Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{1-\cot x}{1+\cot x}\right)dx \\ =\int \left(\frac{1-{\displaystyle \frac{\cos x}{\sin x}}}{1+{\displaystyle \frac{\cos x}{\sin x}}}\right)dx \\ =\int \left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)dx \\ \mathrm{Putting}\sin x+\cos x=t \\ \Rightarrow \left(\cos x-\sin x\right)dx=dt \\ \Rightarrow \left(\sin x-\cos x\right)dx=-dt \\ \therefore I=\int \frac{-dt}{t} \\ =-\ln \left|t\right|+C \\ =-\ln \left|\sin x+\cos x\right|+C \end{gathered}$$