Chapter 3 Integration
Chapter 3
Integration
Part I Integration – as the Inverse of Differentiation
Suppose a function has been differentiated.
Then
integration is the process of reversing our steps in order
to find the original function.
Hence if
dy
 ax n
dx
then
i.e. If
y  ax n
then
a n 1
y
x
n 1
a n 1
 ydx   ax dx  n  1 x  c
n
(except where
n  1 )
Example:
3

f
(
x
)

4
x
If
find f (x )
Solution:
4 31
f ( x) 
x  1.x 4  x 4
3 1
But any constant term of
need
to
conclude
an
f (x ) would be lost so we
arbitrary
constant
such
1
Chapter 3 Integration
that
f ( x)  x 4  c .
So the family of solutions to
n
x
c.
our problem are functions of the form
Integration sign
The integral sign is

2
3
6
t
dt

2
t
c
so we say 
2
Here the function to be integrated is 6t . This is placed
in between the integral sign and dt . This may be the
first time you have seen
dt on its own.
It means that
we are integrating with respect to the variable
Example:
If y  3
find
t.
 ydx
Solution:
0
ydx

3
dx

3
.
1
dx

3
x
 

 dx 
3 01
x  c  3x  c
0 1
2
Chapter 3 Integration
 xdx,

TRY Find the integrals:
1
2
 x dx,
 z dz
3
Exact solutions
Exact solutions can be obtained only when some other
conditions are given that help us to obtain a value for c .
Example:
If f ( x)  4 x find f (x )
3
given
that
f ( x)  20 when x  2
Solution: As before
f ( x)  x 4  c
We also now know that
f ( x)  20
when
x2
i.e.
f (2)  20
Thus we can say that
2 4  c  20,
16  c  20,
c  20  16  4
Thus
f ( x)  x 4  4
3
Chapter 3 Integration
Rules of Integration
Rule 1
If
k is a constant term then  kf ( x)dx  k  f ( x)dx
Rule 2: Sum Rule
  f ( x)  g ( x)dx   f ( x)dx   g ( x)dx
NB: We only need one arbitrary constant.
Rule 3: Linear Composite Rule

1
f (ax  b)dx  F (ax  b)  c
a
where
c is a constant and a  0
2
100
x
 20 sin xdx
Example: 
Solution:
2
 100 x  20 sin xdx 
100 21 20
100 3
x 
cos x  c 
x  20 cos x  c
2 1
1
3
TRY Find the integrals:
 sin x  2 x  1dx,
 x  2 x dx
2
4
Chapter 3 Integration
Definite and Indefinite Integration
So far we have worked out integrals and our results have
3
4
4
x
dx

x
 c.
been functions e.g. 
There is
another integration process which we shall need which is
required to actually solve a problem. This is known as
definite integration and the result is a number rather than
a function.
Example:

2
Find the definite integral 1
4 x 3 dx
Solution: Here the upper limit is 2 and lower limit is 1.
First we integrate as before we get:
x 
4 2
1
Notice how we don’t bother with the constant of
integration in definite integration, but we have written
our limits in square brackets around the function. Now
we work out the value of the function to the upper limit
and subtract form this value of the function at the lower
limit i.e.
x   2   1   16  1  15
4 2
1
4
4
5
Chapter 3 Integration
Hence

2
1
4 x 3 dx  15
Example: Find the definite integral


2
0
3 cos tdt
Solution: Here the upper limit is 0 and the lower limit

is 2 .
1
3
cos
tdt

3
sin t  3 sin t  c
First we integrate 
1
So we want to find the definite integral

3 sin t 02


  3 sin   3 sin 0
2

TRY Find the definite integrals (below):
6
Chapter 3 Integration
1
3
x
 dx,
1
4
x
1
2
dx,
0
 1  3x  2 x dx,
1
2
1
 x
2
3

 x  2 dx,
1
 3x
1
2

 7 dx,
1

2
 cos 2 xdx,
4
 x dx
2
1
1
7
Chapter 3 Integration
Part II Integration – as the Inverse of Differentiation
Substitution
corresponds
in
the
chain
rule
in
differentiation. We will look at two different substitution
methods.
Substitution method for
Example:
 f (ax  b)dx
3
(
3
x

