Answer
$$\frac{1}{8}\left(\frac{5x}{2}-2\sin \left(2x\right)+\frac{3}{4}\sin \left(2x\right)\cos \left(2x\right)+\frac{1}{6}\sin ^3\left(2x\right)+C\right)$$
Work Step by Step
\begin{aligned}
\int \sin ^{6} x d x &=\int\left(\frac{1}{2}(1-\cos 2 x)\right)^{3} d x \\ &=\frac{1}{8} \int(1-\cos 2 x)^{3} d x \\
&=\frac{1}{8} \int\left(1-3 \cos 2 x+3 \cos ^{2} 2 x-\cos ^{3} 2 x\right) d x\\
&=\frac{1}{8} \int\left(1-3 \cos 2 x+\frac{3}{2}(1+ \cos 4x)-\cos ^{2} 2 x\cos 2 x\right) d x\\
&=\frac{1}{8} \int\left(\frac{5}{2}-3 \cos 2 x+\frac{3}{2}\cos 4x-\cos ^{2} 2 x\cos 2 x\right) d x\\
&=\frac{1}{8}\left(\frac{5x}{2}-2\sin \left(2x\right)+\frac{3}{4}\sin \left(2x\right)\cos \left(2x\right)+\frac{1}{6}\sin ^3\left(2x\right)+C\right)
\end{aligned}
