Question

Evaluate $\int {\sin }^{3}xdx$

Answer 1

We know that $\sin 3x=3\sin x-4{\sin }^{3}x$

$\Rightarrow {\sin }^{3}x=\frac{1}{4}(3\sin x-\sin 3x)$

$\int {\sin }^{3}xdx$

$=\frac{1}{4}\int (3\sin x-\sin 3x)dx$

$=\frac{1}{4}[-3\cos x-\left(\frac{-\cos 3x}{3}\right)]+c$

$=\frac{1}{4}[-3\cos x+\frac{\cos 3x}{3}]+c$