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How can one evaluate the integral $$\int\frac{\sqrt{x^2-1}}x\mathrm dx$$?

I tried substituting $x = \cosh t$ but got stuck at $$\int\frac{\sinh^2t}{\cosh t}\mathrm dt$$

Any hints?

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    $\begingroup$ Why not substitute $x = \sec u$? $\endgroup$ – user71641 Jan 14 '15 at 20:45
  • $\begingroup$ Now use $\sinh^{2}t=\cosh^{2}t-1$ $\endgroup$ – user84413 Jan 14 '15 at 20:49
  • $\begingroup$ @user84413: I tried that, but then didn't know how to integrate $\frac1{\cosh x}$... $\endgroup$ – rubik Jan 14 '15 at 20:50
  • $\begingroup$ Here is one way to do it: calc101.com/special_10.html $\endgroup$ – user84413 Jan 14 '15 at 22:13
  • $\begingroup$ @user84413: Thanks for the reference! That's interesting, although I'll never remember that during an exam! $\endgroup$ – rubik Jan 15 '15 at 10:41
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Put $X^2=x^2-1$ then $x^2=X^2+1$.

$2x\mathrm dx=2X\mathrm dX$.

$$\int\frac{\sqrt{x^2-1}}{x}\mathrm dx=\int\frac{X^2}{x^2}\mathrm dX=\int\frac{X^2}{X^2+1}\mathrm dX$$

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  • $\begingroup$ Very nice. I actually thought about this substitution, but didn't see that the $x$ would become $x^2$. Damn! $\endgroup$ – rubik Jan 14 '15 at 21:04
  • $\begingroup$ Thanks. I also though it wouldn't work but who knows let's try! $\endgroup$ – Scientifica Jan 14 '15 at 21:06
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Let $ x = \sec u $. Then, $ \mathrm{d}x = \sec u \tan u \, \mathrm{d}u $. Then, the integral becomes $$ \int \tan^2 u \, \mathrm{d}u = \int \left( \sec^2 u - 1 \right) \, \mathrm{d}u = \tan u - u + \mathcal{C}. $$ Then, you can substitute back and finish.

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  • $\begingroup$ Interesting! I am a bit inexperienced with trig integrals. I just know the basic ones, and always forget about, for example, $\int\tan^2x\mathrm dx$! $\endgroup$ – rubik Jan 14 '15 at 21:15
  • $\begingroup$ @rubik No problem; that happens to everybody! $\endgroup$ – Ahaan S. Rungta Jan 14 '15 at 21:50
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I'll use the hyperbolic substitution you made. (Why not?) Of importance is the hyperbolic dual of the Pythagorean identity, $\cosh^2 x - \sinh^2 x = 1$. Then, one can see that: $$\frac{\sinh^2 t}{\cosh t} = \frac{\cosh^2 t - 1}{\cosh t} $$

This makes your integral: $$\int \cosh t - \operatorname{sech} t\,dt$$

If you know your hyperbolic trig integrals as well as most people know their "normal" trig integrals, you're home free.

Hint: $$\int\operatorname{sech} t\,dt = 2\arctan\left(\tanh\left(\frac{t}{2}\right)\right) + C$$ (According to Wolfram.)

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