Question

By using the properties of definite integrals, evaluate the integral ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^{2}xdx$

Answer 1

Let $I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^{2}xdx$
As ${\sin }^2(-x)=\left(\sin \right(-x){)}^2=(-\sin x{)}^2={\sin }^{2}x$,
Therefore, ${\sin }^{2}x$ is an even functions.
It is known that if $f\left(x\right)$ is an even function, then ${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$
$\therefore I=2{\int }_0^{\frac{\pi }{2}}{\sin }^{2}xdx$
$=2{\int }_0^{\frac{\pi }{2}}\frac{1-\cos 2x}{2}dx$
$={\int }_0^{\frac{\pi }{2}}(1-\cos 2x)dx$
$={[x-\frac{\sin 2x}{2}]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}$