Exercise asked in the math group by Deepak Verma

Exercise:

Integrate,$\int{\sin x\sin 2x\sin 3x\,dx}$

Solution: we know that $\sin x\sin 3x=\frac{1}{2}\left[ \cos \left( x-3x \right)-\cos \left( x+3x \right) \right]=\frac{1}{2}\left[ \cos 2x-\cos 4x \right]$

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

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

We know that $\sin 2x\cos 2x=\frac{1}{2}\left[ \sin 4x+0 \right]=\frac{1}{2}\sin 4x$

So $\int{\sin 2x\cos 2x}\,dx=\frac{1}{2}\int{\sin 4x\,dx}=-\frac{1}{8}\cos 4x+c$

Thus $\frac{1}{2}\int{\sin 2x\cos 2x\,dx}=-\frac{1}{16}\cos 4x+c$

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

Thus $\int{\sin 2x\cos 4x\,dx}=\frac{1}{2}\int{\sin 6x\,dx}-\frac{1}{2}\int{\sin 2x\,dx}=-\frac{1}{12}\cos 6x+\frac{1}{4}\cos 2x+c$

Hence $\frac{1}{2}\int{\sin 2x\cos 4x}\,dx=-\frac{1}{24}\cos 6x+\frac{1}{8}\cos 2x+c$

Therefore $\int{\sin x\sin 2x\sin 3x}\,dx=-\frac{1}{16}\cos 4x+\frac{1}{24}\cos 6x-\frac{1}{8}\cos 2x+c$

$=-\frac{1}{8}\left( \cos 2x+\frac{1}{2}\cos 4x-\frac{1}{3}\cos 6x \right)+c$