Question

Evaluate:${\int }_0^{\frac{\pi }{2}}{\sin }^{3}xdx$

Answer 1

${\int }_0^{\frac{\pi }{2}}{\sin }^{3}xdx$

$=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}4{\sin }^{3}xdx$

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

$=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}(3\sin x-\sin 3x)dx$

$=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}3\sin xdx-\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\sin 3xdx$

$=\frac{3}{4}{\int }_0^{\frac{\pi }{2}}\sin xdx-\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\sin 3xdx$

$=\frac{3}{4}{[-\cos x]}_0^{\frac{\pi }{2}}-\frac{1}{4}{\left[\frac{-\cos 3x}{3}\right]}_0^{\frac{\pi }{2}}$

$=\frac{-3}{4}[\cos \frac{\pi }{2}-\cos 0]+\frac{1}{12}[\cos \frac{3\pi }{2}-\cos 0]$

$=\frac{-3}{4}[0-1]+\frac{1}{12}[\cos (\pi +\frac{\pi }{2})-1]$

$=\frac{3}{4}+\frac{1}{12}[\cos \frac{\pi }{2}-1]$

$=\frac{3}{4}+\frac{1}{12}[0-1]$

$=\frac{3}{4}-\frac{1}{12}$

$=\frac{3\times 3}{4\times 3}-\frac{1}{12}$

$=\frac{9}{12}-\frac{1}{12}$

$=\frac{8}{12}=\frac{2}{3}$