$\int \frac{1+\sin \mathrm{x}}{1-\sin \mathrm{x}}\mathrm{dx}=?$

We will solve the given integral by using following formulas,  $\int {\mathrm{x}}^{\mathrm{n}}\mathrm{dx}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c};\int {\sec }^2 \mathrm{xdx}=\tan \mathrm{x}$+C

Therefore,

$$\begin{gathered} \int \frac{1+\sin x}{1-\sin x}dx \\ =\int \frac{1+\sin x(1+\sin x)}{1-\sin x(1+\sin x)}dx \end{gathered}$$

$\Rightarrow \int \frac{(1+\sin \mathrm{x}{)}^2}{1-{\sin }^2 \mathrm{x}}\mathrm{dx}=\int \frac{1+{\sin }^2 \mathrm{x}+2\sin \mathrm{x}}{{\cos }^2 \mathrm{x}}\mathrm{dx}$

$\Rightarrow \int \frac{1}{{\cos }^2 \mathrm{x}}\mathrm{dx}+2\int \frac{\sin \mathrm{x}}{{\cos }^2 \mathrm{x}}\mathrm{dx}+\int \frac{{\sin }^2 \mathrm{x}}{{\cos }^2 \mathrm{x}}\mathrm{dx}$

$\Rightarrow \int {\sec }^2 \mathrm{xdx}+2\int \frac{\sin \mathrm{x}}{{\cos }^2 \mathrm{x}}\mathrm{dx}+\int {\tan }^2 \mathrm{xdx}$

$\Rightarrow \int {\sec }^2 \mathrm{xdx}+2\int \frac{\sin \mathrm{x}}{{\cos }^2 \mathrm{x}}\mathrm{dx}+\int \left(-1+{\sec }^2 \mathrm{x}\right)\mathrm{dx}$$\Rightarrow 2\int {\sec }^2 \mathrm{xdx}+2\int \frac{\sin \mathrm{x}}{{\cos }^2 \mathrm{x}}\mathrm{dx}-\int 1\mathrm{dx}$

Put  $\cos \mathrm{x}=\mathrm{t}$

Therefore $\Rightarrow \sin \mathrm{xdx}=-\mathrm{dt}$

$\Rightarrow 2\tan \mathrm{x}-2\int \frac{\mathrm{dt}}{{\mathrm{t}}^2}-\mathrm{x}+\mathrm{c}$

$\Rightarrow 2\tan x+2\frac{1}{t}-x+c$

$\Rightarrow 2\tan \mathrm{x}+2\sec \mathrm{x}-\mathrm{x}+\mathrm{c}$

Which is option (a)