Chapter I
Indefinite integral
Problem 1: Integrate the following integrals
i)  sin 4 xdx ii)  sin 2 x cos2 xdx iii)  sin 2 x cos 2 xdx
1
1
(2 sin 2 x) 2 dx   (1  cos 2 x) 2 dx

4
4
1
1
1
  (1  2 cos 2 x  cos2 2 x)dx   (1  2 cos 2 x)dx   2 cos2 2 xdx
4
4
8
1
1
1
1
sin 4 x
)c
= ( x  sin 2 x)   (1  cos 4 x)dx  ( x  sin 2 x)  ( x 
4
8
4
8
4
3 x sin 2 x sin 4 x



c
4
4
32
1
1
ii)  sin 2 x cos2 xdx   2 sin 2 x.2 cos2 xdx   (1  cos 2 x)(1  cos 2 x)dx
4
4
1
x 1
x 1
x x sin 4 x
  (1  cos2 2 x)dx    2 cos2 2 xdx    (1  cos 4 x)dx   
c
4
4 8
4 8
4 8
32
x sin 4 x
 
c
8
32
1
1
1
iii)  sin 2 x cos 2 xdx   2 sin 2 x cos 2 xdx   (1  cos 2 x) cos 2 xdx   (cos2 x  cos2 2 x)dx
2
2
2
sin 2 x 1
sin
2
x
1
sin
2
x
x
sin
4x

  2 cos2 2 xdx 
  (1  cos 4 x)dx 
 
c
4
4
4
4
4
4
16
Solution: i)  sin 4 xdx 
Chapter IIA
Method of substitution
Problem 1: Integrate the following integrals
ax
x
sin 2 xdx
tan xdx
sin 2 xdx
dx v) 
dx
i) 
ii) 
iii)  2
iv) 
2
2
2
2
2
2
2
a sin x  b cos x
a  b tan x
ax
ax
(a cos x  b sin x)

dx
1 x 
dx vii) 
vi)  cos 2 cot 1

1 x 
x x 4 1

sin 2 xdx
2 sin x cos xdx
2 sin x cos xdx
Solution: i) 
=

2
2
2
2
a sin x  b cos x
a sin x  b(1  sin x)
(a  b) sin 2 x  b
1
dz
1
1


ln z  c 
ln[( a  b) sin 2 x  b]  c
Put

( a  b) z ( a  b)
( a  b)
(a  b) sin 2 x  b  z
sin x
(
)dx
 2(a  b) sin x cos xdx  dz
tan xdx
sin x cos xdx
cos x
ii) 
=

a  b tan 2 x 
b sin 2 x  a cos2 x  b sin 2 x
a
cos2 x
1 2 sin x cos xdx
1
 

ln[(b  a) sin 2 x  a]  c similar as ( i)
2
2 (b  a) sin x  a 2(b  a)
sin 2 xdx
(a cos2 x  b 2 sin 2 x) 2
2 sin x cos xdx
2 sin x cos xdx
= 2 2
 2
2
2
2
[a sin x  b (1  sin x)]
[( a  b 2 ) sin 2 x  b 2 ]2
1
dz
1
1
 2
 2
( )  c
2  2
2
(a  b ) z
(a  b ) z
1
1
 2
c
2
2
2
(a  b ) (a  b ) sin 2 x  b 2
iii) 
iv) 
2
Put
(a 2  b 2 ) sin 2 x  b 2  z
 2(a 2  b 2 ) sin x cos x  dz
ax
a  a cos 2t
dx =  
2a sin 2tdt
ax
a  a cos 2t
2a cos2 t
cost
2a sin 2tdt  
2a.2 sin t costdt
2
sin t
2a sin t
 2a  2 cos2 tdt  2a  (1  cos 2t )dt  2at  a sin 2t  c
 
 a cos1
Put
x  a cos 2t
 dx  2a sin 2tdt
x
 a2  x2  c
a
x
a sin 2 t
dx = 
v) 
2a sin t costdt
ax
a  a sin 2 t
a sin 2t
  2a sin 2 tdt  a  (1  cos 2t )dt  at 
c
2
x
x ax
 at  a sin t cost  c  a sin 1
a
c
a
a
a
x
 x(a  x)  c
a


