Question

Solve ${\int }^{}\sin x\sin 2x\sin 3xdx.$

Answer 1

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

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

We have multiplying and dividing by $2$ to get it in addition form

$\frac{1}{2}\int \left[\cos \right(2x-x)-\cos (2x+x\left)\right]\sin 3xdx$

$=\frac{1}{2}\int (\sin 3x\cos x-\sin 3x\cos 3x)dx$

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

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

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