Question

$\int \frac{2+\sin 2x}{1+\cos 2x}{e}^{x}dx$

Answer 1

Consider the given integral.

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

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

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

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

$I=\int e^x{\sec }^{2}xdx+\int \left(\frac{2\sin x\cos x}{2{\cos }^{2}x}\right)e^{x}dx$

$I=\int e^x{\sec }^{2}xdx+\int \tan x{e}^{x}dx$

$I=\int e^x{\sec }^{2}xdx+\tan x{e}^x-\int {\sec }^{2}x{e}^{x}dx$

$I=e^x\tan x+C$

Hence, this is the answer.