Question

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

Answer 1

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

$=\frac{1}{2}\int \sin x(\cos x-\cos 5x)dx$

$=\frac{1}{2}\int (\sin x\cos x-\sin x\cos 5x)dx$

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

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

Hence, the answer is $\frac{1}{4}[\frac{-\cos 2x}{2}+\frac{\cos 6x}{6}-\frac{\cos 4x}{4}+c].$