Question

Solve :
${\int }_0^{\pi }{\sin }^{3}xdx$

Answer 1

$\begin{array}{c}I={\int }_0^{\pi }{\sin }^{3}xdx\\ =2{\int }_0^{\frac{\pi }{2}}{\sin }^{3}xdx\\ =2\left[{\sin }^{2}x\left(-\cos x\right)+{\int }^{}\cos x{\sin }^{2}xdx\right]\\ =2\left[-{\sin }^2\cos x+{\int }^{}2\sin {\cos }^{2}dx\right]\\ =2{\int }^{}2\sin x{\cos }^{2}xdx=-{\int }^{}2t^{2}dt\\ \cos x=t\\ =-\frac{2}{3}{t}^3=\frac{-2}{3}{\left(\cos x\right)}^3\\ \frac{dt}{dx}=-\sin x\\ =2{\left[-{\sin }^{2}x\cos x+\frac{-2}{3}{\left(cosx\right)}^3\right]}_0^{\frac{\pi }{2}}\\ hence,\\ I=2\left[0-\left(-\frac{2}{3}\right)\right]=\frac{4}{3}\end{array}$