© Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules 26: Integration by Substitution...

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules

26: Integration by 26: Integration by SubstitutionSubstitution

Part 1 Part 1

Integration by Substitution Part 1

"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

Module C3

AQA Edexcel

OCR

Module C4

MEI/OCR


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.


4)21( x dx

e.g. 1 dxx 4)21(

Let xu 21

Method:

We must substitute for x and

dx.

2dx

du• Differentiate: dxdu

2

Find dx by treating like a

fraction dx

du

• Define u as the inner function

• Substitute for the inner function . . .


4)21( x

Let xu 21

dx

e.g. 1 dxx 4)21(

2dx

dudx

du

2

4u

• Differentiate:

Method:


dx.



and dx


dxdu

2

dx 4)21( x dx 2

du4u

Let xu 21

e.g. 1 dxx 4)21(

2dx

du• Differentiate:

Method:


dx.



and dx


dx

dxdu

2

4)21( x 2

du4u

Let xu 21

e.g. 1 dxx 4)21(

2dx


Method:


dx.



and dx

• Integrate:

duu

2

4

• Replace u:

C10

5u


C10

5u

dx

dxdu

2

4)21( x 2

du4u

Let xu 21

e.g. 1 dxx 4)21(

2dx


Method:


dx.



and dx

• Integrate:

duu

2

4

• Replace u: x21 C10

)( 5


C10

5u

Let xu 21

dx

dxdu

2

4)21( x 2

du4u

e.g. 1 dxx 4)21(

2dx


Method:


dx.



and dx

• Integrate:

duu

2

4

• Replace u: x21 C10

)( 5

Integration by Substitution Part 1Exercis

esUse substitution to integrate the following. (Where possible, you could also use a 2nd method.)

dxx 8)1(1. dxe x32.

Integration by Substitution Part 1Solution

s: dxx 8)1(1. Let xu 1

1dx

dudxdu

Cu

9

9

Cx

9

)1( 9

duudxx 88)1(So,

Integration by Substitution Part 1Solution

s: dxe x32. Let xu 3

3dx

dudx

du

3

Ce u

3

Ce x

3

3

33 du

edxe uxSo,

due u

3


Definite integrationWe 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.


e.g. 1

1

02)1(

dxe

ex

x

Let xeu 1xe

dx

du dx

e

dux

Limits: 010 eux 2eux 11

So,

1

02)1(

dxe

ex

x

e1

2x

x

e

du

u

e2

duu

e

1

22

1


where xeu 1

So,

1

02)1(

dxe

ex

x

duu

e

1

22

1

You will often see

this written as 2u

du

eduu

1

2

2

eu

1

2

1

1

2

1

e

u

1

2

1

e1

1

2

1

e1

1

We leave answers in the exact form.


In the next examples, the extra x doesn’t

conveniently cancel so we need to substitute for it.


dxdu

duxu6

Let xu 2

e.g. 3 dxxx 6)2(

1dx

du• Differentiate:• Substitute for the inner function

and dx

• Define u as the inner

function:

dxxx 6)2(

The extra x doesn’t cancel so we must substitute for it.

Using xu 2 xu 2

So, duxu6 duuu 6)2(


duuu 6)2(

Can you spot the important difference between these?Ans: We can easily multiply out the brackets in the 2nd

duuu 67 2

( where )xu 2

Cuu

7

2

8

78

Cxx

7

)2(2

8

)2( 78

• Integrate:

• Replace u:

dxxx 6)2(So,


Tip: Don’t be tempted to substitute for the

extra x . . . until you’ve checked to see if it cancels.

dxdu dx

x

x4)1(

4e.g. 4

Let xu 11

dx

du


dxdu dx

x

x4)1(

4e.g. 4

Let xu 11

dx

du

xu 1

dx

x

x4)1(

4So, du

u

x4

4

xu 1

du

u

u4

)1(4

duuu

u44

14

x doesn’t cancel so now substitute:

duu

x4

4

A multiplying constant . . .can be taken outside the integral.

1

3u


duuu 434

dxx

x 4)1(4So,

where xu 1 du

uu 43

114

Cuu

324

32


duuu 434

dxx

x 4)1(4So,

where xu 1 du

uu 43

114

Cuu

324

32

Cuu

32 3

1

2

14


duuu 434

dxx

x 4)1(4So,

Cxx

32 )1(3

4

)1(

2

where xu 1 du

uu 43

114

Cuu

324

32

Cuu

32 3

1

2

14

Remove the brackets and substitute for u:

Integration by Substitution Part 1Exercis

eUse substitution to integrate the following:

dxxx 5)1(


Let xu 11

dx

dudxdu

duxu5 dxxx 5)1(So,xu 1 ux 1

duuu 5)1(

duuu 65

Cuu

76

76

Cxx

7

)1(

6

)1( 76

Solution:

dxxx 5)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 1SUMMAR

Y

e.gs. dxxx 21 dxx 4)21(

• Differentiate the substitution expression and rearrange to find dx

Method:

• Substitute for the inner function

and dx


• If there’s an extra x, cancel itIf x won’t cancel, rearrange the substitution expression to find x and substitute for it

Substitution can be used for a variety of integrals

• Integrate• Substitute back


dx

dxdu

2

4)21( x 2

du4u

Let xu 21

e.g. 1 dxx 4)21(

2dx


Method:


dx.



and dx

• Integrate:

Cu

10

5

duu

2

4

• Replace u: Cx

10

)21( 5


dxx

du

2

x

duux2

Let 21 xu

e.g. 2 dxxx 21

xdx

du2• Differentiate:


and dx


function:

dxxx 21

duu

2

21

Cancel the extra

x

Sometimes x won’t cancel and we have to

make an extra substitution


where

21 xu

• Integrate:

• Replace u:

Cu

23

23

2

duu

2

21

Cx

3

)1( 23

2

So,

dxxx 21

Cu

3

23


dxdu

duxu6

Let xu 2e.g. 3 dxxx 6)2(

1dx

du• Differentiate:• Substitute for the inner function

and dx


function:

dxxx 6)2(

duuu 6)2(

The extra x doesn’t cancel so we must substitute for it.

Using xu 2 xu 2

dxxx 6)2(So,


duuu 6)2(

Now we can easily multiply out the brackets

duuu 67 2

( where )xu 2

Cuu

7

2

8

78

Cxx

7

)2(2

8

)2( 78

• Integrate:

• Replace u:

dxxx 6)2(So,


Definite integrationWe 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.


e.g. 1

1

02)1(

dxe

ex

x

Let xeu 1xe

dx

du dx

e

dux

Limits: 010 eux 2eux 11

So,

1

02)1(

dxe

ex

x

e1

2x

x

e

du

u

e2

duu

e

1

22

1


where xeu 1

So,

1

02)1(

dxe

ex

x

duu

e

1

22

1

You will often see

this written as 2u

du

eduu

1

2

2

eu

1

2

1

1

2

1

e

u

1

2

1

e1

1

2

1

e1

1

Date post:	31-Mar-2015
Category:	Documents
Upload:	bailey-gascoyne
View:	240 times
Download:	5 times

© Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules 26: Integration by Substitution...

Documents