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

1 Answer
May 6, 2018

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