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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
$\endgroup$ 14 Answers
$\begingroup$$$\sec^3(x)=\frac{1}{\cos^3(x)}=\frac{\cos(x)}{(1-\sin^2(x))^2} $$
$u=\sin(x)$.
$\endgroup$ 2 $\begingroup$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 $\int\sec^3(x)\, dx$.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$$\int \sec^3x \,dx=\int \sec x (\sec^2 x \, dx)$
Let $\tan x=t\implies \sec^2 x \,dx=dt$ and $\sec x=\sqrt{1+t^2}$ which changes our integral to $\int \sqrt{1+t^2}\,dt$ which is a standard integral which evaluates to $\frac{t\sqrt{1+t^2}}{2}+\frac{\log(t+\sqrt{1+t^2})}{2}$. Now back substitute $t=\tan x$ in answer
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