Find the integrals of the functions $\sin x\;\sin2x\;\sin3x$

Question

$\begin{array}{1 1} \frac{[\frac{1}{3}\cos 6x-\frac{1}{2}\cos 4x-\cos 2x]}{8}+c. \\\large \frac{[\frac{1}{3}\cos 6x+\frac{1}{2}\cos 4x-\cos 2x]}{8}+c. \\ \large \frac{[\frac{1}{3}\cos 6x-\frac{1}{2}\cos 4x+\cos 2x]}{8}+c. \\ \frac{[\frac{1}{3}\cos 6x+\frac{1}{2}\cos 4x+\cos 2x]}{8}+c.\end{array} $