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The value of $\int \frac{1 + \sin x}{1 - \sin x} d x$ is

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a

2 \tan (\frac{x}{2} + \frac{π}{4}) + C

b

2 \tan (\frac{x}{2} + \frac{π}{4}) + x + C

c

2 \tan (\frac{x}{2} + \frac{π}{4}) - x + C

d

2 \tan^{2} (\frac{x}{2} + \frac{π}{4}) - x + C

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detailed solution

Correct option is C

$\begin{array}{l} \int \frac{1 + \sin x}{1 - \sin x} d x = \int \frac{(1 + \sin x)^{2}}{1 - \sin^{2} x} d x \\ = \int (\sec x + \tan x)^{2} d x \\ = \int \sec^{2} x + \int (\sec^{2} x - 1) d x + 2 \int \sec x \tan x d x \\ = 2 \tan x - x + 2 \sec x + C \\ = 2 (\frac{1 + \sin x}{\cos x}) - x + C \\ = 2 (\frac{1 - \cos (x + \frac{π}{2})}{\sin (x + \frac{π}{2})}) - x + C \\ = 2 \frac{2 \sin^{2} (\frac{x}{2} + \frac{π}{4})}{2 \sin (\frac{x}{2} + \frac{π}{4}) \cos (\frac{x}{2} + \frac{π}{4})} - x + C \\ = 2 \tan (\frac{x}{2} + \frac{π}{4}) - x + C \end{array}$

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