Question

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

Answer 1

We can write integral as

$\left(\frac{2+\sin 2x}{1+\cos 2x}\right)e^x=\left[\frac{2+2\sin x\cos x}{1+2{\cos }^{2}x-1}\right]e^x$

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

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

$=[{\sec }^{2}x+\tan x]e^x$

$=[\tan x+{\sec }^{2}x]e^x$

$\int (\tan x+{\sec }^{2}x)e^{x}dx$ is of the form $\int e^x[f\left(x\right)+f^{\prime }\left(x\right)]dx=e^{x}f\left(x\right)+c$

where $f\left(x\right)=\tan x\Rightarrow f^{\prime }\left(x\right)={\sec }^{2}xdx$

So, our equation becomes

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

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

$=e^x\tan x+c$