<s>[INST] Finding fomulas for hyperbolic functions I'm trying to find formulas for hyperbolic functions, starting with this image

Knowing that the area between the origin, vertex and a point on hyperbola (enclosed by x-axis and hyperbola itself) is equal to $\frac{\alpha}{2}$, I wrote
$$
x^2 - y^2 = 1
$$
$$
x = \sqrt{1 + y^2}
$$
... and started integrating this expression for $x$.
$$
\begin{align} 
\frac{\alpha}{2} 
        & = \int^{y}_0{\sqrt{1 + y^2}\,\textrm{d}y}
     \\ & = \int^{\arctan y}_0{\sqrt{1 + \tan^2\theta}\sec^2\theta\,\textrm{d}\theta}  && \text{let }y = \tan\theta \text{ and }\textrm{d}y = \sec^2\theta\,\textrm{d}\theta
     \\ & = \int^{\arctan y}_0{\sec\theta\sec^2\theta\,\textrm{d}\theta}
\end{align}
$$
Here I used integration by parts to find $\int{sec^3\theta\,\textrm{d}\theta}$
$$
\begin{align} 
\int{\sec^3\theta\,\textrm{d}\theta}
        & = \sec\theta\tan\theta - \int{\sec\theta\tan^2\theta \,\textrm{d}\theta}
     \\ & = \sec\theta\tan\theta - \int{\sec\theta(\sec^2\theta - 1) \,\textrm{d}\theta}
     \\ & = \sec\theta\tan\theta - \int{\sec^3\theta \,\textrm{d}\theta} + \int{\sec\theta \,\textrm{d}\theta}
     \\ & = \frac{1}{2}\bigg(\sec\theta\tan\theta + \int{\sec\theta \,\textrm{d}\theta}\bigg)
     \\ & = \frac{1}{2}\bigg(\sec\theta\tan\theta + \ln{\bigg|\sec\theta + \tan\theta \,\bigg|}\bigg)
\end{align}
$$
... and by substituting $\tan\theta = y$, $\sec\theta = \sqrt{1 + y^2}$ back I got
$$
\alpha = y\sqrt{1 + y^2} + \ln{\bigg|\sqrt{1 + y^2} + y \,\bigg|}
$$
Which doesn't make much sense, because this is not what an inverse hyperbolic sine looks like. Where have I made mistake? What should I do different? 
Thanks.
 [/INST] The integral you have calculated $\int_0^y\sqrt{1+y^2}dy$ does not represent the red area alone, but includes the area of the right angled triangle immediately above it. Therefore you need to subtract $\frac 12 xy$ and then you will get the answer you expected
</s>