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LECTURE 11OF 12

TOPIC : 2.0 TRIGONOMETRIC FUNCTIONS

SUBTOPIC : 2.7 Integration Of TrigonometricFunction

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LEARNING OUTCOMES:

At the end of the lesson,students should be able to:

b. Discuss the following integrals:

iii.

c. Evaluate integrals using integration by parts; ,

where and and

.

dx,bxcosaxsin dx,bxcosaxcos

dxbxsinaxsin dxaxsinxn

dxaxcosx n 2en dxbxsineax

dxbxcose ax

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b. iii. To solve

by using;

? A

? A

? AQPQPQP

QPQPQP

QPQPQP

!

!

!

coscos2

1coscos

coscos21sinsin

sinsin2

1cossin

dx,bxcosaxsin dx,bxcosaxcos dxbxsinaxsin

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Example 1 :

Solve

dx2xcosxcos2)a

dx2xcos3xcos)b dx2xsin4xcos)c

dx2xsinsin x)d

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Solution:

dx2xcosxcos2)a

? A

? A

csinx3xsin

3

1

dxxcos3xcos

dxx)-2x(cos2x)(xcos

!

!

!

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dx2xcos3xcos)b

? A

? A

csin x215xsin

101

csin5

sin5x

2

1

dxxcos5xcos2

1

dx2x)-3x(cos2x)(3xcos2

1

!

-

!

!

!

x

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dx2xsin4xcos)c

? A

? A

c7xsin14

1-xcos2

1

ccos7

cos7x-

2

1

dxsin x7xsin2

1

dx3x)-4x(sin-3x)(4xsin2

1

!

-

!

!

!

x

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dx2xsinsin x)d

? A

? A

cxx

c

!

-

!

!

!

6

3sin

2

sin

sinx-3xsin3

1

2

1-

dxsin x-3xcos2

1

dxx)-(2xcos-x)2(xcos2

1

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Exercise:

dxsin x3xcos2)1 Answer: ? A c4xcos-2xcos241

dx4xcos2xcos)2

Answer:

? A c 2xsin36xsin121

dxsin x3xsin2)3

Answer:

? A c4xsin-2xsin2

4

1

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dx2xcos4xsin)4 Answer: ? A c 2xcos36xcos

12

1

4

0dsin5sin)5 UUU Answer:

12

1

4

6

dx2xcos3xsin)6 T Answer: 0.13

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c. To evaluate integrals using integration by

parts;

,dxaxsinxi. n dxaxcosxn where 2en

! duvvudvu

By Parts Method

Formula:

(The choice must be suitable)

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Example 1:

Solve the following integration:

dxxcosx)a

dx2xsinx)b

dxxcosx) 2c

dx2xsinx)2d

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Solution:

dxxcosx)a

cxcossin xx

cxcos--sin xx

dxsin x-sin xxsin xv

d

xdu

dxxcosdvxu

!

!

!

!!

!!

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dx2xsinx)b

!

!!

!!

dx2

2xcos--

2

2xcos-x

2xcos

2

1-vdxdu

dx2xsindvxu

c

2

2xcosx-

4

2xsin

c22xsin

21

22xcosx-

!

!

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dxxcosx) 2c

? Acsin x2-xcos2xsin xx

dxxcos--xcosx-2-sin xx

xcos-vdxdu

dx2xsindvxu

dxsin xx2-sin xx

dx2xsin x-sin xx

sin xvdx2xdu

dxxcosdvxu

2

2

2

2

!

!

!!

!!

!

!

!!

!!

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dx2xsinx) 2d

c4

2xcos

2

2xsinx

2

2xcosx-

dx22xsin-

22xsinx

22xcosx-

2

2xsinvdxdu

dx2xcosdvxu

dx2x

2

2xcos--

2

2xcos-x

2

2xcos-vdx2xdu

dx2xsindvxu

2

2

2

2

!

-

!

!!

!!!

!

!!

!!

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Exercise:

dx3xsinx)1 Answer: cxxx 3cos33sin91

dx2xcosx)2 Answer: cxxx 2cos4

12sin

2

1

dxxsinx)3 2 Answer: cxxxx 2cos2sin228

1 2

0sin)4 xdxx Answer: T

T

0

2 sin)5 xdxx Answer: 42 T

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T

04sin)6 xdxx Answer: 4

T

7)Prove that

!2

0

224

16

1sin

T

Txdxx.

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dxbxsineii. ax and dxbxcoseax

Example 2:

Solve the following integration:

dxsin xe)x

a

dx2xcose) 2xb

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cxcos-sin xe21

dxsin xe

cxcose-sin xedxsinxe2therefore;

xx

xxx

!

!

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dx2xcose)

2xb

2xsin

2

1vdx2edu

dx2xcosdveu

2x

2x

!!

!!

partsbyagainanddx2xsine-2xsine2

1

dx2e2xsin2

1-2xsin

2

1edx2xcose

2x2x

2x2x2x

!

!

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2xcos2

1-vdx2edu

dx2xsindveu

2x

2x

!!

!!

therefore;

questiontheassamedx2xcose-2xcose212xsine

21

dx2e2xcos2

1--)2xcos

2

1(-e-2xsine

2

1dx2xcose

2x2x2x

2x2x2x2x

n!

!

c2xcos2xsine4

1dx2xcose

c2xcos2xsine2

1dx2xcose2

2x2x

2x2x

!

!

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Exercise:

dxsin xe)13x

Answer: cx)cos-sin x3(e10

1

3x

dx3xcose)2 -x Answer: c3x)cos-3xsin3(e101

x-

dxxcose)3x Answer: cx)cossin x(e

21 x

dxsin xe)4 -x

Answer:

cx)cossin x(e2

1- x-

dx2xsine)5 xx

Answer:

c2x)cos2-2xsin2(e5

1

x-

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Note:

Whenever we integrate functions of the form

bxcoseorbxsine axax , we get similar results after

applying the rule twice.

Summary:1. From the above examples, we can deduce the following;

a) If one factor is a power of x, it must be taken as u

b) If there is no power of x, then the exponential function is

taken as u.

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