The document discusses indefinite integration, defining it as the reverse process of differentiation and providing key theorems and standard integrals. It includes examples of various methods for solving integrals, such as integration by substitution, parts, and forms for irrational and trigonometric functions. A total of 103 questions on the topic are outlined, with 50 solved examples to illustrate the concepts.

1. Total No. of questions in Indefinite Integration are-
In Chapter Examples..................................................... 53
Solved Example............................................................ 50
Total No. of questions...............................................103
INDEFINITEINTEGRATION

2. 1. INTEGRATION OF A FUNCTION
Integration is a reverse process of differentiation.
The integral or primitive of a function f(x) with
respect to x is that function  (x) whose derivative
with respect to x is the given function f(x). It is
expressed symbolically as -
f x dx x
( ) ( )

z 
Thus
f x dx x
( ) ( )

z   
d
dx
x f x
[ ( )] ( )

The process of finding the integral of a function
is called Integration and the given function is
called Integrand. Now, it is obvious that the
operation of integration is inverse operation of
differentiation. Hence integral of a function is
also named as anti-derivative of that function.
Further we observe that-
d
dx
x x
d
dx
x x
d
dx
x k x
( )
( )
( )
2
2
2
2
2 2
2

 
 
U
V
|
|
|
W
|
|
|
  
 t
tan
cons
x
xdx
2 2
So we always add a constant to the integral of
function, which is called the constant of
Integration. It is generally denoted by c. Due to
presence of this constant such an integral is
called an Indefinite integral.
2. BASIC THEOREMS ON INTEGRATION
If f(x), g(x) are two functions of a variable x and
k is a constant, then-
(i) z
k f(x) dx = k z
f(x) dx.
(ii) z
[f(x)  g(x)] dx = z
f(x)dx ± z
g(x) dx
(iii) d/dx ( z
f(x) dx) = f(x)
(iv) zd
dx
f x
( )
F
H
G I
K
Jdx = f(x)
3. STANDARD INTEGRALS
The following integrals are directly obtained from
the derivatives of standard functions.
i. z
0. dx = c
ii. z
1.dx = x + c
iii. z
k.dx = kx + c (k  R)
iv. z
xn dx =
x
n
n

1
1
+ c (n  –1)
v.
1
x
zdx = loge x + c
vi. z
ex dx = ex + c
vii. z
ax dx =
a
a
x
e
log
+ c = ax loga e + c
viii. z
sin x dx = – cos x + c
ix. z
cos x dx = sin x + c
x. z
tan x dx = log sec x +c = – log cos x + c
xi. z
cot x dx = log sin x + c
xii. z
secx dx =log(secx +tanx) + c
=–log (sec x –tan x) + c
= log tan

4 2

F
H
G I
K
J
x
+c
xiii. z
cosec x dx = – log (cosec x + cot x) + c
= log (cosec x – cot x) + c = log tan
x
2
F
H
GI
K
J+ c
xiv. z
sec x tan x dx = sec x + c
xv. z
cosec x cot x dx = – cosec x + c
xvi. z
sec2 x dx = tan x + c
xvii. z
cosec2 x dx = – cot x + c
xviii. z
sinh x dx = cosh x + c
xix. z
cosh x dx = sinh x + c
xx. z
sech2 x dx = tanh x + c
xxi. z
cosech2 x dx = – coth x + c
xxii. z
sech x tanh x dx = – sech x + c
xxiii. z
cosech x coth x = – cosech x + c

