Higher Engineering Mathematics, Sixth Edition

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