1 x 
1  cost 
dx =   cos 2 cot 1
 sin tdt
vi)  cos 2 cot 1

1  x 
1  cost 


sin t / 2 

1
  cos 2 cot1
 sin tdt   cos(2 cot tan t / 2) sin tdt
cost / 2 

1
  cos(2 cot cot( / 2  t / 2)) sin tdt   cos(2( / 2  t / 2)) sin tdt
Put
x  a sin 2 t
 dx  2a sin t cos dt
 at  a sin t cost  c  a sin 1
cos2 t
x2
   cos(  t ) sin tdt   cost sin tdt  
c   c
2
2
dx
sect tan tdt
sect tan tdt

vii) 
=
4
2
2 sect tan t
x x 1
2 sect sec  1

dt t
sec1 x 2
 c 
c
2 2
2
Put
x  cost
 dx   sin tdt
Put
x 2  sect
 2 xdx  sect tan tdt
sect tan tdt
 dx 
2x
Chapter IIB
Method of substitution with some formulae
Problem 1: Integrate the following integrals
ax
xa
dx
dx
dx iv) 
ii) 
iii) 
dx
x
x
( x   )(  x)
(1  x) 1  2 x  x 2
dx
2 zdz
Solution: i) 
=
Put
( x   )(  x)
z 2 (    z 2 )
x   z2
2dz
dz
z
 dx  2 zdz

 2
 2 sin 1
c
2
2
 
   z
   z2
i) 

 2 sin 1
ii)

x 
c
 
Put
dx
 (1  x) 1  2x  x   1/ z 1  2(1/ z  1)  (1/ z  1)
 
1  x  1/ z
 dx  1 / z 2 dz
 1 / z dz
2
2
dz
z 1  2 / z  2  1/ z 2  2 / z  1
dz
 
2
dz
z 4 / z  2  1/ z 2
dz
 
z
4z  2z 2 1
z2
1
dz
1
dz



2
2
2
4z  2z 1
2 z  z  1/ 2
 1  2 z  z 2  1/ 2  1
1
dz
1
dz




2
2
2
2
 (1  2 z  z )  1 / 2
1 / 2  ( z  1) 2
 

2


1
1
1
z

1
1
1
 2x
1
2x
1
1 1  x

sin
c  
sin
c  
sin 1
c 
sin 1
c
1 x
1 x
2
1/ 2
2
1/ 2
2
2
ax
ax ax
ax
1 a  2x
1
a
dx  
dx  
dx  
dx  
dx
2
2
x
x ax
x 2  ax
x 2  ax
x 2  ax
iii) 
1
1
 .2 x 2  ax  
2
2
a
x 2  ax 
2
2
dx  x 2  ax 
a
a

4
4
1
adx

2
2
2 
a a
x    
2 2

2
2
a 
a a 
a  a
 a



 x  ax  ln x    x        c  x 2  ax  ln x   x 2  ax   c
2  2
2 2 
2  2




2
a  2x  a  2 x x  a 
a x  x  a  2 x x  a 
 x 2  ax  ln
  c  x( x  a)  ln
c
2 
2
2
2



