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?
 A: \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}
A: This answer is based on Feynman's trick. Put
\begin{equation*}
 I(a) = \int_{0}^{\pi/4}\arctan\left(a\sqrt{\dfrac{1-\tan^2 x}{2}}\right)\, dx .
\end{equation*}
Then
\begin{gather*}
I'(a) = \int_{0}^{\pi/4}\dfrac{1}{1+a^2\dfrac{1-\tan^2 x}{2}}\cdot \sqrt{\dfrac{1-\tan^2 x}{2}} \, dx = \\[2ex]\int_{0}^{\pi/4}\dfrac{1}{1+a^2\dfrac{\cos 2x}{1+\cos 2x}}\cdot \sqrt{\dfrac{\cos 2x}{1+\cos 2x}} \, dx =
 [y=\cos 2x]\\[2ex]
 = \dfrac{1}{2}\int_{0}^{1}\dfrac{1}{1+(a^2+1)y}\cdot\sqrt{\dfrac{y}{1-y}}\,dy= \left[z=\sqrt{\dfrac{y}{1-y}}\right] =\\[2ex]
\dfrac{1}{2}\int_{-\infty}^{\infty}\dfrac{z^2}{(1+(a^2+2)z^2)(z^2+1)}\, dz = [\mbox{ residue calculus }]=\\[2ex]
\dfrac{\pi}{2}\left(\dfrac{1}{a^2+1}-\dfrac{1}{(a^2+1)\sqrt{a^2+2}}\right)
\end{gather*}
Finally we get
\begin{gather*}
I(1)=I(1)-I(0)=\int_{0}^{1}I'(a)\, da =\dfrac{\pi}{2}\left[\arctan a -\arctan\dfrac{a}{\sqrt{a^2+2}}\right]_{0}^{1} =\dfrac{\pi^2}{24}.
\end{gather*}
A: 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}
