Calculus Integration.pdf

Integration
1 | I n t e g r a t i o n
ADVANCED MATHEMATICS
INTEGRATION
10
BARAKA LO1BANGUT1
V = π ∫ (r2
- x2
)dx
r
-r
=
4
3
πr3

Integration
The author
Name: Baraka Loibanguti
Email: barakaloibanguti@gmail.com
Tel: +255 621 842525 or +255 719 842525

Integration
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Integration
To myself and I

Integration
INTEGRATION
Integration is process of reversing differentiation. Previously, we
learned differentiation of different relations. Here we are going to
find out the relation/function whenever the derivative is given.
The required relation/function is called anti – derivative or definite integrals.
The origin of integral calculus can be traced back
to the ancient Greeks. They were motivated by
the need to measure the length of curves, the
area of a surface, or the volume of a solid.
Archimedes used techniques very similar to
actual integration to determine the length of a
segment of a curve. Democritus 410 B.C. had the
insight to consider that a cone was made up of
infinitely many planes cross sections parallel to
the base.
The theory of integration received very little
stimulus after Archimedes’ remarkable
achievements. It was not until the beginning of the 17th
century that the interest in
Archimedes’ ideas began to develop. Johann Kepler 1571 – 1630 A.D. was the first among
European mathematicians to develop the ideas of infinitesimals in connection with
integration. The use of the term ‘‘integral’’ is due to the Swiss mathematician Johann
Bernoulli 1667 – 1748. In the present chapter, we shall study integration of real-valued
functions of a single variable according to the concepts put forth by the German
mathematician Georg Friedrich Riemann 1826 – 1866. He was the first to establish a
rigorous analytical foundation for integration, based on the older geometric approach.
Suppose we are given the function 3
4
3 2
−
+
= x
x
y , it is the derivative 4
6 +
= x
dx
dy
. The
question is can do reverse to find y? It is possible to get y , the rule that guide integration
is  +
+
=
+
A
n
x
dx
x
n
n
1
1
where A is a constant of integration and 1
−

n . So for the
derivative 4
6 +
= x
dx
dy
multiply both sides dx
Chapter
10
BARAKA LO1BANGUT1

Integration
( )dx
x
dx
dy
dx 4
6 +
=  ( )dx
x
dy 4
6 +
=
Put the integral both sides, as ( )

 +
= dx
x
dy 4
6
  
+
= dx
xdx
dy 4
3 thus

 +
= dx
dx
x
y 4
3 therefore A
x
x
y +
+
= 4
2
3 2
Indefinite integrals
The indefinite integrals: These are integrals of the form
 dx
x
f )
( thus if
( )
  

=
 dx
x
g
dx
x
f
dx
x
g
x
f )
(
)
(
)
(
)
(
Evaluate the integral ( )
 +
+ dx
x
x 7
4
2 2
3
Solution
From ( )
 +
+ dx
x
x 7
4
2 2
3
( ) 


 +
+
=
+
+ dx
dx
x
dx
x
dx
x
x 7
4
2
7
4
2 2
3
2
3


 +
+ dx
dx
x
dx
x 7
4
2 2
3
thus A
x
x
x
+
+








+








= 7
3
4
4
2
3
4
( )
 +
+
+
=
+
+
 A
x
x
x
dx
x
x 7
3
4
2
7
4
2
3
4
2
3
Evaluate ( )
 − dx
x 3
2
Solution
From ( )  
 −
=
− dx
xdx
dx
x 3
3
3
2
 
− dx
xdx 3
2 thus ( )
 +
−
=
− A
x
x
dx
x 3
3
2 2
Example 1
Example 2

Integration
Evaluate  





−
+
− dx
x
x
x
x
1
6
5
3 3
2
Solution
( )  
 −
=
− xdx
dx
x
dx
x
x 5
3
5
3 2
2
thus A
x
x
+








−








=
2
5
3
3
2
3
( )
 +
−
=
−
 A
x
x
dx
x
x
2
5
5
3
2
3
2
Evaluate ( )
 +
− dx
x
x 3
5
3
2
Solution
From ( ) 


 +
−
=
+
− dx
x
xdx
dx
dx
x
x 3
3
5
3
2
5
3
2
A
x
x
x +








+








−
=
4
5
2
3
2
4
2
Therefore ( ) A
x
x
x
dx
x
x +
+
−
=
+
−
 4
5
2
3
2
5
3
2
4
2
3
Evaluate ( )
 − dx
x
x 2
3
3
Solution
Given ( )
 − dx
x
x 2
3
3 thus ( ) 

 −
=
− dx
x
dx
x
dx
x
x 2
3
2
3
3
3
Example 3
Example 4
Example 5
( )
 +
−
=
−
 A
x
x
dx
x
x 3
4
2
3
4
3

Integration
EXERCISE 1
1. ( )
 +
− dx
x
x 5
2 2
3
2. ( )( )
 −
− dx
x
x
x 2
2 2
3
3. ( )
 +
− dx
x
x 2
5
4 3
4.
 





+
+
− dx
x
x
x
12
5
3
3 4
3
5. ( )
 −
+
− dx
x
x
x 3
4
2
10
3
4 6.
 





+
− dx
x
x
x
2
1
3
6
1 2
3
7. ( )
 − dx
x
x 4
3 2
8.
 





− dx
x
x 2
3 1
9. ( )
 −
+
− dx
x
x
x
x 2
3
4
2 10.
 





+
+ dx
x
x
x 3
1
11.
 





− dx
x
x 5
3
1
2 12.
 





− dx
x
x
x
3
4
5
Substitution / Change of variable
Integrals of the form
 dx
x
f
x
f )
(
'
)
( are easily evaluated by letting
)
(
'
)
( x
f
dx
du
x
f
u =

= hence
)
(
' x
f
du
dx =
Evaluate the integral ( )( )
 −
+
+ dx
x
x
x 2
3
3
2 2
Solution
From ( )( )
 −
+
+ dx
x
x
x 2
3
3
2 2
Let 2
3
2
−
+
= x
x
u then 3
2 +
= x
dx
du
Make dx the subject and substitute
( )
3
2 +
=
x
du
dx then ( )
( )
 +
+
3
2
3
2
x
du
u
x
Example 6

Integration
 +








= A
u
udu
2
2
But 2
3
2
−
+
= x
x
u
( )( ) ( )
 +
−
+
=
−
+
+ A
x
x
dx
x
x
x
2
2
2
2
3
2
1
2
3
3
2
Evaluate ( )( )
 −
− dx
x
x
x 12
2
6
2 2
Solution
From ( )( )
 −
− dx
x
x
x 12
2
6
2 2
, Let x
x
u 12
2 2
−
= and 12
4 −
= x
dx
du
( )
6
2
2 −
=
x
du
dx and ( )
( )
 −
−
6
2
2
6
2
x
du
u
x
A
u
udu +








=
 2
2
1
2
1 2
But x
x
u 12
2 2
−
=
( )( ) ( )
 +
−
=
−
−
 A
x
x
dx
x
x
x
2
2
2
12
2
4
1
12
2
6
2
Evaluate ( ) ( )
 +
+ dx
x
x
x 5
5
1
3 3
2
Solution
Let x
x
u
x
x
u 5
5
5
5 3
2
3
+
=

+
=
5
15
2 2
+
= x
dx
du
u Substituting
( )
1
3
5
2
2
+
=
x
udu
dx and ( ) ( )
 +
+
1
3
5
2
1
3 2
2
x
udu
u
x
A
u
du
u +
=

3
2
15
2
5
2
But ( )
x
x
u 5
5 3
+
=
( ) ( )
 +





 +
=
+
+ A
x
x
dx
x
x
x
3
3
3
2
5
5
15
1
5
5
1
3
Therefore, ( ) ( ) ( ) ( )
 +
+
+
=
+
+ A
x
x
x
x
dx
x
x
x 5
5
5
5
15
1
5
5
1
3 3
3
3
2
Example 7
Example 8

Integration
Evaluate ( )( )
 −
+
+ dx
x
x
x
5
2
1
2
7
1
7
Solution
Given ( )( )
 −
+
+ dx
x
x
x
5
2
1
2
7
1
7 Let 1
2
7 2
−
+
= x
x
u therefore
( )
1
7
2 +
=
x
du
dx thus ( )
( )
 +
+
1
7
2
1
7 5
x
du
u
x
 +
= A
u
du
u 6
5
12
1
2
1
But 1
2
7 2
−
+
= x
x
u
Therefore, ( )( ) ( ) A
x
x
dx
x
x
x +
−
+
=
−
+
+

6
2
5
2
1
2
7
12
1
1
2
7
1
7
Evaluate ( )( )
 −
− dx
x
x
x
9
2
4
3
3
2
Solution
( )( )
 −
− dx
x
x
x
9
2
4
3
3
2 Let x
x
u 4
3 2
−
= then 4
6 −
= x
dx
du
( )
x
du
du
3
2
2 −
−
= then ( )
( )
 −
−
−
x
du
u
x
3
2
2
3
2 9
, But x
x
u 4
3 2
−
=
( )
 +
−
−
=
− A
x
x
du
u
10
2
9
4
3
20
1
2
1
( )( ) ( ) A
x
x
dx
x
x
x +
−
−
=
−
−

10
2
9
2
4
3
20
1
4
3
3
2
EXERCISE 2
Evaluate the following
1. ( )( )
 +
−
− dx
x
x
x 2
3
1
6 2
2. ( )
 +
−
− dx
x
x
x 3 2
3
4
4
1
2
3.
 dx
x
x 3
cos
3
sin 4. xdx
x cos
sin7

5.
 xdx
3
cos 6.
 xdx
x 3
cos
5
sin
7.
 xdx
x 3
sin
cos 8.
 xdx
x 4
cos
4
sin2
Example 9
Example 10

Integration
9. ( )
 + dx
x
x
4
2
2
3 10. ( )
 + dx
x 4
1
2
11.
 dx
x
x 2
tan 12. ( )
 + dx
x 1
3
cos
13.
 x
x
x d
tan
sec 3
14. ( )( )
 −
− dx
x
x
x 5
3
2
2
1
15. dx
x
x

2
3
cos
sin 16. dx
x
x
 cos
sin
17. dx
x

5
sin 18.
 xdx
x 2
2
sec
tan
19.  − dx
x
a
x 2
2
20.  + dx
x
x 2
1
Integration of natural logarithm
From, ( )
x
x
dx
d 1
=
ln ( ) dx
x
x
d
1
=
ln
( )
 
= dx
x
x
d
1
ln 
= dx
x
x
1
ln
 +
=
 A
x
x
dx
ln thus  +
=
+
b
ax
A
a
dx
b
ax
ln
1
1
Integral of the form
( )
( )
 dx
x
f
x
f'
Suppose you have
( )
( )
 dx
x
f
x
f'
Let ( ) ( )dx
x
f
du
x
f
u '
=

=
 +
= A
u
u
du
ln But ( )
x
f
u =
( )
( )
( ) A
x
f
dx
x
f
x
f
+
=
 ln
'
Evaluate
 + 1
2x
dx
Example 11

Integration
Solution
From
 + 1
2x
dx
Let 1
2 +
= x
u
2
=
dx
du
2
du
dx =
 





2
1 du
u
thus
 +
= A
u
u
du
ln
2
1
2
1
, But 1
2 +
= x
u
 +
+
=
+
 A
x
x
dx
1
2
ln
2
1
1
2
Evaluate  +
dx
x
x
9
3 2
Solution
 +
dx
x
x
9
3 2
Let 9
3 2
+
= x
u
x
dx
du
6
=
x
du
dx
6
=
Substituting 











 x
du
u
x
6
 +
+
= A
x
u
du
9
3
ln
6
1
6
1 2
But 9
3 2
+
= x
u
 +
+
=
+
 A
x
dx
x
x
9
3
ln
6
1
9
3
2
2
Evaluate 
−
+
dx
x
x
x 1
3
2
Solution

−
+
dx
x
x
x 1
3
2
simplifying  









−
+ dx
x
x
x
x
x
2
1
2
1
2
1
2
1
3
( ) 
 
−
−
+ dx
x
dx
x
dx
x 5
.
0
5
.
0
5
.
1
3
Example 12
Example 13

Integration
A
x
x
x
+
+
+
= 2
1
2
3
2
5
2
2
2
5
A
x
x
x
dx
x
x
x
+
+
+
=







 −
+
 2
1
2
3
2
5
2
2
2
5
2
1
3
Evaluate
 +
dx
x
x
2
sin
cos
Solution
From
 +
dx
x
x
2
sin
cos
Let 2
sin +
= x
u
xdx
du
x
dx
du
cos
cos =

=
 +
= A
u
u
du
ln But 2
sin +
= x
u
 +
+
=
+
 A
x
dx
x
x
2
sin
ln
2
sin
cos
Find  







+
−
−
−
dx
e
e
e
e
x
x
x
x
2
2
2
2
Solution
Let x
x
e
e
u 2
2 −
+
= thus ( )
x
x
e
e
dx
du 2
2
2 −
−
=
( )dx
e
e
du x
x 2
2
2
−
−
=

 +
=
= A
u
u
du
u
du
ln
2
1
2
1
2
But x
x
e
e
u 2
2 −
+
=
 +
+
=








+
−
 −
−
−
A
e
e
dx
e
e
e
e x
x
x
x
x
x
2
2
2
2
2
2
ln
2
1
Example 14
Example 15

Integration
Find  −
+
+
dx
x
x
x
2
3
3
2
2
Solution
Let 2
3
2
−
+
= x
x
u then
3
2
3
2
+
=

+
=
x
du
dx
x
dx
du
 





+
+
3
2
3
2
x
du
u
x
A
u
u
du
+
=
 ln , But 2
3
2
−
+
= x
x
u
 +
−
+
=
−
+
+
 A
x
x
dx
x
x
x
2
3
ln
2
3
3
2 2
2
Find  +
−
−
dx
x
x
x
7
5
3
5
3
2
Solution
From  +
−
−
dx
x
x
x
7
10
3
5
3
2
Let 7
10
3 2
+
−
= x
x
u
10
6 −
= x
dx
du
( )
5
3
2 −
=
x
du
dx
( )
  +
=
=
−
−
A
u
u
du
x
du
u
x
ln
2
1
2
1
5
3
2
5
3
But 7
10
3 2
+
−
= x
x
u
Therefore,  +
+
−
=
+
−
−
A
x
x
dx
x
x
x
7
10
3
ln
2
1
7
10
3
5
3 2
2
Example 16
Example 17

Integration
Evaluate

+
−
dx
x
x
x 3
2
2
Solution
Separate the integrals as 
+
−
dx
x
x
x 3
2
2
thus
  
+
−
x
dx
dx
xdx 3
2
A
x
x
x
x
x
x
+
+
−
=
+
−
 ln
3
2
2
3
2 2
2
Find
 dx
x
x
sin
ln
cot
Solution
 dx
x
x
sin
ln
cot
Let x
u sin
ln
=
x
dx
du
cot
=
x
du
dx
cot
=
 





x
du
u
x
cot
cot
thus
 +
= A
u
u
du
ln but x
u sin
ln
=
A
x
dx
x
x
+
=
 sin
ln
ln
sin
ln
cot
EXERCISE 3
Evaluate the following integrals.
1.  





