Higher Engineering Mathematics

H

Integral calculus

40

Integration using trigonometric and

hyperbolic substitutions

40.1 Introduction

Table 40.1 gives a summary of the integrals that
require the use oftrigonometric and hyperbolic
substitutionsand their application is demonstrated
in Problems 1 to 27.

40.2 Worked problems on integration

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

Problem 1. Evaluate

∫ π

4

0

2 cos^24 tdt.

Since cos 2t=2 cos^2 t−1 (from Chapter 18),

then cos^2 t=

1

2

(1+cos 2t) and

cos^24 t=

1

2

(1+cos 8t)

Hence

∫ π

4

0

2 cos^24 tdt

= 2

∫π

4

0

1

2

(1+cos 8t)dt

=

[

t+

sin 8t

8

]π

4

0

=

⎡

⎢

⎣

π

4

+

sin 8

(π

4

)

8

⎤

⎥

⎦−

[

0 +

sin 0

8

]

=

π

4

or 0. 7854

Problem 2. Determine

∫

sin^23 xdx.

Since cos 2x= 1 −2 sin^2 x(from Chapter 18),

then sin^2 x=

1

2

(1−cos 2x) and

sin^23 x=

1

2

(1−cos 6x)

Hence

∫

sin^23 xdx=

∫

1

2

(1−cos 6x)dx

=

1

2

(

x−

sin 6x

6

)

+c

Problem 3. Find 3

∫

tan^24 xdx.

Since 1+tan^2 x=sec^2 x, then tan^2 x=sec^2 x− 1

and tan^24 x=sec^24 x−1.

Hence 3

∫

tan^24 xdx= 3

∫

( sec^24 x−1) dx

= 3

(

tan 4x

4

−x

)

+c

Problem 4. Evaluate

∫π

3

π

6

1

2

cot^22 θdθ.

Since cot^2 θ+ 1 =cosec^2 θ, then cot^2 θ=cosec^2 θ− 1

and cot^22 θ=cosec^22 θ−1.

Hence

∫π

3

π

6

1

2

cot^22 θdθ

=

1

2

∫ π

3

π

6

(cosec^22 θ−1) dθ=

1

2

[

−cot 2θ

2

−θ

]π

3

π

6

=

1

2

⎡

⎢

⎣

⎛

⎜

⎝

−cot 2

(π

3

)

2

−

π

3

⎞

⎟

⎠−

⎛

⎜

⎝

−cot 2

(π

6

)

2

−

π

6

⎞

⎟

⎠

⎤

⎥

⎦

=

1

2

[(0. 2887 − 1 .0472)−(− 0. 2887 − 0 .5236)]

= 0. 0269