Question

Integrate:
${\int }_{-\pi /2}^{\pi /2}{\sin }^{2}x$

Answer 1

$\begin{array}{c}I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\sin }^{2}xdx\\ I=2{\int }_0^{\frac{\pi }{2}}\frac{1-\cos 2x}{2}dx\\ =2\times \frac{1}{2}{\left[x-\frac{\sin 2x}{2}\right]}_0^{\frac{\pi }{2}}\\ =\frac{\pi }{2}-\frac{\sin 2\times \frac{\pi }{2}}{2}-0\\ =\frac{\pi }{2}-0=\frac{\pi }{2}\end{array}$

Hence, this is the answer.