Answer
$$-\frac{\sin ^{5} x \cos x}{6}+\frac{5}{6}\left( -\frac{\sin ^{3} x \cos x}{4}+\frac{3}{4} \left[-\frac{\sin x \cos x}{2}+\frac{1}{2}x\right]\right) $$
Work Step by Step
Given $$ \int \sin ^{6} x d x$$
Use
$$ \int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x$$
For $n=6 $
\begin{align*}
\int \sin ^{6} x d x&=-\frac{\sin ^{5} x \cos x}{6}+\frac{5}{6} \int \sin ^{4} x d x\\
&=-\frac{\sin ^{5} x \cos x}{6}+\frac{5}{6}\left( -\frac{\sin ^{3} x \cos x}{4}+\frac{3}{4} \int \sin ^{2} x d x\right)\\
&=-\frac{\sin ^{5} x \cos x}{6}+\frac{5}{6}\left( -\frac{\sin ^{3} x \cos x}{4}+\frac{3}{4} \left[-\frac{\sin x \cos x}{2}+\frac{1}{2}x\right]\right)
\end{align*}
