Is this correct? I thought it would be but when I entered it into wolfram alpha, I got a different answer.
$\int (\cos^3x)(\sin^2x)dx = \int(\cos x)(\cos^2x)(\sin^2x)dx = \int (\cos x)(1-\sin^2x)(\sin^2x)dx.$
let $u = \sin x$, $du = \cos xdx$
$\int(1-u^2)u^2du = \int(u^2-u^4)du = \frac{u^3}{3} - \frac{u^5}{5} +C$
Plugging in back $u$, we get $\displaystyle\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5}$ + C