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Nirma University
Institute of Technology
Mathematics and Humanities department
B.Tech.(All), Semester-3 (D2D)
Handouts
Course Name: Foundation mathematics-I
Course Code: 2MA004
Integration: Integration is inverse (opposite) process of differentiation.
e.g: if
d
(sin x)  cos x , Then
dx
But we also know that
Similarly
 cos x
dx = sin x
d
 sin x  5  cos x   cos xdx  sin x  5
dx
d
 sin x  100   cos x   cos xdx  sin x  100
dx
So In general   cos xdx  sin x  c
where ‘c’ is arbitrary constant, It is known as integration
constant and it work for all above cases.
Notation of integration:
Integrand function

Symbol of integration   f  x dx  Variable w.r.t integration taken.
Standard formulae of Integration.
x n 1
1).  x dx =
+c
n 1
n
2).
1
 x dx
= log |x| + c.
3).  cos x dx = sin x + c.
4).  sin x dx = - cos x + c.
5).  sec 2 x dx = tan x + c.
6).  cos ec 2 x dx = - cot x + c.
7).  sec x  tan x dx = sec x + c.
8).  cos ecx  cot x dx = - cosec x + c.
9).
x
2
1
x
dx
= tan 1 + c.
2
a
a
a
dx
10).

11).
| x|
a x
2
2
dx
x2  a2
dx

14).
x
15).
a
a x
2
=
x
+ c.
a
1
x
1
x
sec 1
+ c =  cos ec 1 + c.
a
a
a
a
ax
 c.
log e a
12).  a x dx =
13).
= sin 1
2
= log | x +
x 2  a 2 | + c.
1
xa
dx
log
=
+ c.
2
2a
ax
 a2
2
1
xa
dx
log
=
+ c.
2
2a
xa
x
16).  tan x dx = log |sec x | + c.
17).  cot x dx = log | sin x | + c.
18).  cos ecx dx = log cos ecx  cot x  c = log | tan x 2 | + c.
19).

secx dx = log | sec x + tan x | + c.
e.g.
 x 3  x 2 1
 dx.
 
x

Sol:

=
x3
x

dx +
x3
x

x

dx +
1
2
3
1
2
x
=
 x 2 dx +
dx +
5
1

1
dx +

dx +
x2
x
=
5
x2
1
2
x
2
1
x
1
2
3

dx
1
2
x
dx +
 x 2 dx +
3
dx
x

1
2
dx
1
 x 2 dx
1
1
 1
5
1
x 2
x2
x2
=
+
+
+c
5
3
1
 1
1
1
2
2
2
7
2
2
= x 2 + x 2 + 2  x 2 + c.
7
5
Method of Substitution:
e.g 
x3
dx
1  x8
x3
I= 
dx
1  x8
Taking x 4  t
4 x 3 dx  dt
x 3 dx 

1
dt
4
I=
=
1
 1 t
2
1
dt
4
1
tan 1 t  c
4
=
1
tan 1 ( x 4 )  c
4
Method of Trigonometric Substitution:
For following type of function we have to take following
trigonometric substitution.
Function
Substitution

a2  x2
x = a sin 

x2  a2
x = a sec 

x2  a2
x = a tan 

ax & ax
x = a cos 2

ax
x = a sin 2 

ax
x = a tan 2 

2ax  x 2
x = 2a sin 2 

e.g
1
1
1
,
or
a  b cos x a  b sin x a  b sin x  c cos x
x
1
x a4
4
dx
Solution : I =
x
1
x4  a 4
dx

Taking x 2 a 2 sec  ,    0, 


2
2x dx = a 2 sec  tan  d 
tan
x
t
2
Now ,
I=
 2x
2 xdx
=
 2a
=
 2a
=
 2a
=
x 4 a 4
2
a 2 sec  tan d
2
sec  a 4 sec 2 a 4
tan  d
2
tan 
1
2
d
1
 c
2a 2
But, x 2  a 2 sec    = sec 1

I=
x2
a2
2
1
1 x
sec
c
2a 2
a2
Integration of Function of Type:   2
and
 ax bx  c
dx
e.g.
 4x
2
dx
 3x  1
dx
3
1

4 x 2  x  
4
4

dx
= 
2

3
9 1
4 x     
8  64 4 

1
dx
= 
2
4 
3  7 
x   
8   64 

 8x  3 
2
 + c.
=
tan 1 
7
 7 
=




ax 2 bx  c 
dx
Ax  B

dx and
Integration of Function of Type:   2
 ax bx  c


dx 
ax 2 bx  c 
Ax  B
e.g.  32x  1 dx
2x  x  1
3
1
(4 x  1) 
4
4 dx
2
2x  x 1
Solution: I =