+
dx
x
x
cos
5
sin
2.  +
−
dx
x
x
x
x
sin
cos
sin
cos
3.  





+
dx
x
x
3
sin
1
3
cos
4.  





−
+
+
dx
x
x
x
1
3
3
2
2
Example 18
Example 19

Integration
5.  







dx
x
x
tan
3
sec2
6.  +
dx
x 3
2
1
7.
( )
 +
dx
x
x
4
3
2
8
8.  







+
dx
x
x
8
3
2
9.
 +
dx
x
x
sin
cos
1
10.  −
dx
x
x
3
11.  −
+
dx
x
x
3
2
12.  −
+
dx
x
x
2
4
3
2
13.  −
−
dx
x
x
2
1
3
14.  −
dx
x
x
2
2
1
15.  − 2
2
1 x
x
dx
16.  −1
2
2
x
x
dx
17. dx
x
x
 − 4
4
18.  +
dy
y
y
20
9
3
Standard integrals
1. A
n
x
dx
x
n
n
+
+
=
+
 1
1
2.
 +
−
= A
x
xdx cos
sin
3.
 +
= A
x
xdx sin
cos
4. A
x
x
xdx +
=
−
=
 sec
ln
cos
ln
tan
5.
 +
= A
x
xdx sin
ln
cot 6.
 +
+
= A
x
x
xdx tan
sec
ln
sec
7.
 +
−
= A
x
xdx cot
cosec 8.
 +
= A
x
xdx
x sec
tan
sec
9.
 +
= A
x
xdx tan
sec2
9. A
x
xdx +
−
=
 cot
cosec2
10.
 +
= A
e
dx
e x
x
11.
 +
= A
a
a
dx
a
x
x
ln
, 0

a
12.
 +
=
−
−
A
x
x
dx
)
(
cos 1
2
1
13.
 +
=
+
−
A
x
x
dx
)
(
tan 1
2
1

Integration
14.
 +
=
−
−
A
x
x
x
dx
)
(
sec 1
2
1
15.
 +
=
−
−
A
x
x
dx
)
(
tanh 1
2
1
16.
 +
= A
x
xdx sinh
cosh 17.
 +
= A
x
xdx cosh
sinh
18.
 +
= A
x
xdx cosh
ln
tanh 19. ( )
 +
=
+
−
A
x
dx
x
1
2
1
1
sinh
20. A
x
x
xdx +
−
=
 cot
cosec
ln
cosec
 dx
x
f )
(
sin
INTEGRAL OF TRIGONOMETRIC RATIOS OF THE FORM or  dx
x
f )
(
cos
or  dx
x
f )
(
tan
Integrate
 xdx
3
sin
Solution
From
 xdx
3
sin Let x
u 3
= ,
3
du
dx =
 





3
sin
du
u thus
 +
−
= A
u
udu cos
3
1
sin
3
1
But x
u 3
=
 +
−
=
 A
x
xdx 3
cos
3
1
3
sin
Find ( )
 + dx
x
x 2
tan
5
cos
Solution
( )
 + dx
x
x 2
tan
5
cos thus A
x
x
xdx
xdx +
−
=
+
  2
cos
ln
2
1
5
sin
5
1
2
tan
5
cos
Integrate
 xdx
x 5
cos
4
sin
Solution
Example 20
Example 21
Example 22

Integration
From
 xdx
x 5
cos
4
sin
Recall: ( ) ( ) B
A
B
A
B
A sin
cos
2
sin
sin =
−
−
+
( )
  −
= dx
x
x
xdx
x sin
9
sin
2
1
5
cos
4
sin
2
2
1
A
x
x
xdx
x +
+
−
=
 cos
2
1
9
cos
18
1
5
cos
4
sin
Evaluate the integral dx
x

7
cos
Solution
 
= xdx
x
xdx cos
cos
cos 6
7
thus ( )
 xdx
x cos
cos
3
2
( )
 − xdx
x cos
sin
1
3
2
Let x
u sin
= thus
xdx
du cos
= substituting ( )
 − du
u
3
2
1 then ( )
 −
+
− du
u
u
u 6
4
2
3
3
1
( ) A
u
u
u
u
u +
−
+
−
=
−

7
5
3
3
2
7
1
5
3
1 But x
u sin
=
Therefore A
x
x
x
xdx +
−
+
−
=

7
5
3
7
sin
7
1
sin
5
3
sin
sin
cos
INTEGRATION BY PART
Refer product rule of differentiation ( )
dx
du
v
dx
dv
u
uv
dx
d
+
=
Then vdu
udv
uv
d +
=
)
(
Integrating both sides ( )
  
+
= vdu
udv
uv
d thus  
+
= vdu
udv
uv
Therefore
 
−
= vdu
uv
udv
This is the formula of integrating by part
To get the function u consider the word ILATE where,
I – Inverse of trigonometric functions
L – Logarithmic functions
Example 23

Integration
A – Algebraic functions
T – Trigonometric functions (common and natural logarithm)
E – Exponential functions
The function that comes first is u any other function is dv.
Evaluate
 xdx
x cos
Solution
Consider ILATE, therefore, Algebra comes before Trigonometric
We let x
u =  dx
du= and xdx
dv cos
=

 = xdx
dv cos  x
v sin
=
 
−
= xdx
x
x
xdx
x sin
sin
cos thus  +
+
= A
x
x
x
xdx
x cos
sin
cos
Evaluate
 dx
e
x x
2
Solution
Let 2
x
u = xdx
du 2
=
 and

 = dx
e
dv x x
e
v =

 
−
= dx
xe
e
x
dx
e
x x
x
x
2
2
2
But
 dx
xex need by part integration
dx
e
xe
dx
xe x
x
x

 −
=
A
e
xe
dx
xe x
x
x
+
−
=
 thus for ( ) A
e
xe
e
x
dx
e
x x
x
x
x
+
−
−
=
 2
2
2
Therefore, A
e
xe
e
x
dx
e
x x
x
x
x
+
+
−
=
 2
2
2
2
Find
 xdx
ex
cos
Solution
Let xdx
du
x
u sin
cos −
=

= and dx
e
dv x
= thus
x
x
e
v
dx
e
dv =

= 

Example 24
Example 25
Example 26

Integration
Substituting

 +
= xdx
e
x
e
dx
x
e x
x
x
sin
cos
cos
dx
x
e
x
e
xdx
e x
x
x

 −
= cos
sin
sin
dx
x
e
x
e
x
e
dx
x
e x
x
x
x

 −
+
= cos
sin
cos
cos
A
x
e
x
e
xdx
e x
x
x
+
+
=
 sin
cos
cos
2
( ) A
x
x
e
xdx
e x
x
+
+
=
 sin
cos
2
1
cos
Evaluate dx
x
x
 ln
4
Solution
From dx
x
x
 ln
4
, Let dx
x
du
x
u
1
=

=ln and dx
x
dv 4
=
5
4
5
1
x
v
dx
x
dv =

= 
 thus 
 





−
= dx
x
x
x
x
dx
x
x
1
5
1
ln
5
1
ln 5
5
4
A
x
x
x
dx
x
x +
−
=

5
5
4
25
1
ln
5
1
ln
The tabular method or tic – tac – toe method
So far we saw the process of evaluating integrals by using by part
technique, but what will happen when you have the integral like
 dx
e
x x
4 ? This cause a repetition of a by part method. To avoid repetition
of by part process in integration, we introduce this method to simplify the
repeating of the by part integration process.
Suppose we want to evaluate
 dx
e
x x
4 if we use by part method we will
repeat the process 4 – times. To deal with this, we introduce the so called
tabular method or tic – tac – toe method.
Integrate
 dx
e
x x
4
Solution
Example 27
Example 28

Integration
From  dx
e
x x
4
Let 4
x
u = and dx
e
dv x
=
Consider the table below:-
u sign dv
4
x + x
e
3
4x – x
e
2
12x + x
e
x
24 – x
e
24 + x
e
0 x
e x
e
We multiply by following an arrow direction and the appropriate sign to
connect each product is shown. Therefore,
( ) A
e
x
x
x
x
dx
e
x x
x
+
+
−
+
−
=
 24
24
12
4 2
3
4
4
Integrate ( )dx
x
 ln
Solution
dx
x
ln Let dx
x
du
x
u
1
=

=ln and x
v
dx
dv =

=

 −
= dx
x
x
dx
x ln
ln therefore A
x
x
x
dx
x +
−
=
 ln
ln
Integrate

−
dx
x)
(
tan 1
Solution

−
dx
x)
(
tan 1 , Let 2
1
1 x
dx
du
x
u
+
=

= −
tan and x
v
dx
dv =

=
( ) 
 +
−
= −
−
2
1
1
1
tan
tan
x
xdx
x
x
dx
x therefore,
( ) ( )
 +
+
−
= −
−
A
x
x
x
dx
x 2
1
1
1
ln
2
1
tan
tan
Example 29
Example 30

Integration
EXERCISE 4
Evaluate the following integrals by using by part method
1.
 xdx
x sin 2.
 xdx
x cos
3
3.

−
xdx
3
sin 1
4.
 xdx
x sin
4
5.
 xdx
ln 6.  −
−
dx
x
x
x
2
1
1
sin
7.
 dx
e
x x
3 8.
 xdx
2
sin
9.  xdx
x 3
sin
10.  






 −
dx
x
x
2
1
sin
11. ( )
 dx
x
e x
sin
2
12.
 







+
−
−
−
−
−
dx
x
x
x
x
1
1
1
1
cos
sin
cos
sin
13. ( )
 + xdx
x ln
2
2
14.
( )
 +
dx
x
x
2
1
ln
15.  +
−
dx
x
x
x
2
1
2
1
tan 16.
 dx
e
x x
3
2
17.
 xdx
x ln
2
18.

−
xdx
e x
cos
19. ( )
 dx
x
x 2
3
ln
20.
( )
 −
−
dx
x
x
x
2
3
2
1
2
1
sin

Integration
Integration by partial fractions
Case 1: When the denominator of the rational function contain linear
factors, thus;
( )( ) q
kx
Z
d
cx
B
b
ax
A
q
kx
d
cx
b
ax
x
f
+
+
+
+
+
+
=
+
+
+
+
+
...
)
(
...
)
(
where A, B,
Z, a, b, c, d, k, q are constants.
Integrate
( )( )
 +
−
+
dx
x
x
x
1
2
2
3
Solution
From
( )( )
 +
−
+
dx
x
x
x
1
2
2
3
Decompose the rational function into small rational fractions
( )( ) 1
2
2
1
2
2
3
+
+
−
=
+
−
+
x
B
x
A
x
x
x
Find the values of A and B
( )( ) 1
2
2
1
2
2
3
+
+
−
=
+
−
+
x
B
x
A
x
x
x
By comparison; ( ) ( )
2
1
2
3 −
+
+
=
+ x
B
x
A
x
When 2
=
x then 1
=
A and when 2
1
−
=
x then 1
−
=
B
( )( ) ( )
1
2
1
2
1
1
2
2
3
+
−
−
=
+
−
+
x
x
x
x
x
( )( ) ( )
 
 +
−
−
=
+
−
+
dx
x
dx
x
dx
x
x
x
1
2
1
2
1
1
2
2
3
Therefore
( )( )
A
x
x
dx
x
x
x
+
+
−
−
=
+
−
+
 1
2
ln
2
1
2
ln
1
2
2
3
Evaluate
( )( )
 +
+
−
dx
x
x
x
1
2
1
3
4
Solution
From
( )( )
 +
+
−
dx
x
x
x
1
2
1
3
4
Example 31
Example 32

Integration
Let
( )( ) 1
2
1
3
1
2
1
3
4
+
+
+
=
+
+
−
x
B
x
A
x
x
x
then
( ) ( )
1
3
1
2
4 +
+
+
=
− x
B
x
A
x
Compare the coefficients,



−
=
+
=
+
4
1
3
2
B
A
B
A
Solve for A and B, using a calculator, 13
−
=
A and 9
=
B
( )( ) 1
2
9
1
3
13
1
2
1
3
4
2
+
+
+
−
=
+
+
−
x
x
x
x
x
x
( )( )
   +
+
+
−
=
+
+
−
dx
x
dx
x
dx
x
x
x
x
1
2
9
1
3
13
1
2
1
3
4
2
therefore
( )( )
 +
+
+
+
−
=
+
+
−
A
x
x
dx
x
x
x
x
1
2
ln
2
9
1
3
ln
3
13
1
2
1
3
4
2
Solve  −
+
dx
x
x
4
3
2
Solution
From  −
+
dx
x
x
4
3
2
let
( )( )
2
2
3
4
3
2 +
−
+
=
−
+
x
x
x
x
x
Writing into partial fractions as
( )( ) 2
2
2
2
3
+
+
−
=
+
−
+
x
B
x
A
x
x
x
Thus ( ) ( )
2
2
3 −
+
+
=
+ x
B
x
A
x , when 2
=
x then
4
5
=
A and when
2
−
=
x then
4
1
−
=
B thus,
( )( ) ( ) ( )
2
4
1
2
4
5
2
2
3
+
−
−
=
+
−
+
x
x
x
x
x
 
