Question

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

Answer 1

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