Question:

Let $f(x) = \frac{1}{1 + 3 \cos^2 x} \Rightarrow f(\pi - x) = f(x) \Rightarrow \int_{0}^{\pi} \frac{dx}{1 + 3 \cos^2 x} = 2 \int_{0}^{\pi/2} \frac{dx}{1 + 3 \cos^2 x}$ $= 2 \int_{0}^{\pi/2} \frac{\sec^2 x \, dx}{1 + \tan^2 x + 3} = 2 \int_{0}^{\pi/2} \frac{\sec^2 x \, dx}{4 + \tan^2 x} = \left[ \tan^{-1} \left( \frac{\tan x}{2} \right) \right]_0^{\pi/2}$ $\because \tan \frac{\pi}{2}$ is undefined, we take limit $= \lim_{x \to \pi/2^-} \tan^{-1} \left( \frac{\tan x}{2} \right) - \tan^{-1} \left( \frac{\tan 0}{2} \right) = \pi/2 - 0 = \pi/2$