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$\int \frac{1 + \cos x}{1 - \cos x} d x =$

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$- 2 \cot (\frac{x}{2}) - x + c$

$- 2 \cot (\frac{x}{2}) + x + c$

$2 \cot (\frac{x}{2}) + x + c$

$2 \cot (\frac{x}{2}) - x + c$

answer is B.

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Detailed Solution

$\int \frac{1 + \cos x}{1 - \cos x} d x$

$= \int \frac{2 \cos^{2} \frac{x}{2}}{2 \sin^{2} \frac{x}{2}} d x$

$= \int \cot^{2} \frac{x}{2} d x$

$= \int (\cos e c^{2} \frac{x}{2} - 1) d x$

$= - \frac{\cot x / 2}{1 / 2} - x + c$

$= - 2 \cot \frac{x}{2} - x + c$

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