Evaluate $\int \frac{dx}{1 + \cos x}$.

$\tan(x/2) + C$

The correct answer is Option (1) → $\tan(x/2) + C$

Let $I = \int \frac{dx}{1 + \cos x} = \int \frac{dx}{1 + 2\cos^2 \frac{x}{2} - 1}$

$= \frac{1}{2} \int \frac{1}{\cos^2 \frac{x}{2}} dx = \frac{1}{2} \int \sec^2 \frac{x}{2} dx$

$= \frac{1}{2} \cdot \tan \frac{x}{2} \cdot 2 + C = \tan \frac{x}{2} + C \quad [∵\int \sec^2 x dx = \tan x]$