Forum TI-Planet.org

$mathjax$\int_0^{\ln(2+\sqrt{3})} \frac{dt}{ch(t)}=\int_0^{\ln(2+\sqrt{3})}\frac{2e^t}{1+(e^t)^2}dt=\left[2\arctan(e^t)\right]_0^{\ln(2+\sqrt{3})}=2(\arctan(2+\sqrt{3})-\arctan(1))=2\left(\frac{5\pi}{12}-\frac{\pi}{4}\right)=\frac{\pi}{3}$mathjax$

$mathjax$\arctan(2+\sqrt{3})$mathjax$

$mathjax$\displaystyle{\tan\left(\dfrac{5\pi}{12}\right)=\tan\left(\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)=\frac{\tan\left(\dfrac{\pi}{6}\right)+\tan\left(\dfrac{\pi}{4}\right)}{1-\tan\left(\dfrac{\pi}{6}\right)\tan\left(\dfrac{\pi}{4}\right)}=\frac{\dfrac{1}{\sqrt{3}}+1}{1-\dfrac{1}{\sqrt{3}}\times 1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}=\dfrac{(\sqrt{3}+1)^2}{3-1}=2+\sqrt{3}}$mathjax$

$mathjax$\dfrac{5\pi}{12}\in\left]-\dfrac{\pi}{2},\dfrac{\pi}{2}\right[$mathjax$

$mathjax$\arctan(2+\sqrt{3})=\dfrac{5\pi}{12}$mathjax$