402 Higher Engineering Mathematics
1
3
cos 5x sin 2x dx
=
1
3
1
2
[sin(5x + 2x) − sin (5x − 2x)] dx,
from 7 of Table 40.1
=
1
6
(sin 7x − sin 3x) dx
=
1
6
−cos 7x
7
+
cos 3x
3
+ c
Problem 11. Evaluate
1
0
2 cos 6θ cos θ dθ,
correct to 4 decimal places.
1
0
2 cos 6θ cos θ dθ
= 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 6θ cos θ dθ
= (0.09386 + (−0.19178)) − (0)
= −0.0979, correct to 4 decimal places.
Problem 12. Find 3
sin 5x sin 3x dx.
3
sin 5x sin 3x dx
= 3
−
1
2
[ cos (5x + 3x) − cos (5x − 3x)] 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 157 Further problems on
integration of products of sines and cosines
In Problems 1 to 4, integrate with respect to the
variable.
1. sin 5t cos 2t
−
1
2
cos 7t
7
+
cos 3t
3
+ c
2. 2 sin 3x sin x
sin 2x
2
−
sin 4x
4
+ c
3. 3 cos 6x cos x
3
2
sin 7x
7
+
sin 5x
5
+ c
4.
1
2
cos 4θ sin 2θ
1
4
cos 2θ
2
−
cos 6θ
6
+ c
In Problems 5 to 8, evaluate the definite integrals.
5.
π
2
0
cos 4x cos 3x dx
(a)
3
7
or 0.4286
6.
1
0
2 sin7t cos 3t dt
[0.5973]
7. −4
π
3
0
sin 5θ sin 2θ dθ
[0.2474]
8.
2
1
3 cos 8t sin 3t dt
[−0.1999]
40.5 Worked problems on integration
using the sin θ substitution
Problem 13. Determine
1
(a 2 − x 2 )
dx.
Let x = a sin θ, then
dx
dθ
= a cos θ and dx = a cos θ dθ.
Hence
1
(a 2 − x 2 )
dx
=
1
(a 2 − a 2 sin
2
θ)
a cos θ dθ
=
a cos θ dθ
[a 2 (1 − sin 2 θ)]