2
)
dx
Find the integral 
Solution: Firstly let
u  3x  2 . Now
the integral is
3
u
 dx
Unfortunately we cannot just integrate this, instead we
must firstly transform
dx
to du .
How do we do this?
Well if
u  3x  2
du
3
.
dx
then
This can then be rearranged such that
hence
du  3dx
and
1
dx  du
3 .
We can now rewrite
3
u
 dx
as
8
Chapter 3 Integration
1  1 3
 u  3 du   3  u du .
3
Now we can integrate and so
1 3
1  u4 
1 4


u
du


c

u c



.
3
3 4 
12
Of course we cannot finish here; we must substitute for u
to get our answer in terms of x.

1 4
1
u  c  (3x  2) 4  c
.
12
12
Example: Find the integral
Solution: Rewrite
Let
dx
 2x 1
dx
1
 2 x  1   2 x  1 dx
u  2x  1
Now the integral is
Transform
1
 u dx
dx to du
du
2
dx
9
Chapter 3 Integration
Therefore
1
du  2dx  dx  du
2
Rewrite
1
11  1 1
 u dx   u  2 du   2  u du
Integrate
1 1
1
1
du

ln
u

c

ln 2 x  1  c

2 u
2
2
Example:
Find the integral  sin( 3 x  2)dx
Solution: Let u  3x  2
Now the integral is
Transform
Therefore
 sin udx
dx to du
du
3
dx
1
du  3dx  dx  du
3
1  1
sin
udx

sin
u
  3 du   3  sin udu
Rewrite 
10
Chapter 3 Integration
Integrate
1
1
1


sin
udu


cos
u

c


cos u  c

3
3
3
Substitute for u:
1
1
  cos u  c   cos(3x  2)  c
3
3
Example: Find the integral
 cos(10 x)dx
Solution: Let u  10x
Now the integral is
Transform
Therefore
 cos udx
dx to du
du
 10
dx
1
du  10dx  dx  du
10
Rewrite
1
 1
cos
udx

cos
u
du


  10   10  cos udu
Integrate
1
1
sin u   c
cos
udu

10 
10
Substitute for u:

1
1
sin u  c  sin 10 x   c
10
10
11
Chapter 3 Integration
7
(
5
x

8
)
dx
Find the integral 
Example:
Solution: Let
u  5x  8 .
7
u
 dx
Now the integral is
Transform
Therefore
dx to du
du
5
dx
du  5dx  dx 
1
du
5
 1 7
7
71
u
dx

u
du
  5   5  u du
Rewrite 
Integrate
1 7
1 u 7 1 1 u 8 u 8
u du  .
 . 
c

5
5 7  1 5 8 40

u8
5 x  8
c 
c
40
40
8
Substitute for u:
12
Chapter 3 Integration
Example:
dx
Find the integral  9x  6
Solution: Let
u  9x  6 .
Now the integral is
Transform
Therefore
dx
to du
1
 u dx
du
9
dx
du  9dx  dx 
1
du
9
Rewrite
1
11  1 1
 u dx  u  9 du   9  u du
Integrate
1 1
1
du

ln u  c

9 u
9
1
1
ln
u

c

ln 9 x  6   c
Substitute for u: 9
9
13
Chapter 3 Integration
TRY Find the integrals:
 sin 7 x  1dx,
 cos7 x  1dx,
 e dx,
 t cos2t dt
11x  5
2
14
Chapter 3 Integration
Substitution method for
Example:
2
xf
(
ax
 b)dx

Find the integral
Solution: Firstly let
 xe
x2
dx
u  x2
with the aim of simplifying the unfamiliar term
e
x2
du
 2x
Now we must transform dx to du , so dx
This can then be rearranged such that
and hence
du  2xdx
1
xdx  du
2 .
(Note we do not divide through be x as we want all the x
terms on one side of the = sign and all the u terms on the
other)
We can now rewrite
 xe
x2
dx
 1 u
u1
e
(
xdx
)

e
du

  2   2  e du
x2
as
1 u
1 u
e du  e  c
Now we can integrate and so 2 
2
15
Chapter 3 Integration
Of course we cannot finish here. We must substitute for u
1 u
1 x2

e

c

e c.
to get our answer in terms of x 2
2
Example:
Find the integral
Solution: Let
u  3x 2  2
xdx
 3x 2  2
xdx
x
 2
  dx
Now the integral is
u
3x  2
du
1

