Question

$\int e^{2x}\left(\frac{1+sin2x}{1+cos2x}\right)dx=$

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

Answer 1

The correct option is B $\frac{e^{2x}}{2}\tan 2x+c$

Given,

$\int e^{2x}\left(\frac{\sin 2x}{\cos 2x}\right)dx$

Put $2x=t\to dx=\frac{dt}{2}$

$\int e^t\left(\frac{\sin t}{\cos t}\right)\frac{dt}{2}$

$=\frac{1}{2}\int e^t\left(\frac{1+\sin t}{2{\cos }^2\frac{t}{2}}\right)dt$

$=\frac{1}{2}\int e^t(\frac{1}{2}{\sec }^2\frac{t}{2}+\frac{2\sin \frac{t}{2}\cos \frac{t}{2}}{2{\cos }^2\frac{t}{2}})dt$

$=\frac{1}{2}\int e^t(\frac{1}{2}{\sec }^2\frac{t}{2}+\tan \frac{t}{2})dt$

$=\frac{1}{2}{e}^t\tan \frac{t}{2}+c$

$=\frac{1}{2}{e}^{2x}\tan 2x+c$