The value of ∫1+sin⁡x1−sin⁡xdx is

# The value of $\int \frac{1+\mathrm{sin}x}{1-\mathrm{sin}x}dx$ is

1. A

$2\mathrm{tan}\left(\frac{x}{2}+\frac{\pi }{4}\right)+C$

2. B

$2\mathrm{tan}\left(\frac{x}{2}+\frac{\pi }{4}\right)+x+C$

3. C

$2\mathrm{tan}\left(\frac{x}{2}+\frac{\pi }{4}\right)-x+C$

4. D

$2{\mathrm{tan}}^{2}\left(\frac{x}{2}+\frac{\pi }{4}\right)-x+C$

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### Solution:

$\begin{array}{l}\int \frac{1+\mathrm{sin}x}{1-\mathrm{sin}x}dx=\int \frac{\left(1+\mathrm{sin}x{\right)}^{2}}{1-{\mathrm{sin}}^{2}x}dx\\ =\int \left(\mathrm{sec}x+\mathrm{tan}x{\right)}^{2}dx\\ =\int {\mathrm{sec}}^{2}x+\int \left({\mathrm{sec}}^{2}x-1\right)dx+2\int \mathrm{sec}x\mathrm{tan}xdx\\ =2\mathrm{tan}x-x+2\mathrm{sec}x+C\\ =2\left(\frac{1+\mathrm{sin}x}{\mathrm{cos}x}\right)-x+C\\ =2\left(\frac{1-\mathrm{cos}\left(x+\frac{\pi }{2}\right)}{\mathrm{sin}\left(x+\frac{\pi }{2}\right)}\right)-x+C\\ =2\frac{2{\mathrm{sin}}^{2}\left(\frac{x}{2}+\frac{\pi }{4}\right)}{2\mathrm{sin}\left(\frac{x}{2}+\frac{\pi }{4}\right)\mathrm{cos}\left(\frac{x}{2}+\frac{\pi }{4}\right)}-x+C\\ =2\mathrm{tan}\left(\frac{x}{2}+\frac{\pi }{4}\right)-x+C\end{array}$