a 

 x( x  a)  ln
2 


 x  x  a    c  x( x  a)  a ln x  x  a   a ln 2  c
2
2
2



 x( x  a)  a ln x  x  a  c0
2
2
xa
xa xa
xa
dx  
dx  
dx
x
x xa
x xa
x
a
dx
a

dx  
dx  

dx
x xa
x xa
xa
x xa
2azdz
2adz
 2 xa 
 2 xa 
a  z2
(a  z 2 ) z 2
iv) 
 2 x  a  2a 
dz
 a  z
 2 x  a  2 a tan 1
x  a  z2
 dx  2 zdz
1
z
tan 1
c
a
a
 2 x  a  2a.
2
Put (2nd integral)
2
xa
xa
 c  2 x  a  2 a tan 1
c
a
a
Chapter III
Integration by parts
Problem 1: Integrate the following integrals
ex
x
x2 1
x 1
dx ii)  (1  x log x)dx iii)  e x
i)  sin 1
dx
dx iv)  e x
2
x
xa
( x  1) 3
( x  1)
v)
 2ax  x dx vi)  ( x   )(  x)dx vii)  ( x  2) x  2x  10dx
2
2
x
dx
xa
Solution: i)  sin 1
√
∫
∫
∫
6
∫
7
6
,
(
(
iv) ∫
(
)
∫
)
(
)
(
)
∫
√
6
∫
.
(
)
(
)
(
)
)
)√*
∫ √*
(
) +
√ 7
/
)
(
7
√
∫
0
∫
0(
(
)
)
v) ∫ √(
(
))
(
iii) ∫
)
∫(
(
ii) ∫
Put
(
) +
(
)
)
(
1
)
1
(
)√(
vi) ∫ √(
)(
(
)
∫√ (
)
∫ √(
)
∫ √(
)
.
/ √.
)
∫√
(
)
(
)
∫ √(
(
8
)
)
(
)
(
) 9
)
/
.
/
.
/
.
/
.
(
)√(
)(
)
(
(
)
)
(
vii) ∫(
∫(
)√
∫ √(
∫√
(
)
)√
(
)
∫√
∫√
(
)√
)
√
)
2(
(
)
)√
∫
/
3
)
2(
√
3
Chapter IV
Integration of trigonometric function
Problem 1: ) ∫
)∫
)∫
)∫
)∫
Solution:
)∫
∫
∫
∫
(
∫
)
)∫
∫
(
)
(
(
∫
∫
(
(
)
)
)
)
(
(
)
)
)∫
∫
∫
(
)
∫
∫
(
)
(
)
(
)∫
)
∫
∫
Put
∫
∫
∫
(
∫
)
∫
(
)
.
.
)∫
/
. /
/
∫
.
∫
/
∫
Put
∫
Chapter V
Rational function
Problem 1: ) ∫ (
)(
)
) ∫(
) (
)
)∫
)∫
)∫
Solution:
)∫
(
)∫
)(
(
∫{
)
) (
)
)
(
)
(
)
(
)
(
)
(
}
)
(
)
Now,
(
) (
(
)(
)
(
)
(
)
(
)
Put
Put
Equating the coefficient of
∫
(
) (
∫{
)
(
)
(
)∫
∫
(
)
)∫
)
(
)
(
)
)(
)(
)
(
)
(
)
(
(
)
}
)
(
∫
Put
(
∫{
(
)
(
)
(
)
}
)
,
∫
(
)∫
∫{
)
∫
,
}
(
)(
∫{
)
(
(
)
)
-
,
(
)
}
(
)
[
]
CHAPTER VI
Definite integral
Problem 1: ) ∫
(
)
)∫
(
) 7
)∫
Solution:
)∫ (
)
6
∫
⁄
(
-
)
∫
∫
∫
,
-
∫
,
-
[
)∫
∫
(
,(
]
,
∫
,
(
-
)-
(
⁄
)∫
) -
∫
)
⁄
(
∫
)
(
)
Put
⁄
(
)
∫
∫
⁄
(
⁄
(
(
)
)
(
⁄
)
(
∫
⁄
∫
)
∫
)
⁄
(
)
∫
(
))
(
)
⁄
(
∫
(
)
⁄
[
(
)
(
First principle of integral calculus:
)]
(
)
Area of region by the curve f(x) with the x-axis bounded by x=a and x=b is
, ( )
(
)
(
)
(
) )(
∑ (
)
( )
Symbolically this area is represented as ∫
∫
( )
If
∑ (
and
)
implies that
∑ (
)
then
∫ ( )
∑ ( )
Problem 2: Evaluate from first principle ∫
Solution:
∫
∑
(
[
(
[
(
(
)
)
]
)
)
(
)
]
Problem 3: Evaluate from first principle ∫
Solution:
∫
∑(
)
,
,
(
)(
)
(
(
Problem 4: Evaluate
)
[
]
)
0
1
)
6
7
)
[4
5
4
5
4
5
4
Solution:
)
[
]
[
]
∑
(
)
∫
[
(
)]
)
0
1
[
]
5 ]
∑
)
,
∫
-
6
7
[
]
∑
)
[4
,
∫
5
∑4
4
5
4
(
)-
5
4
5 ]
5
∑
5
Let
∑4
5
∑4
∫
(
5
∑
4
)
Put
∫
,
-
Chapter VII
General properties and reduction fomula
Some properties of definite integral:
4
5
)∫
( )
( )
)∫
( )
∫
)∫
( )
∫ ( )
)∫
( )
∫
)∫
( )
∫
( )
(
)∫
( )
∫
( )
(
∫
( )
( )
∫
(
)
(
Problem 1: Integrate ) ∫
)
)
( )
( )
( )
⁄
√
√
(
) show that ∫
)
) show that ∫
√
∫
)
Solution:
⁄
∫
√
⁄
√
√
√
⁄
∫
⁄
∫
.
⁄
√
∫
√
/
√
.
/
⁄
√
√
∫
√
√
√
√
√
, - ⁄
∫
∫
∫
∫
∫
∫
Put
∫
/
∫
√
√
∫
