© Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules 27: Integration by Substitution Part 2 Part 2

© Christine Crisp

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

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

27: Integration by 27: Integration by SubstitutionSubstitution

Part 2 Part 2

Integration by Substitution Part 2

dxxtan

A useful example of integration by substitution is to find

dxx

x

cos

sinWe write dxxtan

Let xu cos xdx

dusin dx

x

du

sin

So, dxx

x

cos

sin

x

du

u

x

sin

sin

duu

1Cu ln

Cx cosln


Cx 1)(cosln

Using the 3rd law of logs,

Cx

cos

1ln

Cx secln

Cx cosln


duucos1

u2cos duucos

1

u2sin1

e.g. 1 dx

x 21

1

udu

dxcos duudx cos

ux sinLet

duu

ucos

sin1

12

1sincos 22 uu uu 22 sin1cos

N.B. Instead of defining u as a function of x

we have defined x as a function of u. So,

dx

x 21

1

Can you spot what to do next?

Use the identity:


duuu

coscos

1

du1

Cu

Cx 1sin

ux sin

dxx 21

1

duuu

coscos

12

So,

dxx 21

1

We need u from the

substitution expression:

xu 1sin

ux sinwhere

Integration by Substitution Part 2Exercis

e

1. Find dx

x 29

1

ux sin3

using the substitution

2. Show that Cx

dxx

2

tan2

1

4

1 12

ux tan2using the substitution this is an example of the general result

Integration by Substitution Part 2Solution

s:

udu

dxcos3 duudx cos3

ux sin3Let

dx

x 29

1

duuu

cos3sin99

12

duu

ucos3

)sin1(9

12

duuu

cos3cos9

12

1. dxx 29

1


duuu

cos3cos3

1

du1

Cu

Cx

3sin 1

ux sin3where

duuu

cos3cos9

12

dxx 29

1

ux

ux sin3

sin3

3sin 1 x

uTo subs. back:

So, dx

x 29

1


2. Show that Cx

dxx

2

tan2

1

4

1 12

ux tan2using the substitution Solution

:ux tan2

udu

dx 2sec2 duudx 2sec2

duuu

dxx

222

sec2tan44

1

4

1

So,

duuu

22

sec2)tan1(4

1

uu 22 tan1sec Use the identity:


ux

ux tan2

tan2

duuu

22

sec2)tan1(4

1 duu

u2

2sec2

sec4

12

du2

1

Cu 2

1

Cx

2tan

2

1 1 dx

x 24

1So,

2tan 1 x

u


Show that x = sin2θ transforms

dsin to dxx

x 221

x = sin2θ 2)(sin

cossind

dx2 Using rule for brackets

dcossin

sin

sin dx

x

xI So 2

11 2

2

dcossindx 2


dcossin

sin

sin dx

x

xI 2

11 2

2

dcossincos

sinI 22

2

dcossincos

sinI 2

dsinI 22 Proven

221 cossin


dsinI 22

This can be integrated using cos2 = 1–2sin2

2sin2 = 1–cos2

dcos2–1dsinI 22

22

1sin


boundaries the change dxx

x find To

4

1

0 1

x = sin2θ So if x= ¼ sin2 = ¼

sin = ½ = /6

So if x= 0 sin2 = 0

sin = 0 =


22

12

1

6

0

24

1

0

sin dsin dxx

x oS

= /6

6

2

2

1

6sinI

= 0I

4

3

6

I So

4

3

62

3

2

1

6



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.


e.g. 1 dx

x 21

1

udu

dxcos duudx cos

ux sinLet

duu

ucos

sin1

12

1sincos 22 uu uu 22 sin1cos

duu

ucos

sin1

12 duu

ucos

cos

12

N.B. Instead of defining u

as a function of x we

have defined x as a

function of u.

So,

dx

x 21

1

Use the identity:


duuu

coscos

1

du1

Cu

Cx 1sin

ux sin

dxx 21

1

duuu

coscos

12

So,

dxx 21

1

We need u from the

substitution expression:

xu 1sin

ux sinwhere


2. Show that2

tan2

1

4

1 12

xdx

x

ux tan2using the

substitution Solution:

ux tan2

udu

dx 2sec2 duudx 2sec2

duuu

dxx

222

sec2tan44

1

4

1

So,

duuu

22

sec2)tan1(4

1

uu 22 tan1sec Use the identity:


duuu

22

sec2)tan1(4

1 duu

u2

2sec2

sec4

12

du2

1

Cu 2

1

Cx

2tan

2

1 1 dx

x 24

1So,

ux

ux tan2

tan2

2tan 1 x

u

Documents

© Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules 27: Integration by Substitution Part 2 Part 2