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.
 A: By setting $x=\arctan t$ we have:
$$\int_{0}^{\pi/4}\log^2(\cos x)\,dx = \frac{1}{4}\int_{0}^{1}\frac{\log^2(1+t^2)}{1+t^2}.$$
Attack plan: get the Taylor series of $\log^2(1+t^2)$ and integrate it termwise.
Since
$$-\log(1-z)=\sum_{n=1}^{+\infty}\frac{z^n}{n}$$
it follows that
$$[z^n]\log^2(1-z)=\sum_{k=1}^{n-1}\frac{1}{k(n-k)}=2\frac{H_{n-1}}{n},$$
$$\log^2(1+t^2)=\sum_{n=2}^{+\infty}2\frac{H_{n-1}}{n}(-1)^n t^{2n}.\tag{1}$$
If now we set
$$\mathcal{J}_m = \int_{0}^{1}\frac{t^{2m}}{t^2+1}\,dt $$
we have $\mathcal{J}_0=\frac{\pi}{4}$ and $\mathcal{J}_{m+1}+\mathcal{J}_m = \frac{1}{2m+1}$, hence:
$$\mathcal{J}_m = (\mathcal{J}_m+\mathcal{J}_{m-1})-(\mathcal{J}_{m-1}+\mathcal{J}_{m-2})+\ldots\pm(\mathcal{J}_1+\mathcal{J}_0)\mp\mathcal{J}_0,$$
$$\mathcal{J}_m = \sum_{j=0}^{m-1}\frac{(-1)^j}{(2m-2j-1)}+(-1)^m\frac{\pi}{4}=(-1)^m \sum_{j\geq m}\frac{(-1)^j}{2j+1}.\tag{2}$$
From $(1)$ and $(2)$ it follows that:
$$\int_{0}^{\pi/4}\log^2(\cos x)\,dx=\frac{1}{2}\sum_{n=2}^{+\infty}\frac{H_{n-1}}{n}\sum_{r\geq n}\frac{(-1)^r}{2r+1},\tag{3}$$
and summation by parts gives:

$$\int_{0}^{\pi/4}\log^2(\cos x)\,dx=\frac{1}{4}\sum_{n=2}^{+\infty}(H_n^2-H_n^{(2)})\frac{(-1)^n}{2n+1}.\tag{4}$$

UPDATE: the question is now set in an answer to another question. This site (many thanks to @gammatester) is devoted to the evaluation of sums like the one appearing in the RHS of $(4)$. Through Euler-Landen's identity (see the line below $(608)$ in the linked site) it is not too much difficult to see that the RHS of $(4)$ depends on $\operatorname{Li}_3\left(\frac{1+i}{2}\right)$ as stated in the @Lucian's answer.
A: The strategy in this post will be included in another paper.

A solution (in large steps) by Cornel Ioan Valean
In my opinion, this is a very magical & powerful way that manages to circumvent the necessity of using the already famous method proposed by Random Variable which I think most posts on MSE use it for such integrals. It's time for a new way to come in place and join the existing one!
In this post, we magically prove that
$$\int_0^1\frac{\log x\log(1+x^2)}{1+x^2}\textrm{d}x=-\frac{\pi}{16}  \log ^2(2) - \log (2)G-\frac{\pi ^3}{64}+2\Im\biggr \{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr \},$$
by wisely combining a result from the book, (Almost) Impossible Integrals, Sums, and Series, namely the special Fourier series (see eq. 3.284, page 244, and eq. 3.288, page 247),
\begin{equation}
\begin{aligned}
\small \sum_{n=1}^{\infty} (-1)^{n-1}\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(\frac{n}{2}\right)-\frac{1}{n}\right)\sin(2nx)&\small=\sum_{n=1}^{\infty} (-1)^{n-1}\left(\int_0^1 t^{n-1}\frac{1-t}{1+t} \textrm{d}t\right)\sin(2nx)\\ 
&=-\cot(x)\log(\cos(x)), 
\end{aligned}
\end{equation}
where $\displaystyle 0< x<\frac{\pi}{2}$, and the Cornel's integral,
$$\int_0^{\pi/2} x\frac{\log(\cos x)}{\sin x}\textrm{d}x=2\log(2)G-\frac{\pi}{8}\log^2(2)-\frac{5}{32}\pi^3+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\},$$
already calculated in this post How can you approach $\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx$.