6
x

6
xdx

du

xdx

du .
Transform dx
6
Rewrite
Integrate
x
1
11  1 1
 u dx   u xdx   u  6 du   6  u du
1 1
1
du

ln u  c

6 u
6
Substitute for u:


1
1
ln u  c  ln 3x 2  2  c .
6
6
16
Chapter 3 Integration
Example: Find the integral
Solution: Let
 
2
x
cos
7
x
dx

u  7x 2
Now the integral is
 
  x cos 7 x 2 dx   x cos udx   cos uxdx
Transform
xdx to function of du
du
1
 14 x  14 xdx  du  xdx  du .
dx
14
1
 1
cos
uxdx

cos
u
du
  14   14  cos udu
Rewrite 
Integrate
1
1
cos udu  sin u  c

14
14
Substitute for u:
 
1
1
sin u  c  sin 7 x 2  c .
14
14
17
Chapter 3 Integration
 
2
x
sin
2
x
dx
Example: Find the integral 
Solution: Let
u  2x 2
Now the integral is
 
  x sin 2 x 2 dx   x sin udx   sin uxdx
Transform
xdx to function of du
du
1
 4 x  4 xdx  du  xdx  du .
dx
4
Rewrite
Integrate
1  1
sin
uxdx

sin
u

  4 du   4  sin udu
1
1
sin udu   cos u  c

4
4
 
1
1
2

cos
u

c


cos
2
x
c.
Substitute for u:
4
4
18
Chapter 3 Integration
Example: Find the integral
Solution: Let
u  9x 2  6

Now the integral is 
Transform
xdx
 9x 2  6
xdx
x
1

dx

xdx
2


u
u
9x  6
xdx to function of du
du
1
 18 x  18 xdx  du  xdx  du
dx
18
Rewrite
1
1 1
 1 1
xdx

du
u
 u  18   18  u du
Integrate
1 1
1
du  ln u  c

18 u
18


1
1
2
ln
u

c

ln
9
x
 6  c.
Substitute for u: 18
18
19
Chapter 3 Integration
10 xdx
Find the integral  5 x 2  4
Example:
u  5x 2  4
Solution: Let
Now the integral is :
10 xdx
10 x
1
 2

dx  10  xdx
u
u
5x  4
Transform
xdx to function of du
du
1
 10 x  10 xdx  du  xdx  du .
dx
10
1
1 1
1 1
1

10
xdx

10
du

10
du



 u  10  10  u  u du
Rewrite  u
Integrate
1
 u du  ln u  c
Substitute for u:


ln u  c  ln 5x 2  4  c .
20
Chapter 3 Integration
TRY Find the integrals:
9 x 2 dx
 3x 3  4 ,
6 x 2  2 xdx
 2 x 3  x 2  1,
4x
 2 x 2  1 dx,


2
x
cos(
3
x
)dx,



2
x
x
  3 dx
7
21
Chapter 3 Integration
Substitution method for
m
n
cos
ax
sin
axdx

with
(m or n odd)
Example: Find the integral
3
sin
 x cos xdx
u  sin x
Solution: Firstly let
with the aim of
simplifying the worst term.
Begin by transforming
So, if
dx to du .
du
u  sin x then dx  cos x .
This can then be rearranged such that
We can now rewrite
du  cos xdx
3
sin
 x cos xdx
as
3
3
sin
x
(cos
xdx
)

