1.36 Hyperbolic triangles

The area of the triangle $OPQ$ equals ${{1}\over{2}}|OQ||QP|=\,{{{a^{2}}\over{2}}\sinh v\cosh v.}$ The area of the region $APQ$ below hyperbola equals $\int_{a}^{a\cosh v}\sqrt{x^{2}-a^{2}}\,dx.$ Now change variables to $x=a\cosh u$, so ${{dx}\over{du}}=\,{a\sinh u}$ and $\sqrt{x^{2}-a^{2}}={a\sqrt{\cosh^{2}u-1}}\,{=a\sinh u}$; then

Hence