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1.36 Hyperbolic triangles

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

aacoshvx2-a2dx=0va2sinh2udu=a220v(1+cosh2u)du\int_{a}^{a\cosh v}\sqrt{x^{2}-a^{2}}\,dx=\,{\int_{0}^{v}a^{2}\sinh^{2}u\,du}% \,{=\frac{a^{2}}{2}\int_{0}^{v}(1+\cosh 2u)\,du}
=a22[u+12sinh2u]0v=a2v2+a22sinhvcoshv.{=\frac{a^{2}}{2}\left[u+\frac{1}{2}\sinh 2u\right]_{0}^{v}}\,{=\frac{a^{2}v}{% 2}+\frac{a^{2}}{2}\sinh v\cosh v.}

Hence

Area ofAOP=(Area of OPQOPQ)-(Area of APQAPQ)=a2v2.\mbox{Area of}\;AOP\,=\,\mbox{(Area of $OPQ$)}-\mbox{(Area of $APQ$)}=\,{{{a^{% 2}v}\over{2}}.}