.
√
∫
)
√
⁄
√
,
-
∫
(
)
∫
(
)
∫
)
(
∫
)
Put
∫
(
∫
[
∫
[
)
∫
0
]
∫
]
∫
∫
,
(
.
/1
[
]
∫
)
[
]
∫
-
Theorem: state and prove walli’s formula for definite integral or show that
If be positive integer
∫
∫
Or
According as
Proof:
∫
is even or odd.
∫
(
)
(
(
)∫
(
)
)
(
)∫
(
)∫
(
)∫
(
)
(
)∫
(
)∫
6
(
(
7
)
∫
)
Similarly if we proceed we get
according as
∫
is even or odd.
, ,
∫
∫
-
∫
Or
according as
is even or odd.
Theorem: If both m and n are even
(
∫
If m is even and n is odd
(
)(
)
(
)(
If both m and n are odd
(
)(
)
(
)(
Problem 2: Evaluate ∫
)(
(
(
)
)(
(
)(
)
)(
)
)(
)
)
(
)
√
Solution: Put
∫
√
√
∫
Problem 3: Evaluate ∫
(
∫
) ⁄
)
Solution: Put
∫
) ⁄
(
∫
(
)
Problem 4: Evaluate ) ∫
∫
)∫
)∫
Solution:
)∫
)∫
)∫
(
Problem 5: show that ∫ (
)
)
Solution:
∫
∫
∫
(
)
(
)
(
)
(
∫
∫
(
(
(
(
(
∫
)
∫
(
)
(Solution See problem 1 (iii) chapter Vi)
))
(
)
(
(
)∫
Solution:
∫
∫
∫
∫
⁄
6
)∫
⁄
hence deduce
)∫
∫
7
⁄
∫
)
)
Problem 6: Obtain reduction formulae for ) ∫
)
)
)
∫
)
)
(
)
)
∫
∫
∫
∫
)
∫
∫
∫
∫(
)
∫
Problem 7: Obtain reduction formulae for ) ∫
)∫
Solution:
)
∫
(
∫
)
)
∫
∫
∫
∫
)
∫
∫
hence deduce ) ∫
∫(
)
∫(
)
(
)∫
(
)
(
)
(
(
)
(
)
)∫
∫
(
)
∫
(
Problem 8: Obtain reduction formulae for ∫
)
hence deduce ∫
Solution:
∫
∫
[
*(
∫
)
(
∫
(
+
)
)
*(
)
(
)
)
(
)
(
]
)
∫
∫
∫
∫
*(
+
)
+
(
4
(
5
(
)
)
(
(
)
Now the reduction formula for ∫
(
6
(
)
is
)
7
(
)
)
∫
Chapter VIII
Gama beta function
Gama function: The second Eulerian integral is called gama function and is defined as
( )
and
but need not be integer.
∫
Beta function: The first Eulerian integral is called beta function and is defined as
(
)
(
∫
)
and
Problem 1: Show that ) (
)
(
but need not be integer.
) ) ( )
) (
)
( )
( )
)∫
Solution:
) (
)
∫
∫
(
) ( )
∫
) (
)
,
-
When
(
)
∫ (
(
)
(
)
)
∫
∫
∫
( )
∫
is positive integer (
) ( )
)
)
∫
Put
( )
( )
∫
(
)
∫
∫
Problem 2: show that ) ( )
) ( )
∫ .
/
hence integrate
Solution:
) we know
( ) ∫
Put
( )
∫
∫
) we know
( ) ∫
Put
( )
∫
(
)
( )
∫
) (
∫
∫
)∫
)
)∫
∫
)∫ .
/
) we know
( ) ∫
Put
( )
∫ .
/
∫ .
/
) we know
( )
∫
Put
( )
∫
∫
( )
√
) we know
(
)
∫
Put
( )
∫
( )
∫
) we know
( )
∫ .
/
∫ .
/
Put
( )
∫ .
/
)
∫
Problem 3: Show that (
( )
(
∫
)
(
)
Solution: we know
(
)
∫
(
)
Put
(
(
)
∫ (
∫ (
)
)
(
(
)
)
(
)
)
(
∫
)
(
Similarly
(
)
∫
Since (
)
(
∫
)
(
)
(
(
)
∫
)
We know
(
)
∫
(
)
(
)
(proved)
)
hence integrate ∫
(
)
Put
∫
(
(
)
( ) ( )
(
)
)
Problem 4: Show that
( ) ( )
)
) (
(
)
.
)∫
/ .
.
/
/
Solution:
) we know
( )
∫
And
( )
∫
( )
∫
( ) ( )
(
(
(
∫ 8∫
)∫
(
)
9
(
)
) (
∫
)
( ) ( )
(
)
)
) we know
(
)
∫
)
∫ (
(
)
Put
(
∫ (
)
)
(
(
)
)
∫
Or
(
∫
Put
and
)
( ) ( )
(
)
(
(
)
)
.
∫
/ .
/
.
.
/ .
/
.
/
( ) . /
/
/
Example:
.
∫
/ .
.
Problem 5: Show that ( ) (
Solution: We know
( ) ( )
(
)
(
)
Let
( ) (
)
∫
/
. /
)
(
∫
.
/
. /
)
∫
∫
∫
Here
∫
Put
. /
. /
∫
. /
∫
∫
. /
∫
∫
∫
( ) (
)
∫ (
∫
∫
∫
)(
)
∫ (
)
6
7
[
]
∫ (