3. xxiv. z 1
x + a
2 2 dx =
1
a
tan
an–1
x
a
F
H
GI
K
J+ c
xxv.
1
2 2
x a

z dx =
1
2a
log
x a
x a


F
H
G I
K
J+ c
xxvi.
1
2 2
a x

z dx =
1
2a
log
a x
a x


F
H
G I
K
J+ c
xxvii.
1
2 2
a x

z dx = sin–1
x
a
F
H
GI
K
J+ c
= – cos–1
x
a
F
H
GI
K
J+ c
xxviii.
1
2 2
x a

z dx = sinh–1
x
a
F
H
GI
K
J+ c
= log (x + x a
2 2
 ) + c
xxix.
1
2 2
x a

z dx = cosh–1
x
a
F
H
GI
K
J+ c
= log (x + x a
2 2
 ) + c
xxx. a x
2 2

z dx
=
x
2 a x
2 2
 +
a2
2
. sin–1
x
a
+ c
xxxi. x a
2 2

z dx
=
x
2 x a
2 2
 +
a2
2
. sinh–1
x
a
+ c
xxxii. x a
2 2

z dx
=
x
2 x a
2 2
 –
a2
2
. cosh–1
x
a
+ c
xxxiii.
1
2 2
x x a

z dx =
1
a
sec–1
x
a
+ c
xxxiv. zeax sin bx dx
=
e
a b
ax
2 2

(a sin bx – b cos bx) + c
=
e
a b
ax
2 2

sin bx
b
a

F
H
GI
K
J
R
S
T
U
V
W

tan 1
+ c
xxxv. zeax cos bx dx
=
e
a b
ax
2 2

(a cos bx + b sin bx) + c
=
e
a b
ax
2 2

cos bx
b
a

R
S
T
U
V
W

tan 1
+ c
Examples
based on Integration of Function
Ex.1 Evaluate : zx–55 dx
Sol. z
x–55 dx
=
x

54
54
+ c Ans.
Ex.2 Evaluate :
x
x
2
2
1

z
e j dx
Sol.
x x
x
4 2
2 1
 
z dx
= x x
x
3
2
1
 
F
H
G I
K
J
z dx
=
x4
4
+ x2 + log x + c Ans.
Ex.3 Evaluate : x
x

F
H
G I
K
J
z 1
2
dx
Sol. Here I = x
x
 
F
H
G I
K
J
z 1
2 dx
=
x2
2
+ log x – 2x + c Ans.
Ex.4 Evaluate : z
tan2 x. dx
Sol. Here I = z
(sec2 x – 1) dx
= z
sec2 x dx – z
dx
= tan x – x + c Ans.
Ex.5 Evaluate z
cos2 x/2 dx
Sol. Here I =
1
2

z cos x
b g
dx

4. =
1
2 z
dx +
1
2 z
cos x dx
=
1
2
(x + sin x) + c Ans.
Ex.6 Evaluate : e x
log 6
z dx
Sol. Let I = e x
log 6
z dx
 I = x dx
6
z
=
x7
7
+ c Ans.
Ex.7 Evaluate : sec .cos
2 2
x ec x
e j
z dx
Sol. Here I =
sin cos
cos sin
2 2
2 2
x x
x x

z dx
=
1 1
2 2
cos sin
x x

F
H
G I
K
J
z dx
= sec2
x dx 
z cosec x dx
2
z
= tan x – cot x + c Ans.
Ex.8 Evaluate : a e dx
x x
.
z
Sol. I = a e dx
x x
.
z
= ( )
ae dx
x
z
=
ae
ae
x
b g
log ( )
+ c
=
a e
a
c
x x
log 

1
Ans.
Ex.9 Evaluate : sin . cos
2x x dx
z
Sol. Let I = sin . cos
2x x dx
z
=
1
2 z
(sin 3x + sin x) dx
= –
1
2
cos
cos
3
3
x
x

L
N
M O
Q
P
+ c Ans.
Ex.10 Evaluate z dx
x x
4 4 2
2
 
Sol. Let I = z dx
x x
4 4 2
2
 
= z 1
2 1 1
2
x  
b g dx
=
1
2
tan–1 (2x + 1) + c Ans.
Ex.11 Evaluate : zx2
9
 dx
Sol. I = zx2
9
 dx
=
x
2 x2
9
 +
9
2
sinh–1
x
3
F
H
GI
K
J+ c Ans.
Ex.12 Evaluate : z
e3x. cos 4x dx
Sol. Here I = z
e3x . cos 4x dx
=
e x
3
5
cos (4x – tan–1 4/3) + c Ans.
Ex.13 Evaluate : z
sin–1 (cos x) dx
Sol. Here I = z
sin–1 (cos x) dx
= z
sin–1 sin

2

F
H
G I
K
J
x dx
= z
2

F
H
G I
K
J
x dx
=

2
2
2

F
H
G I
K
J

x
+ c Ans.
4. METHODS OF INTEGRATION
When integration can not be reduced into some
standard form then integration is performed using
following methods-
(i) Integration by substitutions
(ii) Integration by parts
(iii) Integration of rational functions
(iv) Integration of irrational functions
(v) Integration of trigonometric functions
4.1 INTEGRATION BY SUBSTITUTION:
Generally we apply this method in the following
two cases.
(i) When Integrand is a function of function-
i.e. f [  (x)]  ' (x) dx
Here we put  (x) = t so that  '(x) dx = dt
and in that case the integrand is reduced
to f(t) dt.
Note:
In this method the integrand is broken into two
factors so that one factor can be expressed in
terms of the function whose differential coefficient
is the second factor.

