# Tough Calculus Integral problem

$\int\sec^3x~dx = \int\sec^2x \sec x~dx = \int\frac{1+\tan^2 x}{ \cos x}~dx = \int\frac{1}{ \cos x}~dx+ \int\frac{\tan^2 x}{ \cos x}~dx$
Remember $\tan x = \frac{\sin x }{\cos x}$ and $\frac{d}{dx}(\sin x) = \cos x$