Question

Please find the value for this integration. What is the answer?

`int_0^(pi/2)sin^3xcosxdx`

Answer 1

`u = sin(x)`

`du = cos(x)dx`

`int_0^(1)u^3du`

...

Answer 2

`= 1/4`

Explanation:

Using the Integral

`intsin^n axcdot cos axcdot dx = 1/[a(n+1)]sin^{n+1} ax +C`
for `(n!=-1)`

Given expression is

`int _0^(pi/2)sin^3 xcdot cos xcdot dx` ,

here `n=3 and a=1`
We obtain `=[ 1/[1(3+1)]sin^{3+1} x +C] _0^(pi/2)`
`=[ 1/4sin^4 x +C] _0^(pi/2)`
Inserting the value of `sin(pi/2)=1 and sin 0=0`
`=[ 1/4xx 1^4 +C] -[ 1/4xx 0^4 +C]`
`= 1/4`