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

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

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

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

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

$= 3\int d x - 2\int d x + 2\int {\cos } xdx$

$= \int d x + 2\int {\cos } xdx = x + 2\sin x + C = 2\sin x + x + C$