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I am stuck trying to find $$\int \sec^3{x} \ dx.$$

Here is my attempt using integration by parts: $$\int \sec^3{x} \ dx = \sec{x}\tan{x} - \int \tan^2{x}\sec{x} \ dx.$$

At this point, I am stuck. How can I continue from here?

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marked as duplicate by Aditya Hase, okarin, user98602, user147263, Hanul Jeon Dec 15 '14 at 2:07

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  • $\begingroup$ @MathNoob That looks like the answer I was searching for. I didn't find anything when I first searched the site. $\endgroup$ – okarin Dec 15 '14 at 1:39
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Hint: $\sec^2(x)=1+\tan^2(x)$.

So $\int \sec^3(x)dx= \int \sec(x)[1+\tan^2(x)]dx = \int \sec(x)dx +\int \sec(x)\tan^2(x)dx$

Can you take it from here?

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  • $\begingroup$ Thanks. This is very helpful. $\endgroup$ – okarin Dec 15 '14 at 1:38
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You can use hyperbolic substitution.

Let take $\tan x = \sinh u$ and $\sec^2 x dx=\cosh u du$ then you get $$\int \sec^3xdx = \int \cosh^2 udu$$

and latter integral is (relatively) easy to calculate.

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