Higher Engineering Mathematics

428 INTEGRAL CALCULUS

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 deter-

mine

∫

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 4 sin

(^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
  4 sin^5 tdt
  = 4
  [
  −
  1
  5
  sin^4 tcost
  −
  4
  15
  sin^2 tcost−
  8
  15
  cost
  ] 1
  0
  = 4
  [(
  −
  1
  5
  sin^4 1 cos 1−
  4
  15
  sin^2 1 cos 1
  −
  8
  15
  cos 1
  )
  −
  (
  − 0 − 0 −
  8
  15
  )]