 +
−
−
=
−
+
dx
x
dx
x
dx
x
x
2
1
4
1
2
1
4
5
4
3
2
therefore
A
x
x
dx
x
x
+
+
+
−
=
−
+
 2
ln
4
1
2
ln
4
5
4
3
2
Find  −
−
−
dx
x
x
x
2
5
3
7
2
Example 33
Example 34

Integration
Solution
Partialize
2
5
3
7
2
−
−
−
x
x
x
by factorize the denominator (if possible)
( )( )
2
1
3
2
5
3 2
−
+
=
−
− x
x
x
x
( )( )
2
1
3
7
2
5
3
7
2 −
+
−
=
−
−
−
x
x
x
x
x
x
( )( ) 2
1
3
2
1
3
7
−
+
+
=
−
+
−
x
B
x
A
x
x
x
thus B
Bx
A
Ax
x +
+
−
=
− 3
2
7
Compare coefficients and solve for A and B



−
=
+
−
=
+
7
2
1
3
B
A
B
A
7
22
=
A and 7
5
−
=
B thus
( )( ) ( ) ( )
2
7
5
1
3
7
22
2
1
3
7
−
−
+
=
−
+
−
x
x
x
x
x
dx
x
dx
x
dx
x
x
x


 −
−
+
=
−
−
−
2
1
7
5
1
3
1
7
22
2
5
3
7
2
A
x
x
dx
x
x
x
+
−
−
+
=
−
−
−
 2
ln
7
5
1
3
ln
21
22
2
5
3
7
2
EXERCISE 5
By partial fraction, integrate the following
1.
( )( )
 +
−
−
dx
x
x
x
2
3
2
1
2
2.
( )( )
 +
+
dx
x
x
x
2
1
3.
( )( )
 −
−
+
dx
x
x
x
2
3
2
2
4.  − 2
2
8 x
dx
5.
( )( )
 −
+
dx
x
x
x
3
3
2
1
2
6.
( )( )
 −
−
dx
x
x
x
sin
2
sin
1
cos
7.
( ) ( )

 −
=
− 1
1 4
4
3
4
x
x
dx
x
x
x
dx 8.
( )( )( )
 −
+
−
dx
x
x
x
x
3
1
2
9.
( )( )
 +
− 1
1 x
x
x
dx
10.  +
dx
x
x 2
sin
sin
1

Integration
Case 2: When the denominator contain quadratic factor(s)
( )( ) ( ) q
px
E
d
cx
D
Cx
b
ax
B
Ax
q
px
d
cx
b
ax
x
f
+
+
+
+
+
+
+
+
=
+
+
+
+
+
...
...
)
(
2
2
2
2
where
A, B, C, D, E, a, b, c, d, p, q are constants.
Evaluate
( )( )
 +
−
+
dx
x
x
x
2
3
2
2
Solution
Partialize
( )( ) 2
3
2
3
2
2
2
+
+
+
−
=
+
−
+
x
C
Bx
x
A
x
x
x
( ) ( )( )
3
2
2 2
−
+
+
+
=
+ x
C
Bx
x
A
x
C
Cx
Bx
Bx
A
Ax
x 3
3
2
2 2
2
−
+
−
+
+
=
+
Compare the coefficients





=
+
=
−
=
+
−
0
2
3
2
1
3
B
A
C
A
C
B
Then,
11
5
=
A ,
11
5
−
=
B and
11
4
−
=
C thus
( )( ) 2
3
2
3
2
2
11
4
11
5
11
5
2
+
+
−
−
=
+
−
+
x
x
x
x
x
x
Therefore
( )( )
   +
+
−
−
=
+
−
+
dx
x
x
dx
x
dx
x
x
x
2
4
5
11
1
3
1
11
5
2
3
2
2
2
( )( )
 +








−
+
−
=
+
−
+ −
A
x
x
x
dx
x
x
x
2
tan
11
2
2
2
3
ln
11
5
2
3
2 1
2
2
Example 35
Example 36

Integration
Evaluate
( )
 +
+
dx
x
x
x
1
3
2
Solution
Partialize
( ) 1
1
3
2
2
2
+
+
+
=
+
+
x
C
Bx
x
A
x
x
x
( ) ( )
C
Bx
x
x
A
x +
+
+
=
+ 1
3 2
2
Cx
Bx
A
Ax
x +
+
+
=
+ 2
2
2
3 then







=
−
=
=
=
+
0
2
3
1
C
B
A
B
A
( )
 
 +
−
=
+
+
dx
x
x
dx
x
dx
x
x
x
1
2
3
1
3
2
2
2
( )
 +
+
−
=
+
+
A
x
x
dx
x
x
x
1
ln
ln
3
1
3 2
2
2
( )
 +
+
=
+
+
A
x
x
dx
x
x
x
1
ln
1
3
2
3
2
2
Evaluate
 −
+
−
+
−
dx
x
x
x
x
x
9
18
2
3
4
2
3
2
Solution
( )( )
1
2
9
3
4
9
18
2
3
4
2
2
2
3
2
−
+
+
−
=
−
+
−
+
−
x
x
x
x
x
x
x
x
x
Partialize
( )( ) 1
2
9
1
2
9
3
4
2
2
2
−
+
+
+
=
−
+
+
−
x
C
x
B
Ax
x
x
x
x
( )( ) ( )
9
1
2
3
4 2
2
+
+
−
+
=
+
− x
C
x
B
Ax
x
x
C
Cx
B
Bx
Ax
Ax
x
x 9
2
2
3
4 2
2
2
+
+
−
+
−
=
+
−
Example 37

Integration
Compare the coefficients





=
+
−
−
=
+
−
=
+
3
9
4
2
1
2
C
B
B
A
C
A
Then,
37
16
=
A ,
37
66
−
=
B and
37
5
=
C
  −
+
+
−
dx
x
dx
x
x
1
2
1
37
5
9
66
16
37
1
2
   −
+
+
−
+
dx
x
dx
x
dx
x
x
1
2
1
37
5
9
1
37
66
9
37
16
2
2
A
x
x
x +
−
+






−
+
= −
1
2
ln
74
5
3
tan
37
22
9
ln
37
8 1
2
EXERCISE 6
Evaluate the following integrals
1.
( )( )
 +
+
dx
x
x
x
3
1 2
2.
( )
 +1
6
x
x
dx
3.  







−
+
−
dx
x
x
x
x
1
2
3
4.  −1
4
x
dx
5.
( )( )
 −
+
−
dx
x
x
x
4
1
3
2
6.
( )
 +
dx
x
x 1
1
2
7.
( )( )
 −
+
+
dx
x
x
x
2
2
5
2
8.
( )( )
 −
+
dx
x
x
x
1
3
1
2
9.
( )( )
 −
+ 4
2
1 2
x
x
dx
10.
( )
 + 2
2
x
x
dx

Integration
Case 3: When the denominator contains the repeated root(s).
( ) ( ) ( ) ( ) ( ) q
px
D
b
ax
C
b
ax
B
b
ax
A
q
px
b
ax
x
f
r
n
n
n +
+
+
+
+
+
+
+
=
+
+ −
−
...
...
)
(
1
where A, B, C, D, a, b, p, q are constants and n is a positive integer, where
r is less than or equal to n.
Evaluate
( )
 +
+
dx
x
x
x
2
2
3
1
Solution
Partialize
( ) ( )2
2
2
2
3
2
3
1
+
+
+
+
=
+
+
x
C
x
B
x
A
x
x
x
( ) ( )2
2
2
2
3
2
3
1
+
+
+
+
=
+
+
x
C
x
B
x
A
x
x
x
( ) ( ) Cx
x
Bx
x
A
x 3
2
3
2
1 2
+
+
+
+
=
+
When 0
=
x then 4
1
=
A , when 2
−
=
x then 6
1
=
C and when 1
=
x
then 12
1
−
=
B
( ) ( )2
2
2
6
1
)
2
(
12
1
12
1
2
3
1
+
+
+
−
=
+
+
x
x
x
x
x
x
( ) ( )



 +
+
+
−
=
+
+
dx
x
dx
x
dx
x
dx
x
x
x
2
2
2
1
6
1
2
1
12
1
1
12
1
2
3
1
Therefore,
( ) ( )
A
x
x
x
dx
x
x
x
+
+
−
+
−
=
+
+
 2
6
1
2
ln
12
1
ln
12
1
2
3
1
2
Evaluate
( )( )
 −
+
−
dx
x
x
x
2
1
2
5
7
3
Solution
Partialize
( )( ) ( )2
2
1
2
1
2
5
1
2
5
7
3
−
+
−
+
+
=
−
+
−
x
C
x
B
x
A
x
x
x
( )( ) ( )2
2
1
2
1
2
5
1
2
5
7
3
−
+
−
+
+
=
−
+
−
x
C
x
B
x
A
x
x
x
( ) ( )( ) ( )
5
1
2
5
1
2
7
3 2
+
+
−
+
+
−
=
− x
C
x
x
B
x
A
x
Example 38
Example 39

Integration





−
=
−
=
+
−
=
−
=
+
+
=
−
=
+
−
1
when
10
4
12
9
1
when
4
6
6
0
when
7
5
5
x
C
B
A
x
C
B
A
x
C
B
A
thus
11
2
−
=
A ,
11
4
=
B and 1
−
=
C
( )( ) ( ) ( ) ( )2
2
1
2
1
1
2
11
4
5
11
2
1
2
5
7
3
−
−
−
+
+
−
=
−
+
−
x
x
x
x
x
x
Then
( ) ( ) ( )


 −
−
−
+
+
− dx
x
dx
x
dx
x 2
1
2
1
1
2
1
11
4
5
1
11
2
A
x
x
x +






−
+
−
+
+
−
=
1
2
1
2
1
1
2
ln
11
2
5
ln
11
2
Therefore
( )( )
A
x
x
x
x
x
x
+
−
+
+
−
=
−
+
−
 2
4
1
5
1
2
ln
11
2
1
2
5
7
3
2
EXERCISE 7
1.
( )
 +1
2
x
x
dx
2.
( )
 + 2
1
x
x
dx
3.
( )( )
dx
x
x
x
 +
+
+
2
3
1
2
4.
( ) ( )
dx
x
x
x
 −
+ 1
2
1
3 2
5.
( )( )
dx
x
x
x
 −
−
+
3
1
2
3
6.
( )( )
dx
x
x
x
x
 −
+
+
+
2
2
1
3
3
2
3
7.
( )
 −
−
dx
x
x
3
2
1
3
8.
( ) ( )
 +
−
+
−
dx
x
x
x
x
1
5
3
1
3
2
2
2
9.
( ) ( )
 +
−
dx
x
x
x
2
2
1
1
4
10.
( )( )( )
 −
+
−
+
dx
x
x
x
x
1
3
1
2
4
2
Case 4: When the denominator has no real factors or cannot be
factorized easily k
n
r
n
n
n
dx
cx
bx
ax
x
f
−
−
−
+
+
+ 2
)
(
Find (a)  +
+
dx
x
x 2
2
4
2
(b)  +
+ 3
11
2 2
x
x
dx
Solution
Example 40

Integration
(a)  +
+
dx
x
x 2
2
4
2
Consider ( ) 1
1
2
2 2
2
+
+
=
+
+ x
x
x therefore,
( ) 1
1
4
2
2
4
2
2
+
+
=
+
+ x
x
x
( )
  +
+
=
+
+
dx
x
dx
x
x 1
1
4
2
2
4
2
2
thus
( ) ( )

 +
+
=
+
+
dx
x
dx
x 1
1
1
4
1
1
4
2
2
Let 
tan
1 =
+
x
Hence 
d
dx 2
sec
= thus
 +
=
+
A
d




4
1
tan
sec
4 2
2
Therefore, ( )
 +
+
=
+
+
−
A
x
dx
x
x
1
tan
4
2
2
4 1
2
(b)  +
+ 3
11
2 2
x
x
dx
The expression
8
97
4
11
2
3
11
2
2
2
−






+
=
+
+ x
x
x
therefore,  
−






+
=
+
+
8
97
4
11
2
3
11
2 2
2
x
dx
x
x
dx
 












+
−
−
=






+
−
− 2
2
4
11
97
4
1
97
8
4
11
97
16
1
97
8
x
dx
x
dx
(See the hyperbolic substitution section)
Let 





+
=
4
11
97
4
tanh x
 then
4
11
tanh
4
97
+
= x
 hence
dx
d =


2
sech
4
97
Substituting 
 
−
=
−
− 



d
d
4
97
97
8
tanh
1
sech
4
97
97
8
2
2
C
+
− 
97
2
replacing  , then C
x
+





 +
− −
97
11
4
tanh
97
2 1

Integration
Therefore,  +





 +
−
=
+
+
−
A
x
x
x
dx
97
11
4
tanh
97
2
3
11
2
1
2
EXERCISE 8
1.  +
− 3
2 2
x
x
dx
2.  +
+ 1
7
5 2
x
x
dx
3.  +8
2 2
x
dx
4.  +
+ 1
2
2 2
x
x
dx
5.
 +
− 1
2
2
x
x
dx
6.
 +
− 5
2
2
x
x
dx
7.  +
+ 11
4
2 2
x
x
dx
8.  +
− 7
8
4 2
x
x
dx
9. dx
x
x
 −
+ 2
3
2
2
10. dx
x
x
 +
− 2
3
3
5
2
Case 5: The use of ( ) B
D
dx
d
A
N +
= , where N is Numerator, D is
denominator, A and B are constants,
)
(
)
(
x
g
x
f
then B
dx
dg
A
x
f +
=
)
(
Evaluate  +
+
−
dx
x
x
x
5
3
3
2
2
Solution
From
5
3
3
2
2
+
+
−
x
x
x
Let ( ) B
D
dx
d
A
N +
= then ( ) B
x
A
x +
+
=
− 3
2
3
2
B
A
Ax
x +
+
=
− 3
2
3
2 hence 1
=
A and 6
−
=
B
6
3
2
3
2 −
+
=
− x
x
5
3
6
3
2
5
3
3
2
2
2
+
+
−
+
=
+
+
−
x
x
x
x
x
x
Example 41

Integration
5
3
6
5
3
3
2
5
3
6
3
2
2
2
2
+
+
−
+
+
+
=
+
+
−
+
x
x
x
x
x
x
x
x
 
 +
+
−
+
+
+
=
+
+
−
+
dx
x
x
x
x
x
dx
x
x
x
5
3
6
5
3
3
2
5
3
6
3
2
2
2
2
A
x
x
x
dx
x
x
x
+







