The area of the triangle OPQOPQ equals 12|OQ||QP|=a22sinhvcoshv.{{1}\over{2}}|OQ||QP|=\,{{{a^{2}}\over{2}}\sinh v\cosh v.} The area of the region APQAPQ below hyperbola equals ∫aacoshvx2-a2dx.\int_{a}^{a\cosh v}\sqrt{x^{2}-a^{2}}\,dx. Now change variables to x=acoshux=a\cosh u, so dxdu=asinhu{{dx}\over{du}}=\,{a\sinh u} and x2-a2=acosh2u-1=asinhu\sqrt{x^{2}-a^{2}}={a\sqrt{\cosh^{2}u-1}}\,{=a\sinh u}; then
Hence