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

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