430 Higher Engineering Mathematics
du
dx
= (n − 1) sin
n−2 x cos x and
du = (n − 1) sin
n−2 x cos x dx
and let dv = sin x dx, from which,
v =
sin x dx =−cos x. Hence,
I n =
sin
n−1 x sin x dx
= (sin
n−1 x)(−cos x)
−
(−cos x)(n − 1) sin
n−2 x cos x dx
= −sin
n−1 x cos x
+ (n − 1)
cos
2 x sin
n−2 x dx
= −sin
n−1 x cos x
+ (n − 1)
(1 − sin
2 x) sin
n−2 x dx
= −sin
n−1 x cos x
+ (n − 1)
sin
n−2 x dx −
sin
n x dx
i.e.
I n = −sin
n−1 x cos x
+ (n − 1)I n−2 −(n − 1)I n
i.e.
I n + (n − 1)I n
= −sin
n−1 x cos x + (n − 1)I n−2
and
n I n = −sin
n−1 x cos x + (n − 1)I n−2
from which,
sin
n x dx =
I n = −
1
n
sin
n−1 xcos x +
n − 1
n
I n−2
(4)
Problem 8. Use a reduction formula to determine
sin 4 x dx.
Using equation (4),
sin
4 x dx = I 4 = −
1
4
sin 3 x cos x +
3
4
I 2
I 2 = −
1
2
sin 1 x cos x +
1
2
I 0
and
I 0 =
sin
0 x dx =
1 dx = x
Hence
sin
4 x dx = I 4 = −
1
4
sin
3 x cos x
+
3
4
−
1
2
sin x cos x +
1
2
(x)
= −
1
4
sin
3 x cos x −
3
8
sin x cos x
+
3
8
x + c
Problem 9. Evaluate
1
0 4 sin
5 t dt , correct to 3
significant figures.
Using equation (4),
sin
5 t dt = I 5 = −
1
5
sin
4 t cos t +
4
5
I 3
I 3 = −
1
3
sin
2 t cos t +
2
3
I 1
and
I 1 = −
1
1
sin 0 t cos t + 0 = −cos t
Hence
sin
5 t dt = −
1
5
sin
4 t cos t
+
4
5
−
1
3
sin
2 t cos t +
2
3
(−cos t )
= −
1
5
sin
4 t cos t −
4
15
sin
2 t cos t
−
8
15
cos t + c
and
t
0
4 sin
5 t dt
= 4
−
1
5
sin
4 t cos t
−
4
15
sin
2 t cos t −
8
15
cos t
1
0
= 4
−
1
5
sin
4 1 cos1 −
4
15
sin
2 1 cos1
−
8
15
cos 1
−
−0 − 0 −
8
15