Proving the area of circle and hyperbolic function is the same

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

$$A = \frac{1}{2}a^2\phi$$