“Teach A Level Maths”
Vol. 2: A2 Core Modules
26: Integration by Substitution
Part 1
© Christine Crisp
Integration by Substitution Part 1
Module C3
AQA
Module C4
Edexcel
MEI/OCR
OCR
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Integration by Substitution Part 1
Integration by substitution can be used for a variety
of integrals: some compound functions, some products
and some quotients.
Sometimes we have a choice of method.
Integration by Substitution Part 1
e.g. 1
 (1  2 x ) dx
4
We have already met this type of integral.
See if you can do it.
Solution:
Reversing the chain rule gives
 (1  2 x ) dx 
4

(1  2 x )
5
C
5( 2)
(1  2 x )
5
C
10
We’ll use this example to illustrate integration by
substitution but if you got it right you can continue to
use the reverse chain rule ( also called inspection ).
Integration by Substitution Part 1
e.g. 1
 (1  2 x ) dx
4
Method: We must substitute for x and dx.
•
Define u as the inner function
Let u  1  2 x
•
Differentiate:
•
du
dx
2

du
2
 dx
Substitute for the inner du
function . . .
Find dx by treating
like a fraction
4
 (1  2 x ) dx dx
Integration by Substitution Part 1
e.g. 1
 (1  2 x ) dx
4
Method: We must substitute for x and dx.
•
Define u as the inner function
Let u  1  2 x
•
Differentiate:
•
du
2

du
dx
2
 dx
Substitute for the inner function . . . and dx

4
dx 
(
1

2
x
)


u
4
Integration by Substitution Part 1
e.g. 1
 (1  2 x ) dx
4
Method: We must substitute for x and dx.
•
Define u as the inner function
Let u  1  2 x
•
Differentiate:
•
du
dx
2

du
2
 dx
Substitute for the inner function . . . and dx
4
4 du
 (1  2 x ) dx   u
2
Integration by Substitution Part 1
e.g. 1
 (1  2 x ) dx
4
Method: We must substitute for x and dx.
•
Define u as the inner function
Let u  1  2 x
•
Differentiate:
•
du
2

dx
du
2
 dx
Substitute for the inner function . . . and dx
4
u
4
4 du
 (1  2 x ) dx   u
 
du
2
2
5
•
Integrate:
•
Replace u:

u
10
C
Integration by Substitution Part 1
e.g. 1
 (1  2 x ) dx
4
Method: We must substitute for x and dx.
•
Define u as the inner function
Let u  1  2 x
•
Differentiate:
•
du
2

dx
du
 dx
2
Substitute for the inner function . . . and dx
4
u
4
4 du
 (1  2 x ) dx   u
 
du
2
2
5
•
Integrate:

•
Replace u:

u
C
10
5
( 1  2x )
10
C
Integration by Substitution Part 1
e.g. 1
 (1  2 x ) dx
4
Method: We must substitute for x and dx.
•
Define u as the inner function
Let u  1  2 x
•
Differentiate:
•
du
2

dx
du
 dx
2
Substitute for the inner function . . . and dx
4
u
4
4 du
 (1  2 x ) dx   u
 
du
2
2
5
•
Integrate:

•
Replace u:

u
C
10
5
( 1  2x )
10
C
Integration by Substitution Part 1
Students studying the OCR specification should skip
the next slide.
SKIP
Integration by Substitution Part 1
e.g. 2

x 1  x dx
2
This is a product but we can’t use integration by
parts. Why not?
ANS: 1  x 2 is a compound function with inner
function non-linear. We can’t integrate it.
If we chose to integrate x instead, at the next
stage we would have a more complicated integral
than the one we started with.
We substitute as before, but using the inner
function of the 2nd factor in the product.
Integration by Substitution Part 1
e.g. 2
•
•
•

x 1  x dx
2
Define u as the inner function:
2
Let
u  1 x
du
du
Differentiate:
 2x 
 dx
dx
2x
Substitute for the inner function and dx
 x 1  x dx 
2

x u
1
du
2x
Cancel the
extra x
u 2
If x won’t cancel we will have
 2todumake an extra
substitution. We’ll do an example later.
Integration by Substitution Part 1
So,
1
 x 1  x dx 
2
2
u

du
where
2
3
•
Integrate:

2
u
2
C
3
2
3

•
u
2
C
3
Substitute back:
(1  x )
2

3
3
2
C
u  1 x
2
Integration by Substitution Part 1
SUMMARY
Substitution can be used for a variety of integrals
4
2
e.gs.
(1  2 x ) dx
x 1  x dx
Method:
• Define u as the inner function
• Differentiate the substitution expression and
rearrange to find dx
• Substitute for the inner function and dx
• If there’s an extra x, cancel it
If x won’t cancel, rearrange the substitution
expression to find x and substitute for it
• Integrate
• Substitute back


