Question

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

Answer 1

Given that ,

$y=\int \frac{1}{1-\tan x}dx$ ……(1)

Put $t=1-\tan x$, we get

$\Rightarrow \tan x=t-1\Rightarrow {\tan }^{2}x={(t-1)}^2\Rightarrow {\sec }^{2}x={(t-1)}^2-1$

${\sec }^{2}xdx=dt$

$dx=\frac{dt}{{\sec }^{2}x}$

$y=\int \frac{1}{t}\frac{dt}{{\sec }^{2}x}$

$y=\int \frac{1}{t}\frac{dt}{{(t-1)}^2-1}$

$=\int \frac{1}{t(t^2-2t)}dt$

$=\int \frac{1}{t^2(t-2)}dt$

$=\int (\frac{-t-2}{4t^2}+\frac{1}{4(t-2)})dt$

$=-\int \frac{t+2}{4t^2}dt+\frac{1}{4}\int \frac{1}{t-2}dt$

$y=-\frac{1}{8}\int \frac{2t+4}{t^2}dt+\frac{1}{4}\int \frac{1}{t-2}dt$

$=-\frac{1}{8}\log t^2+\frac{1}{4}\log (t-2)+C$

$=-\frac{1}{8}\log (1+\tan x)+\frac{1}{4}\log (1+\tan x-2)+C$

$=-\frac{1}{8}\log (1+\tan x)+\frac{1}{4}\log (\tan x-1)+C$