Answer
$$\frac{1}{32}(12 x-8 \sin 2 x+\sin 4 x) +C=-\frac{1}{4}\sin^3 x\cos x-\frac{3}{8} \sin x\cos x+\frac{3}{8}x+C$$
Work Step by Step
Since
\begin{align*}
\sin 2x& =2\sin x\cos x\\
\cos 2x &=cos^2x-\sin ^2 x=1-2\sin^2x
\end{align*}
Then
\begin{align*}
\frac{1}{32}(12 x-8 \sin 2 x+\sin 4 x)+C&=\frac{1}{32}(12 x-16 \sin x\cos x+2\sin 2 x\cos 2x)+C\\
&=\frac{1}{32}\left(12 x-16 \sin x\cos x+4\sin x\cos x[1-2\sin^2 x]\right)+C\\
&=\frac{1}{32}\left(12 x-16 \sin x\cos x+4\sin x\cos x -8\sin^3 x\cos x \right)+C\\
&=-\frac{1}{4}\sin^3 x\cos x-\frac{3}{8} \sin x\cos x+\frac{3}{8}x+C
\end{align*}
