The value of ∫1+sin⁡x1−sin⁡xdx is

∫1+sin⁡x1−sin⁡xdx=∫(1+sin⁡x)21−sin2⁡xdx=∫(sec⁡x+tan⁡x)2dx=∫sec2⁡x+∫sec2⁡x−1dx+2∫sec⁡xtan⁡xdx=2tan⁡x−x+2sec⁡x+C=21+sin⁡xcos⁡x−x+C=21−cos⁡x+π2sin⁡x+π2−x+C=22sin2⁡x2+π42sin⁡x2+π4cos⁡x2+π4−x+C=2tan⁡x2+π4−x+C

The value of ∫1+sin⁡x1−sin⁡xdx is

∫1+sin⁡x1−sin⁡xdx=∫(1+sin⁡x)21−sin2⁡xdx=∫(sec⁡x+tan⁡x)2dx=∫sec2⁡x+∫sec2⁡x−1dx+2∫sec⁡xtan⁡xdx=2tan⁡x−x+2sec⁡x+C=21+sin⁡xcos⁡x−x+C=21−cos⁡x+π2sin⁡x+π2−x+C=22sin2⁡x2+π42sin⁡x2+π4cos⁡x2+π4−x+C=2tan⁡x2+π4−x+C

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

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

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

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

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

answer is C.

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Detailed Solution

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