TITLE: "Discovering" the hyperbolic functions $\cosh(x)$ and $\sinh(x)$
QUESTION [5 upvotes]: I'm trying to derive the definitions of hyperbolic functions with this image in mind, where $a := \cosh u$, $b := \sinh u$, and $u = 2A$.
I have 
$$2A = 2\int_{0}^b \sqrt{1+y^2} \ \mathrm{d}y = b\sqrt{b^2 + 1} + \log\left(b + \sqrt{b^2 + 1}\right)$$
and therefore 
$$\begin{align}
\cosh u &= \cosh\left(\;b\sqrt{b^2 + 1} + \log\left(b + \sqrt{b^2 + 1}\right)\;\right) = \sqrt{b^2 + 1} \\
\sinh u &= \sinh\left(\;b\sqrt{b^2 + 1} + \log\left(b + \sqrt{b^2 + 1}\right)\;\right) = b
\end{align}$$
The goal here is to solve for $u$ in terms of $b$ and then derive the usual formulas, but that's proven to be impossible (unless I made a mistake, which I think might be the case here). Where can I go from here? 
Also, I'm using the derivation shown in this guide.

EDIT: The correct integral is
$$2A = 2\int_{0}^b \left( \sqrt{1+y^2} - \frac{a}{b}y \ \mathrm{d}y \right)  = \log\left(b + \sqrt{b^2 + 1}\right)$$
so 
$$b = \cfrac{e^{2u} - 1}{2e^u}$$ 
and from here the rest follows trivially.

REPLY [5 votes]: Don't forget that the region $A$ is sliced off by the line $x=\frac ab y$, so you should amend your integral to be
$$
2A = 2\int_{0}^b \left(\sqrt{1+y^2}-\frac ab y\right) dy.\tag1$$
But since $(a,b)$ lies on the hyperbola, we have $a=\sqrt{b^2+1}$. Plugging this in to (1) and evaluating the integral, you'll end up with
$$u=2A=\log(b+\sqrt{b^2+1}).\tag2$$
(Note that this amendment amounts to subtracting off double the area of a right triangle with legs $a$ and $b$.)
Your next step is to solve (2) for $b$ in terms of $u$.