$\sin x\sin 2x\sin 3x$

: Let $I = \int {\sin x\sin 2x\sin 3xdx}$
$= \cfrac{1}{2}\int {\left( {2\sin x\sin 2x} \right)\sin 3x} dx = \cfrac{1}{2}\int {\left( {\cos x - \cos 3x} \right)\sin 3x\,dx}$

$= \cfrac{1}{4}\int {2\sin 3x\,cosx\,} dx - \cfrac{1}{4}\int {2sin3x\,cos3x\,dx}$

$= \cfrac{1}{4}\int {\left( {\sin 4x + \sin 2x} \right)dx - \cfrac{1}{4}\int {\sin 6x\,dx} }$

$= - \cfrac{1}{{16}}\cos 4x - \cfrac{1}{8}\cos 2x + \cfrac{1}{{24}}\cos 6x + C$

$= \cfrac{1}{4}\left[ {\cfrac{1}{6}\cos 6x - \cfrac{1}{4}\cos 4x - \cfrac{1}{2}\cos 2x} \right] + C$