Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int e^{2x}\left(\frac{1+\sin 2x}{1+\cos 2x}\right)dx \\ =\int e^{2x}\left(\frac{1}{1+\cos 2x}+\frac{\sin 2x}{1+\cos 2x}\right)dx \\ =\int e^{2x}\left(\frac{1}{2{\cos }^{2}x}+\frac{2\sin x\cos x}{2{\cos }^{2}x}\right)dx \\ =\int e^{2x}\left(\frac{{\sec }^{2}x}{2}+\tan x\right)dx \\ \mathrm{Let}{e}^{2x}\tan x=t \\ \Rightarrow \left(e^{2x}{\sec }^{2}x+2e^{2x}\tan x\right)dx=dt \\ =\left[\frac{e^{2x}{\sec }^{2}x}{2}+e^{2x}\tan x\right]dx=\frac{dt}{2} \\ \therefore I=\int \frac{dt}{2} \\ =\frac{t}{2}+\mathrm{C} \\ =\frac{e^{2x}\tan x}{2}+\mathrm{C} \end{gathered}$$