 +
−
+
+
=
+
+
− −
 11
3
2
tan
11
12
5
3
ln
5
3
3
2 1
2
2
Evaluate
 +
−
−
+
dx
x
x
x
x
8
6
3
3
2
2
Solution
The rational function involved in this integral is an improper rational
function. So divide first to lower the numerator degree.
Use long division method,
1
11
9
8
6
3
3
8
6
2
2
2
−
+
−
−
+
+
−
x
x
x
x
x
x
x
8
6
11
9
1 2
+
−
−
+
x
x
x
  +
−
−
+ dx
x
x
x
dx
8
6
11
9
2
Now Partialize,
( )( )
4
2
11
9
8
6
11
9
2 −
−
−
=
+
−
−
x
x
x
x
x
x
( )( ) 4
2
4
2
11
9
−
+
−
=
−
−
−
x
B
x
A
x
x
x
( ) ( )
2
4
11
9 −
+
−
=
− x
B
x
A
x
When 4
=
x ,
2
25
=
B and when 2
=
x ,
2
7
−
=
A
Example 42

Integration
   −
−
−
+
2
2
7
3
2
25
x
dx
x
dx
dx
Therefore, ( ) ( ) A
x
x
x
dx
x
x
x
x
+
−
−
−
+
=
+
−
−
+
 2
ln
2
7
4
ln
2
25
8
6
3
3
2
2
Evaluate
 +
+
dx
x
x
5
2
2
4
Solution
From
 +
+
dx
x
x
2
5
2
4
Dividing by long division
2
9
4
2
5
2
2
5
0
2
2
2
2
2
4
2
4
2
−
−
−
+
−
+
−
+
+
+
x
x
x
x
x
x
x
x
2
9
2
2
5
2
2
4
+
+
−
=
+
+
x
x
x
x
( )
   +
+
−
=
+
+
dx
x
dx
x
dx
x
x
2
9
2
2
5
2
2
2
4
 +








+
−
=
+
+ −
A
x
x
x
dx
x
x
2
tan
2
9
2
3
2
5 1
3
2
4
EXERCISE 9
1.  +
+
−
dx
x
x
x
2
3
2
1
2
2
2.  +
+
+
dx
x
x
x
1
4
2
Example 43

Integration
3.  +
+
+
dx
x
x
x
2
2
2
3
2
4.  −
+
+
dx
x
x
x
7
4
5
2
2
5.  +
+
dx
x
x
4
5
2
2
6.  +
+
dx
x
x
6
2
3
2
2
7.
( )
 +
+
dx
x
x
x
4
2
2
8.  +
−
+
dx
x
x
x
1
2
3
2
9.  +
+
−
dx
x
x
x
4
3
3
5
2
10.  −
−
+
dx
x
x
x
3
2
1
2
11.  +
−
−
dx
x
x
x
3
1
2
12.  +
+
−
dx
x
x
x
1
9
3
2
2
INTEGRATION OF TRIGONOMETRIC FUNCTIONS
Case 6: The use of ( ) BD
D
dx
d
N +
= (for trigonometric) where N is
Numerator, D is denominator and A and B are constants
 +
+
dx
x
b
c
x
b
c
x
b
c
x
b
c
3
4
3
3
1
2
1
1
cos
sin
cos
sin
Evaluate dx
x
x
x
 − sin
2
cos
sin
Solution
From dx
x
x
x
 − sin
2
cos
sin
Let ( ) BD
D
dx
d
N +
=
( ) ( )
x
x
B
x
x
dx
d
A
x sin
2
cos
sin
2
cos
sin −
+
−
=
x
B
x
B
x
A
x
A
x sin
2
cos
cos
2
sin
sin −
+
−
−
=
Compare



=
+
−
=
−
−
0
2
1
2
B
A
B
A
Solve for A and B,
5
1
−
=
A and
5
2
−
=
B
Example 44

Integration
( ) ( )
 
 −
−
−
−
+
=
−
dx
x
x
x
x
dx
x
x
x
x
dx
x
x
x
sin
2
cos
sin
2
cos
5
2
sin
2
cos
cos
2
sin
5
1
sin
2
cos
sin
Consider
 −
+
dx
x
x
x
x
sin
2
cos
cos
2
sin
5
1
, Let x
x
u sin
2
cos −
=
( )dx
x
x
du cos
2
sin +
−
=
 −
= u
u
du
ln
5
1
5
1
, But x
x
u sin
2
cos −
=
x
x
dx
x
x
x
x
sin
2
cos
ln
5
1
sin
2
cos
cos
2
sin
5
1
−
−
=
−
+

Also consider
x
dx
dx
x
x
x
x
5
2
5
2
sin
2
cos
sin
2
cos
5
2
−
=
−
=
−
−
 
Therefore, A
x
x
x
dx
x
x
x
+
−
−
−
=
−
 5
2
sin
2
cos
ln
5
1
sin
2
cos
sin
Evaluate dx
x
x
x
 + sin
cos
3
cos
Solution
From dx
x
x
x
 + sin
cos
3
cos
( ) ( )
x
x
B
x
x
dx
d
A
x sin
cos
3
sin
cos
3
cos +
+
+
=
x
B
x
B
x
A
x
A
x sin
cos
3
sin
3
cos
cos +
+
−
=
Compare coefficients



=
+
−
=
+
0
3
1
3
B
A
B
A
Solve for A and B then
10
1
=
A and
10
3
=
B
 
 +
+
+
+
−
=
+
dx
x
x
x
x
dx
x
x
x
x
dx
x
x
x
sin
cos
3
sin
cos
3
10
3
sin
cos
3
sin
3
cos
10
1
sin
cos
3
cos
 +
+
+
=
+
A
x
x
x
dx
x
x
x
10
3
sin
cos
3
ln
10
1
sin
cos
3
cos
Example 45

Integration
EXERCISE 10
1.
 +
dx
x
x
x
2
sin
2
cos
2
sin
2.
 −
+
dx
x
x
x
x
cos
5
sin
7
cos
sin
2
3.
 +
+
dx
x
x
x
x
cos
4
sin
3
cos
3
sin
2
4.  +
dx
x
x
x
cos
4
sin
3
cos
3
5.  dx
x
x
cos
sin
5
6. dx
x
x
 + 5
cos
2
sin
4
7.  +
dx
x
x
x
sin
cos
sin
8.  +
dx
x
x
x
sin
4
cos
cos
3
9.  +
−
dx
x
x
x
x
sin
cos
sin
cos
10.  dx
x
x
cos
tan
(a)  + x
b
a
dx
cos
(b)  + x
b
a
dx
sin
(c)  + x
b
x
a
dx
sin
cos
Use of t formula
Refer, 2
2
1
1
cos
t
t
x
+
−
= and 2
1
2
sin
t
t
x
+
= where 





=
2
tan
x
t
Solve  + x
dx
cos
5
3
Solution
 + x
dx
cos
5
3
Let 





=
2
tan
x
t then 





=
2
sec
2
1 2 x
dx
dt
Thus dx
x
dt
=






2
sec
2
2
Example 46

Integration
But 2
2
1
1
cos
t
t
x
+
−
=









+
−
+
+
2
2
2
1
1
5
3
1
2
t
t
t
dt
then
( ) ( ) 
 −
=
−
+
+ 2
2
2
2
8
2
1
5
1
3
2
t
dt
t
t
dt
( )

 −
=
− 2
2
2
1
4
1
4 t
dt
t
dt
, Let 
tanh
2
=
t
thus 
d
dt 2
sech
2
=
 +
=
−
A
d
se




2
1
tanh
1
ch
2
4
1
2
2
But 





= −
2
tanh 1 t
 and 





=
2
tan
x
t
A
x
x
dx
+














=
+
−
 2
tan
2
1
tanh
cos
5
3
1
Solve  − x
dx
cos
3
5
Solution
From  − x
dx
cos
3
5
Let 





=
2
tan
x
t then






=
2
sec
2
2 x
dt
dx therefore
2
1
2
t
dt
dx
+
=

 +
=








+
−
−
+
2
2
2
2
8
2
2
1
1
3
5
1
2
t
dt
t
t
t
dt
thus
( )

 +
=
+ 2
2
2
1
8
2
2
t
dt
t
dt
Example 47

Integration


 d
dt
t 2
sec
2
1
tan
2 =

= therefore
( ) 
 +
=
+
=
+
A
d
t
dt




2
1
tan
1
sec
2
1
2
1 2
2
2
But ( )
t
2
tan 1
−
=
 and






=
2
tan
x
t thus A
x
x
dx
+














=
−
−
 2
tan
2
tan
2
1
cos
3
5
1
Integrate  + x
dx
sin
1
Solution
From  + x
dx
sin
1
, Let 





=
2
tan
x
t then 





=
2
sec
2
1 2 x
dx
dt
thus
2
2 1
2
2
sec
2
t
dt
x
dt
dx
+
=






= substituting 
 +
+
=
+
+
+ dt
t
t
dt
t
t
t
2
2
2
2
1
2
1
2
1
1
2
( )
dt
t
dt
t
t 
 +
=
+
+ 2
2
1
2
2
1
2
Let 1
+
= t
u then dt
du=
 +
−
= A
u
u
du 2
2
2
but 1
+
= t
u
( )
 +
+
−
=
+
A
t
t
dt
1
2
1
2
2
but 





=
2
tan
x
t
 +
+






−
=
+
A
x
x
dx
1
2
tan
2
sin
1
Evaluate  +6
sin
4
cos
3 x
x
dx
Solution
Example 48
Example 49

Integration
From  +6
sin
4
cos
3 x
x
dx
, Let dx
x
dt
x
t 





=







=
2
sec
2
1
2
tan 2
then

+






+
+








+
−






+
6
1
2
4
1
1
3
2
tan
1
2
2
2
2
2
t
t
t
t
x
dt

+
+
+
+
−
+
2
2
2
2
1
6
6
8
3
3
1
2
t
t
t
t
t
dt
 +
+ 9
8
3
2
2
t
t
dt
 +
+ 3
3
8
3
2
2
t
t
dt

+






+
9
11
3
4
3
2
2
t
dt
A
t
+







 +
−
11
4
3
tan
11
2 1
But 





=
2
tan
x
t
( )
A
x
+







 +
−
11
4
tan
3
tan
11
2 2
1
Evaluate
( )
 + 2
cos
sin
2 x
x
dx
Solution
From
( )
 + 2
cos
sin
2 x
x
dx
Divide by x
cos
( )
 + 2
1
tan
2
sec
x
xdx
thus,
( )
1
tan
2
2
1
+
−
x
Example 50

Integration
EXERCISE 11
1.  +
− 5
sin
4
cos
3 x
x
dx
2.  + x
dx
sin
5
4
3.  − x
x
dx
cos
sin
1
4.  +
− 13
sin
5
cos
12 x
x
dx
5.  +
+ 3
cos
sin
2 x
x
dx
6.  + x
dx
tan
3
2
7.  +
dx
x
x
x
sin
3
cos
2
sin
8.  +
+
dx
x
x
x
x
cos
sin
2
cos
sin
9.  +
+
dx
x
x
x
sin
cos
1
cos
10.  +
dx
x
2
sin
2
1
R– FORMULA IN INTEGRATION
( ) 




 sin
sin
cos
cos
cos r
r
r 
=
 or
( ) 




 sin
cos
cos
sin
sin r
r
r 
=

Evaluate  + x
x
dx
sin
4
cos
3
Solution
 + x
x
dx
sin
4
cos
3
Let ( )

+
=
+ x
r
x
x sin
sin
4
cos
3

 sin
cos
cos
sin
sin
4
cos
3 x
r
x
r
x
x +
=
+
x
x
r cos
3
sin
cos =

3
sin =

r (1)
x
x
r sin
4
cos
sin =

4
cos =

r (2)
Divide equation (1) by equation (2)






= −
4
3
tan 1

Square equation (1) and (2) and sum then
5
4
3 2
2
=
+
=
r
Example 51

Integration














+
=
+ −
4
3
tan
sin
5
sin
4
cos
3 1
x
x
x















+ −
4
3
tan
sin
5 1
x
dx
 













+ −
dx
x
4
3
tan
cosec
5
1 1
A
x
x +














+
−














+ −
−
4
3
tan
cot
4
3
tan
cosec
ln
5
1 1
1
EXERCISE 12
1.  − x
x
dx
sin
12
cos
5
2.  + x
x
dx
sin
cos
3.  + x
x
dx
sin
7
cos
24
4.
( )
 − x
x
dx
cos
sin
2
5.  + 3
tan
sec2
x
xdx 6.
 +
dx
x
x
4
cot
cos
7.  −
+
dx
x
x
x
x
sin
cos
cos
2
sin
8.  +
dx
x
sin
3
4
9.  +
dx
x
x 2
cos
3
2
sin
1
10.  +
dx
x
x
4
tan
sec
Integral of the form  xdx
x n
m
sin
cos
Evaluate  xdx
2
sin
Solution
Example 52

Integration
Given
 xdx
2
sin from x
x
x 2
2
sin
cos
2
cos −
= thus
( )
x
x 2
cos
1
2
1
sin2
−
=
( )

 −
= dx
x
dx
x 2
cos
1
2
1
sin2
therefore A
x
xdx +
−
=
 2
sin
4
1
2
1
sin2
Integrate dx
x
x
 sin
cos3
Solution
Let
x
du
dx
x
u
sin
cos −
=

=
A
u
du
u
x
du
x
u +
−
=
−
=
− 

4
3
3
4
1
sin
sin But x
u cos
= then
A
x
xdx
x +
−
=

4
3
cos
4
1
sin
cos
Evaluate dx
x
x

2
3
sin
cos
Solution

 = xdx
x
x
dx
x
x 2
2
2
3
sin
cos
cos
sin
cos
( )