Integration by Substitution Part 1
Exercises
Use substitution to integrate the following. (Where
possible, you could also use a 2nd method.)
1.
 (1  x )
8
dx
3.
 x(1  x
2.
2 4
e
) dx
3 x
dx
Integration by Substitution Part 1
Solutions:
1.
 (1  x )
So,
8
Let
dx
 (1  x )
8
dx 
u  1 x
du
 1   du  dx
dx
u


u
8
du
9
C
9
9
(1  x )
9
C
Integration by Substitution Part 1
Solutions:
2.
So,
e
3 x

3 x
e
Let
dx
du 
dx  e  

 3 
u
e
 
du
3
u
3 x
e
e

C 
C
3
3


u
u  3 x
du
du
 dx
 3  
3
dx
Integration by Substitution Part 1
Solutions:
2 4
3.
x (1  x ) dx

So,
Let
u  1 x
du
du
 dx
 2x 
2x
dx
2 4
4 du
 x(1  x ) dx  xu 2 x
4
u

du
2
5
2 5
u
(1  x )

C 
C
10
10


2
Integration by Substitution Part 1
Definite integration
We work in exactly the same way BUT we must also
substitute for the limits, since they are values of x
and we are changing the variable to u.
A definite integral gives a value so we never return
to x.
Integration by Substitution Part 1
e.g. 1

1
0
x
(1  x )
2 3
Let
u  1 x
du
du
 2x 
 dx
dx
2x
dx
Limits:
So,
2
x  0  u  1 x  1
2
x  1  u  1 x  2
1
2 x du
x
dx 
3
2 3
2x
0 (1  x )
1 u
2




2
1
1
2u
3
du


2
1
u
3
2
du
Integration by Substitution Part 1
So,

1
0
x
(1  x )
2 3
dx 

2
1
u
3
du
2
2
2
u 
 

41
2
1 

 
2 
4
u

1
 1   1
 
  

 16   4 
3

16
Integration by Substitution Part 1

e.g. 2
1
0
e
x
(1  e )
x
2
dx
Let
u  1 e
du
du
x
e

 dx
x
dx
e
x
Limits: x  0  u  1  e 0  2
x  1  u  1 e
So,

1
0
e
x
(1  e )
x
2
dx 



1e
2
1 e
2
e
x
du
2
e
u
1
u
2
x
du
Integration by Substitution Part 1
So,

1
0
e
x
(1  e )
x
2
dx 
u
We leave answers in
the exact form.
1
u
2
You will often see this 
written as du
2

1 e
1 e
2
u
1
2
du
where
2
du
1 e
u 


 1  2
u  1 e
1 e
 1
  
 u 2
1 

 1
 
  

 1 e
 2
1
1
 
2 1 e
x
Integration by Substitution Part 1
Exercises
1.
2
0
xe
x
2
dx
Give exact answers.
2.

1
0
2x
4 x
2
dx
Integration by Substitution Part 1
Solutions:
2
1.
0
xe
x
2
Let
dx
u x
du
2

 2x
du
 dx
2x
dx
Limits: x  0  u  0
x2  u4
So
N.B.
2
0
xe
x
2
dx 

4
0
e 1
0
xe
u
du

2x
e4 
e0 
    
 2 
 2 

4
0
e 1
4

4
e 
du  

2
 2 0
e
u
2
u
Integration by Substitution Part 1

2.
1
0
2x
4 x
2
Let
dx
du
u  4 x
 2x
2
du

2x
dx
Limits: x  0  u  4
x 1  u5
So,

1
0
2x
4 x
2

dx 
 ln u
5
4
4
5
2 x du
u 2x


5
4
1
du
u
 ln 5  ln 4  ln
We can use the log laws to simplify this.
5
4
 dx
Integration by Substitution Part 1
In the next examples, the extra x doesn’t
conveniently cancel so we need to substitute for
it.
Integration by Substitution Part 1
e.g. 3
 x(2  x ) dx
6
This function:
time we could
• Another
Define product.
u as the inner
Letuse uintegration
 2 x
by parts but instead we’ll use substitution.
du
 1  du  dx
• Differentiate:
dx
• Substitute for the inner function and dx
 x( 2  x )
6
dx 
 xu
6
du
Using
u  2xdoesn’t
x  u
 2  xso we must substitute
The extra
cancel
for it.
So,
 xu du
6


( u  2)u du
6
Integration by Substitution Part 1
So,
 x( 2  x )
6
dx 

( u  2)u du
6
( where u  2  x )
Can you spot the important difference between these?
Ans: We can easily multiply out the brackets in the 2nd
  u  2u du
7
•
Integrate:

u
8

6
2u
8
•
Replace u:

7
C
7
(2  x )
8
8

2( 2  x )
7
7
C
Integration by Substitution Part 1
e.g. 4
Let
4x
 (1  x )
4
dx
u  1 x
du
 1  du  dx
dx
Tip: Don’t be tempted to substitute for the extra
x . . . until you’ve checked to see if it cancels.
Integration by Substitution Part 1
e.g. 4
Let
4x
 (1  x )
So,
4
dx
4x
 (1  x )
4

dx
x doesn’t cancel so
u  1 x  u 1  x
now substitute:
4x
u

u  1 x
du
 1  du  dx
dx
4x
du
4
u
4
du 
A multiplying constant . . .
can be taken outside the
integral.

