Question

$\int e^{\tan x}\cdot \tan x\cdot {\sec }^{2}xdx$.

Answer 1

Putting $\tan x=m$

${\sec }^{2}xdx=dm$

$\int e^{m}mdm$

Integrating by parts, we get

$m{e}^m-e^m+C$

$\tan x{e}^{\tan x}-e^{\tan x}+C$