5. Ex.14 Evaluate : z
x 1 2
 x dx
Sol. Let 1 + x2 = t
then 2x dx = dt
or x dx = 1/2 dt.
  x 1 2
 x dx = 
1
2 t dt
=
t
c
3 2
3
/
 =
1
3
(1 + x2)3/2 + c Ans.
Ex.15 Evaluate : x2
ex3
dx.
Sol. Let x3 = t then
3x2 dx = dt
 I =
1
3 et dt =
1
3
et + c
=
1
3
ex3
+ c Ans.
Ex.16 Evaluate : x tan x2 sec x2 dx.
Sol. Let x2 = t
2x dx = dt  x dx = 1/2 dt
 I =
1
2 tan t sec t dt
=
1
2
sec t + c =
1
2
sec x2 + c .Ans.
(ii) When integrand is the product of two
factors such that one is the derivative of
the other i.e. I = zf'(x) f(x) dx.
In this case we put f(x) = t and convert it
into a standard integral.
Ex.17 Evaluate : z
log x
x
dx.
Sol. Let log x = t

1
x
dx = dt
 I = z
t.dt =
1
2
t2 + c
=
1
2
(log x)2 + c Ans.
Ex.18 Evaluate : z
tan

1
2
1
x
x
dx.
Sol. Let tan–1 x = t then
1
1 2
 x
dx = dt
 I = z
t dt =
t2
2
+ c
=
1
2
(tan–1 x)2 + c Ans.
Ex.19 Evaluate : z
tan x.sec2 x dx.
Sol. Let tan x = t
sec2 x. dx = dt
 I = z
tan x . sec2 x dx = z
t.dt
=
t2
2
+ c =
tan2
2
x
+ c Ans.
(iii) Integral of a function of the form f (ax+b).
Here we put ax + b = t and convert it
into standard integral. Obviously if
z
f(x) dx =  (x), then
z
f(ax + b) dx =
1
a
 (ax + b)
Ex.20 Evaluate : z 1
3 2
 x
dx
Sol. Here I = z 1
3 2
 x
dx
= –
1
2
log (3 – 2x) + c Ans.
Ex.21 Evaluate : z
cos 3x cos 5x dx
Sol. I = z
cos 3x cos 5x.dx
=
1
2 z
(cos 8x + cos 2x) dx.
=
1
2
1
8
8
1
2
2
sin sin
x x

L
N
M O
Q
P
+ c Ans.
(iv) Standard form of Integrals:
(a) z
f x
f x
'( )
( )
dx = log [f(x)] + c

6. (b) z
[f(x)]n f'(x) dx =
f x
n
n
( )


1
1
+ c
(provided n  – 1)
(c) zf x
f x
'( )
( )
dx = 2 f x
( ) + c
Ex.22 Evaluate : z
tan3 x. sec2 x dx
Sol. Let tan x = t
sec2 x dx = dt
 z
tan3 x. sec2 x dx = z
t3 dt
=
t4
4
+ c
=
tan4
4
x
+ c Ans.
Ex.23 Evaluate : zsec
tan
2
x
x
dx.
Sol. Let t = tan x
dt = sec2 x dx
 I = zdt
t
= 2 t1/2 + c
= 2 tan x c
 Ans.
Ex.24 Evaluate za b x
x



tan 1
3
2
1
e j dx.
Sol. Let a + b tan–1 x = t
 b
1
1 2

F
H
G I
K
J
x
dx = dt
or
1
1 2

F
H
G I
K
J
x
dx =
dt
b
 za b x
x



tan 1
3
2
1
e j dx =
1
b z
t3 dt
=
1
b
t4
4
+ c

1
4b
(a + b tan–1 x)4 + c Ans.
Ex.25 Evaluate z
e e
e e
x x
x x



 dx.
Sol. Let ex + e–x = t
(ex – e–x).dx = dt
 I = z
dt
t
= log t + c
= log (ex + e–x) + c Ans.
(v) Integral of the form
z dx
a x b x
sin cos

putting a = r cos  and b = r sin  . we get
I = z dx
r x
sin ( )
 
