How to integrate $\sec^3 x \, dx$? [duplicate]
Possible Duplicate:
Indefinite integral of secant cubed
How to integrate $\sec^3 x \, dx$? Can someone please give a method, I tried separating $\sec^3 x$ as $\sec x(\sec^2 x)$ then applying by-parts method but it didn't yield anything useful
Solution 1:
$$\sec^3(x)=\frac{1}{\cos^3(x)}=\frac{\cos(x)}{(1-\sin^2(x))^2} $$
$u=\sin(x)$.
Solution 2:
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 $\int\sec^3(x)\, dx$.