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$?

Since I mentioned it in comments, there is a general approach to such problems, the Weierstrass half-angle substitution.
Let $u=\tan \frac{x}{2}$ then $\cos x = \frac{1-u^2}{1+u^2}$ and $dx=\frac{2du}{1+u^2}$. 
So:
$$\int \sec^3 x \,dx = \int \frac{2(1+u^2)^2du}{(1-u^2)^3}$$
You are still going to need to do a partial fraction decomposition.
It's messy. but this trick is worth knowing because it can be used in so many questions.
Cheating to keep this brief, Wolfram Alpha gives the partial fraction decomposition as:
$$\frac{2(1+u^2)^2}{(1-u^2)^3} = \frac{1}{2(u+1)} -\frac{1}{2(u+1)^2} + \frac{1}{(u+1)^3}  - \frac{1}{2(u-1)} - \frac{1}{2(u-1)^2} - \frac{1}{(u-1)^3} $$
