∫sin x sin 2x sin 3x dx is equal to

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

The product is turned into a sum with the product-to-sum formulas:

    \(2\sin A\sin B = \cos(A - B) - \cos(A + B)\)
    \(2\sin A\cos B = \sin(A + B) + \sin(A - B)\)

With A = 3x, B = x:

    \(\displaystyle \sin x\sin 3x = \frac{1}{2}(\cos 2x - \cos 4x)\)

So

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

    \(\displaystyle \sin 2x\cos 2x = \frac{1}{2}\sin 4x\)

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

Hence

    \(\displaystyle \sin x\sin 2x\sin 3x = \frac{1}{4}\left(\sin 2x + \sin 4x - \sin 6x\right)\)

Integrating term by term:

    \(\displaystyle \int\sin x\sin 2x\sin 3x\,dx = \frac{1}{4}\left(-\frac{\cos 2x}{2} - \frac{\cos 4x}{4} + \frac{\cos 6x}{6}\right) + C\)

    \(\displaystyle = -\frac{\cos 2x}{8} - \frac{\cos 4x}{16} + \frac{\cos 6x}{24} + C\)

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

NCERT (2026-27 edition):
Integration Using Trigonometric Identities