Higher Engineering Mathematics, Sixth Edition

402 Higher Engineering Mathematics

∫

1

3

cos5xsin2xdx

=

1

3

∫

1

2

[sin( 5 x+ 2 x)−sin( 5 x− 2 x)]dx,

from 7 of Table 40.1

=

1

6

∫

(sin7x−sin3x)dx

=

1

6

(

−cos 7x

7

+

cos 3x

3

)

+c

Problem 11. Evaluate

∫ 1

0

2cos6θcosθdθ,

correct to 4 decimal places.

∫ 1

0

2cos6θcosθdθ

= 2

∫ 1

0

1

2

[cos( 6 θ+θ)+cos( 6 θ−θ)]dθ,

from 8 of Table 40.1

=

∫ 1

0

(cos 7θ+cos 5θ)dθ=

[

sin7θ

7

+

sin5θ

5

] 1

0

=

(

sin7

7

+

sin5

5

)

−

(

sin0

7

+

sin0

5

)

‘sin7’ means ‘the sine of 7radians’ (≡ 401 ◦ 4 ′)and

sin 5≡ 286 ◦ 29 ′.

Hence

∫ 1

0

2cos6θcosθdθ

=( 0. 09386 +(− 0. 19178 ))−( 0 )

=− 0. 0979 ,correct to 4 decimal places.

Problem 12. Find 3

∫

sin5xsin3xdx.

3

∫

sin5xsin3xdx

= 3

∫

−

1

2

[cos( 5 x+ 3 x)−cos( 5 x− 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

(4sin2x−sin 8x)+c

Now try the following exercise

Exercise 157 Further problems on

integration of products of sines and cosines

In Problems 1 to 4, integrate with respect to the

variable.

1. sin5tcos 2t

[

−

1

2

(

cos 7t

7

+

cos 3t

3

)

+c

]

2. 2sin3xsinx

[

sin2x

2

−

sin4x

4

+c

]

3. 3cos6xcosx [
   3
   2

(

sin7x

7

+

sin5x

5

)

+c

]

4.

1

2

cos 4θsin2θ

[

1

4

(

cos2θ

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

2sin7tcos 3tdt [0.5973]

7. − 4

∫ π

3

0

sin5θsin2θdθ [0.2474]

8.

∫ 2

1

3cos8tsin3tdt [−0.1999]

40.5 Worked problems on integration

using thesinθsubstitution

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)]