Question

The value of integral $\int e^x\left(\frac{2 \tan x}{1+\tan x}+\cot ^2\left(x+\frac{\pi}{4}\right)\right) d x$ is equal to
[Note: Where C is integration constant.]

• 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)

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