Answer
$$\int \sin ^{4} x \cos ^{2} x d x =\frac{-1}{6}\cos ^{3} x \sin ^{3} x -\frac{1}{8}\cos ^{3} x \sin x + \frac{ 1}{16} x+\frac{1}{32}\sin 2 x +C$$
Work Step by Step
$$
\int \sin ^{4} x \cos ^{2} x d x
$$
Here $ m=2,\ n=4$, Use the formula $$
\begin{aligned} \int \cos ^{m} x \sin ^{n} x d x=&-\frac{\cos ^{m+1} x \sin ^{n-1} x}{m+n}+\\ & \frac{n-1}{m+n} \int \cos ^{m} x \sin ^{n-2} x d x \end{aligned}
$$
Then
\begin{align*}
\int \sin ^{4} x \cos ^{2} x d x&= -\frac{\cos ^{3} x \sin ^{3} x}{6} + \frac{3}{6} \int \cos ^{2} x \sin ^{ 2} x d x \\
&=-\frac{\cos ^{3} x \sin ^{3} x}{6} + \frac{1}{2}\left(-\frac{\cos ^{3} x \sin x}{4}+ \frac{ 1}{4} \int \cos ^{2} x d x \right)\\
&=-\frac{\cos ^{3} x \sin ^{3} x}{6} + \frac{1}{2}\left(-\frac{\cos ^{3} x \sin x}{4}+ \frac{ 1}{8} \int( 1+\cos 2 x ) d x \right)\\
&=-\frac{\cos ^{3} x \sin ^{3} x}{6} + \frac{1}{2}\left(-\frac{\cos ^{3} x \sin x}{4}+ \frac{ 1}{8} ( x+\frac{1}{2}\sin 2 x ) \right)+C\\
&=\frac{-1}{6}\cos ^{3} x \sin ^{3} x -\frac{1}{8}\cos ^{3} x \sin x + \frac{ 1}{16} x+\frac{1}{32}\sin 2 x +C
\end{align*}
