<s>[INST] 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.
 [/INST] 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 $$\int_{0}^{\pi /4} \log^{2}(\sin x) \ dx = \frac{\pi^{3}}{192} + G\frac{  \log(2)}{2} + \frac{3 \pi}{16} \log^{2}(2) - \text{Im} \  \text{Li}_{3}(1-i).$$
</s>