 −
= xdx
x
x
xdx
x
x 2
2
2
2
sin
sin
1
cos
sin
cos
cos
Let xdx
du
x
u cos
sin =

=
( ) A
u
u
du
u
u +
−
=
−

5
3
2
2
5
1
3
1
1 But x
u sin
=
A
x
x
dx
x
x +
−
=

5
3
2
3
sin
5
1
sin
3
1
sin
cos
Example 53
Example 54

Integration
EXERCISE 13
1. dx
x
x

2
5
cos
sin 2. dx
x
x

4
3
cos
sin
3. dx
x

7
sin 4. xdx
x 5
8
cos
sin

5. Show that
 +
+
+
= A
x
x
x
x
xdx tan
sec
ln
2
1
tan
sec
2
1
sec3
6.
 xdx
x 7
5
sin
cos 7.
 xdx
11
cos
8. dx
x
x
 tan
sec4
9.  xdx
x 2
7
sin
cos
10. Show that
4
1
16
π
sin
2
0
2
2
π
+
=
 dx
x
x
11.  xdx
x 7
3
cos
sin 12.  xdx
4
cos
Integrals of the form  +
dx
x
b
x
a 2
2
sin
cos
1
Integrate  + x
x
dx
2
2
sin
5
cos
3
Solution
From  + x
x
dx
2
2
sin
5
cos
3
Divide throughout by x
2
cos
 +
dx
x
x
2
2
tan
5
3
sec
Let x
u tan
= xdx
du 2
sec
=

  +
=
+ 2
2
3
5
1
3
1
5
3 u
du
u
du
Let 
tan
3
5
=
u

d
du 2
sec
3
5
= therefore 
d
du 2
sec
5
3
=
Example 55

Integration
 
=
+
= 



d
d
5
3
3
tan
1
sec
5
3
3
1
2
2
But








= −
u
3
5
tan 1

A
u +








−
3
5
5
3
3 1
tan But x
u tan
=
A
x +








−
tan
3
15
tan
15
15 1
Therefore, A
x
x
x
dx
+








=
+
−
 tan
3
15
tan
15
15
sin
5
cos
3
1
2
2
Solve  − x
dx
2
cos
1
Solution
From  − x
dx
2
cos
1
x
x
x 2
2
sin
cos
2
cos −
=
  
=
= xdx
x
dx
x
dx 2
2
2
cosec
2
1
sin
2
1
sin
2
A
x
xdx
x
dx
+
−
=
=
−
 
2
2
2
cot
2
1
cosec
2
1
cos
1
 +
−
=
−
A
x
x
dx 2
2
cot
2
1
cos
1
Evaluate  +
dx
x
x
2
sin
4
cos
Solution
From  +
dx
x
x
2
sin
4
cos
Let x
u sin
= thus xdx
du cos
=
Example 56
Example 57

Integration
 






+
=
+ 2
2
2
1
4
1
4 u
du
u
du
A
x
u
u
du
+






=






=
+
−
−
 sin
2
1
tan
2
1
2
tan
2
1
4
1
1
2
Therefore,  +






=
+
−
A
x
dx
x
x
sin
2
1
tan
2
1
sin
4
cos 1
2
Evaluate  +
dx
x
x
2
cos
1
sin
Solution
From  +
dx
x
x
2
cos
1
sin

 +
=
+
dx
x
x
x
dx
x
x
x
2
2
2
tan
2
tan
sec
sin
cos
2
sin
 +
dx
x
x
x
2
sec
1
tan
sec
Let x
u sec
= xdx
x
du tan
sec
=

( )
 +
=
+
−
A
u
u
du 1
2
tan
1
But x
u sec
=
( ) A
x
dx
x
x
+
=
+
−
 sec
tan
cos
1
sin 1
2
Example 58

Integration
EXERCISE 14
1.  + x
x
dx
2
2
sin
4
cos
2.  +3
2
cos x
dx
3. dx
x
x
 + 2
2
sin
cos
9
1
4. dx
x
x
 + sin
cos
2
1
5. dx
x
x
 − cos
3
sin
1
2
6.  −
dx
x
x
x
sin
cos
sin
7.  +
dx
x
x
x
cos
sin
1
tan
8.  +
dx
x
2
cos
2
1
9.  −
dx
x
x sec
cos
4
2
10.  −
+
dx
x
x
x
x
sin
cos
sin
cos
Integrals of the form ( ) dx
r
qx
px
b
ax
 +
+
+ 2
Use ( ) B
r
qx
px
dx
d
A
b
ax +
+
+
=
+ 2
if substitution technique is not
working.
Evaluate ( )
 +
+
+ dx
x
x
x 1
2
2
2 2
Solution
Let ( ) B
x
x
dx
d
A
x +
+
+
=
+ 1
2
2
2 2
( ) B
x
A
x +
+
=
+ 2
4
2



=
=
+
1
4
2
2
A
B
A
then
4
1
=
A and
2
3
=
B thus ( )
2
3
2
4
4
1
2 +
+
=
+ x
x
( ) 
 +
+
+
+
+
+ dx
x
x
dx
x
x
x 1
2
2
2
3
1
2
2
2
4
4
1 2
2
Consider ( )
 +
+
+ dx
x
x
x 1
2
2
2
4
4
1 2
, Let 1
2
2 2
+
+
= x
x
u 
( )dx
x
udu 2
4
2 +
= thus  du
u2
4
2








=
3
2
1 3
u
Example 59

Integration
But 1
2
2 2
+
+
= x
x
u thus
3
2
1
2
2
6
1





 +
+ x
x
Consider ( )
 +
+ dx
x
x 1
2
2
3 2
 





+






+ dx
x
2
2
2
1
2
1
2
2
3
dx
x
 







+






+






 1
2
1
2
2
1
2
2
3
2
dx
x
 







+






+ 1
2
1
2
4
2
3
2
, let 1
2
2
1
2
sinh +
=






+
= x
x
 thus


d
dx
2
cosh
=
 

 d
cosh
sinh
8
2
3 2
 
d
2
cosh
8
2
3
thus ( )
 + 
 d
1
2
cosh
16
2
3
A
+
+ 

16
2
3
2
sinh
32
2
3
But ( )
1
2
sinh 1
+
= −
x

( )
( ) ( ) A
x
x +
+
+
+ −
−
1
2
sinh
16
2
3
1
2
sinh
2
sinh
32
2
3 1
1
Combining the two solutions ( )
 +
+
+ dx
x
x
x 1
2
2
1
2 2
( )
( ) ( ) A
x
x
x
x +
+
+
+
+





 +
+
= −
−
1
2
sinh
16
2
3
1
2
sinh
2
sinh
32
2
3
1
2
2
6
1 1
1
3
2
Evaluate
 +
−
+
dx
x
x
x
1
1
2
4
2
Solution
From
 +
−
+
dx
x
x
x
1
1
2
4
2
Divide throughout by 2
x
Example 60

Integration







+
−






+
dx
x
x
x
x
x
2
2
2
2
2
1
1
1
1
complete the square













+
−
+
−
+
dx
x
x
x
2
2
1
1
1
1
2
2
2

+






−
+
+
dx
x
x
x
1
2
1
1
1
2
2
2
thus 
+






−
+
dx
x
x
x
1
1
1
1
2
2
and let
dx
x
du
x
x
u 





+
=

−
= 2
1
1
1
( ) A
u
u
du
+
=
+
−

1
2
tan
1
but
x
x
u
1
−
=
Therefore
 +






−
=
+
−
+ −
A
x
x
dx
x
x
x 1
tan
1
1 1
2
4
2
EXERCISE 15
1. ( )
 −
+ dx
x
x 5
2
3 2
2. ( )
 +
+
+ dx
x
x
x 1
1 2
3. ( ) ( )
 −
+
+ dx
x
x
x
5
2
10
2
2
1
2
4.
 −
+
−
dx
x
x
x
1
1
2
4
2
5. ( )
 +
+
− dx
x
x
x 8
4
2
5 2
6. ( )
 +
−
− dx
x
x
x 2
1
3 2
7. ( )
 +
−
− dx
x
x
x 3
5
2
5
7 2
8. ( )( )
 −
+
− dx
x
x
x
3
2
1
3
1
2 9. ( )( )
 −
− dx
x
x
x
3
2
3
2
3
4
10.  −
+
dx
x
x
2
2
1
2 11.  +
+
−
dx
x
x
x
4
2
2
4
2
12. ( ) ( ) ( )
 +
+
+
=
+
+ A
x
x
dx
x
x 7
3
3
15
2
4
3 2
3

Integration
SUBSTITUTION OF INVERSE OF TRIGONOMETRIC FUNCTIONS
Integrals of the form dx
x
f
 − 2
)
(
1
1
or
( )
 +
dx
x
f 2
)
(
1
1
Refer the derivatives of inverse of trigonometric functions
1. If )
(
sin 1
x
y −
= then
2
1
1
x
dx
dy
−
=
2. If )
(
cos 1
x
y −
= then
2
1
1
x
dx
dy
−
−
=
3. If )
(
tan 1
x
y −
= then 2
1
1
x
dx
dy
+
=
4. If )
(
sec 1
x
y −
= then
1
1
2
−
=
x
x
dx
dy
5. If )
(
csc 1
x
y −
= then
1
1
2
−
−
=
x
x
dx
dy
6. If )
(
cot 1
x
y −
= then 2
1
1
x
dx
dy
+
−
=
Evaluate dx
x
 − 2
9
4
1
Solution
This integral resembles, cosine or sine inverse, any of the function can be used.
Given dx
x
 − 2
9
4
1
, this can expressed as
( ) ( )
  −
=
−
dx
x
dx
x 2
2
3
2
4
9
1
1
2
1
1
4
1
let x
2
3
sin =
 , then dx
d =


cos
3
2
  +
=
=
−
 A
d
d 




3
1
3
1
sin
1
cos
3
2
2
1
2
, but 





= −
x
2
3
sin 1
 therefore
A
x
x
dx
+






=
−

−
2
3
sin
3
1
9
4
1
2
Example 61

Integration
EXERCISE 16
1.  − 2
3
5 x
dx
2.  −
+
dx
x
x 2
9
2
1
1
3.  +
dx
x2
49
4
1
4.  −
+
dx
x
x 2
6
1
5.  −
−
dx
x
x
x
4
2
1
1
sin 6.  −
+ 2
3
3 x
x
dx
7.  +
+
dx
x
x 2
1
1
8.
( )
 −
−
2
1
2
4
2 x
x
dx
9.  −
−
dx
e
e
e
x
x
x
2
4
5
10.  −
+
dx
x
x
2
4
3
2
SUBSTITUTION OF INVERSE OF HYPERBOLIC FUNCTIONS
Refer the derivatives of inverse of hyperbolic functions, for further studies
of these integrals of these form, check the hyperbolic functions topic in
book 2.
1. If )
(
cosh 1
x
y −
= then
1
1
2
−
=
x
dx
dy
2. If )
(
sinh 1
x
y −
= then
2
1
1
x
dx
dy
+
=
3. If )
(
tanh 1
x
y −
= then 2
1
1
x
dx
dy
−
=
4. If )
(
sech 1
x
y −
= then
2
1
1
x
x
dx
dy
−
=
5. If )
(
cosech
y -1
x
= then
1
1
2
+
=
x
x
dx
dy
6. If )
(
coth 1
x
y −
= then
1
1
2
−
=
x
dx
dy

Integration
Integrate dx
x
x
 −
+
2
3
1
1
6
Solution
From dx
x
x
 −
+
1
3
1
6
2
, thus   −
+
− 1
3
1
3
6
2
2
x
dx
dx
x
x
INTEGRAL OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Suppose we have  dx
ax
, for all 0

a
Let x
a
y = then a
x
y
a
y x
ln
ln
ln
ln =

=
a
dx
dy
y
ln
1
= thus 
 =

= adx
y
dy
adx
y
dy ln
ln
dx
y
a
y 
=ln but x
a
y = thus dx
a
a
a x
x

=ln
Therefore A
a
a
dx
a
x
x
+
=
 ln
, where 0

a
Integrate dx
x x
 




 2
3
)
(
Solution
From dx
x x
 




 2
3
)
( , let x
dx
dy
x
y 2
2
=

=
( ) 
 = dy
x
dy
x y
y
3
2
1
2
3 , let
3
ln
2
1
=
dx
dy
y
thus 
 =

= ydx
dy
ydx
dy
3
ln
2
1
3
ln
2
replacing y
Therefore, A
dx
x
x
+
=
 9
ln
3
3
2
2
Example 63
Example 62

Integration
Integrate ( )
 + dx
e x
x
2
3
Solution
( )
  
+
=
+ dx
dx
e
dx
e x
x
x
x
2
3
2
3 thus ( ) A
e
dx
e
x
x
x
x
+
+
=
+
 2
ln
2
3
2
3
Integrate 
−
dx
x 1
2
4
Solution
Let ( ) 4
ln
1
2
ln
4 1
2
−
=

= −
x
y
y x
dx
y
dy
dx
dy
y
16
ln
4
ln
2
1
=

=


 =

= ydx
y
ydx
dy 16
ln
16
ln but 1
2
4 −
= x
y

−
−
= dx
x
x 1
2
1
2
4
16
ln
4 therefore A
dx
x
x
+
=
−
−
 16
ln
4
4
1
2
1
2
Integrate ( )
 −
+ dx
e x
x
x 2
3
10
Solution
Using a general observation ( )
 −
+ dx
e x
x
x 2
3
10
Can be expressed as 
  −
+ dx
e
dx
dx x
x
x 2
3
10 thus
( ) A
e
dx
e x
x
x
x
x
x
+
−
+
=
−
+

2
2
2
1
3
ln
3
10
ln
10
3
10
Example 64
Example 65
Example 66
Question
Show that A
dx
x
x
+
=

+
+
9
ln
3
3
2
2
2
2

Integration
Integrate
 +
+
dx
x
x
x
x
5
2
5
ln
5
10
5
4
Solution
From
 +
+
dx
x
x
x
x
5
2
5
ln
5
10
5
4
, let x
x
u 5
2 5
+
= thus 5
ln
5
10 4 x
x
dx
du
+
=
A
u
du
u
+
=
 ln
1
thus ( ) A
x
dx
x
x x
x
x
+
+
=
+
+
 5
2
ln
5
2
5
ln
5
10 5
5
4
DEFINITE INTEGRALS
The fundamental theorem of calculus
The integrals of the form ( ) ( ) ( )
a
F
b
F
dx
x
f
b
a
−
=
 are called definite
integrals. If b
a = then  =
−
=
−
=
b
a
a
F
a
F
b
F
b
F
dx
x
f 0
)
(
)
(
)
(
)
(
)
(
Evaluate the value of ( )
 −
+
2
1
3
3
2 dx
x
x
Solution
From ( )
2
1
2
4
2
1
3
3
4
3
2 A
x
x
x
dx
x
x +
−
+
=
−
+









