# Hyperbolic integral

1. Jun 20, 2015

What exactly is the integral of $\frac{1}{\sqrt{x^2 - 1}}$?
I know that the derivative of $\cosh^{-1}{x}$ is $\frac{1}{\sqrt{x^2 - 1}}$, but $\cosh^{-1}{x}$ is only defined for $x \geq 1$, whereas $\frac{1}{\sqrt{x^2 - 1}}$ is defined for all $|x| \geq 1$. How do I take that into account? Do I write:
$$\int \frac{1}{\sqrt{x^2 - 1}} dx = \cosh^{-1}{|x|} + c$$
UPDATE: I tried differentiating $\cosh^{-1}{|x|}$ as a piecewise function and I ended up with $-\frac{1}{\sqrt{x^2 - 1}}$ for $x \leq 1$. Now I'm more confused.
Should I write the integral as:
$$\int \frac{1}{\sqrt{x^2 - 1}} dx = S(x) \cosh^{-1}{|x|}$$
where $S$ is the signum function?

Last edited: Jun 20, 2015
2. Jun 20, 2015

### Matternot

Consider a comparison to ln(x)

notice how:
$\frac{d}{dx}\ln(x)$ is the same as $\frac{d}{dx}\ln(-x)$

This is inconsistent with your $\cosh^{-1}{|x|}$

The idea that resolves this is that you can use the idea that $\frac{1}{\sqrt{x^2 - 1}}$ is an even function.

There could be some quirks using complex numbers... I'll try it and get back.

Last edited: Jun 20, 2015
3. Jun 20, 2015

### Matternot

one idea is to consider $\cosh(x) = cos(ix)$

i.e.
$\cosh^{-1}(x) = -i cos^{-1}(x)$
this leads to simply $±\frac{1}{\sqrt{x^{2}-1}}$

Most obvious solution would be your last line

Last edited: Jun 20, 2015
4. Jun 21, 2015

I think I figured it out. It's $\ln|x + \sqrt{x^2 - 1}|$.
Which is identical to the inverse hyperbolic cosine, except it's defined for both positive and negative $x + \sqrt{x^2 - 1}$.