=
1
r z
cosec (x +  ) dx.
=
1
r
log tan (x/2 +  /2) + c
=
1
2 2
a b

log tan (x/2 + 1/2 tan–1 b/a) + c
Ex.26 Evaluate : z 1
sin cos
x x

dx
Sol. Here a = 1 & b = 1
So z 1
sin cos
x x

dx
=
1
1 1

log tan
x
2
1
2
1
1

F
H
G I
K
J

tan + c
=
1
2
log tan
x
2 8

F
H
G I
K
J

+ c Ans.
Ex.27 Evaluate : z 1
3 sin cos
x x

dx
Sol. By putting a = 3 & b = 1
z 
1
3 sin cos
x x
dx
=
1
1 3

log tan
x
2
1
2
1
3
1

F
H
G I
K
J

tan + c
=
1
2
log tan
x
2 12

F
H
G I
K
J

+ c. Ans.
(vi) Standard Substitutions : Following standard
substitutions will be useful-
Integrand form Substitution
(i) a x
2 2
 or x = a sin  or
1
2 2
a x

x = a cos 
(ii) x a
2 2
 or x = a tan  or
1
2 2
x a

x = a cot  or x = a sinh 
(iii) x a
2 2
 or x = a sec  or
1
2 2
x a

x = a cosec orx = a cosh 

7. (iv)
x
a x

or
a x
x

x = a tan2

or x a x

b gor
1
x a x

b g
(v)
x
a x

or
a x
x

or x a x

b g x = a sin2

or
1
x a x
( )

(vi)
x
x a

or x = a sec2

x a
x

or
x x a

b gor
1
x x a
( )

(vii)
a x
a x


or x = a cos 2
a x
a x


(viii)
x
x




or x =  cos2
 +  sin2

x x
 
 
b g
b g
 

b g
Ex.28 Evaluate : z
1
1


sin
sin
x
x
dx.
Sol. I = z
1
1


sin
sin
x
x
dx.
= zcos / sin /
cos / sin /
x x
x x
2 2
2 2
2
b g b g
b g b g


L
N
M
M
O
Q
P
Pdx
= z
tan2

4 2

F
H
G I
K
J
x
dx
= z
[sec2

4 2

F
H
G I
K
J
x
–1 ] dx
= 2 tan

4 2

F
H
G I
K
J
x
– x + c Ans.
Ex.29 Evaluate : z dx
x a x

b g
Sol. Let x = a sin2
 then
dx = 2a sin  cos  d 
 I = z2
2 2
a
a a
sin . cos
sin . cos
 
 
d
= 2 z
d  = 2  + c
= 2 sin–1 x a
/
e j+ c Ans.
4.2 INTEGRATION BY PARTS :
4.2.1 If u and v are two functions of x, then
z
(u.v) dx = u v dx
z
e j– zdu
dx
F
H
G I
K
J. v dx
z
e jdx.
i.e. Integral of the product of two functions =
first function × integral of second function –
z
[(derivative of first) × ( Integral of second)]
Note :
(i) From the first letter of the words inverse
circular, logarithmic, Algebraic, Trigonometric,
Exponential functions, we get a word ILATE.
Therefore first arrange the functions in the
order according to letters of this word and
then integrate by parts.
(ii) For the integration of Logarithmic or Inverse
trigonometric functions alone, take unity (1)
as the second function.
4.2.2 If the integral is of the form
z
ex [f(x) + f'(x)] dx, then by breaking this
integral into two integrals, integrate one
integral by parts and keep other integral
as it is, By doing so, we get-
z
ex [f(x) + f'(x)]dx = ex f(x) + c
Ex.30 Evaluate : z
x sin x dx.
Sol. Here integrating by parts by taking x as the
first function and sin x as the second function.
I = z
x. sin x dx.
= x (–cos x) – z
1. (–cos x) dx.
= – x cos x + z
cos x. dx
= – x cos x + sin x + c Ans.

