Question

$\int e^x\left[\frac{2+\sin 2x}{1+\cos 2x}\right]dx=$

• A. $e^x\tan x+c$
• B. $e^x+\tan x+e^y$
• C. $e^{2x}\tan x+c$
• D. $e^x\tan 2x+c$

Answer 1

The correct option is A $e^x\tan x+c$

$\int e^x\left[\frac{2+\sin 2x}{1+\cos 2x}\right]dx$

$\int e^x\left[\frac{2+2\sin x\cos x}{2{\cos }^{2}x}\right]dx$

$=\int e^x({\sec }^{2}x+\tan x)dx$

And , $\int e^x\left(f\right(x)+f^{{}^{\prime }}(x\left)\right)dx=e^{x}f\left(x\right)+c$

Here also, $\frac{d}{dx}\tan x={\sec }^{2}x$

Hence, integration results to:-

$=e^x\tan x+c$

Hence, answer is option-(A).