Answer
$$ \frac{\sin ^{4} x \cos x}{5}+\frac{1}{5}\left( -\cos x+\frac{1}{3}\cos^3x\right)+C$$
Work Step by Step
Since $$\int \sin ^{m} x \cos ^{n} x d x=\frac{\sin ^{m+1} x \cos ^{n-1} x}{m+n}+\frac{n-1}{m+n} \int \sin ^{m} x \cos ^{n-2} x d x$$
Then for $m=3,\ n= 2$, we get
\begin{align*}
\int \sin ^{3} x \cos ^{2} x d x&=\frac{\sin ^{4} x \cos x}{5}+\frac{1}{5} \int \sin ^{3} x d x\\
&= \frac{\sin ^{4} x \cos x}{5}+\frac{1}{5}\left( \int (1-\cos^2 x)\sin xdx\right)\\
&= \frac{\sin ^{4} x \cos x}{5}+\frac{1}{5}\left( -\cos x+\frac{1}{3}\cos^3x\right)+C
\end{align*}
