Question

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

Answer 1

Solve the given integration

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

We know that,

$\tan x=\frac{\sin x}{\cos x}$

Now ,

$\int \frac{1+\tan x}{1-\tan x}dx$ $=\int \left\{\frac{\sin x+\cos x}{\cos x-\sin x}\right\}dx$

Put , $t=\cos x-\sin x$

On differentiating we get,

$-dt=(\sin x+\cos x)dx$

Now,

$$\begin{gathered} I=\int \left\{\frac{\sin x+\cos x}{\cos x-\sin x}\right\}dx \\ I=\frac{-dt}{t} \\ I=-\log t+C \\ I=-\log \left(\cos x+\sin x\right)+C \end{gathered}$$

Hence , Integral $\int \frac{1+\tan x}{1-\tan x}dx$ is $-\log \left(\cos x+\sin x\right)+C$.