Question

Find the integrals of the functions $\sin x\sin 2x\sin 3x$

Answer 1

It is known that, $\sin A\sin B=\frac{1}{2}\left\{\cos \right(A-B)-\cos (A+B\left)\right\}$
$\therefore \int \sin x\sin 2x\sin 3xdx=\int [\sin x\cdot \frac{1}{2}\left\{\cos \right(2x-3x)-\cos (2x+3x\left)\right\}]dx$
$=\frac{1}{2}\int \left(\sin x\cos \right(-x)-\sin x\cos 5x)dx$
$=\frac{1}{2}\int (\sin x\cos x-\sin x\cos 5x)dx$
$=\frac{1}{2}\int \frac{\sin 2x}{2}dx-\frac{1}{2}\int \sin x\cos 5xdx$
$=\frac{1}{4}\left[\frac{-\cos 2x}{2}\right]-\frac{1}{2}\int \left\{\frac{1}{2}\sin \right(x+5x)+\sin (x-5x\left)\right\}dx$
$=\frac{-\cos 2x}{8}-\frac{1}{4}\int (\sin 6x+\sin (-4x\left)\right)dx$
$=\frac{-\cos 2x}{8}-\frac{1}{4}[\frac{-\cos 6x}{3}+\frac{\cos 4x}{4}]+C$
$=\frac{-\cos 2x}{8}-\frac{1}{8}[\frac{-\cos 6x}{3}+\frac{\cos 4x}{2}]+C$
$=\frac{1}{8}[\frac{\cos 6x}{3}-\frac{\cos 4x}{2}-\cos 2x]+C$