I'm having a difficult time evaluating the integral
$$\mathcal{J} = \int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$$
This is integral arose after simplifying the integral $\displaystyle \int_{0}^{\pi/4 } \arctan \sqrt{\frac{1-\tan^2 x}{2}} \, \mathrm{d}x$;
\begin{align*}
\require{cancel.js}
\int_{0}^{\pi/4} \arctan \sqrt{\frac{1-\tan^2 t}{2}}\, \mathrm{d}t &\overset{1-\tan^2 t \mapsto 2t^2}{=\! =\! =\! =\! =\! =\!=\!=\!} \int_{0}^{\sqrt{2}/2} \frac{t \arctan t}{\sqrt{1-2t^2} \left ( 1-t^2 \right )} \, \mathrm{d}t \\
&=\cancelto{0}{\left [ – \arctan \sqrt{1-2t^2} \arctan t \right ]_0^{\sqrt{2}/2}} + \int_{0}^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t
\end{align*}
My main guess is that differentiation under the integral sign is the way to go here. Any ideas?
Best Answer
\begin{align}J&=\int_0^{\frac{1}{\sqrt{2}}} \frac{\arctan\left(\sqrt{1-2x^2}\right)}{1+x^2}\,dx\\ &\overset{x=\frac{1}{\sqrt{2}}\sin u}=\frac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\frac{\cos u\arctan(\cos u)}{1+\frac{1}{2}\sin^2 u}\,du\\ &=\sqrt{2}\int_0^{\frac{\pi}{2}}\frac{\cos u\arctan(\cos u)}{2+\sin^2 u}\,du\\ &=\left[\arctan\left(\frac{1}{\sqrt{2}}\sin u\right)\arctan(\cos u)\right]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}\frac{\arctan\left(\frac{1}{\sqrt{2}}\sin u\right)\sin u}{1+\cos^2 u}\,du\\ &=\int_0^{\frac{\pi}{2}}\frac{\arctan\left(\frac{1}{\sqrt{2}}\sin u\right)\sin u}{1+\cos^2 u}\,du\\ &=\int_0^{\frac{\pi}{2}}\int_0^{\frac{1}{\sqrt{2}}}\left(\frac{\sin^2 u}{(1+\cos^2 u)(1+a^2\sin^2 u)}\,da\right)\,du\\ &=\int_0^{\frac{1}{\sqrt{2}}}\left[\frac{\sqrt{2}\arctan\left(\frac{1}{\sqrt{2}}\tan u\right)}{2a^2+1}-\frac{\arctan\left(\sqrt{1+a^2}\tan u\right)}{(2a^2+1)\sqrt{1+a^2}}\right]_{u=0}^{u=\frac{\pi}{2}}\,da\\ &=\frac{\pi}{2}\int_0^{\frac{1}{\sqrt{2}}}\frac{\sqrt{2}}{2a^2+1}\,da-\frac{\pi}{2}\int_0^{\frac{1}{\sqrt{2}}}\frac{1}{(2a^2+1)\sqrt{1+a^2}}\,da\\ &=\frac{\pi}{2}\Big[\arctan\left(\sqrt{2}a\right)\Big]_0^{\frac{1}{\sqrt{2}}}-\frac{\pi}{2}\left[\arctan\left(\frac{a}{\sqrt{1+a^2}}\right)\right]_0^{\frac{1}{\sqrt{2}}}\\ &=\frac{\pi}{2}\times \frac{\pi}{4}-\frac{\pi}{2}\times \frac{\pi}{6}\\ &=\boxed{\frac{\pi^2}{24}} \end{align}