8. Ex.31 Evaluate : z
x2 ex dx.
Sol. Here integrating by parts by taking x2 as the
first function and ex as the second function
I = z
x2 ex dx
= x2 ex – z
2x.ex dx
= x2 ex – 2 [x.ex – z
1.ex dx]
(taking x as first function)
= x2 ex – 2xex + 2ex + c Ans.
Ex.32 Evaluate : z
tan–1 x dx.
Sol. Integrating it by parts by taking tan–1 x as
the first function and 1 as the second
function, we get
I = z
tan–1 x. 1 dx
= tan–1 x. x – z1
1 2
 x
. x dx
= x. tan–1 x –
1
2
log (1 + x2) + c Ans.
Ex.33 Evaluate : z
ex (sin x + cos x) dx
Sol. I = z
ex (sin x + cos x) dx
This is of the form
z
ex [f(x) + f'(x)] dx = ex f(x) + c
Now here f(x) = sin x
 I = ex sin x + c Ans.
Ex.34 Evaluate : z
ex (log x + 1/x) dx.
Sol. I = z
(ex log x + ex. 1/x) dx
= ex log x + c
Here f x x
f x x
( ) log
& '( ) /


L
N
M O
Q
P
1
Ans.
4.2.3 If the integral is of the form
z
[x f'(x) + f(x)]dx then by breaking this
integral into two integrals integrate one
integral by parts and keep other integral
as it is, by doing so, we get
z
[x f'(x) + f(x) ]dx = x f (x) + c
Ex.35 Evaluate : z
(x sec2 x + tan x) dx
Sol. Here I = z
(x sec2 x + tan x) dx
= z
[x f'(x) + f(x)]dx where f(x) = tanx
= x f(x) + c
= x. tan x + c Ans.
4.3 Integration of Rational functions :
4.3.1 When denominator can be factorized
(using partial fraction) :
Let the integrand is of the form
f x
g x
( )
( )
, where both
f(x) and g(x) are polynomials. If degree of f(x) is
greater than degree of g(x) then first divide f(x) by
g(x) till the degree of the remainder becomes
less than the degree of g(x). Let Q(x) is the
quotient and R(x) , the remainder then
f x
g x
( )
( )
= Q(x) +
R x
g x
( )
( )
Now in R(x)/g(x), factorize g(x) and then write
partial fractions in the following manner-
(i) For every non repeated linear factor in the
denominator, write
1
x a x b
 
b g
b g
=
A
x a

+
B
x b

(ii) For repeated linear factors in the denominator,
write-
1
3
( )
x a x b
 
b g
=
A
x a
( )

+
B
x a
( )
 2
+
C
x a
( )
 3
+
D
x b
( )

(iii) For every non repeated quadratic factor in
the denominator, write
1
2
( )
ax bx c x d
  
b g
=
Ax B
ax bx c
C
x d

 


2
Note :
(i) If integrand is of the form
1
( ) ( )
x a x b
 
then
use the following method for obtaining partial
fractions-
Here
1
( ) ( )
x a x b
 
=
1
a b

b g
a b
x a x b

 
L
N
M
M
O
Q
P
P
b g
b g
=
1
a b

b g
x a x b
x a x b
  
 
L
N
M
M
O
Q
P
P
b gb g
b g
b g
=
1
a b

b g
1 1
x b x a



L
N
M O
Q
P
(ii) If integrand is of the form
x
x a x b
 
b g
b gthen

9. x
x a x b
 
b g
b g
=
1
b a

b a x
x a x b

 
L
N
M
M
O
Q
P
P
b g
( )( )
=
1
b a

b x a a x b
x a x b
  
 
L
N
M
M
O
Q
P
P
b g ( )
( ) ( )
=
1
b a

b
x b
a
x a



L
N
M O
Q
P
Ex.36 Evaluate : zx
x
2
2
1

dx
Sol. Given integral
I = z1
1
1
2


F
H
G I
K
J
x
dx
= z
dx + z dx
x x
 
1 1
b g
b g
= x +
1
2 z 1
1
1
1
x x



F
H
G I
K
Jdx
= x +
1
2
log
x
x


F
H
G I
K
J
1
1
+ c Ans.
Ex.37 Evaluate : z x
x x
2
2
 
dx.
Sol. Here I = z x
x x
 
2 1
b g
b g
dx
=
1
3 z 2
2
1
1
x x



F
H
G I
K
Jdx
=
1
3
[2 log (x – 2) + log (x + 1)] + c
=
1
3
log [(x – 2)2 (x + 1)] + c Ans.
Ex.38 Evaluate : z 2
3 2
2
x
x x
 
