Find: $\int \sin^{3} x \, dx$

We use the trigonometric identity $\sin 3x = 3 \sin x - 4 \sin^{3} x$.
Rearranging this,we get $\sin^{3} x = \frac{3 \sin x - \sin 3x}{4}$.
Now,integrate both sides:
$\int \sin^{3} x \, dx = \int \frac{3 \sin x - \sin 3x}{4} \, dx$
$= \frac{3}{4} \int \sin x \, dx - \frac{1}{4} \int \sin 3x \, dx$
$= \frac{3}{4} (-\cos x) - \frac{1}{4} \left(-\frac{\cos 3x}{3}\right) + C$
$= -\frac{3}{4} \cos x + \frac{1}{12} \cos 3x + C$.
Alternatively,using substitution:
$\int \sin^{3} x \, dx = \int \sin^{2} x \cdot \sin x \, dx = \int (1 - \cos^{2} x) \sin x \, dx$.
Let $t = \cos x$,then $dt = -\sin x \, dx$,or $\sin x \, dx = -dt$.
Substituting these into the integral:
$\int (1 - t^{2}) (-dt) = \int (t^{2} - 1) \, dt = \frac{t^{3}}{3} - t + C$.
Substituting back $t = \cos x$:
$= \frac{1}{3} \cos^{3} x - \cos x + C$.