Question

What is the integral of `int sin^3xcos^2x dx` from `0` to `pi/2`?

Answer 1

`2/15`

Explanation:

Rewrite the integrand as `int_0^(pi/2)sin^2xcos^2xsinxdx`

Recalling the identity `sin^2x=1-cos^2x,` we get

`int_0^(pi/2)(1-cos^2x)cos^2xsinxdx`

We can now solve this with a simple substitution.

`u=cosx`
`du=-sinxdx`

Calculate the new bounds:

Upper: `u=cos(pi/2)=0`
Lower: `u=cos0=1`

`-int_1^0u^2(1-u^2)du=-int_1^0(u^2-u^4)du`

`=-(1/3u^3-1/5u^5)|_1^0`
`-(-1/3+1/5)=2/15`