Question

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

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

Answer 1

The correct option is A $e^x\tan (x-\frac{\pi }{4})+C$

Let $I=\int e^x(\frac{2\tan x}{1+\tan x}+{\cot }^2(x+\frac{\pi }{4}))dx$
$=\int e^x(\frac{2\tan x}{1+\tan x}+{\tan }^2(x-\frac{\pi }{4}))dx$
$=\int e^x(\frac{2\tan x}{1+\tan x}+{\sec }^2(x-\frac{\pi }{4})-1)dx$
$=\int e^x(\frac{\tan x-1}{1+\tan x}+{\sec }^2(x-\frac{\pi }{4}))dx$
$=\int e^x(\tan (x-\frac{\pi }{4})+{\sec }^2(x-\frac{\pi }{4}))dx$
Use the property $\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)+C$
So,
$I=e^x\tan (x-\frac{\pi }{4})+C$