Evaluate $\int \sin^{3} x\,dx$.

Explanation

1. 01

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

2. 02

   $\int = \frac{1}{4}\left(-3\cos x + \frac{\cos 3x}{3}\right) = -\frac{3\cos x}{4} + \frac{\cos 3x}{12} + C$.