Evaluate $\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos 2x}{1 + \cos 2x} dx$.

$\frac{\pi}{2} – 1$

The correct answer is Option (1) → $\frac{\pi}{2} – 1$

$\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos 2x}{1 + \cos 2x} dx$

$f(x) = \frac{\cos 2x}{1 + \cos 2x}$

$f(-x) = \frac{\cos 2x}{1 + \cos 2x}$

Hence, f(x) is even function.

$∴I = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos 2x}{1 + \cos 2x} dx = 2 \int\limits_{0}^{\frac{\pi}{4}} \frac{\cos 2x}{1 + \cos 2x} dx$

$= 2 \int\limits_{0}^{\frac{\pi}{4}} \left[ 1 - \frac{1}{1 + \cos 2x} \right] dx$

$= 2 \left[ \int\limits_{0}^{\frac{\pi}{4}} 1 dx - \int\limits_{0}^{\frac{\pi}{4}} \frac{1}{2} \sec^2 x dx \right]$

$= 2 \left[ x - \frac{1}{2} \tan x \right]_0^{\frac{\pi}{4}}$

$I = 2 \left[ \frac{\pi}{4} - \frac{1}{2} - 0 - 0 \right] = \frac{\pi}{2} – 1$