Question

$\int \frac{1+\cos x}{1-\cos x}dx$

Answer 1

$$\begin{gathered} \int \left(\frac{1+\cos x}{1-\cos x}\right)dx \\ =\int \left(\frac{2{\cos }^2{\displaystyle \frac{x}{2}}}{2{\sin }^2{\displaystyle \frac{x}{2}}}\right)dx\left[\therefore 1+\cos x=2{\cos }^2\frac{x}{2}\&1-\cos x=2{\sin }^2\frac{x}{2}\right] \\ =\int {\cot }^2\frac{\mathrm{x}}{2}dx \\ =\int \left({\mathrm{cosec}}^2\frac{x}{2}-1\right)dx \\ =\frac{-\cot \left({\displaystyle \frac{x}{2}}\right)}{{\displaystyle \frac{1}{2}}}-x+\mathrm{C} \\ =-2\cot \left(\frac{x}{2}\right)-x+\mathrm{C} \end{gathered}$$