Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x$.

(C) Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x$.
Divide numerator and denominator by $\cos^{2} x$:
$I = \int_{0}^{\frac{\pi}{2}} \frac{1}{1+4 \tan ^{2} x} d x$.
Let $2 \tan x = t$,then $2 \sec^{2} x d x = d t$,so $d x = \frac{d t}{2(1+\tan^{2} x)} = \frac{d t}{2(1+t^{2}/4)} = \frac{2 d t}{4+t^{2}}$.
When $x = 0, t = 0$; when $x = \frac{\pi}{2}, t \to \infty$.
$I = \int_{0}^{\infty} \frac{1}{1+t^{2}} \cdot \frac{2 d t}{4+t^{2}} = 2 \int_{0}^{\infty} \frac{d t}{(1+t^{2})(4+t^{2})}$.
Using partial fractions: $\frac{1}{(1+t^{2})(4+t^{2})} = \frac{1}{3} \left( \frac{1}{1+t^{2}} - \frac{1}{4+t^{2}} \right)$.
$I = \frac{2}{3} \left[ \int_{0}^{\infty} \frac{d t}{1+t^{2}} - \int_{0}^{\infty} \frac{d t}{4+t^{2}} \right]$.
$I = \frac{2}{3} \left[ \tan^{-1}(t) \Big|_{0}^{\infty} - \frac{1}{2} \tan^{-1}(\frac{t}{2}) \Big|_{0}^{\infty} \right]$.
$I = \frac{2}{3} \left[ \frac{\pi}{2} - \frac{1}{2} \cdot \frac{\pi}{2} \right] = \frac{2}{3} \left[ \frac{\pi}{2} - \frac{\pi}{4} \right] = \frac{2}{3} \cdot \frac{\pi}{4} = \frac{\pi}{6}$.