Using properties of definite integral, calculate the value of:

Let $I = \int_{0}^{\pi/2} \frac{\sin^2 x}{1 + \sin x \cos x} \, dx$. Using property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$: $I = \int_{0}^{\pi/2} \frac{\sin^2(\pi/2 - x)}{1 + \sin(\pi/2 - x)\cos(\pi/2 - x)} \, dx = \int_{0}^{\pi/2} \frac{\cos^2 x}{1 + \cos x \sin x} \, dx$. Adding both equations: $2I = \int_{0}^{\pi/2} \frac{\sin^2 x + \cos^2 x}{1 + \sin x \cos x} \, dx = \int_{0}^{\pi/2} \frac{1}{1 + \sin x \cos x} \, dx$. Dividing numerator and denominator by $\cos^2 x$: $2I = \int_{0}^{\pi/2} \frac{\sec^2 x}{\sec^2 x + \tan x} \, dx = \int_{0}^{\pi/2} \frac{\sec^2 x}{1 + \tan^2 x + \tan x} \, dx$. Let $t = \tan x$, so $dt = \sec^2 x \, dx$: $2I = \int_{0}^{\infty} \frac{dt}{(t + 1/2)^2 + (\sqrt{3}/2)^2} = \left[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2t + 1}{\sqrt{3}}\right) \right]_{0}^{\infty} = \frac{2}{\sqrt{3}} \left( \frac{\pi}{2} - \frac{\pi}{6} \right) = \frac{2\pi}{3\sqrt{3}}$. Hence, $I = \frac{\pi}{3\sqrt{3}}$.