$\int\sin x \sin 2x \sin 3x\, dx$ is equal to

$-\frac{1}{48}(6 \cos 2x + 3 \cos 4x - 2 \cos 6x) + C$, Where C is constant of integration

The correct answer is Option (3) → $-\frac{1}{48}(6 \cos 2x + 3 \cos 4x - 2 \cos 6x) + C$, Where C is constant of integration

$\int \sin x \sin 2x \sin 3x \, dx$

Use identity

$\sin x \sin 3x=\frac{1}{2}\left[\cos 2x-\cos 4x\right]$

So integrand becomes

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

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

Use identities

$\sin A\cos B=\frac{1}{2}\left[\sin(A+B)+\sin(A-B)\right]$

$\sin 2x\cos 2x=\frac{1}{2}\sin 4x$

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

Hence integrand

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

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

Integrate termwise

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

$=-\frac{1}{16}\cos 4x+\frac{1}{24}\cos 6x-\frac{1}{8}\cos 2x+C$

Taking LCM $48$

$=-\frac{1}{48}\left(6\cos 2x+3\cos 4x-2\cos 6x\right)+C$