$\int\limits^{\pi}_{0}e^x(tan\, x + sec^2\, x)dx$

0

Evaluate $\int_{0}^{\pi} e^x (\tan x + \sec^2 x) \, dx$

Split integral:

$\int_{0}^{\pi} e^x \tan x \, dx + \int_{0}^{\pi} e^x \sec^2 x \, dx$

Use integration by parts for $\int e^x \tan x \, dx$:

Let $u = \tan x$, $dv = e^x dx \Rightarrow du = \sec^2 x dx$, $v = e^x$

$\int e^x \tan x \, dx = e^x \tan x - \int e^x \sec^2 x \, dx$

Adding $\int e^x \sec^2 x \, dx$ gives:

$e^x \tan x - \int e^x \sec^2 x \, dx + \int e^x \sec^2 x \, dx = e^x \tan x$

Evaluate from $0$ to $\pi$:

$[e^x \tan x]_0^\pi = e^\pi \tan \pi - e^0 \tan 0 = 0 - 0 = 0$

Answer: $0$