TITLE: Difficult Integral $\int_0^{1/\sqrt{2}}\frac{\arcsin({x^2})}{\sqrt{1+x^2}(1+2x^2)}dx=$
QUESTION [7 upvotes]: I have a difficult integral to compute.I know the result, but need to know the method of calculation.
How prove this result? $$\int_0^{1/\sqrt{2}}\frac{\arcsin({x^2})}{\sqrt{1+x^2}(1+2x^2)}dx=\frac{\pi^2}{144}$$

REPLY [4 votes]: $\displaystyle J=\int_0^{\tfrac{1}{\sqrt{2}}}\dfrac{\arcsin (x^2)}{\sqrt{1+x^2}
(1+2x^2)}dx$
Perform change of variable $y=x^2$, one obtains:
$\displaystyle J=\int_0^{\tfrac{1}{2}}\dfrac{\arcsin x}{2\sqrt{x(1+x)}(1+2x)}dx$
$\displaystyle J=\left[\arcsin(x)\arctan\left(\sqrt{\dfrac{x}{1+x}}\right)\right]_0^{\frac{1}{2}}-\int_0^{\tfrac{1}{2}}\dfrac{\arctan\left(\sqrt{\dfrac{x}{1+x}}\right)}{\sqrt{1-x^2}}dx$
$\displaystyle J=\dfrac{\pi^2}{36}-\int_0^{\tfrac{1}{2}}\dfrac{\arctan\left(\sqrt{\dfrac{x}{1+x}}\right)}{\sqrt{1-x^2}}dx$
Define on $[0,1]$, $\displaystyle F(a)=\int_0^{\tfrac{1}{2}}\dfrac{\arctan\left(a\sqrt{\dfrac{x}{1+x}}\right)}{\sqrt{1-x^2}}dx$
For all $a$ in $[0,1]$, $\displaystyle F'(a)=\int_0^{\tfrac{1}{2}}\dfrac{x}{\sqrt{x(1-x)}(a^2x+x+1)}dx$
For all $a$ in $[0,1]$, $F'(a)=2\left[\dfrac{\sqrt{2+a^2}\arctan\left(\dfrac{\sqrt{x}}{\sqrt{1-x}}\right)-\arctan\left(\dfrac{\sqrt{2+a^2}\sqrt{x}}{\sqrt{1-x}}\right)}{(1+a^2)\sqrt{2+a^2}}\right]_0^{\tfrac{1}{2}}$
For all $a$ in $[0,1]$, $F'(a)=\dfrac{\pi}{2}\dfrac{1}{1+a^2}-\dfrac{2\arctan\left(\sqrt{2+a^2}\right)}{(1+a^2)\sqrt{2+a^2}}$
Therefore,
$\displaystyle \int_0^{\tfrac{1}{2}}\dfrac{\arctan\left(\sqrt{\dfrac{x}{1+x}}\right)}{\sqrt{1-x^2}}dx=\int_0^1 F'(a)da=\dfrac{\pi}{2}\int_0^1\dfrac{1}{1+a^2}da-2\int_0^1\dfrac{\arctan\left(\sqrt{2+a^2}\right)}{(1+a^2)\sqrt{2+a^2}}da$
Therefore,
$\displaystyle J=\dfrac{\pi^2}{36}-\dfrac{\pi^2}{8}+2\int_0^1\dfrac{\arctan\left(\sqrt{2+a^2}\right)}{(1+a^2)\sqrt{2+a^2}}da$
The last integral is the Ahmed's integral (see How to evaluate Ahmed's integral? )
Finally,
$J=\dfrac{\pi^2}{36}-\dfrac{\pi^2}{8}+2\times \dfrac{5\pi^2}{96}=\dfrac{\pi^2}{144}$