Answer
We use:
$$\int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x$$
Work Step by Step
Given $$\int \sin ^{6} x \cos ^{2} x d x$$
Since
\begin{align*}
\int \sin ^{6} x \cos ^{2} x d x&= \int \sin^6x (1-\sin^2 x)dx\\
&=\int (\sin^6 x-\sin^8 x)dx
\end{align*}
Then use
$$\int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x $$
