Question:
$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x d x$
Solution:
Let $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x d x$
As $\sin ^{2}(-x)=(\sin (-x))^{2}=(-\sin x)^{2}=\sin ^{2} x$, therefore, $\sin ^{2} x$ is an even function.
It is known that if $f(x)$ is an even function, then $\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$
$I=2 \int_{0}^{\frac{\pi}{2}} \sin ^{2} x d x$
$=2 \int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2} d x$
$=\int_{0}^{\frac{\pi}{2}}(1-\cos 2 x) d x$
$=\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}$
$=\frac{\pi}{2}$