Looking for closed-forms of $\int_0^{\pi/4}\ln^2(\sin x)\,dx$ and $\int_0^{\pi/4}\ln^2(\cos x)\,dx$

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


$$\int_0^\frac\pi4\Big(\ln\sin x\Big)^2~dx~=~\dfrac{23}{384}\cdot\pi^3~+~\dfrac9{32}\cdot\pi\cdot\ln^22~+~\underbrace{\beta(2)}_\text{Catalan}\cdot\dfrac{\ln2}2~-~\Im\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg].$$

$$\int_0^\frac\pi4\Big(\ln\cos x\Big)^2~dx~=~\dfrac{-7}{384}\cdot\pi^3~+~\dfrac7{32}\cdot\pi\cdot\ln^22~-~\underbrace{\beta(2)}_\text{Catalan}\cdot\dfrac{\ln2}2~+~\Im\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg].$$


$$S=\int_0^\frac\pi4\Big(\ln\sin x\Big)^2~dx~+~\int_0^\frac\pi4\Big(\ln\cos x\Big)^2~dx=I+J.$$

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, $~K=\displaystyle\sum_{n=1}^\infty\frac{(-1)^n~H_n}{(2n+1)^2}$.


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.