Proof: We differentiate both sides of the Fourier series that leads to
$$2 \sum_{n=1}^{\infty} (-1)^{n-1}\left(\int_0^1 t^{n-1}\frac{1-t}{1+t} \textrm{d}t\right)n\cos(2nx)=1+\frac{\log(\cos(x))}{\sin^2(x)},$$
and if we multiply both side by $x \sin(x)$ and integrate from $x=0$ to $x=\pi/2$, we arrive at
$$\int_0^{\pi/2} x\sin(x)\textrm{d}x+\int_0^{\pi/2}x\frac{\log(\cos(x))}{\sin(x)}\textrm{d}x$$
$$=2 \log (2)-1+2 \log (2)\underbrace{\int_0^1 \frac{\log (x)}{1+x^2}\textrm{d}x}_{\displaystyle \text{Trivial}}+\frac{1}{2}\underbrace{\int_0^1 \log (x) \log \left(1-x^2\right)\textrm{d}x}_{\displaystyle \text{Trivial}}$$
$$+\frac{1}{2}\underbrace{\int_0^1\frac{\log (x) \log \left(1-x^2\right)}{x^2}\textrm{d}x}_{\displaystyle \text{Trivial}}-2\underbrace{\int_0^1\frac{ \log (x) \log \left(1-x^4\right)}{1-x^4}\textrm{d}x}_{\displaystyle \text{Beta function in disguise}}$$
$$+2\underbrace{\int_0^1\frac{x^2 \log (x) \log \left(1-x^4\right)}{1-x^4}\textrm{d}x}_{\displaystyle \text{Beta function in disguise}}+2\color{blue}{\int_0
^1 \frac{\log (x) \log(1+x^2)}{1+x^2}\textrm{d}x},$$
from which the desired result follows.
Note the following values of the Beta function forms in disguise:
$$\int_0^1 \frac{\log (x) \log \left(1-x^4\right)}{1-x^4} \textrm{d}x=\frac{1}{16}\int_0^1 \frac{\log(x)\log (1-x)}{ x^{3/4}(1-x) } \textrm{d}x$$
$$=\frac{7 }{4}\zeta (3)+\frac{\pi ^3}{32}-\frac{3}{16}\log (2)\pi ^2-\frac{\pi }{4}G-\frac{3}{2}\log(2)G,$$
and
$$\int_0^1 \frac{x^2\log (x) \log \left(1-x^4\right)}{1-x^4} \textrm{d}x=\frac{1}{16}\int_0^1 \frac{\log(x)\log (1-x)}{x^{1/4}(1-x)} \textrm{d}x$$
$$=\frac{7}{4} \zeta (3)+\frac{3}{2} \log (2)G-\frac{1}{4} \pi G-\frac{3}{16}\log(2)\pi^2-\frac{\pi ^3}{32}.$$
A note: this method can also be adjusted to extract other very difficult integrals, which is possible by further exploiting and developing ideas like the ones in the paper A symmetry-related treatment of two fascinating sums of integrals by C.I. Valean.
