# Relationship integration math problem

1. Jul 9, 2014

### MisterX

This is about attempting to solve $\left( y'\right)^2 = y^2 - 1$.

$$\int\frac{dy}{\sqrt{y^2 -1}} = \pm \int dx$$

using a trig. substitution and another trick,

$$\int\frac{dy}{\sqrt{y^2 -1}} = \ln\left(y \pm \sqrt{y^2 - 1} \right) + C$$

I'm not sure about that $\pm$ sign. It came in when doing $\tan (\sec^{-1} y)$ since I had $$\sec \theta = y \longrightarrow\tan^2 \theta = y^2 -1\longrightarrow \tan \theta = \pm \sqrt{y^2 -1}$$
The solution is supposed to be something like $y = \cosh x$. But I am not sure how I'm supposed to be getting that from separating and integrating.

$\ln\left(y + \sqrt{y^2 - 1} \right) = \pm x + C$
$y + \sqrt{y^2 - 1} = C_1e^{\pm x}$

2. Jul 9, 2014

### Mr-R

Hello Misterx

Please see the attached picture. Not sure if I am correct. I suggest waiting for someone to verify things

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3. Jul 9, 2014

### pasmith

You can do $$\int \frac{1}{\sqrt{y^2 - 1}}\,dy$$ by substituting $y = \cosh u$ so that $dy = \sinh u\,du$ and using the relationship $$\cosh^2 u - \sinh^2 u = 1.$$
Alternatively, if you solve $$y = \cosh x = \frac{e^{x} + e^{-x}}{2}$$ for $x = \mathrm{arccosh}(y) \geq 0$ you will find that $$\mathrm{arccosh}(y) = \ln\left(y + \sqrt{y^2 - 1}\right).$$

Note that substituting $y = \cosh u$ gets you to the result with considerably less effort than substituting $y = \sec \theta$.