Answer
$$ -\frac{1}{8} \cos ^{7} x \sin x+\frac{1}{48} \cos ^{5} x \sin x+\frac{5}{192} \cos ^{3} x \sin x+\frac{5}{128} \cos x \sin x+\frac{5}{128} x+C$$
Work Step by Step
Since
\begin{align*}
\int \sin ^{2} x \cos ^{6} x d x&=\int \cos ^{6} x\left(1-\cos ^{2} x\right) d x\\
&=\int \cos ^{6} x d x-\int \cos ^{8} x d x
\end{align*}
Use
$$\int \cos ^{n} u d u=\frac{1}{n} \cos ^{n-1} u \sin u+\frac{n-1}{n} \int \cos ^{n-2} u d u $$
Then
\begin{align*}
\int \sin ^{2} x \cos ^{6} x d x &=\int \cos ^{6} x d x-\int \cos ^{8} x d x\\
&=\int \cos ^{6} x d x-\left(\frac{1}{8} \cos ^{7} x \sin x+\frac{7}{8} \int \cos ^{6} x d x\right)\\
&=-\frac{1}{8} \cos ^{7} x \sin x+\frac{1}{8} \int \cos ^{6} x d x\\
&=-\frac{1}{8} \cos ^{7} x \sin x+\frac{1}{8}\left(\frac{1}{6} \cos ^{5} x \sin x+\frac{5}{6} \int \cos ^{4} x d x\right)\\
&=-\frac{1}{8} \cos ^{7} x \sin x+\frac{1}{48} \cos ^{5} x \sin x+\frac{5}{48} \int \cos ^{4} x d x\\
&=-\frac{1}{8} \cos ^{7} x \sin x+\frac{1}{48} \cos ^{5} x \sin x+\frac{5}{48}\left(\frac{1}{4} \cos ^{3} x \sin x+\frac{3}{4} \int \cos ^{2} x d x\right)\\
&=-\frac{1}{8} \cos ^{7} x \sin x+\frac{1}{48} \cos ^{5} x \sin x+\frac{5}{192} \cos ^{3} x \sin x+\frac{5}{64} \int \cos ^{2} x d x\\
&=-\frac{1}{8} \cos ^{7} x \sin x+\frac{1}{48} \cos ^{5} x \sin x+\frac{5}{192} \cos ^{3} x \sin x+\frac{5}{64} \int \cos ^{2} x d x\\
&=-\frac{1}{8} \cos ^{7} x \sin x+\frac{1}{48} \cos ^{5} x \sin x+\frac{5}{192} \cos ^{3} x \sin x+\frac{5}{64}\left(\frac{1}{2} \cos x \sin x+\frac{1}{2} x\right)\\
&=-\frac{1}{8} \cos ^{7} x \sin x+\frac{1}{48} \cos ^{5} x \sin x+\frac{5}{192} \cos ^{3} x \sin x+\frac{5}{128} \cos x \sin x+\frac{5}{128} x+C
\end{align*}
