Integration using trigonometric and hyperbolic substitutions 401
π
4
0
4 cos
4
θ dθ = 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
2 2θ)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
=
3θ
2
+ sin 2θ +
sin 4θ
8
π
4
0
=
3
2
π
4
+ sin
2π
4
+
sin 4(π/4)
8
− [0]
=
3π
8
+ 1 = 2.178,
correct to 4 significant figures.
Problem 8. Find
sin 2 t cos 4 t dt.
sin
2 t cos
4 t dt =
sin
2 t (cos
2 t )
2 dt
=
1 − cos 2t
2
1 + cos 2t
2
2
dt
=
1
8
(1 − cos 2t )(1 + 2 cos 2t + cos
2 2t ) dt
=
1
8
(1 + 2 cos 2t + cos
2 2t − cos 2t
− 2 cos 2 2t − cos 3 2t ) dt
=
1
8
(1 + cos 2t − cos
2 2t − cos
3 2t ) dt
=
1
8
1 + cos 2t −
1 + cos 4t
2
− cos 2t (1 − sin
2 2t )
dt
=
1
8
1
2
−
cos 4t
2
+ cos 2t sin
2 2t
dt
=
1
8
t
2
−
sin 4t
8
+
sin
3 2t
6
+ c
Now try the following exercise
Exercise 156 Further problems on
integration of powers of sines and cosines
In Problems 1 to 6, integrate with respect to the
variable.
1. sin 3 θ
(a)−cos θ +
cos 3 θ
3
+ c
2. 2 cos 3 2x
sin 2x −
sin 3 2x
3
+ c
3. 2 sin
3 t cos
2 t
−2
3
cos
3 t +
2
5
cos
5 t + c
4. sin 3 x cos 4 x
− cos 5 x
5
+
cos 7 x
7
+ c
5. 2 sin 4 2θ
3θ
4
−
1
4
sin 4θ +
1
32
sin 8θ + c
6. sin 2 t cos 2 t
t
8
−
1
32
sin 4t + c
40.4 Worked problems on integration
of products of sines and cosines
Problem 9. Determine
sin 3t cos 2t dt.
sin 3t cos 2t dt
=
1
2
[sin(3t + 2t ) + sin (3t − 2t )] dt,
from 6 of Table 40.1, which follows from Section 17.4,
page 170,
=
1
2
(sin 5t + sin t ) dt
=
1
2
−cos 5t
5
− cos t
+ c
Problem 10. Find
1
3
cos 5x sin 2x dx.