End of story
A: we can prove, using the same strategy of Random Variable, the following equality:
$$\int_0^{\pi/4}\ln^2(\cos x)\ dx=\frac7{192}\pi^3+\frac5{16}\pi\ln^22-\frac12\ln2G-\text{Im}\operatorname{Li_3}(1+i)$$
proof :
\begin{align*} \ln(1+e^{2ix}) &=  \ln (e^{-ix}+e^{ix}) + \ln(e^{ix}) \\ &=  \ln(2\cos x) + ix 
\end{align*}
squaring both sides and integrating, we get
$$\int_0^{\pi/4}\ln^2(1+e^{2ix})\ dx=\int_0^{\pi/4}(\ln(2\cos x)+ix)^2\ dx$$
equating the real parts on both sides and rearranging the terms, we have:
$$
\int_0^{\pi/4}\ln^2(\cos x)\ dx=\int_0^{\pi/4}(x^2-\ln^22)\ dx-2\ln2\int_0^{\pi/4}\ln(\cos x)\ dx$$
$$+\text{Re}\int_0^{\pi/4}\ln^2(1+e^{2ix})\ dx$$
$$=\frac{\pi^3}{192}-\frac{\pi}{4}\ln^22-2\ln2\left(\frac12G-\frac{\pi}{4}\ln2\right)+\text{Re}\int_0^{\pi/4}\ln^2(1+e^{2ix})\ dx$$
$$=\frac{\pi^3}{192}+\frac{\pi}{4}\ln^22-\ln2G+\text{Re}\int_0^{\pi/4}\ln^2(1+e^{2ix})\ dx \tag{1}$$
Evaluating the last integral:
\begin{align*}
I&=\text{Re}\int_0^{\pi/4}\ln^2(1+e^{2ix})\ dx=\frac12\text{Im}\int_1^i\frac{\ln^2(1+x)}{x}\ dx\\
&=\frac12\text{Im}\left(\ln(-i)\ln^2(1+i)+2\ln(1+i)\operatorname{Li_2}(1+i)-2\operatorname{Li_3}(1+i)\right)\\
&=\frac{\pi^3}{32}+\frac{\pi}{16}\ln^22+\frac12\ln2G-\text{Im}\operatorname{Li_3}(1+i)\tag{2}
\end{align*}
Plugging $(2)$ in $(1)$ we get our result.
note that we used:
$$\ln(-i)=-\frac{\pi}{2}i$$
$$\ln(1+i)=\frac12\ln2+\frac{\pi}{4}i$$
$$\operatorname{Li_2}(1+i)=\frac{\pi^2}{16}+\left(\frac{\pi}{4}\ln2+G\right)i$$
which give us:
$$\ln(-i)\ln^2(1+i)=\frac{\pi^2}{8}\ln2+\left(\frac{\pi^3}{32}-\frac{\pi}{8}\ln^22\right)i$$
$$\ln(1+i)\operatorname{Li_2}(1+i) =-\frac{\pi}{4}G-\frac{\pi^2}{32}\ln2+\left(\frac12\ln2G+\frac{\pi^3}{64}+\frac{\pi}{8}\ln^22\right)i$$
A: my approach to problem $(3)$: 
\begin{align}
I&=\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx=-2\int_0^{\pi/4}\ln(\tan x)\ln(\cos x)\ dx\\
&=-2\int_0^{\pi/4}\ln(\sin x)\ln(\cos x)\ dx+2\int_0^{\pi/4}\ln^2(\cos x)\ dx\\
&=-\int_0^{\pi/2}\ln(\sin x)\ln(\cos x)\ dx+2\int_0^{\pi/4}\ln^2(\cos x)\ dx\\
&=-\left(\frac{\pi}{2}\ln^22-\frac{\pi^3}{48}\right)+2\left(\frac7{192}\pi^3+\frac5{16}\pi\ln^22-\frac12\ln2~G-\text{Im}\operatorname{Li_3}(1+i)\right)\\
&=\frac3{32}\pi^3+\frac{\pi}8\ln^22-\ln2~G-2\text{Im}\operatorname{Li_3}(1+i)
\end{align}
note that we evaluated the first integral using the derivative of beta function and as follows:
\begin{align}
J&=\int_0^{\pi/2}\ln(\sin x)\ln(\cos x)\ dx=\frac18\frac{\partial^2}{\partial{a}\partial{b}}\beta(a,b)\Bigg\rvert_{a\to1/2,~b\to1/2}\\
&=\frac18\beta(a,b)\left(\left(\psi(a)-\psi(a+b)\right)\left(\psi(b)-\psi(a+b)\right)-\psi^{(1)}(a+b)\right)\Bigg\rvert_{a\to1/2,~b\to1/2}\\
