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

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Solution

Now,
$\int \frac{cos x - cos 2 x}{(1 - cos x)} d x$
$= \int \frac{cos x - 2 {cos}^{2} x + 1}{(1 - cos x)} d x$ [ Since $cos 2 x = 2 {cos}^{2} x - 1$ ]
$= \int \frac{(1 - cos x) (1 + 2 cos x)}{(1 - cos x)} d x$ [ Since $- 2 {cos}^{2} x + cos x + 1 = (1 - cos x) (1 + 2 cos x)$
$\int (1 + 2 cos x) d x$
$= x + 2 sin x + c$ [ Where $c$ is integrating constant]

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