u

 du
Now we can integrate and so
3
u
 du 
1 4
u c
.
4
Of course we cannot finish here; we must substitute for u
to get our answer in terms of x.
1 4
1
 u  c  sin 4 x  c
.
4
4
22
Chapter 3 Integration
Part III Integration by Parts
Integration by Parts
If u and v are functions of x then by the product rule
d
dv
du
(uv)  u
v
dx
dx
dx .
Integrating both sides with respect to x gives
dv
du
uv   u dx   v dx
dx
dx .
Rearranging the terms in the equation above we have
dv
du
 u dx dx  uv   v dx dx
So if we have a product of two functions to integrate, one
factor is chosen as u and the other is the differential
coefficient of some function v. Knowing u and v, we can
substitute them into the equation above and we end up
with another product to integrate but usually this is
simpler than the original one.
23
Chapter 3 Integration
Example: Find
 x cos( x)dx
Solution: We need to rewrite as
u
dv
du
dx  uv   v dx
dx
dx
So we define the separate parts:
du
u  x;
 1,
dx
dv
dv
 cos( x), v   dx   cos( x)dx  sin x  c
dx
dx
Now we combine our terms to make up the full function
and find the solution
u
dv
du
dx  uv   v dx
dx
dx
  x cos xdx  x. sin x   sin x.1dx
 x. sin x   sin x.dx  x sin x  ( cos x)  c  x sin x  cos x  c
24
Chapter 3 Integration
Example:
2
x
ln xdx
Find 
Solution: We need to rewrite as
dv
du
 u dx dx  uv   v dx dx
So we define the separate parts:
du 1
u  ln x,
 ,
dx x
3
dv
dv
x
 x 2 , v   dx   x 2 dx   c
dx
dx
3
Now we combine our terms to make up the full function
and find the solution
dv
du
dx  uv   v dx
dx
dx
x3 x3 1
2
  x ln xdx  ln x.   .. dx
3
3 x
x3
x2
x3
1 2
x3
1  x3 
x3
x3
 ln x   dx  c  ln x   x dx  c  ln x     c  ln x   c
3
3
3
3
3
3 3 
3
9
u
25
Chapter 3 Integration
How do we decide which part of the function is u and
dv
which part is dx ?
The priority for u is as follows:
(1) ln x , (2)
xn ,
kx
e
(3)
(in that order)
Example: Use integration by parts to find
 t sin( t )dt
Solution: We need to rewrite as
dv
du
 u dt dt  uv   v dt dt
So we define the separate parts:
u  t,
du
 1,
dt
dv
dv
 sin t , v   dt   sin tdt   cos t  c
dt
dt
Now we combine our terms to make up the full function
and find the solution
u
dv
du
dt  uv   v dt
dt
dt
  t sin tdt  t.( cos t )   cos t..1dt  t cos t   cos tdt  t cos t  sin t  c
26
Chapter 3 Integration
Example: Use integration by parts to find
t
te
 dt
Solution: We need to rewrite as
dv
du
 u dt dt  uv   v dt dt
So we define the separate parts:
du
u  t,
 1,
dt
dv
dv
t
 e , v   dt   e t dt   e t  c
dt
dt
Now we combine our terms to make up the full function
and find the solution
dv
du
 u dt dt  uv   v dt dt
  te t dt  t.(e t )    e t .1dt


 te t   e t dt  te t   e t  c  te t  e t  c
27
Chapter 3 Integration
3
x
 ln xdx,
TRY Find the integrals:
 x
x
sin
  2 dx
TRY Find the definite integral:
2
 x ln xdx
1
28
Chapter 3 Integration
Using the Integration by Parts rule twice
Sometimes to get the result integration by parts has to be
applied again to evaluate the integral on the left hand
side.
Example:
2
x
sin xdx
Find 
Solution: We need to rewrite as
dv
du
 u dx dx  uv   v dx dx
So we define the separate parts:
u  x2 ,
du
 2 x,
dx
dv
dv
 sin x, v   dx   sin xdx   cos x  c
dx
dx
Now we combine our terms to make up the full function
and find the solution
u
dv
du
dx  uv   v dx
dx
dx
  x 2 sin xdx  x 2 .( cos x)   cos x.2 xdx   x 2 cos x  2 x cos xdx  c
29
Chapter 3 Integration
So now we need to find
 x cos xdx (see
above
repeated here….)
Define the separate parts:
du
u  x;
 1,
dx
dv
dv
 cos( x), v   dx   cos( x)dx  sin x  c
dx
dx
Combine terms to make up the full function and find the
solution
u
dv
du
dx  uv   v dx   x cos xdx  x. sin x  sin x.1dx
dx
dx
 x. sin x   sin x.dx  x sin x  ( cos x)  c  x sin x  cos x  c
Put everything together:
dv
du
u
dx

uv

v
 dx
 dx dx
  x 2 sin xdx   x 2 cos x  2 x cos xdx  c   x 2 cos x  2x sin x  cos x   c
TRY Find the integral:
2
t
 cos tdt,
30
Chapter 3 Integration
Integration by Partial Fractions
x
How do we integrate something like ( x  2)( x  1) ?
We can’t use any of the techniques discussed previously.
Therefore we have to express
x
( x  2)( x  1)
as a partial fraction and then integrate the result.
Consider the following example:
7 1 1
 
