Prove

It is known that, $\sin A \sin B=\frac{1}{2}\{\cos (A-B)-\cos (A+B)\}$

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

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

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

$=\frac{1}{2} \int \frac{\sin 2 x}{2} d x-\frac{1}{2} \int \sin x \cos 5 x d x$

$=\frac{1}{2} \int \frac{\sin 2 x}{2} d x-\frac{1}{2} \int \sin x \cos 5 x d x$

$=\frac{1}{4}\left[\frac{-\cos 2 x}{2}\right]-\frac{1}{2} \int\left\{\frac{1}{2} \sin (x+5 x)+\sin (x-5 x)\right\} d x$

$=\frac{-\cos 2 x}{8}-\frac{1}{4} \int(\sin 6 x+\sin (-4 x)) d x$

$=\frac{-\cos 2 x}{8}-\frac{1}{4}\left[\frac{-\cos 6 x}{3}+\frac{\cos 4 x}{4}\right]+\mathrm{C}$

$=\frac{-\cos 2 x}{8}-\frac{1}{8}\left[\frac{-\cos 6 x}{3}+\frac{\cos 4 x}{2}\right]+\mathrm{C}$

$=\frac{1}{8}\left[\frac{\cos 6 x}{3}-\frac{\cos 4 x}{2}-\cos 2 x\right]+\mathrm{C}$