430 Higher Engineering Mathematics
du
dx
=(n− 1 )sinn−^2 xcosx and
du=(n− 1 )sinn−^2 xcosxdx
and let dv=sinxdx, from which,
v=
∫
sinxdx=−cosx. Hence,
In=
∫
sinn−^1 xsinxdx
=(sinn−^1 x)(−cosx)
−
∫
(−cosx)(n− 1 )sinn−^2 xcosxdx
=−sinn−^1 xcosx
+(n− 1 )
∫
cos^2 xsinn−^2 xdx
=−sinn−^1 xcosx
+(n− 1 )
∫
( 1 −sin^2 x)sinn−^2 xdx
=−sinn−^1 xcosx
+(n− 1 )
{∫
sinn−^2 xdx−
∫
sinnxdx
}
i.e. In=−sinn−^1 xcosx
+(n− 1 )In− 2 −(n− 1 )In
i.e. In+(n− 1 )In
=−sinn−^1 xcosx+(n− 1 )In− 2
and nIn=−sinn−^1 xcosx+(n− 1 )In− 2
from which,
∫
sinnxdx=
In=−
1
n
sinn−^1 xcosx+
n− 1
n
In− 2 (4)
Problem 8.∫ Use a reduction formula to determine
sin^4 xdx.
Using equation (4),
∫
sin^4 xdx=I 4 =−
1
4
sin^3 xcosx+
3
4
I 2
I 2 =−
1
2
sin^1 xcosx+
1
2
I 0
and I 0 =
∫
sin^0 xdx=
∫
1dx=x
Hence
∫
sin^4 xdx=I 4 =−
1
4
sin^3 xcosx
+
3
4
[
−
1
2
sinxcosx+
1
2
(x)
]
=−
1
4
sin^3 xcosx−
3
8
sinxcosx
+
3
8
x+c
Problem 9. Evaluate
∫ 1
0 4sin
(^5) tdt, correct to 3
significant figures.
Using equation (4),
∫
sin^5 tdt=I 5 =−
1
5
sin^4 tcost+
4
5
I 3
I 3 =−
1
3
sin^2 tcost+
2
3
I 1
and I 1 =−
1
1
sin^0 tcost+ 0 =−cost
Hence
∫
sin^5 tdt=−
1
5
sin^4 tcost
- 4
5
[
−
1
3
sin^2 tcost+
2
3
(−cost)
]
=−
1
5
sin^4 tcost−
4
15
sin^2 tcost
−
8
15
cost+c
and
∫t
0
4sin^5 tdt
= 4
[
−
1
5
sin^4 tcost
−
4
15
sin^2 tcost−
8
15
cost
] 1
0
= 4
[(
−
1
5
sin^4 1cos1−
4
15
sin^2 1cos1
−
8
15
cos1
)
−
(
− 0 − 0 −
8
15
)]