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
[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)

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