Proving the area of circle and hyperbolic function is the same I am an A-level student and due to the schools being closed down, I am reading ahead in Further Maths. I am stuck on a question regarding integrating an area due to a hyperbolic function. The question is below:

I know how to find the area of the circle but I am stuck on the rectangular hyperbola. I have tried to integrate the function 
$$
y=\sqrt{a^2+x^2}
$$
by substituting in:
$$
\frac{x}{a}=\sinh{u}
$$
However, that doesn't seem to prove it. I then tried to also tried to integrate it parametrically by using 
$$
x=\cosh{\theta}
$$
$$
y=\sinh{\theta}
$$
But I still haven't been able to prove it. Could you help?
 A: If you want to set up the integral, you need the area between the hyperboloid and the line. Right now you are getting the area below the hyperboloid. But notice that
$$A = \frac{a^2\cosh \phi \sinh \phi}{2} - \int_a^{a\cosh\phi}\sqrt{x^2-a^2}\:dx$$
where the first term is the area of the whole triangle with the side lengths of the point $P_2$. Then using $x = a\cosh t$ we have that 
$$A = \frac{a^2\cosh \phi \sinh \phi}{2} - a^2 \int_0^\phi \sinh^2 t\:dt $$
Some hyperbolic identity magic gives us that 
$$\sinh^2 t = \frac{1}{2}\cosh 2t - \frac{1}{2}$$
$$\cosh \phi \sinh\phi = \frac{1}{2}\sinh2\phi$$
simplifying the formula to
$$A = \frac{1}{2}a^2\phi$$
A: In the second diagram, regard the blue region as made up of a bunch of "infinitesimal"
triangles with vertices $(0,0)$, $(a\cosh t,a\sinh t)$ and $(a\cosh(t+\delta t)
,a\sinh(t+\delta t))$ for $t$ between $0$ and $\phi$. The area of this triangle is
$$\frac{a^2}2\left(\cosh t\sinh(t+\delta t)-\sinh t\cosh(t+\delta t)\right)
=a^2\frac{\sinh(\delta t)}2=a^2\frac{\delta t}2+\text{higher order infinitesimals}.$$
Therefore the blue area is
$$a^2\int_0^\phi\frac{dt}{2}$$
etc.
