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$\int (1 - \cos x) {cosec}^{2} xdx$ is equal to

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a

\tan x + C

b

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

c

\frac{1}{2} \tan \frac{x}{2} + C

d

None of these

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detailed solution

Correct option is B

$\begin{array}{l} \int (1 - \cos x) {cosec}^{2} xdx \\ \begin{aligned} = \int {cosec}^{2} xdx - \int {cosec}^{2} xcos xdx = \cot x - \int cosec xcot xdx \\ \begin{array}{ll} = & - \cot x + cosec x + C = \frac{1 - \cos x}{\sin x} + C \\ = \frac{2 \sin^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} + C \\ = \tan \frac{x}{2} + C \end{array} \end{aligned} \end{array}$

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