Answer
Rewrite as:
$$\sin ^{3} x \cos ^{2} x =\sin ^{2} x \cos ^{2} x \sin x=[ \cos ^{2} x - \cos ^{4} x ]\sin x $$
and substitute:
$u=\cos x\ \ \ \ \ \ \ \ du =-\sin x$
Work Step by Step
We rewrite as:
\begin{align*}
\int \sin ^{3} x \cos ^{2} x d x&=\int \sin ^{2} x \cos ^{2} x \sin xd x\\
&= \int[1- \cos ^{2} x ] \cos ^{2} x \sin xd x\\
&=\int[ \cos ^{2} x - \cos ^{4} x ]\sin xd x
\end{align*}
and then use the substitution:
$u=\cos x\ \ \ \ \ \ \ \ du =-\sin x$
