Answer
$$\frac{\sin ^{3} x \cos ^{3} x}{6}+\frac{3}{6}\left( \frac{\sin ^{3} x \cos x}{4}+\frac{ 1}{8} [x-\frac{1}{2}\sin2x] \right)+C$$
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 $m=2,\ n=4$, we get
\begin{align*}
\int \sin ^{2} x \cos ^{4} x d x&=\frac{\sin ^{3} x \cos ^{3} x}{6}+\frac{3}{6} \int \sin ^{2} x \cos ^{2} x d x\\
&= \frac{\sin ^{3} x \cos ^{3} x}{6}+\frac{3}{6}\left( \frac{\sin ^{3} x \cos x}{4}+\frac{ 1}{4} \int \sin ^{2} x d x \right)\\
&= \frac{\sin ^{3} x \cos ^{3} x}{6}+\frac{3}{6}\left( \frac{\sin ^{3} x \cos x}{4}+\frac{ 1}{8} \int (1-\cos 2x) d x \right)\\
&=\frac{\sin ^{3} x \cos ^{3} x}{6}+\frac{3}{6}\left( \frac{\sin ^{3} x \cos x}{4}+\frac{ 1}{8} [x-\frac{1}{2}\sin2x] \right)+C
\end{align*}
