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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

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Please show the rest of your work so we can help. –  Code-Guru Aug 7 '12 at 14:41
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marked as duplicate by Hans Lundmark, Noah Snyder, Martin Sleziak, Norbert, Hagen von Eitzen Oct 17 '12 at 17:06

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4 Answers

up vote 13 down vote accepted

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

$u=\sin(x)$.

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Is this one your own or can something in the literature be cited? –  Michael Hardy Aug 8 '12 at 15:15
    
@MichaelHardy Well it is my idea, I also calculate $\int \sec(x)dx$ this way.... BUT, the theory of trig integrals says that if $\cos(x)$ appears at an odd power, then the sub $u=\sin(x)$ works. Typically people use this for products, but it also works for ratios....So in some sense it is a pretty standard method, is just that people in general fail to realize that it can be applied to more general settings... –  N. S. Aug 8 '12 at 15:22
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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 $\int\sec^3(x)\, dx$.

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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.

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$\int \sec^3x dx=\int \sec x (\sec^2 x dx)$

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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