Find $\int\sin^3x\,dx$

$-\frac{3}{4} \cos x + \frac{1}{12} \cos 3x + C$

The correct answer is Option (3) → $-\frac{3}{4} \cos x + \frac{1}{12} \cos 3x + C$

From the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$, we find that

$\sin^3 x = \frac{3 \sin x - \sin 3x}{4}$

Therefore, $\displaystyle \int \sin^3 x \, dx = \frac{3}{4} \int \sin x \, dx - \frac{1}{4} \int \sin 3x \, dx$

$= -\frac{3}{4} \cos x + \frac{1}{12} \cos 3x + C$