How to evaluate $ \int \sec^3x \, \mathrm{d}x$? How to evaluate $ \int \sec^3x \, \mathrm{d}x$ ?
I tried integration by parts and then I got $$\int \sec^3x \, \mathrm{d}x=\sec x \tan x - \int\sec x \tan^2 x \, \mathrm{d}x.$$ Now I'm stuck solving $\int\sec x \tan^2 x \, \mathrm{d}x$? How to solve that and how to solve my initial question? What is the smartest way solving $ \int \sec^3x \, \mathrm{d}x$?

This is a well-known problem. Using $ \left( u, v \right) = \left( \sec x, \tan x \right) $ gives $$ \begin {align*} \displaystyle\int \sec^3  \, \mathrm{d}x &= \displaystyle\int \sec x \tan x - \displaystyle\int \sec x \tan^2 x \, \mathrm{d}x \\&= \sec x \tan x - \displaystyle\int \sec^3 x \, \mathrm{d}x + \displaystyle\int \sec x \, \mathrm{d}x. \end {align*} $$If we let our integral be $I$, then, $ 2I = \sec x \tan x + \displaystyle\int \sec x \, \mathrm{d}x = \sec x \tan x + \ln \left| \sec x + \tan x \right| + \mathcal{C}_0 $, so we have $$ \displaystyle\int \sec^3 x \, \mathrm{d}x = \boxed {\dfrac {\sec x \tan x + \ln \left| \sec x + \tan x \right|}{2} + \mathcal{C}}. $$
