=∫11−sin⁡xdx=∫11−sin⁡x1+sin⁡x1+sin⁡xdx=∫1+sin⁡xcos2⁡xdx=∫1cos2⁡x+sin⁡xcos⁡x⋅1cos⁡xdx=tanx+secx+c

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$\int \frac{1}{1 - \sin x} d x$

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$\cot x + \cos e c x + c$

$\tan x - \sec x + c$

$\tan x + \sec x + c$

$\cot x - \cos e c x + c$

answer is C.

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Detailed Solution

$\begin{array}{l} = \int \frac{1}{1 - \sin x} d x \\ = \int (\frac{1}{1 - \sin x}) (\frac{1 + \sin x}{1 + \sin x}) d x \\ = \int \frac{1 + \sin x}{\cos^{2} x} d x \\ = \int [\frac{1}{\cos^{2} x} + \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}] d x \end{array}$