+

−
+
−








+

−
+
=
+
−
+ A
A
A
x
x
x
1
3
1
1
1
2
3
2
4
2
3
4
2
2
2
4
2
1
2
4
Therefore ( ) 5
3
2
2
1
3
=
−
+
 dx
x
x
By first expressing
3
5
2
17
4
4
2
2
−
+
−
+
x
x
x
x
into partial fractions, show that
 





−
=
−
+
−
+
2
1 2
2
4
45
ln
2
3
5
2
17
4
4
dx
x
x
x
x
Example 68
Example 69
Example 67

Integration
Solution
Use long division method first to simplify the function
3
5
2
17
4
4
2
2
−
+
−
+
x
x
x
x
2
11
6
6
10
4
17
4
4
3
5
2
2
2
2
−
−
−
+
−
−
+
−
+
x
x
x
x
x
x
x
3
5
2
11
6
2
3
5
2
17
4
4
2
2
2
−
+
+
−
=
−
+
−
+
x
x
x
x
x
x
x
Partial fractions
( )( )
3
1
2
11
6
3
5
2
11
2 +
−
+
=
−
+ x
x
x
x
x
( )( ) 3
1
2
3
1
2
11
6
+
+
−
=
+
−
+
x
B
x
A
x
x
x
thus ( ) ( )
1
2
3
11
6 −
+
+
=
+ x
B
x
A
x
When 3
−
=
x , 1
=
B and when
2
1
=
x , 4
=
A
( )( ) ( ) ( )
3
1
1
2
4
3
1
2
11
+
+
−
=
+
− x
x
x
x
Now combine the solutions
 −
+
−
+
2
1 2
2
3
5
2
17
4
4
dx
x
x
x
x


 +
+
−
−
2
1
2
1
2
1 3
1
1
2
1
4
2 dx
x
dx
x
dx





 
−
=
4
5
9
ln
2
Therefore 





−
=
−
+
−
+
 4
45
ln
2
3
5
2
17
4
4
2
1 2
2
dx
x
x
x
x
Hence shown
Evaluate ( )
 +
3
π
tan
sec
0
dx
x
x
Solution
From ( )
 +
3
π
0
tan
sec dx
x
x thus 
 + 3
π
3
π
tan
sec
0
0
xdx
xdx
Example 70

Integration
3
π
sec
ln
tan
sec
ln
0
x
x
x +
+
=
2
3
1
ln
2
ln
3
2
ln +
=
+
+
=
Therefore ( )
2
3
1
ln
tan
sec
3
0
+
=
+


dx
x
x
Express
( )2
3
1
2
−
+
x
x
in to partial fractions and hence show that
( )
49
ln
6
3
1
2
10
4 2
+
=
−
+
 dx
x
x
Solution
From
( ) ( )2
2
3
3
3
1
2
−
+
−
=
−
+
x
B
x
A
x
x
and B
A
Ax
x +
−
=
+ 3
1
2
Compare the coefficients 2
=
A and 7
=
B
( ) ( )2
2
3
7
3
2
3
1
2
−
+
−
=
−
+
x
x
x
x
( ) ( )


 −
+
−
=
−
+ 10
4 2
10
4
10
4 2
3
7
3
2
3
1
2
dx
x
dx
x
dx
x
x
( )
10
4
10
4 2 3
7
3
ln
2
3
1
2
−
−
−
=
−
+
 x
x
dx
x
x
( )
( ) ( )
7
1
ln
2
1
7
ln
2
3
1
2
10
4 2
−
−
−
=
−
+
 dx
x
x
( )
49
ln
6
3
1
2
10
4 2
+
=
−
+
 dx
x
x
Hence shown
Show that
4
π
tan
1
2
π
0
=
+
 x
dx
Solution
Example 71
Example 72

Integration
Let  +
= 2
π
0 tan
1 x
dx
In
 +
= 2
π
0 sin
cos
cos
x
x
xdx
In
  +
=






−
+






−






−
= 2
π
0
2
π
0 sin
cos
sin
2
sin
2
cos
2
cos
x
x
xdx
x
x
dx
x
In



 +
= 2
π
0 sin
cos
sin
x
x
xdx
In
dx
x
x
x
dx
x
x
x
In 
 +
+
+
= 2
π
0
2
π
0 sin
cos
cos
cos
sin
sin
2
 +
+
= 2
π
0 sin
cos
cos
sin
2 dx
x
x
x
x
In
2
π
0
2 x
In = then
2
π
2 =
n
I therefore
4
π
=
n
I , proved
EXERCISE 17
1.  





+
+
8
1 3
4
3
dx
x
x 2. ( )( )
 −
+
+
2
1
4
2
10
2
2
1
2 dx
x
x
x
3.  −
+
+
5
1 2
5
12
3
2
dx
x
x
x
4. ( )
−
2
π
2
π
2
sin dx
x
5.
( )
 +
4
1 2
3
1
dx
x
6.
( )( )
 +
+
+
2
1 3
2
1
dx
x
x
x
7. ( )dx
x
x

3
π
0
cos
3
cos 8.  





+
7
3
7
1 2
49
1
1
dx
x
9.
( )

−






−
2
tan
4
π
1
2
sin
3
2
cos
4
3
dx
x
x
10. ( )dx
x
x

1
.
1
0
3
tan
sec 11.  





3
2
ln
dx
x
x

Integration
APPLICATIONS OF INTEGRATION
There are several applications of integration.
AREA UNDER THE CURVE: 
=
b
a
ydx
Area
Suppose we want to find the area under the curve )
(x
f
y = sketched
below bounded from a
x = to b
x = with the curve.
One way to estimate the area, A, (Figure 10.1) is to divide the area into
rectangular strips, (Figure 10.2) the sum of the areas of strips is
approximated to the area under the curve. The thinner the strips the better
approximation, i.e. as the number of strips increase the better
approximation of the area under the curve.
Since each strip have the same width, δx the length of each strip is y,
therefore, the area of one strip is given as x
y
A 
  , to get the area under
the curve we sum the area of all strips from a
x = to b
x = , this can be
written as 
=
=
b
a
x
A

Area , as the width of strips decrease, the number of
rectangles increase, therefore, as 0
→
x
 ,
dx
dA
x
A
y
x
A
x
=






→

→ 



 0
lim
Then y
dx
dA
= becomes 
=

= ydx
A
ydx
dA , the boundary limit of x is
a
x = to b
x = . This defines the area under the curve as, 
=
b
a
ydx
A .
Procedures to find the area
(a) Draw a sketch of the given region to visualize the nature of the
curve and divide intervals appropriately to avoid getting negative
or zero answer.
(b) Write the defined integral(s) to represent the region
A
Figure 10.1 Figure 10.2
a a
b b

Integration
(c) Evaluate the defined integral.
Determine the area under the curve 3
+
−
= x
y from 0
=
x to 2
=
x
Solution
Sketch
( )
 +
−
=
2
0
3 dx
x
A
2
0
2
3
2 







+
−
= x
x
A thus 4
=
A
The area is four sq. units
Find the area under the curve 4
2
+
−
= x
y from 1
=
x to 2
=
x
Solution

=
b
a
ydx
Area therefore ( )
 +
−
=
2
1
2
4 dx
x
A
2
1
3
4
3
x
x
A +
−
=






+
−
−






+
−
= 4
3
1
8
3
8
A
The area is
3
5
square units.
Evaluate the area of the curve 2
3
2x
x
y +
= from 1
−
=
x to 1
=
x
Solution
( )
−
+
=
1
1
2
3
2 dx
x
x
A
Example 70
Example 71
Example 72
4
2
+
−
= x
y
3
+
−
= x
y

Integration
( ) ( )

 +
+
+
=
−
1
0
2
3
0
1
2
3
2
2 dx
x
x
dx
x
x
A
1
0
3
4
0
1
3
4
3
2
4
3
2
4 







+
+








+
=
−
x
x
x
x
A
3
.
1
3
2
4
1
3
2
4
1
=






+
+






−
−
=
A
The area is 1.3 square units.
Determine the area of the curve 3
3 2
3
+
−
−
= x
x
x
y from 1
−
=
x to 3
=
x
Solution
From the sketch, a part of the
integral lies below the x – axis
and the other part lie above, if
we evaluate the integral direct
from 1
−
=
x to 3
=
x , the result
will be zero, we divide the
integral as follows
( ) ( )

 −
+
−
−
+
+
−
−
=
0
1
2
3
3
0
2
3
3
3
3
3 dx
x
x
x
dx
x
x
x
A
While dividing it, we should make sure the area obtained is always positive,
this can be done by taking the absolute value of each integral above.
0
1
2
3
4
3
0
2
3
4
3
2
4
3
2
4
−








+
−
−
+








+
−
−
= x
x
x
x
x
x
x
x
A
The area is 4.5 square units.
Example 73
3
3 2
3
+
−
−
= x
x
x
y
2
3
2x
x
y +
=

Integration
Find the area of the curve x
y sin
3
= from 0
=
x to 2π
=
x
Solution

=
b
a
ydx
A then ( )

=
2π
0
sin
3 dx
x
A
( ) ( )

 +
=
2π
π
π
0
sin
3
sin
3 dx
x
dx
x
A
( ) ( ) π
2
π
π
0
cos
3
cos
3 x
x
A +
=
π
cos
3
π
2
cos
3
0
cos
3
π
cos
3 −
+
−
=
A
A = 12 Square units.
Find an approximate area of the shaded region below
Solution
)
sin(
3 x
y =
Example 74
2
4 x
x
y −
=
Example 75

Integration
( )
 −
=
4
0
2
4 dx
x
x
A thus
4
0
3
2
3
2
x
x
A −
=
The area is
3
32
=
A square units.
( )
 −
=
b
a
dx
y
y
A 2
1
AREA BETWEEN CURVES: (where a and b are abscissa at a
point of intersection)
If the area under the curve )
(x
f
y = is
given by 
=
b
a
dx
x
f
A )
(
1 and the area
under the curve )
(x
g
y = is given as
Figure 10.3
a b
)
(x
f
y =
)
(x
g
y =
A

Integration

=
b
a
dx
x
g
A )
(
2 , then the area
between the curves )
(x
f
y = and )
(x
g
y = from a
x = to b
x = (Where a and
b are the values of x at the point of intersection of the curves) is given by
 
 −
=
b
a
dx
x
g
x
f
A )
(
)
( , (Figure 10.3)
Find the area between curves 2
4 x
y −
= and x
x
y 2
2
−
=
Solution
First, find the points of intersection of the curves, 2
2
4
2 x
x
x −
=
−
0
4
2
2 2
=
−
− x
x
( )( ) 0
2
1 =
−
+ x
x
1
−
=
x and 2
=
x
( )
 −
=
b
a
dx
y
y
A 1
2
( ) ( )
( )
−
−
−
−
=
2
1
2
2
2
4 dx
x
x
x
A
( )
−
+
−
=
2
1
2
2
2
4 dx
x
x
A
x
x
y 2
2
−
=
2
4 x
y −
=
Example 76

Integration
2
1
2
3
3
2
4
−
+
−
= x
x
x
A therefore the area is 9 square units
Find the area enclosed by the curves 3
7
2 2
+
+
= x
x
y and 2
4
9 x
x
y −
+
=
Solution
The points of intersection of the curve
( )( ) 0
2
1
4
9
3
7
2 2
2
=
+
−

−
+
=
+
+ x
x
x
x
x
x thus 1
=
x or 2
−
=
x
( )
 −
=
b
a
dx
y
y
A 1
2
( ) ( )
( )
−
−
+
−
+
+
=
1
2
2
2
4
9
3
7
2 dx
x
x
x
x
A
( )
−
−
+
=
1
2
2
6
3
3 dx
x
x
A
1
2
2
3
6
2
3
−
−
+
= x
x
x
A
2
27
=
A square units
Find the area between curves 1
2
−
= x
y and x
y −
=1
Solution
Point of intersections
1
1 2
−
=
+
− x
x
2
−
=
x and 1
=
x
( )
 −
=
b
a
dx
y
y
A 1
2
( ) ( )
( )
−
+
−
−
=
2
1
2
1
1 dx
x
x
A
( )
−
−
−
=
2
1
2
2 dx
x
x
A
Example 77
2
4
9 x
x
y −
+
=
3
7
2 2
+
+
= x
x
y
Example 78
1
2
−
= x
y
x
y −
=1

Integration
2
1
2
3
2
2
3
−
−
−
= x
x
x
A
Area is 4.5 square units
Find the area of the curve x
x
x
x
y 6
3
6
3 2
3
4
+
−
−
= with x – axis
Solution
The points of intersection
x
x
x
x
y 6
3
6
3 2
3
4
+
−
−
=
( )( )( ) 0
1
1
3
3 =
+
−
− x
x
x
x
0
=
x , 3
=
x , 1
=
x and 1
−
=
x
( )
−
+
−
−
=
3
1
2
3
4
6
3
6
3 dx
x
x
x
x
A
3
1
2
3
4
5
3
2
3
5
3
−
+
−
−
= x
x
x
x
A
5
2
22
=
A sq. units
Example 79
x
x
x
x
y 6
3
6
3 2
3
4
+
−
−
=

Integration
EXERCISE 18
1. Find the under the curve x
x
x
f 4
)
( 3
−
= from 2
−
=
x to 2
=
x .
2. Find the area enclosed by the x – axis and that part of the curve
2
3
2
)
( x
x
x
f +
−
= .
3. Sketch the graph of ( ) 2
ln x
x
y +
= and hence calculate the area bounded
with the x axis from 1
=
x to 2
=
x .
4. Find the area of the region bounded by the curve 1
2
3 2
+
−
= x
x
y , the
line 0
1 =
+
x , 0
2 =
−
x and 0
=
y
5. The area between the curve 9
3 2
2
=
+ y
x and the y-axis from 3
−
=
y
and 3
=
y is rotated about the y-axis. Find the volume of the solid
generated.
6. Find the area between the curve 1
+
= x
y and 2
3 x
y −
=
7. Find the area between the curves 2
y
x = and 4
3 +
= y
x
8. Find the area bounded by the curve x
y +
=1 and 1
−
= x
y
9. Find the area between the coordinates axes and the curve 2
−
= x
y
10. Find the area under the curve 4
=
y between the lines 0
=
x and 5
=
x
1
=
+
x
y
from 0
=
x to 40
=
x
12. Find the area between curves 2
x
y = and 2
2
+
−
= x
y
13. Find the value of a if the area between the curve 2
2
+
=ax
y and 2
x
y =
from 1
=
x to 2
=
x is 10 square units.
14. Find the area between the curves x
y 4
2
= and x
y
x 8
2
2
=
+ bounded
with the x – axis.
3
2
=
+ y
y
x from 0
=
x to 3
=
x

Integration
VOLUME OF THE SOLID OF REVOLUTION
Suppose the area enclosed by the line ax
y = with x – axis and the line c
x =
is rotated through one revolution, the solid formed is called solid of
revolution, which is cone in this case.
Consider the figure below
Volume of disc A, h
πr
V 2
=
To get the volume of the figure from a
x =
to b
x = is the summation of the
approximated volume of the discs, given
y
r = and δx
h = therefore volume,

=
=
=
b
x
a
x
2
δx
πy
V .

