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Q.

$\int \frac{x + \sin x}{1 - \cos x} d x$ is equal to

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a

$x \tan \frac{x}{2} + C$

b

$\cot \frac{x}{2} + C$

c

$\log (1 + \cos x) + C$

d

$\log (x + \sin x) + C$

answer is A.

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Detailed Solution

Let

$\begin{array}{l} I = \int \frac{x + \sin x}{1 + \cos x} d x = \int \frac{x}{2} \sec^{2} \frac{x}{2} + \tan \frac{x}{2} d x \\ \Rightarrow I = \int x \frac{1}{2} \sec^{2} \frac{x}{2} d x + \int \tan \frac{x}{2} d x \\ \Rightarrow 1 = x \tan \frac{x}{2} - \int \tan \frac{x}{2} d x + \int \tan \frac{x}{2} d x + C = x \tan \frac{x}{2} \end{array}$

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