Question

Find the integral of the function: $\sin x\sin 2x\sin 3x$

Answer 1

To find: $\int \sin x\sin 2x\sin 3xdx$

$=\int \left(\sin x\sin 2x\right)\sin 3xdx$

Multiply and divide by $2$

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

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

$[\because 2\sin A\sin B=\cos (A-B)-\cos (A+B\left)\cos \right(-x)=\cos x]$

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

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

Again, multiply and divide by $2$

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

$=\frac{1}{4}\int (2\sin 3x.\cos x-\sin 6x)dx$

$[\because \sin 2x=2\sin x\cos x]$

$=\frac{1}{4}\int (2\sin 3x.\cos x-\sin 6x)dx$

$[\because 2sinA\cos B=\sin (A+B)+\sin (A-B\left)\right]$

$=\frac{1}{4}\left[\sin \right(3x+x)+\sin (3x-x)-\sin 6x]dx$

$=\frac{1}{4}[\sin 4x+\sin 2x-\sin 6x]dx$

$=\frac{1}{4}[\sin 4xdx+\sin 2xdx-\sin 6xdx]dx$

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

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

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

Where $C$ is constant of integration.