dx
Sol. I = z 2
1 2
x
x x
 
b gb g
dx
= 2 z x
x x
 
1 2
b gb gdx.
=
2
(2 –1) z 2 1
1 2

 
b g
b g
b g
x
x x
dx
= 2 z2 1 2
1 2
x x
x x
  
 
L
N
M
M
O
Q
P
P
b gb g
b g
b g dx
= 2 z 2
2
1
1
x x



L
N
M O
Q
Pdx
= z4
2
x 
dx – z2
1
x 
dx
= 4 log (x + 2) – 2 log (x + 1) + c
= 2 log
x
x


2
1
2
b g
( )
+ c Ans.
4.3.2 When denominator can not be factorised:
In this case integral may be in the form
(i) z dx
ax bx c
2
 
, (ii) z px q
ax bx c

 
b g
2 dx
Method:
(i) Here taking coefficient of x2 common from
denominator, write -
x2 + (b/a) x + c/a = (x + b/2a)2 –
b ac
a
2
2
4
4

Now the integrand so obtained can be
evaluated easily by using standard formulas.
(ii) Here suppose that px + q = A [diff. coefficient
of (ax2 + bx + c) ] + B
= A (2ax + b) + B ...(1)
Now comparing coefficient of x and constant
terms.
we get A = p/2a, B = q – (pb/2a)
 I = P/2a z2
2
ax b
ax bx c

 
dx
+ 






a
2
pb
q
z dx
ax bx c
2
 
Now we can integrate it easily.
Ex.39 Evaluate : z dx
x x
2
1
 
Sol. Here I = z dx
x x
2
1
 
= z dx
x  
1 2 3 4
2
/ /
b g b g
=
2
3
tan–1
x 
F
H
G I
K
J
1 2
3 2
/
/
+ c
=
2
3
tan–1
2 1
3
x 
F
H
G I
K
J+ c Ans.

10. Ex.40 Evaluate : zx
x x

 
1
1
2 dx
Sol. I = zx
x x

 
1
1
2 dx
=
1
2 z2 1 1
1
2
x
x x
 
 
b g dx
=
1
2 z2 1
1
2
x
x x

 
dx +
1
2 z dx
x x
2
1
 
=
1
2 z2 1
1
2
x dx
x x

 
b g +
1
2 z dx
x  
F
H
G
I
K
J
1 2
3
2
2
2
/
b g
=
1
2
log (x2 + x + 1) +
1
3
tan–1
2 1
3
x 
F
H
G I
K
J+ c
Ans.
4.3.3 Integration of rational functions
containing only even powers of x.
To find integral of such functions, first we
divide numerator and denominator by x2, then
express numerator as d(x ± 1/x) and
denominator as a function of (x ± 1/x).
Following examples illustrate it.
Ex.41 Evaluate : zx
x
2
4
1
1


dx
Sol. Dividing numerator and denominator by x2,
we get
I = z
1 1
1
2
2 2


/
/
x
x x
e j
e jdx
= z
1 1
1
2
2
2


F
H
GI
K
J
R
S
T
U
V
W

L
N
M
M
O
Q
P
P
/ x
x
x
e j
dx
Now taking x –1/x = t
 [1 + 1/x2] dx = dt , we get
I = zdt
t2
2

=
1
2
tan–1
t
2
F
H
G I
K
J
+ c
=
1
2
tan–1
x
x
2
1
2

F
H
G
I
K
J+ c Ans.
Ex.42 Evaluate : z x
x x
2
4 2
1
4 1

 
dx.
Sol. Dividing numerator and denominator by x2,
we get
I = z 1 1
4 1
2
2 2

 
/
/
x
x x
e j dx
= z 1 1
1 2
2
2

 
F
H
G
G
G
I
K
J
J
J
/
/
x
x x
e j
l q
dx
Now taking x + 1/x = t  1
1
2

L
N
M O
Q
P
x
dx = dt
we get
I = zdt
t2
2

=
1
2
tan–1
t
2
F
H
G I
K
J+ c
=
1
2
tan–1
x x

L
N
M O
Q
P
1
2
/
+ c
=
1
2
tan–1
x
x
2
1
2

F
H
G
I
K
J+ c Ans.
4.4 Integration of irrational functions :
If anyone term in Nr or Dr is irrational then it is
made rational by suitable substitution. Also if
integral is of the form-
z dx
ax bx c
2
 
