Question

$\int \frac{{\mathrm{cosec}}^{2}x}{1+\cot x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{{\mathrm{cosec}}^{2}x}{1+\cot x}dx \\ \mathrm{Putting}\cot x=t \\ \Rightarrow -{\mathrm{cosec}}^{2}x=\frac{dt}{dx} \\ \Rightarrow {\mathrm{cosec}}^{2}xdx=-dt \\ \therefore I=\int \frac{-dt}{1+t} \\ =-\ln \left|1+t\right|+C \\ =-\ln \left|1+\cot x\right|+C\left[\because t=\cot x\right] \end{gathered}$$