Question:

Let $I = \int_{0}^{\pi} \frac{x dx}{1 + \cos^2 x} = \int_{0}^{\pi} \frac{(\pi - x) dx}{1 + \cos^2 (\pi - x)} = \int_{0}^{\pi} \frac{\pi dx}{1 + \cos^2 x} - I$<br/>$\Rightarrow 2I = \int_{0}^{\pi} \frac{\pi dx}{1 + \cos^2 x} = 2\pi \int_{0}^{\pi/2} \frac{dx}{1 + \cos^2 x} = 2\pi \int_{0}^{\pi/2} \frac{\sec^2 x dx}{2 + \tan^2 x}$<br/>Let $\tan x = t$ so that for $x \to 0, t \to 0$ and for $x \to \pi/2, t \to \infty$. Hence we can write,<br/>$I = \pi \int_{0}^{\infty} \frac{dt}{2 + t^2} = \pi \frac{1}{\sqrt{2}} \left[ \tan^{-1} \frac{t}{\sqrt{2}} \right]_{0}^{\infty} = \frac{\pi^2}{2\sqrt{2}}$