Question

The value of integral $\int e ^{ x }\left(\frac{2 \tan x }{1+\tan x }+\cot ^2\left( x +\frac{\pi}{4}\right)\right) dx$ is equal to
where $C$ is constant of integration.

• A. $e ^{ x } \tan \left(\frac{\pi}{4}- x \right)+ C$
• B. $e ^{ x } \tan \left( x -\frac{\pi}{4}\right)+ C$
• C. $e ^{ x } \tan \left(\frac{3 \pi}{4}- x \right)+ C$
• D. $e ^{ x } \tan \left( x -\frac{3 \pi}{4}\right)+ C$

Answer 1

Answer: (B)

$\text { Let } I=\int e^x\left(\frac{2 \tan x}{1+\tan x}+\tan ^2\left(x-\frac{\pi}{4}\right)\right) d x=\int e^x\left(\frac{2 \tan x}{1+\tan x}+\sec ^2\left(x-\frac{\pi}{4}\right)-1\right) d x$
$=\int e ^{ x }\left(\frac{\tan x -1}{1+\tan x }+\sec ^2\left( x -\frac{\pi}{4}\right)\right) dx =\int e ^{ x }\left(\tan \left( x -\frac{\pi}{4}\right)+\sec ^2\left( x -\frac{\pi}{4}\right)\right) dx $
$= e ^{ x } \tan \left( x -\frac{\pi}{4}\right)+ C$