Question

Solve:
$\int \sin x\sin 2x\sin 3x$

Answer 1

$\frac{1}{2}\left(2\sin x\sin 2x\right)\sin 3x$

$=\frac{1}{2}\left(\cos x-\cos 3x\right)\sin 3x$

$=\frac{1}{4}\left(2\sin 3x\cos x-2\sin 3x\cos 3x\right)$

$=\frac{1}{4}\left[\sin 4x+\sin 2x-\sin 6x\right]$

the given integral becomes

$\int \sin x\sin 2x\sin 3xdx$

$=\frac{1}{4}\int \left\{\sin 4x+\sin 2x-\sin 6x\right\}dx$

$=\frac{1}{4}[-\frac{\cos 4x}{4}-\frac{\cos 2x}{4}+\frac{\cos 6x}{6}]+c$