$\int_0^1\frac{\ln x\ln^2(1-x^2)}{\sqrt{1-x^2}}dx=\frac{\pi}{2}\zeta(3)-2\pi\ln^32$ I'm looking for proof of the following identity
$$\int_0^1\frac{\ln x\ln^2(1-x^2)}{\sqrt{1-x^2}}dx=\frac{\pi}{2}\zeta(3)-2\pi\ln^32$$
I have worked on this problem for quite some time, however since I'm not much comfortable with beta functions and stuff, I was unable to prove the required. I'm looking for an elementary approach, however, any detailed method (including beta function) is most welcomed. Thanks.
 A: Substitute $t=\sin x$
\begin{align}
& \int_0^1\frac{\ln x\ln^2(1-x^2)}{\sqrt{1-x^2}}dx\\
=& \> 4\int_0^{\pi/2} \ln (\sin t)\ln^2(\cos t)dt\\
 =& \> 2\int_0^{\pi/2} [\ln (\sin t)\ln^2(\cos t)+ \ln^2 (\sin t)\ln(\cos t)]dt\\ 
 =& \> \frac23\int_0^{\pi/2} [\ln^3(\sin t\cos t)-2 \ln^3 (\sin t)]dt\\ 
=&\>\frac23 \int_0^{\pi/2} \ln^3\frac{\sin t}2 dt- \frac43 \int_0^{\pi/2} \ln^3 (\sin t)\>dt \\ =&\> - \frac23 \int_0^{\pi/2} \ln^3 (2\sin t)\>dt 
+4\ln^2 2\int_0^{\pi/2} \ln(2\sin t)dt - 2\pi \ln^32\\
=&\>\frac\pi2\zeta(3) -2\pi\ln^32
\end{align}
where $\int_0^{\pi/2} \ln(2\sin t)dt=0$ and  $\int_0^{\pi/2} \ln^3 (2\sin t)dt = -\frac{3\pi}4\zeta(3)$.
A: $$I=\int_0^1\frac{\ln x\ln^2(1-x^2)}{\sqrt{1-x^2}}dx$$ $$=\frac{1}{4}\int_0^1\frac{\ln t\ln^2(1-t)}{\sqrt t\sqrt{1-t}}dt=\frac{1}{4}\frac{d^2}{db^2}\frac{d}{da}\int_0^1t^{a-1/2}(1-t)^{b-1/2}dt|_{a=b=0}$$
$$=\frac{1}{4}\frac{d^2}{db^2}\frac{d}{da}B(a+1/2;b+1/2)|_{a=b=0}$$
$$\frac{1}{4}B(a+1/2;b+1/2)=\frac{1}{4}\frac{\Gamma(a+1/2)\Gamma(b+1/2)}{\Gamma(a+b+1)}=\frac{\pi}{4}\frac{\Gamma(1+2a)}{\Gamma(1+a)}\frac{\Gamma(1+2b)}{\Gamma(1+b)}\frac{4^{-a}4^{-b}}{\Gamma(1+a+b)}$$
$$\Gamma(a)\Gamma\bigl(a+1/2\bigr)=2\sqrt{\pi}\,4^{-a}\,\Gamma(2a);\,\,a\Gamma(a)=\Gamma(1+a)=1-\gamma a+\Gamma''(1)\frac{a^2}{2}+...$$
$$\Psi(t)=\frac{\Gamma'(t)}{\Gamma(t)};\,\,\Psi(1)=-\gamma;\,\,\Gamma''(1)=\gamma^2+\frac{\pi^2}{6};\,\,\Psi'(1)=\zeta(2);\,\,\Psi''(1)=-2\zeta(3)$$
$$I=\frac{\pi}{4}\frac{d^2}{db^2}\frac{d}{da}\biggl(\frac{(1-a\gamma+...)(1-a\ln4+...)}{\Gamma(1+b)+\Gamma'(1+b)a+...}\frac{\Gamma(1+2b)}{\Gamma(1+b)}4^{-b}\biggr)|_{a=b=0}$$
$$=-\frac{\pi}{4}\frac{d^2}{db^2}\biggl(\bigl(1-b\ln4+\frac{b^2}{2}\ln^24-..\bigr)\frac{1-2\gamma b+2\Gamma''(1)b^2+..}{(1-\gamma b+\Gamma''(1)\frac{b^2}{2}+..)^2}\bigl(\gamma+\ln4+\Psi(1+b)\bigr)\biggr)|_{b=0}$$
$$=-\frac{\pi}{4}\frac{d^2}{db^2}\biggl(\bigl(1-b\ln4+\frac{b^2}{2}\ln^24+..\bigr)\bigl(1-2\gamma b+2\Gamma''(1)b^2+..\bigr)\bigl(1+2\gamma b-\Gamma''(1)b^2+3\gamma^2b^2+..\bigr)\bigl(\gamma+\ln4-\gamma+b\zeta(2)-b^2\zeta(3)\bigr)+...\Bigr)|_{b=0}$$
$$=-\frac{\pi}{2}\Bigl(\frac{\pi^2}{6}\ln4-\ln4\zeta(2)-\zeta(3)+\frac{1}{2}\ln^34\Bigr)$$
$$I=\frac{\pi}{2}\zeta(3)-2\pi\ln^32$$