12 4 3 .
1
1
7
4 and 3 are said to be partial fractions of 12 .
x
2
1


If we write:
( x  2)( x  1) x  2 x  1
1
2
we can then say that x  2 and x  1
x
are the partial fractions of ( x  2)( x  1) .
The problem is how do we find these partial fractions?
That is what we need to find out.
31
Chapter 3 Integration
Finding the Partial Fractions
Rules of partial fractions
If we need to write an algebraic fraction
f ( x)
g ( x)
as a sum
of partial fractions, we should
 factorize the denominator g (x )
 then use one of the two rules:
Rule 1
numerator
A
B


(ax  b)(cx  d ) ax  b cx  d
x
Example: Resolve ( x  2)( x  1) into partial fractions.
Solution:
 Let
x
A
B


( x  2)( x  1) ( x  2) ( x  1) .
 Now add the fractions
A( x  1)  B( x  2)
A
B


( x  2)( x  1) .
( x  2) ( x  1)
 Hence we can say
x
A( x  1)  B( x  2)

( x  2)( x  1)
( x  2)( x  1) .
 Since we have a common denominator (‘bottom’) on
both sides of the equal sign, the numerators (‘top’)
32
Chapter 3 Integration
must equal each other, that is
x  A( x  1)  B( x  2) .
 Now we need to find A and B. Expand out
x  A( x  1)  B( x  2)
such that
x  Ax  A  Bx  2B .
 Gather all the x components together
x  Ax  Bx and thus 1  A  B .
 Now gather all the ‘constant’ components together
0  A  2B .
 We now have two equations:

1 A B
0  A  2B
(1)
(2)
that we can solve simultaneously.
 Subtract equation (2) from equation (1)
1 A B
-
0  A  2B
 1 .
If we substitute B  1 into 1  A  B we get
1  A  1 So 2  A .
Hence we now know that A  2 and B  1 .
1

B
thus B
 If we go back to our initial expression:
x
A
B


( x  2)( x  1) ( x  2) ( x  1) ,
33
Chapter 3 Integration
we can substitute for A and B to get

x
2
1


( x  2)( x  1) ( x  2) ( x  1) .
x
Thus ( x  2)( x  1)
expressed in terms of its partial fractions is
2
1

( x  2) ( x  1) .
Example: Resolve
2x
x2  x  2
into partial fractions.
Solution:
 We know that
So we can let
x 2  x  2  ( x  1)( x  2)
2x
A
B


2
x  x  2 ( x  2) ( x  1) .
 Now add the fractions
A( x  1)  B( x  2)
A
B


( x  2)( x  1) .
( x  2) ( x  1)
 Hence we can say
2x
A( x  1)  B( x  2)

( x  2)( x  1)
( x  2)( x  1) .
34
Chapter 3 Integration
 Since we have a common denominator (‘bottom’) on
both sides of the equal sign, the numerators (‘top’)
must equal each other, that is:
2 x  A( x  1)  B( x  2) .
 Now we need to find A and B. Expand out
2 x  A( x  1)  B( x  2)
such that
2x  Ax  A  Bx  2B .
 Gather all the x components together
2x  Ax  Bx
and thus 2  A  B .
 Now gather all the ‘constant’ components together
0  A  2B .
 We now have two equations:
2  A B
0  A  2B
(1)
(2)
that we can solve simultaneously.
 Subtract equation (2) from equation (1)
2  A B
-
0  A  2B
2
Thus
3B
B2
3.
35
Chapter 3 Integration
 If we substitute
B2
3
into
2  A  B we get
2
2  A
3
2 62 4
A 2 

