## Inverting Cosh x

To find
$cosh^{-1}x$
let
$y=coshx=\frac{e^x+e^{-x}}{2}$
the multiplying by
$2e^x$
gives
$2e^xy=e^{2x}+1 \rightarrow 2y(e^x)-e^{2x}-1=0 \rightarrow e^{2x}-2ye^x+1=0$

This is a quadratic equation in
$e^x$
.
$e^x= \frac{2y \pm \sqrt{(-2y)^2- 4 \times 1 \times 1}}{2} = y \pm \sqrt{y^2-1}$
.
Hence
$x= ln(y \pm \sqrt{y^2-1})= ln(y \pm \sqrt{y^2+1})$
.
Only the positive square root will return a valid value since
$x, \; coshx$
, increase together which is impossible with the negative square root so
$x= ln(y + \sqrt{y^2-1})$
.
Then
$cosh^{-1}x= ln(x + \sqrt{x^2-1})$
.