Question:

$\int e^{\tan x} (\sec x - \sin x) dx = \int e^{\tan x} \left( \sec^2 x \cdot \cos x + (-\sin x) \right) dx = e^{\tan x} \cos x + C$. This uses the form $\int e^{g(x)} (g'(x) f(x) + f'(x)) dx = e^{g(x)} f(x) + C$, where $g(x) = \tan x$ and $f(x) = \cos x$.