Looking for closed-forms of $\int_0^{\pi/4}\ln^2(\sin x)\,dx$ and $\int_0^{\pi/4}\ln^2(\cos x)\,dx$ A few days ago, I posted the following problems
Prove that
\begin{equation}
\int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}\\[20pt]
-\int_0^{\pi/2}\ln^3(\cos x)\,dx=\frac{\pi}{2}\ln^3 2+\frac{\pi^3}{8}\ln 2 +\frac{3\pi}{4}\zeta(3)
\end{equation}
and the OP receives some good answers even I then could answer it.

My next question is finding the closed-forms for

\begin{align}
\int_0^{\pi/4}\ln^2(\sin x)\,dx\tag1\\[20pt]
\int_0^{\pi/4}\ln^2(\cos x)\,dx\tag2\\[20pt]
\int_0^1\frac{\ln t~\ln\big(1+t^2\big)}{1+t^2}dt\tag3
\end{align}

I have a strong feeling that the closed-forms exist because we have nice closed-forms for
\begin{equation}
\int_0^{\pi/4}\ln(\sin x)\ dx=-\frac12\left(C+\frac\pi2\ln2\right)\\
\text{and}\\
\int_0^{\pi/4}\ln(\cos x)\ dx=\frac12\left(C-\frac\pi2\ln2\right).
\end{equation}
The complete proofs can be found here.
As shown by Mr. Lucian in his answer below, the three integrals are closely related, so finding the closed-form one of them will also find the other closed-forms. How to find the closed-forms of the integrals? Could anyone here please help me to find the closed-form, only one of them, preferably with elementary ways (high school methods)? If possible, please avoiding contour integration and double summation. Any help would be greatly appreciated. Thank you.

Following the same approach as in this answer,
$$ \begin{align} &\int_{0}^{\pi/4} \log^{2} (2 \sin x) \ dx = \int_{0}^{\pi/4} \log^{2}(2) \ dx + 2 \log 2 \int_{0}^{\pi/4}\log(\sin x) \ dx + \int_{0}^{\pi /4}\log^{2}(\sin x) \ dx \\ &= \frac{\pi}{4} \log^{2}(2) - \log (2) \left(G + \frac{\pi}{2} \log (2) \right) + \int_{0}^{\pi/4} \log^{2}(\sin x) \ dx \\ &= \int_{0}^{\pi /4} \left(x- \frac{\pi}{2} \right)^{2} \ dx + \text{Re} \int_{0}^{\pi/4} \log^{2}(1-e^{2ix}) \ dx \\ &= \frac{7 \pi^{3}}{192} + \frac{1}{2} \text{Im} \int_{{\color{red}{1}}}^{i} \frac{\log^{2}(1-z)}{z} \ dz \\ &= \frac{7 \pi^{3}}{192} + \frac{1}{2} \text{Im} \left(\log^{2}(1-i) \log(i) + 2 \log(1-i) \text{Li}_{2}(1-i) - 2 \text{Li}_{3}(1-i) \right) \\ &= \frac{7 \pi^{3}}{192} + \frac{1}{2} \left(\frac{\pi}{8} \log^{2}(2) - \frac{\pi^{3}}{32} + \log(2)  \ \text{Im} \ \text{Li}_{2}(1-i) - \frac{\pi}{2} \text{Re} \ \text{Li}_{2}(1-i)- 2 \ \text{Im} \ \text{Li}_{3}(1-i)\right) . \end{align}$$
Therefore,
$$ \begin{align}\int_{0}^{\pi/4} \log^{2}(\sin x) \ dx &= \frac{\pi^{3}}{48} + G \log(2)+ \frac{5 \pi}{16}\log^{2}(2) + \frac{\log(2)}{2} \text{Im} \ \text{Li}_{2}(1-i) - \frac{\pi}{4}    \text{Re} \ \text{Li}_{2}(1-i) \\ &- \text{Im} \ \text{Li}_{3}(1-i) \approx 2.0290341368 . \end{align}$$
The answer could be further simplified using the dilogarithm reflection formula $$\text{Li}_{2}(x) {\color{red}{+}} \text{Li}_{2}(1-x) = \frac{\pi^{2}}{6} - \log(x) \log(1-x) $$
and the fact that $$ \text{Li}_{2}(i) = - \frac{\pi^{2}}{48} + i G.$$
EDIT: 
Specifically, $$\text{Li}_{2}(1-i) = \frac{\pi^{2}}{16} - i G - \frac{i \pi}{4} \log(2). $$
So $$\int_{0}^{\pi /4} \log^{2}(\sin x) \ dx = \frac{\pi^{3}}{192} + G\frac{  \log(2)}{2} + \frac{3 \pi}{16} \log^{2}(2) - \text{Im} \  \text{Li}_{3}(1-i).$$