 =








=
=
=
→
b
a
2
b
x
a
x
2
0
δx
dx
π
δx
π
lim
V y
y
Therefore the volume is given by

=
b
a
dx
y
V 2
π
A y
δx
b
a
Rotation

Integration
The volume is given by 
=
b
a
dx
y
V 2
π , if the curve is revolved about x – axis
or dy
x
V
b
a

= 2
π if the curve is revolved about y – axis.
Note that, if the curve is symmetrical about the y – axis or x – axis and the
same curve is made to revolve about the x – axis or y – axis, we revolve only
the part of the curve lying one side of either the x – axis or y – axis. To get
the volume generated by considering the revolution of both sides is given by
dx
y
V
b
a

= 2
π
2 .
Find the volume of the solid of revolution formed by rotating the area
enclosed by the curve x
x
y +
= 2
, the x – axis and the ordinates 1
=
x and
2
=
x through one revolution about the x – axis.
Solution

=
b
a
dx
y
V 2
π
( )
 +
=
2
1
2
2
π dx
x
x
V
( )
 +
+
=
2
1
2
3
4
2
π dx
x
x
x
V
2
1
3
4
5
3
1
2
1
5
1
π 





+
+
= x
x
x
V
π
30
481
=
V
Therefore, the volume is 50.37 cubic units
Example 80
x
x
y +
= 2

Integration
Find the volume of the solid formed when rectangular hyperbola 8
=
xy
rotated once through one revolution with positive x – axis from 2
=
x and
5
=
x
Solution
8
=
xy then
x
y
8
=
 





=
5
2
2
8
π dx
x
V
dx
x
V  





=
5
2 2
64
π
5
2
64
π 





−
=
x
V
π
5
96
=
V
The volume is 60.32 cubic units
Find the volume in the first quadrant bounded by curve 1
2 2
+
= x
y , the y –
axis and the lines 2
=
y and 5
=
y .
Solution
1
2 2
+
= x
y
2
1
−
=
y
x
 






 −
=
5
2
2
2
1
π dy
y
V
( )
 −
=
5
2
1
2
π
dy
y
V
Example 81
Example 82
8
=
xy
1
2 2
+
= x
y

Integration
5
2
2
2
2
π








−
= y
y
V
π
4
15
=
V
The volume is 11.78 cubic units
EXERCISE 19
1. A bowl has a shape that can be generated by revolving the graph of
2
2
x
y = between 0
=
y and 5
=
y about the y-axis. Find the volume
of the bowl.
2. Find the volume of solid revolution when a curve 7
=
xy is rotated
about the x axis from 3
=
x to 5
=
x , leave your answer with π .
3. Calculate the volume generated when the curve 3
4
2
+
= x
y
x is
rotated once about x-axis from 1
=
x to 2
=
x .
4. Find the volume generated when the loop of the curve
( )2
2
5
4 y
y
x −
= revolve about the y-axis.
5. Find the volume generated by the rotation of the curve 5
3
2
=
+ y
y
x
about y – axis, from y = 0.5 to y=1.6
6. Find the volume generated by revolving the curve )
cos(x
y = from
π
2
3
−
=
x to π
2
3
=
x along the x – axis once.
7. Find the volume generated by revolving the curve 4
+
= x
y from
1
=
x to 2
=
x about the x – axis once.
8. The curve 0
2
3
2
=
−
+ x
y is rotated about x – axis once, from 2
=
x
to 4
=
x
9. The bowl in a shape of parabola open upward has the equation
( )
y
x +
= 3
2
2
is filled with water, find the volume of water contained
in a bowl in cubic centimetres if the vertical height of the bowl is 8
units from the bottom and 4 cubic units = 15 cm3
.
10. Find the volume generated by revolving the curve π
4
4
5
=
+
x
y
along
line 1
−
=
y from 0
=
y to 4
=
y .

Integration
11. Find the volume formed when the loop of the curve
2
3
3





 −
=
x
y
y
revolve about the y – axis.
12. Show that the volume formed by the curve 9
2
2
=
+ y
x is a same as
the volume of the sphere with radius 3 units.
ARC LENGTH
There is no simple formula to calculate length of the curve other than a
circle. If the length of the portion of a curve is needed we use the method
of summing small portions of the required curve to the end of the required
portion. Suppose that the arc AB of the length δs is such an element, then
the length s of the curve PB is given by 
=
=
=
b
x
a
x
s s
δ
The length of the curve AB, where A and B are close to each other, then
By Pythagoras theorem
( ) ( ) ( )2
2
2
δy
δx
δs +

2
2
δx
δy
1
δx
δs






+
=






b
a
P
A
B
C
δs δy
δx
c

Integration
δx
δx
δy
1
δs
2






+
=

=
=






+
=
b
x
a
x
2
δx
δx
δy
1
s
As 0
δx → then
dx
dy
δx
δy
→

=
=
→






+
=
b
x
a
x
2
0
δx
δx
δx
δy
1
lim
s
 





+
=
b
a
2
dx
dx
dy
1
s
The length of an arc is given by the formula,  





+
=
b
a
2
dx
dx
dy
1
s
Find the length of the portion of the curve 2
x
y = between 0
=
x and 1
=
x
Solution
From x
dx
dy
x
y 2
2
=

=
( ) dx
x
s  +
=
1
0
2
2
1
Use hyperbolic substitution
Let 
sinh
2 =
x  
d
dx cosh
2 =
 +
=
2
1
2
cosh
sinh
1
2
1


 d
s

=
2
1
2
cosh
2
1

d
s
Remember 

 2
2
sinh
cosh
2
cosh +
= and 1
sinh
cosh 2
2
=
− 

1
cosh
2
2
cosh 2
−
= 

( )
1
2
cosh
2
1
cosh2
+
= 

Example 83

Integration

=
1
0
2
cosh
2
1

d
s
( )
 +
=
1
0
1
2
cosh
4
1

 d
s






+
= 

2
sinh
2
1
4
1
s
( )


 +
= cosh
sinh
4
1
s
( )




 +
+
= −
x
x
x
s 2
sinh
4
1
2
4
1 1
2
( )
1
0
2
2
1
2
2
ln
4
1
2
4
1





 +
+
+
+
= x
x
x
x
s
( )
5
2
ln
5
2
4
1
+
+
=
s
( )
5
2
ln
4
1
5
2
1
+
+
=
s
The length is 1.48 unit length
Find the length of the arc 2
3
x
y = from 1
=
x to 8
=
x correct to 1 decimal
place.
Solution
From 2
1
2
3
2
3
x
dx
dy
x
y =

=
 





+
=
b
a
dx
dx
dy
s
2
1
( )
 +
=
8
1
2
2
1
2
3
1 dx
x
s
Example 84

Integration
 +
=
8
1 4
9
1 dx
x
s thus 22.803
The length is 22.8 units
Find the length of the arc of x
y sec
ln
= from 0
=
x to
3
π
=
x
Solution
From x
x
x
x
dx
dy
x
y tan
sec
tan
sec
sec
ln =
=

=
 +
=
3
π
0
2
tan
1 dx
x
s
3
π
0
3
π
0
tan
sec
ln
sec x
x
xdx
s +
=
= 
32
.
1
3
2
ln 
+
=
s units
Find the length of an arc of the curve x
e
y x
−
= 2
from 3
=
x to 10
=
x
Solution
x
e
y x
−
= 2
then 1
2 2
−
= x
e
dx
dy
( )dx
e
s x
 −
+
=
10
3
2
1
2
1

=
10
3
2
2 dx
e
s x
, dx
e
s x

=
10
3
2
10
3
2 x
e
s = 3
10
2
2 e
e
s −
=
( )
1
2 7
3
−
= e
e
s
The length is 31121.72 units
Find an approximate value of the length of the portion of the curve
x
x
y
2
1
6
1 3
+
= from 1
=
x to 2
=
x
Solution
Example 85
Example 86
Example 87

Integration
From
x
x
y
2
1
6
1 3
+
= then 2
2
2
1
2
1
x
x
dx
dy
−
=
 





−
+
=
2
1
2
2
2
2
1
2
1
1 dx
x
x
s
 +
−
+
=
2
1 4
4
4
1
2
1
4
1
1 dx
x
x
s
 





+
=
2
1
2
2
2
2
1
2
1
dx
x
x
s
 





+
=
2
1 2
2
2
1
2
1
dx
x
x
s
2
1
3
2
1
6 x
x
s −
=
The length is 1.417 unit correct to 4 significant figures.
Find the length of the arc of the curve 2
2
2
a
y
x =
+ between the points
where 
cos
a
x = and 
cos
a
x =
Solution
From 0
2
2
2
2
2
=
+

=
+
dx
dy
y
x
a
y
x therefore
y
x
dx
dy
−
=
 





−
+
=
b
a
dx
y
x
s
2
1
 






 +
=


cos
cos 2
2
2
a
a
dx
y
x
y
s thus
 







=


cos
cos 2
2
a
a
dx
y
a
s
 





=


cos
cos
a
a
dx
y
a
s But 2
2
x
a
y −
=
 







−
=


cos
cos 2
2
a
a
dx
x
a
a
s thus  







−
=


cos
cos 2
2
a
a
dx
x
a
a
s
Example 88

Integration
Now let 
cos
a
x = 
d
a
dx sin
−
=

 −
−
=




cos
cos 2
2
2
2
sin
sin
a
a
d
x
a
a
a
s

−
=



cos
cos
a
a
d
a
s thus then 
a
s −
= and 





= −
a
x
1
cos



cos
cos
1
cos
a
a
a
x
a
s 





−
= −














−
−














−
= −
−
a
a
a
a
a
a
s

 cos
cos
cos
cos 1
1

 a
a
s +
−
=
The length is ( )

 −
a units
PARAMETRIC LENGTH OF THE CURVE
For the parametric equations of the curve
( ) ( )
t
f
dt
dx
t
f
x '
=

= and ( ) ( )
t
g
dt
dy
t
g
y '
=

=
 





+






=
b
a
dx
dt
dy
dt
dx
s
2
2
The length of the curve given in parametric is given by
 





+






=
b
a
dx
dt
dy
dt
dx
s
2
2
Determine the length of the parametric curve given by the following
parametric equations, t
y sin
3
= and t
x cos
3
= from 0
=
t to π
2
=
t
Solution
Example 89

Integration
From t
dt
dy
t
y cos
3
sin
3 =

= and t
dt
dx
t
x sin
3
cos
3 −
=

=
 





+






=
π
2
0
2
2
dt
dt
dy
dt
dx
s
( ) ( )
 +
−
=
π
2
0
2
2
cos
3
sin
3 dt
t
t
s

=
π
2
0
3 dt
s therefore
π
2
0
3t
s =
π
6
=
s , the length of the arc is 18.85 unit length
EXERCISE 20
1. Calculate the arc length of the curve 2
2
4 x
y −
= from 2
−
=
x to
2
=
x
2. Calculate the arc length of the curve x
x
x
f −
= 2
)
( from 1
=
x to
2
=
x
3. Show that the length of the curve x
ey
cos
= from 0
=
x to
4
π
=
x is ( )
2
1
ln +
=
s
4. Show that the length of the curve x
y cosh
= from 0
=
x to 1
=
x
is 1
2
1
ln
2
−







 −
e
5. Find the length of the curve 3
2
x
y = from 0
=
x to 1
=
x correct
to 4 significant figures.
6. Compute the length of the curve ( )
1
2
2
−
= x
y form 1
=
x to
5
=
x
7. Find the length of the curve ( )
)
tan(
ln x
y = from 1
=
x to 3
=
x
8. Find the length of the curve u
x 3
cos
2
= , u
y 3
sin
2
= from π
=
u
to π
2
=
u
9. Find the length of the curve given parametrically as 2
3 2
+
= t
x ,
3
5 2
−
= t
y form the point where 3
=
t to 8
=
t
10. Determine the length of the curve given as r
r
p 4
2
−
= ,
2
2
3 r
r
q −
= from 4
=
r to 10
=
r

Integration
MISCELLANEOUS EXERCISE
1.
 −
−
dx
x
e
e
x
x
1
2.  





+
+
dx
x
x
9
6
2
3.
 +
dx
x 9
3
1
4.
( )
 −
dy
y
y 3
1
5. ( )dx
x x
 2 6.  





dx
x
x
2
cos
7.
 





+
−
dx
x
x
5
4
1
2
2
8.
 +
−
dx
x
x
x
x
sin
4
cos
3
sin
2
cos
9.  −
+
−
dx
x
x
x
2
2
1
4
1 10. ( )
 dx
x
x 2
ln
11. ( )
 − dx
x
x 6
5
1 12.  −
dx
x
2
sin
5
1
1
13.  +
dx
x
x 2
2
sin
9
cos
4
14. 
+
dx
x
x
2
cos
sin
1
15.
 xdx
x sin
3
16.
 dx
e
x x
5
17.  xdx
x cos
5
18.
 dy
y
y ln
1
19.  −
dx
x
x
3
1
20.  xdx
x
x 2
sin
2
cos
21.  xdx
x 5
6
sin
cos 22. ( )dx
x

−1
sin
23.
 +
+
dx
1
cos
2
sin
1

 24. 
4
1
dx
x
e x
25.
( ) ( )
 +
− 1
1 2
2
x
x
dx
26. dx
x
x
x
 −
+ 20
2
2
3
27. x
x
x d
tan 1
2

−
27.
( )
 +
dx
x
x
2
1
ln
28. ( )
 + dx
x
x 4
ln
29.  +
dx
x
x
tan
2
1
sec2
30. dx
e x
x

−
31.  





