Reduction formulae 431
=4[(− 0. 054178 − 0. 1020196
− 0. 2881612 )−(− 0. 533333 )]
= 4 ( 0. 0889745 )= 0. 356
Problem 10. Determine a reduction formula for
∫ π
2
0
sinnxdxand hence evaluate
∫π
2
0
sin^6 xdx
From equation (4),
∫
sinnxdx
=In=−
1
n
sinn−^1 xcosx+
n− 1
n
In− 2
hence
∫ π
2
0
sinnxdx=
[
−
1
n
sinn−^1 xcosx
]π 2
0
+
n− 1
n
In− 2
=[0−0]+
n− 1
n
In− 2
i.e. In=
n− 1
n
In− 2
Hence
∫ π
2
0
sin^6 xdx=I 6 =
5
6
I 4
I 4 =
3
4
I 2 , I 2 =
1
2
I 0
and I 0 =
∫ π
2
0
sin^0 xdx=
∫ π
2
0
1dx=
π
2
Thus
∫ π
2
0
sin^6 xdx=I 6 =
5
6
I 4 =
5
6
[
3
4
I 2
]
=
5
6
[
3
4
{
1
2
I 0
}]
=
5
6
[
3
4
{
1
2
[π
2
]}]
=
15
96
π
(b)
∫
cosnxdx
LetIn=
∫
cosnxdx≡
∫
cosn−^1 xcosxdxfrom laws
of indices.
Using integration by parts, let u=cosn−^1 x from
which,
du
dx
=(n− 1 )cosn−^2 x(−sinx)
and du=(n− 1 )cosn−^2 x(−sinx)dx
and let dv=cosxdx
from which,v=
∫
cosxdx=sinx
Then
In=(cosn−^1 x)(sinx)
−
∫
(sinx)(n− 1 )cosn−^2 x(−sinx)dx
=(cosn−^1 x)(sinx)
+(n− 1 )
∫
sin^2 xcosn−^2 xdx
=(cosn−^1 x)(sinx)
+(n− 1 )
∫
( 1 −cos^2 x)cosn−^2 xdx
=(cosn−^1 x)(sinx)
+(n− 1 )
{∫
cosn−^2 xdx−
∫
cosnxdx
}
i.e.In=(cosn−^1 x)(sinx)+(n− 1 )In− 2 −(n− 1 )In
i.e.In+(n− 1 )In=(cosn−^1 x)(sinx)+(n− 1 )In− 2
i.e.nIn=(cosn−^1 x)(sinx)+(n− 1 )In− 2
Thus In=
1
n
cosn−^1 xsinx+n−n^1 In− 2 (5)
Problem 11. Use a reduction formula to
determine
∫
cos^4 xdx.
Using equation (5),
∫
cos^4 xdx=I 4 =
1
4
cos^3 xsinx+
3
4
I 2
and I 2 =
1
2
cosxsinx+
1
2
I 0
and I 0 =
∫
cos^0 xdx
=
∫
1dx=x