CHAPTER 6 : INTEGRATION
We know that in algebra the operations of addition and subtraction, multiplication and division
form reverse pairs of operations. In calculus the operations of derivative and integration (antiderivative) are also thought of an inverse of one another. Thus they also form a reverse pair of
operations and the rules for finding derivatives will be useful in establishing corresponding rules
for finding anti-derivatives.
6.1 Anti - Derivatives ( Indefinite Integrals )
If a function  exists , for a given function f, such that  ( x ) = f ( x ), then f is called antiderivative of f.
An anti-derivative is also known as primitive or an indefinite integral.
and is read as "  ( x ) is an integral of f ( x )"
Symbolically, this is written as
w.r. to x .
Here f ( x ) is called the integrand and the process of finding the integral is called integration.
The above definition poses the following questions.
(1) Does every function poses an anti-derivative?
(2) Is an anti-derivative unique in case it exists?
(3) Is it always possible to find an anti-derivative?
The following answers are suggested :
(1) There exist a large number of functions possessing an anti-derivative, e.g. all continuous
functions. In finding an anti-derivative, we shall always assume its existence. However every
function need not posses an anti-derivative.
(2) Let
so that
for any constant ‘c’
so that  ( x ) + c is also an anti-derivative of f ( x ). This accounts for the name "Indefinite
Integral".
(3) In the present chapter, we shall study several methods of finding anti-derivatives. This,
however, does not ensure that anti-derivative of any given function can always be obtained.
Note :
i) The word integration literally means to add together small bits. Now the question arises that is
there any connection between the two meanings of integration :
(i) The Inverse of differentiation.
(ii) The Adding together of small bits.
Yes, there is such a connection between the above said two concepts. This, has been
mentioned later in the fundamental theorem of Integral calculus.
ii) Also note that the constant " c " in the above discussion is known as the constant of
integration.
iii) Any integral of f ( x ) is in the form  ( x ) + c. which is known as the general integral of f ( x
).
An integral obtained by giving particular value to ‘ c ’ is called a particular integral.
Some Standard Results
Given below are some standard results, obtained by using the derivatives of some well-known
functions.
(1)
(2)
(3)
(4)
(5)
(For the sake of convenience it is customary to omit the modulus sign)
Important Rule
In any of the standard result if in place of x we have a linear expression of x in the form ( ax + b
), then the same formula is applicable but we must divide the integral by ‘a’ which is the
coefficient of x.
i.e.
Some Rules for Integration
Example 1
Evaluate
Solution : Using rule 2, we find that
Also, using formula ( 1 ), we get
Example 2
Evaluate
Solution : Using rule 2 and formula (4), we get
Example 3
Evaluate
Solution :
Example 4
Evaluate
Solution :
using rules 1, 2 and applying standard results (1), (2), we get
Example 5
Evaluate
Solution :
using (22), with a2 = 49, we get
Example 6
Solution :
applying (24), we get
Example 7
Evaluate
Solution :
By rule (1) and (2), we have
Using standard results (1), (4), (13) and (11) respectively, you find
Example 9
Integrate w. r. to x. the following
Solution :
Using (1), (3), (4), (5) and again (1) respectively we get
Example 10
Integrate w. r. to. x, the following
Solution :
Example 11
Solution :
Example 12
Solution : By actual division, i.e.
then expressing
Example 13
Given that f’ (x) = 3x2 + 4x + a, f ( 0) = 1 and f (2) = 11 Find f(x)
Solution : On integrating w.r. to x, the given relation, we get
Now f (0) = 1 gives f (0) = (0)3 + 2 (0)2 + a (0) + c = 1
i. e. c = 1.
Also, f (2) = 11 gives, f (2) = (2)3 + 2 (2)2 + a (2) + c = 11
i.e. 8 + 8 + 2a + c = 11
i.e. 2a + c = -5
i.e. 2 a + 1 = -5
i.e. 2a = -6
 f (x) = x3 + 2 x2 + 3x + 1
Example 14
Integrate w. r. to x, imgex14.gif
Solution : By actual division
Example 15
Evaluate
Solution :
6.2 Integration Of Some Trigonometric Functions
Here are some important identities for ready reference.
Use Of Trigonometric Identities
Example 16
Evaluate
Solution :
Example 17 Evaluate
Solution :
Example 18
Evaluate
Solution :
Example 19
Evaluate
Solution :
Example 20
Integrate w. r. to. x.
Solution :
Example 21
Evaluate
Solution :
Example 22
Evaluate
Solution :
Example 23
Integrate w.r. to x, cos 2x sin 4x
Solution :
Using cos A sin B = ½ [ sin (A+B) - sin (A-B) ]
i.e. cos 2x sin 4x = ½ [ sin (2x + 4x ) - sin ( 2x - 4x ) ]
= ½ [ sin 6x - sin (-2x) ]
= ½ [ sin 6 x + sin 2x ]
Example 24
Evaluate
Solution :
Example 25
Evaluate
Solution :
Example 26
Evaluate
Solution :
Example 27 Evaluate
Solution :
Example 28
Integrate
Solution :
6.3 Methods Of Integration
There are special methods of reducing an integral to a standard form. The following are the
methods of integration.
(1) Integration by substitution
(2) Integration by parts
(3) Integration by partial fractions.
6.4 Substitution and Change of Variables
In this method, we transform the integral to a standard form by changing the variable x of the
integration to t by means of a suitable substitution of the type x =  ( t ).