=
3
4
=
3
1
log | 2 x 2  x  1 | +
4
4
=
1
4x  1
dx +
2
4
 x 1
 2x
 2x

2
1
dx
 x 1
1
2
2

1  7  
 
2 x    
4   4  



dx
 4x  1 
3
1
 + c.]
log | 2 x 2  x  1 | +
tan 1 
4
2 7
 7 
Integration of Type:  sin m x cos n xdx
In the type [  sin m x cos n xdx ], the substitution is depends on value
of m and n,
Case-I When m, n both are odd then take sin x = t
Case-II When m is odd and n is even then take cosx = t
Case-III When m is even and n is odd then take sinx = t
Case-IV When m and n are even then use sin 2 x =
and
cos 2 x =
1  cos 2 x
and simplify and then integrate.
2
1  cos 2 x
2
e.g.  sin 5 x cos 2 xdx
Solution: taking cosx = t, sinx dx = -dt
I   sin 4 x sin x cos 2 xdx
=  (1  cos 2 x) 2 cos 2 x sin xdx
=  (1  t 2 )2 t 2 (dt )
=  (1  2t 2 t 4 )( t 2 )dt
=  (2t 4 t 6 t 2 )dt
=
2t 5 t 7 t 3
  c
5
7 3
=
2
1
1
cos 5 x  cos 7 x  cos 3 x  c
5
7
3
1
x 1
3
x 3

log
c
.  log
4
x 1 4
x 3
Integration by parts method:
 u v
 du

dx = u  v dx -      vdx dx
 dx 

To choose u and v is very crucial in this case. First one
has to select u in LIATE order and remaining part of function
is v.
Where, L – Logarithemic.
I – Inverse Trigonometric
A – Algebric.
T – Trigonometric.
E – Exponential.
e.g.
x
2
log xdx
x
Solution: I=
2
log xdx
d

dx    log x   x 2 dx  dx
 dx

= log x
x
= log x 
x3
1 x3
   dx
3
x 3
=
2
x3
1
log x   x 2 dx
3
3
x3
1
=
log x  x 3  c
3
9
Definite Integration:
Fundamental principles of definite integration:
b
d (F )
 f ( x)
If
dx
e.g.
1
  4x
3
 f ( x)dx  F (b)  F (a)
a
 3x 2  2 x  1 dx
0
1
1
1
1
0
0
0
0
1
1
Sol: 4 x3dx  3 x 2 dx + 2  xdx   1dx
1
 x4 
 x3 
 x2 
=  4   + 3   +  2   + [x]10
 4 0  3 0  2 0
= 1111 = 4
Definite integration by simplification & substitution
method:
e.g.
1
e
x
0
dx
 e x
1
e
Sol: Here, I =
x
0
=
1

0
=
dx
 e x
dx
ex 
1
ex
1
e x dx
0 e 2 x  1
Taking e x = t
 e x dx = dt
therefore, when x = 0  t  e 0  1 and x = 1  t  e 1  e

e
t
I=
1
dt
1
2
= [tan 1 t ]1e
= tan 1 e  tan 1 1
 tan 1 e 

.
4
Find out the definite integration by integration by parts
method:
b
b
b
a
a
a
 du

 u  vdx  u  vdx    dx  vdx dx
1
e.g.  x tan 1 xdx
0
Sol: I =
1
 x tan
1
xdx
0
1

x2 
I =  tan 1 x   2 0

1
1
x2

0 1  x 2 2 dx
1
2
  1  1 ( x  1)  1
=    0  
dx
2
 4 2  2 0 1 x
=
 1 
1 
  1 
dx
8 2 0  1 x 2 
=
 1
 [ x  tan 1 x]10
8 2
=
 1  
 1
8 2  4 
=
 1
 .
4 2
1
Definite Integration by using properties:
We have following properties of definite integration
1).
b

a
b
2).

c
b
a
c
f  x  dx   f  x  dx   f  x  dx if a  c  b
a
f  x  dx    f  x  dx
a
b
a
3).
If f  x  is odd function
 f  x  dx 0
a
a
4).

a
a
f  x  dx  2  f  x  dx If f  x  is even function
0
a
5).

0
a
f  x  dx   f  a  x  dx
0
b
6).

a
b
f  x  dx   f  a  b  x  dx
a