By using the properties of definite integrals,evaluate the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x \, dx$.

(N/A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2} x \, dx$.
Since $\sin^{2}(-x) = (\sin(-x))^{2} = (-\sin x)^{2} = \sin^{2} x$,the function $f(x) = \sin^{2} x$ is an even function.
Using the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ for an even function:
$I = 2 \int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx$.
Using the trigonometric identity $\sin^{2} x = \frac{1 - \cos 2x}{2}$:
$I = 2 \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \, dx = \int_{0}^{\frac{\pi}{2}} (1 - \cos 2x) \, dx$.
Integrating the terms:
$I = \left[ x - \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}}$.
Evaluating at the limits:
$I = \left( \frac{\pi}{2} - \frac{\sin(\pi)}{2} \right) - (0 - 0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.