Question

$\int e^x(\frac{2\tan x}{1+\tan x}+{\cot }^2(x+\frac{\pi }{4}))dx$ is equal to

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

Answer 1

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

$I=\int e^x(\frac{2\tan x}{1+\tan x}+{\tan }^2(x-\frac{\pi }{4}))dx=\int e^x(\frac{2\tan x}{1+\tan x}-1+{\sec }^2(x-\frac{\pi }{4}))dx=\int e^x(\frac{\tan x-1}{1+\tan x}+{\sec }^2(x-\frac{\pi }{4}))dx=\int e^x(\tan (x-\frac{\pi }{4})+{\sec }^2(x-\frac{\pi }{4}))dx\underset{f\left(x\right)}{\downarrow }\underset{f^{\prime }\left(x\right)}{\downarrow }\{\because \int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+c\}=e^x\tan (x-\frac{\pi }{4})+c$