Question

Solve the integral $I={\int }_0^{\pi }{\sin }^{2}xdx$.

Answer 1

$I={\int }_0^{\pi }{\sin }^{2}xdx$
${\int }_0^{\pi }{\sin }^{2}xdx={\int }_0^{\pi }\left(\frac{1-\cos 2x}{2}\right)dx(\because {\sin }^{2}x=\frac{1-\cos 2x}{2})$
$=\frac{1}{2}[{\int }_0^{\pi }dx-{\int }_0^{\pi }(\cos 2x\left)dx\right]$
$=\frac{1}{2}{[x-\frac{\sin 2x}{2}]}_0^{\pi }$
$=\frac{1}{2}\left[\right(\pi -0)-\frac{(\sin 2\pi -\sin 0)}{2}]=\frac{\pi }{2}$.