Higher Engineering Mathematics

INTEGRATION USING TRIGONOMETRIC AND HYPERBOLIC SUBSTITUTIONS 399

H

Now try the following exercise.

Exercise 156 Further problems on integra-

tion of sin^2 x, cos^2 x, tan^2 xand cot^2 x

In Problems 1 to 4, integrate with respect to the

variable.

1. sin^22 x

[

1

2

(

x−

sin 4x

4

)

+c

]

2. 3 cos^2 t

[

3

2

(

t+

sin 2t

2

)

+c

]

3. 5 tan^23 θ

[

5

(

1

3

tan 3θ−θ

)

+c

]

4. 2 cot^22 t [−(cot 2t+ 2 t)+c]
   In Problems 5 to 8, evaluate the definite integrals,
   correct to 4 significant figures.

5.

∫ π

3

0

3 sin^23 xdx

[π

2

or 1. 571

]

6.

∫ π

4

0

cos^24 xdx

[π

8

or 0. 3927

]

7.

∫ 1

0

2 tan^22 tdt [− 4 .185]

8.

∫ π

3

π

6

cot^2 θdθ [0.6311]

40.3 Worked problems on powers of

sines and cosines

Problem 5. Determine

∫

sin^5 θdθ.

Since cos^2 θ+sin^2 θ=1 then sin^2 θ=(1−cos^2 θ).

Hence

∫

sin^5 θdθ

=

∫

sinθ( sin^2 θ)^2 dθ=

∫

sinθ(1−cos^2 θ)^2 dθ

=

∫

sinθ(1−2 cos^2 θ+cos^4 θ)dθ

=

∫

(sinθ−2 sinθcos^2 θ+sinθcos^4 θ)dθ

=−cosθ+

2 cos^3 θ

3

−

cos^5 θ

5

+c

[Whenever a power of a cosine is multiplied by a

sine of power 1, or vice-versa, the integral may be

determined by inspection as shown.

In general,

∫

cosnθsinθdθ=

−cosn+^1 θ

(n+1)

+c

and

∫

sinnθcosθdθ =

sinn+^1 θ

(n+1)

+c

Problem 6. Evaluate

∫π

2

0

sin^2 xcos^3 xdx.

∫ π

2

0

sin^2 xcos^3 xdx=

∫π

2

0

sin^2 xcos^2 xcosxdx

=

∫ π

2

0

(sin^2 x)(1−sin^2 x)(cosx)dx

=

∫ π

2

0

(sin^2 xcosx−sin^4 xcosx)dx

=

[

sin^3 x

3

−

sin^5 x

5

]π

2

0

=

⎡

⎢

⎣

(

sin

π

2

) 3

3

−

(

sin

π

2

) 5

5

⎤

⎥

⎦−[0−0]

=

1

3

−

1

5

=

2

15

or 0. 1333

Problem 7. Evaluate

∫π

4

0

4 cos^4 θdθ, correct

to 4 significant figures.

∫ π

4

0

4 cos^4 θdθ= 4

∫ π

4

0

(cos^2 θ)^2 dθ

= 4

∫ π

4

0

[

1

2

(1+cos 2θ)

] 2

dθ

=

∫ π

4

0

(1+2 cos 2θ+cos^22 θ)dθ

=

∫ π

4

0

[

1 +2 cos 2θ+

1

2

(1+cos 4θ)

]

dθ

=

∫ π

4

0

(

3

2

+2 cos 2θ+

1

2

cos 4θ

)

dθ