By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x$.

Let $I = \int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x$ ..... $(1)$
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get:
$I = \int_{0}^{\frac{\pi}{2}} \cos ^{2} \left(\frac{\pi}{2} - x\right) d x$
Since $\cos(\frac{\pi}{2} - x) = \sin x$,we have:
$I = \int_{0}^{\frac{\pi}{2}} \sin ^{2} x d x$ ..... $(2)$
Adding $(1)$ and $(2)$,we obtain:
$2I = \int_{0}^{\frac{\pi}{2}} (\sin ^{2} x + \cos ^{2} x) d x$
Since $\sin ^{2} x + \cos ^{2} x = 1$,we have:
$2I = \int_{0}^{\frac{\pi}{2}} 1 d x$
$2I = [x]_{0}^{\frac{\pi}{2}}$
$2I = \frac{\pi}{2} - 0$
$2I = \frac{\pi}{2}$
$I = \frac{\pi}{4}$