Evaluate $\int \frac{\cos x - \cos 2x}{1 - \cos x} dx$.

$2\sin x + x + C$

The correct answer is Option (2) → $2\sin x + x + C$

Let $I = \int \frac{\cos x - \cos 2x}{1 - \cos x} dx = \int \frac{2 \sin \frac{3x}{2} \cdot \sin \frac{x}{2}}{1 - 1 + 2 \sin^2 \frac{x}{2}} dx$

$= 2 \int \frac{\sin \frac{3x}{2} \cdot \sin \frac{x}{2}}{2 \sin^2 \frac{x}{2}} dx = \int \frac{\sin \frac{3x}{2}}{\sin \frac{x}{2}} dx$

$= \int \frac{3 \sin \frac{x}{2} - 4 \sin^3 \frac{x}{2}}{\sin \frac{x}{2}} dx \quad [∵\sin 3x = 3 \sin x - 4 \sin^3 x]$

$= 3 \int dx - 4 \int \sin^2 \frac{x}{2} dx = 3 \int dx - 4 \int \frac{1 - \cos x}{2} dx$

$= 3 \int dx - 2 \int dx + 2 \int \cos x \, dx$

$= \int dx + 2 \int \cos x \, dx = x + 2 \sin x + C = 2 \sin x + x + C$