Evaluate $\int{\sin^3(x)\cos^2(x)}dx$ I'm trying to solve $\int{\sin^3(x)\cos^2(x)}dx$.
I got $-\frac{1}{2}\cos(x)+C$, but the memo says $\frac{1}{5}\cos^5(x)-\frac{1}{3}\cos^3(x)+C$
This is my working:

Your help is appreciated!

This can be rewritten as $ \int \sin^2(x) \cos^2(x) \sin(x) dx $
using the property that $ \sin^2(x) = 1 - \cos^2(x) $, we can rewrite the original integral as 
$ \int (1 - \cos^2(x)) \cos^2(x) \sin(x)dx $
substitute $u = \cos(x)$ gets us 
$ -\int (1 - u^2)(u^2)du $
= $ -\int u^2 - u^4 du$
now we have an easy integrand to work with:
$ -\int u^2 - u^4 du = -(\frac{u^3}{3} - \frac{u^5}{5} + C)  $
= $\frac{u^5}{5} - \frac{u^3}{3} + C$
substitute $ u = \cos(x) $ to obtain the final answer
$\frac{\cos^5(x)}{5} - \frac{cos^3(x)}{3} + C$
