Question

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

• A. $\log \left|\sec (\frac{\pi }{4}-x)\right|+c$
• B. $\log \left|\cos (\frac{\pi }{4}+x)\right|+c$
• C. $\log \left|\sin (\frac{\pi }{4}-x)\right|+c$
• D. $\log \left|\tan \frac{x}{2}\right|+c$

Answer 1

The correct option is A $\log \left|\sec (\frac{\pi }{4}-x)\right|+c$

$\tan (\frac{\pi }{4}-x)=\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\cdot \tan x}=\frac{1-\tan x}{1+\tan x}$

$\therefore \int \frac{1-\tan x}{1+\tan x}\cdot dx=\int \tan (\frac{\pi }{4}-x)dx$

$=log|\sec (\frac{\pi }{4}-x)|+c$

$\because \int \tan x=log\left(\sec x\right)+c$.