)
6
7
[
]
We know
(
)
(
)
If
then
(
)
(
( ) (
( ) (
(
)
(
)
)
)
)
Chapter Ix
Length of curves
Length of curves for Cartesian equation:
√
√
√
(
)
∫√
(
)
(
)
Chapter x
Area and volume of curves and surface of revolution
Problem 1: Find the area of the ellipse
Solution:
between the major and minor axes
Y
b
X
a
Area of the ellipse
∫
∫
√
Put
∫
[
√
∫
∫ (
)
]
Problem 2: Find the area of the parabola
Solution:
bounded by
Y
X=4
X
Area of the bounded region
∫ √
√ ∫
⁄
√
[
]
√
√
(
Problem 3: Find the area of the loop of the curve
Solution:
) (
)
If
implies that the curve cut x axis at points
The area of the loop is
∫
∫
√
(
) (
)
Put
∫ (
)√
Putting
∫
∫
∫ (
,
)
-
∫
∫ (
(
(
)
)
∫
[
]
)
Problem 4: Find the area above the x axis included between the parabola
and the circle
Solution:
The points of intersection of the curves are (0,0) and (
The required area is
∫ (
∫ (√
)
√
)
)
Putting
∫ (√
∫ √
∫
√
∫
∫ (
[
]
√
)
(
[
]
)
Problem 5: Find the area bounded by the cardioide
Solution:
(
)
The curve is symmetric about the initial line then the required area is
∫
,
∫
-
(
∫ (
)
∫ (
)
)
[
]
The solid of revolution, the axis of revolution is x axis:
Let a curve
whose Cartesian equation is given
( ) be rotated about the
a solid of revolution, let us consider the portion
of this solid bounded by
. Consider a circular slices
with coordinates (
) and (
The volume of the slices with thickness
is
Hence total volume of the solid bounded by
and
is
axis and form
and
).
∑
∫
Again the area of the slices is
Hence total surface area of the solid bounded by
∑
∫
∫
√
and
(
is
)
Similarly if the axis of revolution is y axis:
∫
∫
√
(
)
Problem 6: Find the volume and area of the curved surface of a paraboloid of revolution form by
revolving the parabola
about the axis bounded by
Solution:
Here,
√
√
The required volume is
∫
∫
Also the required surface area is
,
-
√
∫
)
√
∫ √
) ]
[(
√
(
√ {(
)
√ ∫ √
}
Problem 7: Find the volume and area of the curved surface of the reel generated by the parabola
bounded by the latus rectum revolves about the tangent at vertex.
Solution:
Here the axis of revolution being
The volume is
∫
axis and the extreme values of
∫
6
7
4
The surface is
∫
√
(
)
∫
√
Putting
∫
∫ (
6
√
∫
)
(
)7
being
5
[ √
]
Problem 8: Find the volume and surface area of the solid generated by revolving the cycloid
(
)
(
) about its base.
Solution:
The extreme values of
The required volume
∫
are given by
∫ (
)
∫
∫ (
∫
)
∫
The required surface area
∫ √
∫ (
∫
∫
)√ (
) √* (
(
)
)
+
*
+
∫
∫
Problem 9: Find the volume and surface area of the solid generated by revolving the cardioide
(
) about the initial line.
Solution:
The extreme points of curve are given by
The required volume
∫
∫
(
∫ (
)
∫ (
)
(
) (
)(
)
((
)
)
)
Putting
∫ (
)
∫ (
)
[
]
[(
)
(
(
)]
)
The required surface area
√
∫
∫
(
∫ (
√(
)
)
)
√ (
)
Putting
√ ∫ √
√
(
[ ]
)