“Teach A Level Maths”
Vol. 2: A2 Core Modules
27: Integration by Substitution
Part 2
© Christine Crisp
Integration by Substitution Part 2
Module C3
AQA
Module C4
Edexcel
OCR
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Integration by Substitution Part 2
A useful example of integration by substitution is
to find
 tan x dx

sin x
We write
 tan x dx  cos x dx
du
du
Let u  cos x 
  sin x 
 dx
dx
 sin x
sin x
sin x
du
So,
dx 

cos x
u
 sin x
1

du   ln u  C
u
  ln cos x  C



Integration by Substitution Part 2
  ln cos x  C
Using the 3rd law of logs,
 ln (cos x ) 1  C
1
 ln
cos x
C
 ln sec x  C
Integration by Substitution Part 2
We have already shown that
y  sin
1
dy
x 

dx
1
1  x2
Since indefinite integration is the reverse of
differentiation,

1
1 x
2
dx  sin 1 x  C
We can show this result directly by using a trig
substitution.
The following examples and exercises are difficult.
You are unlikely to be asked to do them in an exam
but you will find it useful to follow the method.
Integration by Substitution Part 2
e.g. 1

1
1  x2
x  sin u
Let
dx
dx
u  dx
cos u du
N.B. Instead of defining u as acos
function
of 
x
we
du
have defined 1x as a function of 1u.
So,
dx 
cos u du
2
2
1 x
1  sin u


Use the identity:
Can you spot what to do next?
cos u  sin u  1  cos u  1  sin u
1
1
 
cos u du  
cos u du
1  sin 2 u
cos 2 u
2
2
2
2
Integration by Substitution Part 2
So,

1
1 x
2
dx 

1
cos 2 u
cos u du
where

 1 du


We need u from the
substitution expression:


1
1 x
2
x  sin u
1
cos u du
cos u
uC
x  sin u  u  sin 1 x
dx  sin 1 x  C
Integration by Substitution Part 2
We can get a more general result by a similar
method:

1
a2  x2
1 
x
dx  sin    C
a
Check that this is in your formula book. You can
then quote it without proof and use it for any value
of a.
However, you may like to try using substitution for
examples in the next exercise.
Integration by Substitution Part 2
Exercise
1. Find

1
9  x2
using the substitution
x  3 sin u
1
1  x 
dx  tan    C
2
2
4 x
2 
substitution x  2 tan u
2. Show that
using the
dx

1
this is an example of the general result

1
1  x 
dx

tan

C


a
a2  x2
a
1
Integration by Substitution Part 2
Solutions:
1.


1
9  x2
1
9  x2
dx





dx 
Let
x  3 sin u
dx
 3 cos u  dx  3 cos u du
du
1
3 cos u du
2
9  9 sin u
1
3 cos u du
9(1  sin 2 u)
1
3 cos u du
9 cos 2 u
Integration by Substitution Part 2

1
9 x
2
dx 



1
2
9 cos u
So,

where
x  3 sin u
1
3 cos u du
3 cos u
  1 du
To subs.
back:
3 cos u du
uC
x
1  x 
x  3 sin u   sin u  u  sin  
3
3
1
1  x 
dx  sin    C
3
9  x2
Integration by Substitution Part 2
1
1  x 
dx

tan

C


2
4  x2
2
using the substitution x  2 tan u
Solution:
x  2 tan u
dx
 2 sec2 u  dx  2 sec2 u du
du
1
1
2
So,
dx 
2 sec u du
2
2
4 x
4  4 tan u
1
2

2
sec
u
du
4(1  tan 2 u)
2. Show that


Use the identity:
1


sec2 u  1  tan 2 u
Integration by Substitution Part 2
 4(1  tan
1
2 sec u du 
2
2
u)
 4 sec u
1
2
2 sec u du
2


1
du
2
1
 uC
2
x
1  x 
x  2 tan u 
 tan u  u  tan  
2
2
1
x
1



1
So,

tan

C
dx


2
4  x2
2

2
Integration by Substitution Part 2
SUMMARY
The following results can be proved by trig
substitutions:

x
dx  sin    C
2
2
a
a x
1
1
1 x
dx  tan
C
2
2
a
a
a x
1

1 
Integration by Substitution Part 2
Integration by Substitution Part 2
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Integration by Substitution Part 2
e.g. 1
Let

1
1 x
x  sin u
2
dx
N.B. Instead of defining u
as a function of x we have
defined x as a function of u.
dx
 cos u  dx  cos u du
du
1
1
So,
dx 
cos u du
2
1 x
1  sin 2 u


Use the identity:
cos 2 u  sin 2 u  1  cos 2 u  1  sin 2 u
1
1

cos u du 
cos u du
2
1  sin u
cos 2 u


Integration by Substitution Part 2
So,

1
1 x
2
dx 

1
cos 2 u
cos u du
where

 1 du


We need u from the
substitution expression:


1
1 x
2
x  sin u
1
cos u du
cos u
uC
x  sin u  u  sin 1 x
dx  sin 1 x  C
Integration by Substitution Part 2

1
1 x
dx

tan
2
2
4  x2
using the substitution x  2 tan u
Solution:
x  2 tan u
dx
 2 sec2 u  dx  2 sec2 u du
du
1
1
2
So,
dx 
2 sec u du
2
2
4 x
4  4 tan u
1
2

2
sec
u
du
4(1  tan 2 u)
2. Show that

Use the identity:
1


sec2 u  1  tan 2 u
Integration by Substitution Part 2
 4(1  tan
1
2 sec u du 
2
2
u)
 4 sec u
1
2
2 sec u du
2


1
du
2
1
 uC
2
x
1  x 
x  2 tan u 
 tan u  u  tan  
2
2
1
1
1 x
So,
dx  tan
C
2
2
2
4 x

2