Higher Engineering Mathematics

INTEGRATION USING TRIGONOMETRIC AND HYPERBOLIC SUBSTITUTIONS 401

H

Problem 11. Evaluate

∫ 1

0

2 cos 6θcosθdθ,

correct to 4 decimal places.

∫ 1

0

2 cos 6θcosθdθ

= 2

∫ 1

0

1

2

[ cos (6θ+θ)+cos (6θ−θ)] dθ,

from 8 of Table 40.1

=

∫ 1

0

(cos 7θ+cos 5θ)dθ=

[

sin 7θ

7

+

sin 5θ

5

] 1

0

=

(

sin 7

7

+

sin 5

5

)

−

(

sin 0

7

+

sin 0

5

)

‘sin 7’ means ‘the sine of 7 radians’ (≡ 401 ◦ 4 ′) and
sin 5≡ 286 ◦ 29 ′.

Hence

∫ 1

0

2 cos 6θcosθdθ

=(0. 09386 +(− 0 .19178))−(0)

=− 0. 0979 , correct to 4 decimal places

Problem 12. Find 3

∫

sin 5xsin 3xdx.

3

∫

sin 5xsin 3xdx

= 3

∫

−

1

2

[ cos (5x+ 3 x)−cos (5x− 3 x)] dx,

from 9 of Table 40.1

=−

3

2

∫

( cos 8x−cos 2x)dx

=−

3

2

(

sin 8

8

−

sin 2x

2

)

+c or

3

16

(4 sin 2x−sin 8x)+c

Now try the following exercise.

Exercise 158 Further problems on integra-

tion of products of sines and cosines

In Problems 1 to 4, integrate with respect to the

variable.

1. sin 5tcos 2t

[

−

1

2

(

cos 7t

7

+

cos 3t

3

)

+c

]

2. 2 sin 3xsinx

[

sin 2x

2

−

sin 4x

4

+c

]

3. 3 cos 6xcosx [
   3
   2

(

sin 7x

7

+

sin 5x

5

)

+c

]

4.

1

2

cos 4θsin 2θ

[

1

4

(

cos 2θ

2

−

cos 6θ

6

)

+c

]

In Problems 5 to 8, evaluate the definite integrals.

5.

∫π

2

0

cos 4xcos 3xdx

[

(a)

3

7

or 0. 4286

]

6.

∫ 1

0

2 sin 7tcos 3tdt [0.5973]

7.− 4

∫ π

3

0

sin 5θsin 2θdθ [0.2474]

8.

∫ 2

1

3 cos 8tsin 3tdt [−0.1999]

40.5 Worked problems on integration

using the sinθsubstitution

Problem 13. Determine

∫

1

√

(a^2 −x^2 )

dx.

Letx=asinθ, then

dx

dθ

=acosθand dx=acosθdθ.

Hence

∫

1

√

(a^2 −x^2 )

dx

=

∫

1

√

(a^2 −a^2 sin^2 θ)

acosθdθ

=

∫

acosθdθ

√

[a^2 (1−sin^2 θ)]

=

∫

acosθdθ

√

(a^2 cos^2 θ)

, since sin^2 θ+cos^2 θ= 1

=

∫

acosθdθ

acosθ

=

∫

dθ=θ+c