4
4( u  1)
u
1
u
 uu
4
du
4
3

1
u
4
du
Integration by Substitution Part 1
So, 4
x
 (1  x )
4
dx  4 

1
u
4 u
3
3

1
u
4
u
du
where
4
3
 u 2
u
 4
  2  3

du

 C


u  1 x
Integration by Substitution Part 1
So, 4
x
 (1  x )
4
dx  4 

1
u
4 u
3
3

1
u
4
u
du
where
4
u  1 x
du
3
 u 2
u 
 C
 4

 2


3


1
1 

 4 
C


2
3
3u 
 2u
Integration by Substitution Part 1
So, 4
x
 (1  x )
4
dx  4 

1
u
4 u
3
3

1
u
4
u
du
where
4
u  1 x
du
3
 u 2
u 
 C
 4

 2


3


1
1 

 4 
C


2
3
2
u
3
u


Remove the brackets
2
4
and substitute for u:


C
2
3
(1  x )
3(1  x )
Integration by Substitution Part 1
Exercise
Use substitution to integrate the following:
 x(1  x )
5
dx
Solution:
Integration by Substitution Part 1
 x (1 
So,
5
Let
u  1 x
du
 1   du  dx
dx
5
5
x
(
1

x
)
dx


xu
du


u  1 x  x  1 u
5
   (1  u)u du
x ) dx

  6u


u
5

 u du
6
u
7
6
7
6
(1  x )
6
C

(1  x )
7
7
C
Integration by Substitution Part 1
Integration by Substitution Part 1
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 1
SUMMARY
Substitution can be used for a variety of integrals
4
2
e.gs.
(1  2 x ) dx
x 1  x dx
Method:
• Define u as the inner function
• Differentiate the substitution expression and
rearrange to find dx
• Substitute for the inner function and dx
• If there’s an extra x, cancel it
If x won’t cancel, rearrange the substitution
expression to find x and substitute for it
• Integrate
• Substitute back


Integration by Substitution Part 1
e.g. 1
 (1  2 x ) dx
4
Method: We must substitute for x and dx.
•
Define u as the inner function
Let u  1  2 x
•
Differentiate:
•
•
•
du
dx
2

du
2
 dx
Substitute for the inner function and dx
4
u
4
4 du
 (1  2 x ) dx   u
 
du
2
2
5
u

C
Integrate:
10
5
(1  2 x )
Replace u:

C
10
Integration by Substitution Part 1
e.g. 2
•
•
•

x 1  x dx
2
Define u as the inner function:
2
Let
u  1 x
du
du
Differentiate:
 2x 
 dx
dx
2x
Substitute for the inner function and dx
 x 1  x dx 
2

x u
2x
1


u
du
Cancel the
extra x
2
du
2
Sometimes x won’t cancel and we have to make an
extra substitution
Integration by Substitution Part 1
So,
1
 x 1  x dx 
2
2
u

du
where
2
3
•
Integrate:

2
u
2
C
3
2
3

u
2
C
3
•
Replace u:
(1  x )
2

3
3
2
C
u  1 x
2
Integration by Substitution Part 1
e.g. 3
•
•
•
 x(2  x ) dx
Define u as the inner function:
6
Let
u  2 x
du
 1  du  dx
Differentiate:
dx
Substitute for the inner function and dx
 x(2  x ) dx 
6
 xu du
6
The extra x doesn’t cancel so we must substitute
for it.
Using
So,

u  2 x  u2  x
x ( 2  x ) dx 
6

( u  2)u du
6
Integration by Substitution Part 1
So,
 x( 2  x )
6
dx 

( u  2)u du
6
( where u  2  x )
Now we can easily multiply out the brackets
  u  2u du
7
•
Integrate:

u
8

6
2u
8
•
Replace u:

7
C
7
(2  x )
8
8

2( 2  x )
7
7
C
Integration by Substitution Part 1
Definite integration
We work in exactly the same way BUT we must also
substitute for the limits, since they are values of x
and we are changing the variable to u.
A definite integral gives a value so we never return
to x.
Integration by Substitution Part 1

e.g. 1
1
0
e
x
(1  e )
x
2
dx
Let
u  1 e
du
du
x
e

 dx
x
dx
e
x
Limits: x  0  u  1  e 0  2
x  1  u  1 e
So,

1
0
e
x
(1  e )
x
2
dx 



1e
2
1 e
2
e
x
du
2
e
u
1
u
2
x
du
Integration by Substitution Part 1
So,

1
0
e
x
(1  e )
x
2
dx 
You will often see this
written as du
u
2

1 e
u
2

1
1 e
2
u
1
2
du
where
2
du
1 e
u 


 1  2
u  1 e
1 e
 1
  
 u 2
1 

 1
 
  

 1 e
 2
1
1
 
2 1 e
x