<s>[INST] On the integral $\int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$ 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?
 [/INST] Continue with\begin{align}
& \int_{0}^{\pi/4 } \tan^{-1} \sqrt{\frac{1-\tan^2 x}{2}}dx
\>\>\>\>\>\>\> \tan x =\sin t\\
=& \int_0^{\pi/2}\frac{\cos t}{2-\cos^2t}\ \tan^{-1}\frac{\cos t}{\sqrt2} \ dt\\
=& \int_0^{\pi/2}\int_0^{\pi/4} \frac{\sqrt2 \cos^2t}{(2-\cos^2t)(2\cos^2y+\sin^2 y\cos^2t)}dy\ dt\\
=& \ \frac\pi2\int_0^{\pi/4} \bigg(1-\frac{\cos y}{\sqrt{2-\sin^2y}}\bigg)dy=\frac\pi2\bigg(\frac\pi{4}-\frac\pi6\bigg)=\frac{\pi^2}{24}
\end{align}
</s>