objective of taking out the
mystique attached with the calculus problems. Starting with the indefinite integrals, the book
covers the definite integrals, the area bounded, and the differential equations in a completely
new way focusing on the ways to find the answers to the complicated problems tactically.
The Revised Edition contains all types of questions viz., Single Correct Answers, Multiple Correct
Options, Passage-Based, Matching Types, Assertion and Reason, Integer AnswerTypes - with
Complete Solutions.
those who want to confidently tackle calculus problems in JEE & all other Engineering
acted in India...
ABOUTTHEAUTHOR
Amit M. Agarwal is a renowned name and an authority in the field of Mathematics with a number of
accolades to his credit. Immediately after doing his Post Graduation from Meerut University in 1997,
Mr. Agarwal started teaching Mathematics in his own institute at Meerut and simultaneously writing
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BOOK NTHESERES
feeing7
^arihant
Arihant Prakashan (Series), Meerut
^7
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