Evaluate the following integrals:

Let $I=\int e^{x}(\tan x-\log \cos x) d x$

$I=\int e^{x} \tan x d x-\int e^{x} \log \cos x d x$

Integrating by parts,

$=\int e^{x} \tan x d x-\left\{e^{x} \log \cos x-\int e^{x}\left(\frac{d}{d x} \log \cos x\right) d x\right.$

$=\int e^{x} \tan x d x-e^{x} \log \cos x d x-\int e^{x} \tan x d x$

$=-e^{x} \log \cos x d x+c$

$=e^{x} \log \sec x+c$