−
4
0
1
4
sin dx
x

Integration
32. dx
x
x
 +
3
1 3
2
1
33.  +
+
− 3
2 x
x
dx
34. Prove that
( )
( ) 1
2
ln
4
2
ln
1
1
ln
2
1
1
0
−
−
=
+
+
 dx
x
x
x
35. Evaluate (a)  −
−
dx
x
x
x
2
4
4
1
2
(b) ( )
 + dx
x
x tan
sec
36. Express 4
1
1
x
−
in partial fractions and hence show that e






+
=
−
−
 2
1
tan
3
1
3
ln
4
1
1
1 1
2
1
0 4
dx
x
37. Evaluate
( )
 +
3
1 2
1
1
dx
x
x
express your answer in the form of
b
a
ln
where a and b are non-zero positive integers.
38. Prove that 





+
=
+
 3
4
ln
25
4
π
50
3
cos
4
sin
3
sin
2
π
0
dx
x
x
x
39. By making the substitution y
x −
=π , or otherwise, prove that
π
3
2
sin
π
0
3
=
 dx
x
x
40. Use the substitution x
x −
=π show that
4
π
cos
1
sin 2
π
0 2
=
+
 dx
x
x
x
41. Find the area enclosed by the curve x
x
x
x
f 2
3
)
( 2
3
+
−
= and the x –
axis between 0
=
x and 2
=
x
42. Find the area enclosed between the curve x
x
f 3
sin
)
( = and the x –
axis between π
−
=
x and π
=
x
43. Find the area enclosed between the curves x
y 4
2
= and y
x 4
2
=
44. Find the area between curves 1
2
2
=
+ y
x and 3
3 2
2
=
+ y
x
45. Find the area enclosed between curve 2
3 x
x
y +
−
= and the line 4
=
y
46. Find the volume of the solid revolution formed when curve 3
2
−
= x
y
from 1
=
x to 3
=
x
47. Find the length of the arc of the curve x
y 3
sin
3 = between the points
π
=
x and π
2
=
x
48. Find the area of the surface formed by the rotation of the curve
x
y 8
2
= about the x – axis, from the origin to 2
=
x

Integration
49. Find the volume of the solid revolution obtained by rotating the region
bounded by the curve x
y =
2
, 0

x , the x-axis and the line 0
=
x and
2
=
x respectively about the x-axis.
50. Find the volume of solid of revolution obtained by rotating the region
bounded by the line x
y
2
1
= , the x-axis and the line 0
=
x and 4
=
x
respectively about the x-axis.
51. Find the volume of the sphere obtained by rotating the upper-half of
the circle 2
2
2
a
y
x =
+ about the x-axis.
52. Show that the volume of the solid of revolution obtained by rotating
the region bounded by the curve 3
x
y = , the y-axis and the lines 0
=
y
and 2
=
y about the y-axis is
5
π
4
63
=
V cubic units.
53. Show that ( ) ( )
1
1
0
−
=

+ a
b
b
ax
e
a
e
dx
e
54. Find the value of n so that ( ) 1
2
2
1
2
2
=
−
 dx
x
nx
55. Find the area of the region between the curve 2
4 x
y −
= from 0
=
x
to 3
=
x and the x-axis.
56. Evaluate
a) ( )
 + dx
x 1
3
ln
b)
( )( )
 −
−
+ x
x
e
e
dx
2
2
1
1
57. (a) Find the following (i)

−
dx
x
x
2
3
cos
3
cos
5
2
(ii) dx
x
x
 + 2
9
(b) Show that 





+
=
−
+
 3
5
ln
2
1
4
4
4
3 2
2
dx
x
x
(c) Use partial fractions to find  −
dx
x
x 2
1
2
58. (a) Find dx
x
x
 ln and hence evaluate dx
x
x

2
1
ln
(b) Find  −
+
dz
z
z
2
7
6
when 1
)
3
( =
y

Integration
59. (a) Show that
2
1
2
cos
π
0
4
1
=
 dx
x
(b) Express
( )( )2
1
1
3
4
+
+ x
x
x
in partial fractions
(c) Hence show that
( )( )
2
ln
1
1
1
3
4
1
0 2
−
=
+
+
 dx
x
x
x
60. Find the value of ( )dx
x
x

π
π
4
1
6
1
2
cos
3
sin
2
61. If n is a non-zero integer and m any constant not equal to n, show that
( ) ( ) ( )
( ) n
n
m
m
dx
nx
mx
n
π
sin
2
1
2
cos
2
cos 2
2
π
0
2
1
−
−
=
 .
62. Evaluate ( ) ( )dx
x
x

π
5
.
0
0
4
4
cos
cos
63. Show that
(a) A
x
x
dx
x +
−
=
 2
sin
4
1
2
1
sin2
(b) ( )
4
π
16
1
sin 2
π
0
2
2
1
+
=
 dx
x
x
64. Show that
480
47
cos
sin
π
0
2
3
3
1
=
 

 d
65. Integrate  − x
x
dx
2
2
sin
9
cos
4
66. Given that ( ) x
y
dx
dy 2
sec
2
+
= and that 4
π
)
0
( =
y , find an expression
for y in terms of x.
67. (a) The curves x
y sin
3
= and x
y cos
4
= ( )
π
0 2
1

 x intersect at the
point A, and meet the x-axis at the origin O and the point ( )
0
π,
2
1
B
respectively. With a neat graphs prove that the area enclosed by the
arcs OA, AB and the line OB is unit square units.
(b) If N is a foot of the perpendicular from A to the axis of x, find by
integration the volume obtained when the area enclosed by AN, NB
and the arc AB is completely rotated about the x-axis, giving the
answer correct to five significant figures.

Integration
68. Evaluate the following integrals
(a) ( )
 −
+
i
i
dx
x
x
x
3
2
2
2
sinh
2 (b)  





+
5
3 3
2
3
i
dx
x
69. Find the volume of solid of revolution formed by revolving the finite
region bounded by the curve 1
2
−
= x
y and ( )2
1
−
= x
y about the y
– axis.
70. Find the volume of the solid of revolution generated by rotating the
region in the first quadrant of the curve ( )( )2
3
1 −
−
= x
x
y , 0
=
y
about y – axis.
71. Find the volume of the solid generated when the region bounded by
the curve 3
3
3
+
−
= x
x
y , 0
=
x , 0
=
y and 2
=
x is rotated about the
y – axis.
72. Find the volume of the solid obtained by rotating the region enclosed
by the curve 2
x
y = and 2
4 x
x
y −
= about the line 2
=
x .
73. Consider the region enclosed by the curve x
x
x
f +
= 3
)
( , 2
=
x and
the x – axis. Rotate this region about the y – axis and find the resulting
volume.
74. Find the volume of the solid revolution generated when the region
defined by the curve ( )
1
2
2
+
= x
x
y about the y – axis.
75. Find the volume of the solid generated by revolving the region
bounded by
2
1
1
−
+
=
x
y , 0
=
y , 2
=
x and 6
=
x about the y – axis.
76. For the region bounded by 3
1
+
=
x
x
y , 0
=
y , 0
=
x and 7
=
x . Find
the volume of the solid generated by revolving the region about the x
– axis.
77. Evaluate (a) ( )
 dx
x
x sin
ln
)
cos( (b)
 





+
−
dx
x
x
2
1
1
2
sin
78. Show that 






 +
=
−
+
−
−
 25
4
6
6
sin
6
1 1
1
1 2
dx
x
x
79. Find the area between the curves given in the figure below.

Integration
80. The slope of a curve at a point whose abscissa is x is
2
1
2
x
+ and the
curve passes through the point 1
=
x and 0
=
y . Find the equation of
the curve and the area bounded by it, with the x – axis and the abscissa
1
=
x and 3
=
x
81. Find the expression of ( )
 





+
+ dx
x
x
dx
dy
3
sin
3 2
82. Show that
( )
( )
( )
 = x
Af
dx
x
f
x
f
ln
'
, hence show that if
( ) 
= dx
x
f
x
f
x
f
)
(
)
(
'
)
(
ln 2
1
5
then ( )9
)
(x
f
A =
83. (a) Evaluate ( )

2
π
0
)
cos(
)
sin( dx
x
x
x
(b) Find ( )
 dx
x
x
x 6
cos
4
cos
2
cos
84. Show that ( ) ( )
 +
−
+
−
−
=
−
A
bx
a
b
bx
a
b
a
dx
bx
a
x
2
2
1
ln where a, b
and A are real constants.
85. Find (a)  dx
x
xln
1
(b)  −
dx
x
x 1
1
(c)  − dx
x
x 9
4
86. Find (a)  −
+
dx
e
x
x2
1
(b)  +
dx
x
x
2
cos
2
4
sin
87. Show that the volume represented by a solid of revolution of the
curve 2
2
2
r
y
x =
+ from r
x −
= to r
x = is a sphere.

Integration
88. Show that
2
2
π
cos
1
π
0 2
=
+
 dx
x
x
89. If ( ) 5
.
4
4
3
2
2
=
−
+
a
dx
x
x find the possible value(s) of a
90. Show that  





−
−
=
−
−
x
A
dx
x
x
x
x
2
2
π
cos
2
1
cos
sin
2
1
cos
sin
2
2
8
8
91. Find dx
x
x
x
 +
−
8
4
1
3
1
tan
92. Show that A
x
x
dx
x
x
+
−
=
+
−

− 2
1
2
1
4
π
2
sin
1
2
sin
1
tan
93. Show that for all real values of m,  =
+
2
π
0 4
π
cos
sin
sin
dx
x
x
x
m
m
m
hence
integrate  +
2
π
0 cos
sin
sin
dx
x
x
x
.
94. (a) The diagram below shows the shaded region Q enclosed between
the curve 2
2
1 x
x
y −
= , the x-axis, and the line 5
.
0
=
x .
The area of this region is given by  −
=
5
.
0
0
2
2
1 dx
x
x
A , find the area Q
(b) Find the exact volume generated when the region Q is rotated
through four right angles about the x-axis.
Q

Integration
95. Find (a)
 +
dx
x
x
10
4
1
(b) ( )

−
dx
x
3
sin
cos 1
(c)
 dx
x
x
2
2
sec
cosec
96. (a) Show that  +
−
=
−
A
x
x
dx
x
x
sec
8
tan
5
cos
sin
8
5
2
(b) Show that ( ) ( )
b
a
a
b
dx
x
x
b
a
1
1
2
2
tan
tan
2
1
1 −
−
−
−
−
=
+
−

97. (a) Verify that  +
+
−
=
+
A
x
x
x
dx
x
x
2
2
tan
2
2
sec
2
2
sin
1
2
sin
(b) The marginal cost of a firm with fixed cost of 30 is given by
2
4
50
5
)
( x
x
x
f −
+
= , find the cost function.
(c) The slope of the curve is given by
4
9
2
3
2
3
2
+
+
+
x
x
x
x
, if the curve
passes through point ( )
1
,
0 find the equation of the curve.
(d) Find the area bounded by the curve ( )
3
ln 2
+
= x
y with the x – axis
from 4
−
=
x to 4
=
x .
98. (a) Find  +
−
dx
x
x
x
cos
2
1
2
cos
cos
(b) Suppose ( )( )
b
x
x
x
f −
+
= 1
)
(
' , 2
)
0
( =
f and 3
)
1
( =
f find )
(x
f
(c) Oil flows from the bottom of a storage tank at the rate of
t
t
r 5
150
)
( −
= litres per minutes, where 30
0 
 t . Find the
amount of oil that flows from the tank during the first 20 minutes.
(d) A car accelerates along a straight line, so that its acceleration at
time t is ( ) 2
2
3
)
( −
−
= ms
t
t
t
a . Find the displacement of the car
during the time period 3
2 
 t
99. (a) Use the concept of complex numbers or otherwise, find
( )
 + dx
x
x 5
4
cos
sin
(b) Find  +
+
dx
x
x
x
x
sin
cos
3
sin
9
cos
2
(c) Find  +
dx
x
x
x
4
4
cos
sin
2
sin
(d) Determine the volume of the solid obtained by rotating the region
bounded by 2
2
−
= x
y and x
y = about line 4
=
y .
100. (a) Find the area inside the curve 4
2
2
x
x
y −
=
(b) Using calculus show that the perimeter (arc length) of the
circle with radius r is πr
2 .

Integration
(c) The area defined by the inequalities 1
2
+
 x
y , 0

x , 2

y
is rotated completely about the y-axis. Find the volume of
the solid generated.
101. Integrate the following;-
(a)
 +
dx
x
x
1
4
8
3
(b)
 −
dx
x
x
1
8
3
(c)
 +
−
dx
x
x
x
2
3
2
1
(d)
 +
+
+
−
dx
x
x
x
x
x
1
3
3 2
4
6
3
102. Integrate the following:-
(a)
 +
−
dx
x
x
2
1
1
)
(
tan
(b)
 +
dx
x
x
1
)
(
arctan
2
2
(c)
 +
+
dx
x
x
1
)
1
ln(
(d)
( )
 −
+
+
dx
x
x
x
1
)
1
ln(
2
103. Show that
 +





 −
+
+
+
+
=
−
+
−
A
x
x
x
x
x
x
dx
x
x
x
x 1
ln
1
2
1
2
1
2
2
3
4
3
5
104. Show that A
x
x
x
dx
x
x
+
−
−
=
−
−
 1
2
2
sin
1
1
105. Show that 





+
=
−
+
−
+
 45
4
ln
2
3
5
2
17
4
4
2
1 2
2
dx
x
x
x
x

Integration
ADVANCED MATHEMATICS
INTEGRATION
BARAKA LO1BANGUT1

Calculus Integration.pdf

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