3
3
3.
Hence we now know that
A
4
3
2
B

and
3.
 If we go back to our initial expression:
2x
A
B


x 2  x  2 ( x  2) ( x  1) ,
we can substitute for A and B to get

4
2
2x
3 
3

x 2  x  2 ( x  2) ( x  1) .
2x
Thus x 2  x  2
expressed in terms of its partial fractions is
4
2

3( x  2) 3( x  1) .
36
Chapter 3 Integration
TRY Resolve into partial fractions:
 12u  13
,
2u  1u  3
5 x  13
x  3x  2 
Rule 2
numerator
A
B


ax  b (ax  b) 2
(ax  b) 2
Example:
2t
Resolve (t  1) 2 into partial fractions.
Solution:
2t
A
B


 We can let (t  1) 2
t  1 (t  1) 2 .
 Now add the fractions
A(t  1)  B
A
B


2
(t  1) 2 .
t  1 (t  1)
2t
A(t  1)  B

 Hence we can say (t  1) 2
(t  1) 2 .
 Since we have a common denominator (‘bottom’) on
both sides of the equal sign, the numerators (‘top’)
must equal each other, that is 2t  A(t  1) 
 Now we need to find A and B. Expand out
B.
2t  A(t  1)  B
37
Chapter 3 Integration
such that
2t  At  A  B .
 Gather all the t components together
2t  At
and thus 2  A .
 Now gather all the ‘constant’ components together
0  A  B .
and thus
0  2  B  B  2 .
Hence we now know that A  2 and B
 If we go back to our initial expression:
2t
A
B


2
t  1 (t  1) 2
(t  1)
2.
,
we can substitute for A and B to get
2t
2
2


2
t  1 (t  1) 2
(t  1)
 Thus
.
2t
(t  1) 2
expressed in terms of its partial fractions is
2
2

t  1 (t  1) 2
.
x 1
TRY Resolve into partial fractions:
x  12
38
Chapter 3 Integration
Integration by partial fractions
Now that we can express algebraic functions as partial
fractions then we can also integrate such functions.
Example:
x
dx

Find ( x  2)( x  1)
.
x
Solution: We know from above that ( x  2)( x  1)
2
1

expressed in partial fractions is:
( x  2) ( x  1)
So our integral is now
 2
1 
2
1



dx

dx

  ( x  2) ( x  1)   ( x  2)  ( x  1)dx
1
1
 2
dx  
dx
( x  2)
( x  1)
1
2
dx
First we find
( x  2)
du
u  x2
 1  dx  du
Let
dx
1
2 du  2 ln u  c
Our integral is now  u
39
Chapter 3 Integration
So re-substitute for u
Now find
2 ln u  c  2 ln x  2  c
1
 ( x  1)dx
du
u

x

1

 1  dx  du
Let
dx
1
du  ln u  c
Our integral is now  u
ln u  c  ln x  1  c
So re-substitute for u
Now bring our terms back together for our solution:
1
1
2
dx  
dx  2 ln x  2  ln x  1  c
( x  2)
( x  1)
40
Chapter 3 Integration
Example:
2x
dx
Find  x 2  x  2
2x
Solution: We know from above that x 2  x  2
4
2

Expressed in partial fractions is: 3( x  2) 3( x  1)
So our integral is now
 4
2 
4
2
  3( x  2)  3( x  1) dx   3( x  2) dx   3( x  1)dx
4
1
2
1
 
dx  
dx
3 ( x  2)
3 ( x  1)
4
1
dx
First we find 3  ( x  2)
du
u  x2
 1  dx  du
Let
dx
4 1
4
du  ln u  c
Our integral is now 3  u
3
So re-substitute for u
4
4
ln u  c  ln x  2  c
3
3
41
Chapter 3 Integration
2
1
dx

Now find
3 ( x  1)
du
u

x

1

 1  dx  du
Let
dx
2 1
2
du

ln u  c
Our integral is now 3  u
3
2
2
ln u  c  ln x  1  c
So re-substitute for u 3
3
Now bring our terms back together for our solution:
4
1
2 1
4
2
dx

dx

ln
x

2

ln x  1  c


3 ( x  2)
3 ( x  1)
3
3
TRY
Determine
the
following
integrals
(See
solutions from problems above):
3x  4
 ( x  1)( x  2) dx,
2s  7
 s 2  s  2ds
42