What exactly is the integral of ##\frac{1}{\sqrt{x^2 - 1}}##?(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Hyperbolic integral

Loading...

Similar Threads - Hyperbolic integral | Date |
---|---|

I Help with simplifying series of hyperbolic integrals | Nov 19, 2017 |

Are hyperbolic substitutions absolutely necessary? | Sep 6, 2015 |

Are hyperbolic functions used in Calculus 3? | May 19, 2015 |

Hyperbolic integral. | Jan 16, 2014 |

Integral with hyperbolic: cosh x | Oct 12, 2013 |

**Physics Forums - The Fusion of Science and Community**