How to integrate $\sec^3 x \, dx$?

$$\sec^3(x)=\frac{1}{\cos^3(x)}=\frac{\cos(x)}{(1-\sin^2(x))^2} $$

$u=\sin(x)$.

Use integration by parts; $u = \sec(x)$, $dv = \sec^2(x)\, dx$, $v = \tan(x)$ and $du = \sec(x)\tan(x)$. Now use the Pythagorean identity for $\tan$ and $\sec$. You will solve for the $\int\sec^3(x)\, dx$.

There's a whole Wikipedia article about just this integral: Integral of secant cubed.

You're mistaken to think that integration by parts doesn't help.