, zax bx c
2
  dx
then we integrate it by expressing
ax2 + bx + c = (x +  )2 + 
Also for integrals of the form
z px q
ax bx c

 
2
dx, z
(px + q) ax bx c
2
  dx.
First we express px + q in the form
px + q = A
d
dx
ax bx c
2
 
R
S
T
U
V
W
e j + B and then
proceed as usual with standard form.
Ex.43 Evaluate : z dx
x x
2 3 2
 
Sol. I = z dx
x x
2 3 2
 

11. =
1
3 z dx
x x
2 3 3 2
/ /
 
=
1
3 z dx
x
x
2
3 3
1
36
1
36
2
   
F
H
G I
K
J
=
1
3
z dx
x
25
36
1
6
2
 
F
H
G I
K
J
=
1
3
sin–1
6 1
5
x 
F
H
G I
K
J+ c Ans.
Ex.44 Evaluate : zx x
2
2
 .dx
Sol. Let I = zx x
2
2
 .dx
= zx  
1 1
2
b g .dx
=
1
2
(x+1) x x
2
2
 –
1
2
cosh–1 (x+1) + c
Ans.
Ex.45 Evaluate : z x
x x

 
2
2 4
2
dx.
Sol. Let I = z
1
2
2 2 6
2 4
2
( )
x
x x
 
 
dx.
=
1
2 z 2 2
2 4
2
x
x x

 
dx + 3 z dx
x  
1 3
2
b g
= x x
2
2 4
  + 3 sinh–1
x 
F
H
G I
K
J
1
3
+ c Ans.
4.5 Integration of Trigonometric functions :
Here we shall study the methods for
evaluation of following types of integrals.
I. (i) z dx
a b x
 sin2
(ii) z dx
a b x
 cos2
(iii) z dx
a x b x x c x
cos sin cos sin
2 2
 
(iv) z dx
a x b x
sin cos

b g
2
Method :
Divide numerator and Denominator by cos2 x in
all such type of integrals and then put tan x = t.
Ex.46 Evaluate : z dx
x
1 3 2
 sin
Sol. I = z sec
sec tan
2
2 2
3
x dx
x x

.
(Dividing Numr and Denr by cos2 x)
= zsec
tan
2
2
1 4
x dx
x

=
1
2
tan–1 (2 tan x) + c Ans.
Ex.47 Evaluate : z dx
x x
2 3
2
sin cos

b g
Sol. I = z dx
x x
2 3
2
sin cos

b g
Dividing numr and Denr by cos2 x.
= zsec
tan
2
2
2 3
x dx
x 
b g
= –
1
2 2 3
tan x 
b g
+ c Ans.
II. (i) z dx
a b x
 cos
(ii) z dx
a b x
 sin
(iii) z dx
a x b x
cos sin

(iv) z dx
a x b x c
sin cos
 
Method :
In such types of integrals we use following formulae
for sin x and cos x in terms of tan (x/2).
sin x =
2
2
1
2
2
tan
tan
x
x
F
H
GI
K
J

F
H
GI
K
J
, cos x =
1
2
1
2
2
2

F
H
GI
K
J

F
H
GI
K
J
tan
tan
x
x
and then take tan(x/2) = t and integrate another
method for evaluation of integral

12. (iii) put a = r cos  , b = r sin  , then
I =
1
r z dx
x
sin( )
 