&=\frac18\beta(1/2,1/2)\left((\psi(1/2)-\psi(1))^2-\psi^{(1)}(1)\right)\\
&=\frac{\pi}8\left(4\ln^22-\zeta(2)\right)\\
&=\frac{\pi}2\ln^22-\frac{\pi^3}{48}
\end{align}
A: 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 $$\int_{0}^{\pi /4} \log^{2}(\sin x) \ dx = \frac{\pi^{3}}{192} + G\frac{  \log(2)}{2} + \frac{3 \pi}{16} \log^{2}(2) - \text{Im} \  \text{Li}_{3}(1-i).$$
A: 
$$\int_0^\frac\pi4\Big(\ln\sin x\Big)^2~dx~=~\dfrac{23}{384}\cdot\pi^3~+~\dfrac9{32}\cdot\pi\cdot\ln^22~+~\underbrace{\beta(2)}_\text{Catalan}\cdot\dfrac{\ln2}2~-~\Im\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg].$$
$$\int_0^\frac\pi4\Big(\ln\cos x\Big)^2~dx~=~\dfrac{-7}{384}\cdot\pi^3~+~\dfrac7{32}\cdot\pi\cdot\ln^22~-~\underbrace{\beta(2)}_\text{Catalan}\cdot\dfrac{\ln2}2~+~\Im\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg].$$


$$S=\int_0^\frac\pi4\Big(\ln\sin x\Big)^2~dx~+~\int_0^\frac\pi4\Big(\ln\cos x\Big)^2~dx=I+J.$$
But, by a simple change of variable, $t=\dfrac\pi2-x,~J$ can be shown to equal $\displaystyle\int_\frac\pi4^\frac\pi2\Big(\ln\sin x\Big)^2~dx$,
in which case $I+J=\displaystyle\int_0^\frac\pi2\Big(\ln\sin x\Big)^2~dx=\dfrac{\pi^3}{24}+\dfrac\pi2\ln^22.~$ So we know their sum! Now all
that's left to do is to find out their difference, $D=I-J.~$ Then we'll have $I=\dfrac{S+D}2$ and
$J=\dfrac{S-D}2$.

$$D=I-J=\int_0^\frac\pi4\Big(\ln\sin x\Big)^2~dx-\int_0^\frac\pi4\Big(\ln\cos x\Big)^2~dx=\int_0^\frac\pi4\Big(\ln^2\sin x-\ln^2\cos x\Big)~dx$$
$$=\int_0^\frac\pi4\Big(\ln\sin x-\ln\cos x\Big)~\Big(\ln\sin x+\ln\cos x\Big)~dx=\int_0^\frac\pi4\ln\frac{\sin x}{\cos x}~\ln\big(\sin x~\cos x\big)~dx=$$
$$=\int_0^\frac\pi4\ln\tan x\cdot\ln\frac{\sin2x}2~dx=\frac12\int_0^\frac\pi2\ln\tan\frac x2\cdot\ln\frac{\sin x}2~dx=\int_0^1\ln t\cdot\ln\frac t{1+t^2}\cdot\frac{dt}{1+t^2}$$
where the last expression was obtained by using the famous Weierstrass substitution, $t=\tan\dfrac x2$
$$=\int_0^1\frac{\ln t\cdot\Big[\ln t-\ln(1+t^2)\Big]}{1+t^2}dt~=~\int_0^1\frac{\ln^2t}{1+t^2}dt~-~\int_0^1\frac{\ln t~\ln\big(1+t^2\big)}{1+t^2}dt~=~\frac{\pi^3}{16}-K,$$
where $~K=2~\Im\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]-\dfrac{\pi^3}{64}-\dfrac\pi{16}\ln^22-\underbrace{\beta(2)}_\text{Catalan}\ln2.~$ It follows then that our two
definite integrals possess a closed form expression if and only if $~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)$ has one as well. As
an aside, $~\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3).~$ Also, $~K=\displaystyle\sum_{n=1}^\infty\frac{(-1)^n~H_n}{(2n+1)^2}$.