Theorem : If x =  ( t ) is a differentiable function of ‘t’,
Then
Note : Comparing
with , we observe that dx is replaced
by
. Hence supposing dx and dt as if they were separate entities, we have the following
working rule :
This technique is often compared with the chain - rule of derivatives since they both apply to
composite functions.
Example 29
Evaluate
Solution :
Here the inner function of the composite function
Example 30
Evaluate
Solution : Let 4 - 3x = t,
then -3 = dt
Example 31
Evaluate
Solution : Let
Example 32
Evaluate
Solution : Let log x = t
Example 33
Evaluate
Solution : Let
then -2 x dx = dt
 2x dx = -dt
Example 34
Evaluate
Solution : Let arc tan x = t
Example 35
Evaluate
Solution : Let arc sinx = t
Example 36
Evaluate
Solution : Let log x = t
Then
Example 37
Evaluate
Solution : Let cos x = t
then -sin x dx = dt
 sin x dx = -dt
Example 38
Evaluate
Solution : Let x2 = t
then 2x dx = dt
Again put sint t = u, then cos t dt = du
Example 39
Evaluate
Solution : Let 1 + x2 = t
then 2x dx = dt
 x dx = ½ dt
Example 40
Evaluate
Solution :
Example 41
Evaluate
Solution :
Let cos x + sin x = t then (-sin x + cos x) dx = dt
Example 42
Evaluate
Solution :
Example 43
Evaluate
Solution :
To Evaluate
Method : Let N = numerator = a cosx + b sinx
D = denominator = c cosx + d sinx
We find two constants l and m such that N = l (D) + m (D’) where D' =
i.e. a cos x + b sin x = ( l c + m d ) cos x + ( l d - m c ) sin x. Equating co-efficients of cos x and
sin x, we get \ a = (c + md) and b= (d - mc). By solving these equations, simultaneously, we
obtain l and m. Putting these values in (1) we can write the given integral as :
Example 44
Evaluate
Solution : We find constants l and m such that N = lD + mD’ ........(1)
6.5 Some Standard Substitutions
While evaluating integral by the substitution method, there is no definite rule for proper
substitution. We can use some standard substitutions for certain types of integrals such as :
Example 48
Evaluate
The radical is in the form
put x = a sin = 3 sin   dx = 3 cos d
The radical is in the form
put x = a sin = 3 sin   dx = 3 cos d
Solution :
The radical is in the form
put x = a sin = 3 sin   dx = 3 cos d
Example 49
Evaluate
Solution :
The radical is in the form
put x = a sin = 3 sin   dx = 3 cos d
Example 50
Evaluate
Solution :
Now, add the following results to your list of standard results.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Integrals of the Form
non-negative integers
Case I Either Index Odd
Express the integer and as the product of either
(i) sin x and a polynomial in cos x or
(ii) cos x and a polynomial in sin x.
Put cos x or sin x = t and then solve.
Case II Both Indices Even
It is necessary to change the integrand into sum of cosines and sines of multiples of angles using
sin2  = ½ (1 - cos2) and cos2 = ½ ( 1 + cos2) , more than once if needed.
Case II Both Indices Even
It is necessary to change the
integrand into sum of cosines
and sines of multiples of angles
using sin2  = ½ (1 - cos2)
and cos2 = ½ ( 1 + cos2) ,
more than once if needed.
Case II Both Indices Even
It is necessary to change the integrand into sum of cosines and sines of multiples of angles using
sin2  = ½ (1 - cos2) and cos2 = ½ ( 1 + cos2) , more than once if needed.
6.6 Integration by Parts
This method is used to integrate a product, although one of the factors of the product could be
unity. Also note that one of the product of functions can be easily integrable. The factors of the
product will be two different kinds of functions. They will be :
1 Algebraic
2 Circular 3 Inverse circular
4 Logarithmic 5 Exponential etc.
Theorem : If u and v are functions of x then
Here u is called the first function which is to be differentiated and v is the second function which
is to be integrated.
Rule for the proper choice of the first function : Let us denote the various types of functions that
we come across by the first alphabet i.e. L. I. A. T. E. for logarithmic, Inverse circular,
Algebraic, Trigonometric (circular) and Exponential respectively. The first function to be
selected will be the one which comes first in the order of the word " LIATE
6.7 Integration By Partial
Fractions
If an algebraic fraction is expressed in terms of two or more simpler fractions, then this simple
fractions are called partial fractions.
In many problems, if we resolve an algebraic fraction into partial fractions, its integration
becomes much simpler.
Let the fraction to be resolved into partial fraction is
. Then there are two cases
(1) Degree of P (x) < degree of Q (x).
This algebraic fraction is known as the proper fraction.
(2) Degree of P(x) > degree of Q (x).
This type of algebraic fraction is known as an improper fraction.
Consider these cases :
(a) To each linear factor (a1x2 + b1) of Q (x), there will be a corresponding partial fraction in the
form
(b) To each quadratic factor (a1x2 + b1x + c1), there will be a corresponding partial fraction in the
form
(c) To each repeated linear factor (a1x + b1)n of Q (x), there will be ‘n’ corresponding partial
fractions of the form
(d) To each repeated Quadratic factor (a1x2 + b1 + c1 )n there will be corresponding ‘ n ’ partial
fractions as
terms