=
1
r z
cosec (x +  ) dx.
=
1
r
log tan (x/2 +  /2) + c
=
1
2 2
a b

log tan
x b
a
2
1
2
1

F
H
G I
K
J

tan + c
Ex.48 Evaluate : z dx
x
5 4
 cos
Sol. I = z dx
x
x
5 4
1 2
1 2
2
2



L
N
M
M
O
Q
P
P
tan ( / )
tan ( / )
= zsec ( / )
tan ( / )
2
2
2
9 2
x
x

dx
= 2 z dt
t
32 2

where tan (x/2) = t
=
2
3
tan–1
t
3
+ c
=
2
3
tan–1
1
3 2
tan
x
F
H
G I
K
J+ c Ans.
Ex.49 Evaluate : z dx
x x
sin cos
 3
.
Sol. I = z dx
x x
sin cos
 3
.
=
1
1 3

log tan
x
2 6

F
H
G I
K
J

+ c
=
1
2
log tan
x
2 6

F
H
G I
K
J

+ c Ans.
Ex.50 Evaluate : z dx
x
1 sin
Sol. Here I = z dx
x
1 sin
 I = z dx
x x
sin ( / ) cos ( / )
2 2

=
2
2
log tan
x
4 8

F
H
G I
K
J

+ c
= 2 log tan
x
4 8

F
H
G I
K
J

+ c. Ans.
III. zp x q x
a x b x
sin cos
sin cos


dx
z p x
a x b x
sin
sin cos

dx
z q x
a x b x
cos
sin cos

dx
For their integration, we first express Nr. as
follows-
Nr = A (Dr) + B (derivative of Dr.)
Then integral = Ax + B log (Dr) + C
Ex.51 Evaluate : z sin cos
sin cos
x x
x x


2 3
dx
Sol. Let sin x + cos x = A (2 sinx + 3 cosx)
+ B(2 cos x – 3 sin x)

2 3 1 0
3 2 1 0
A B
A B
  
  
U
V
W
 A = 5/13 , B = –1/13
 I = (5/13) x – (1/13)log(2 sinx + 3cos x) +c
Ans.
5. SOME INTEGRATES OF DIFFERENT
EXPRESSIONS OF ex
(i) zae
b ce
x
x

dx [put ex = t]
(ii) z1
1 ex dx [Multiply and divide
by e–x and put e–x =t]
(iii) z1
1 ex dx [Multiply and divide
by e–x and put e–x =t]
(iv) z 1
e e
x x
  dx [Multiply and divide
by ex]
(v) z
e e
e e
x x
x x



 dx
f x
f x
form
'( )
( )
L
N
M O
Q
P
(vi) z
e
e
x
x


1
1
dx [Multiply anddividebye–x/2]
(vii) ze e
e e
x x
x x


F
H
G
I
K
J

 dx [Integrand = tanh2 x]
(viii) ze
e
x
x
2
2
2
1
1


F
H
G
I
K
Jdx [Integrand = coth2 x]
(ix) z 1
2
e e
x x
 
e j
dx [Integrand = 1/4 sech2 x]

13. (x) z 1
2
e e
x x
 
e j
dx [Integrand =1/4 cosech2 x]
(xi) z 1
1 1
  
e e
x x
e je j
dx [Multiply and divide by ex
and put ex = t]
(xii) z 1
1 ex
dx [Multiply and divide by e–x/2]
(xiii) z 1
1 ex
dx [Multiply and divide by e–x/2]
(xiv) z 1
e
1
x

dx [Multiply & divide by e–x/2]
(xv) z 1
e
2
1
x

dx [Multiply and divide by
2 e–x/2]
(xvi) z x
e
1 dx [Integrand
= (1 – ex)/ x
e
1 ]
(xvii) z x
e
1 dx [Integrand
= (1 + ex) / x
e
1 ]
(xviii) z 1
ex
 dx [Integrand
= (ex – 1) / 1
ex
 ]
(xix) z a
e
a
e
x
x


dx [Integrand
= (ex + a)/ 2
x
2
a
e 
Examples
based on
Some integrates of different
expressions e
x
Ex.52 Evaluate :
1
1
ex

z dx.
Sol. Here I =
1
1
ex

z dx
=
e
e
x
x



z
1
dx = log (1 – e–x) + c Ans.
Ex.53 Evaluate : ex

z 1 dx .
Sol. Here I = ex

z 1 dx
=
e
e
x
x


z 1
1
dx
=
e
e
x
x

z 1
dx –
1
1
ex

z dx
Let ex – 1 = t2, then ex dx = 2t dt
 I = 2 dt
z–
2
1
2
t 
z dt
= 2t – 2 tan–1 (t) + c
= 2 e e
x x
  
L
N
M O
Q
P

1 1
1
tan + c Ans.