<s>[INST] Evaluating $\int_0^{\frac{\pi}{2}}\ln\left(\frac{\ln^2\sin\theta}{\pi^2+\ln^2\sin\theta}\right)\,\frac{\ln\cos\theta}{\tan\theta}\,d\theta$ Prove
$$\int_0^{\frac{\pi}{2}}\ln\left(\frac{\ln^2\sin\theta}{\pi^2+\ln^2\sin\theta}\right)\,\frac{\ln\cos\theta}{\tan\theta}\,d\theta = \frac{\pi^2}{4}$$
 [/INST] I am also a proponent of the opinion that the proposed rules are against the way how we communicate ideas in this community. At the same time, however, the mathematical part of OP is something worth it to be dealt with. So here is a solution:

1. Preliminary
Before the calculation we make some preliminary results:

Lemma 1. For any $u > 0$ and $n > 0$, we have
  $$\frac{1}{n^{2}} \log \left(1 + \frac{4\pi^{2}n^{2}}{u^{2}} \right) = \pi^{2} \int_{u/2}^{\infty} \frac{2}{s^{2} + n^{2}\pi^{2}} \, \frac{ds}{s}.$$

Proof. Differentiating both sides with respect to $u$, we check that they must equal up to a constant. Taking $u \to \infty$, we find that this constant should equal zero. ////

Lemma 2. For any real $x$, we have
  $$ \sum_{n=1}^{\infty} \frac{2}{s^{2} + n^{2}\pi^{2}} = \frac{s \coth s - 1}{s^{2}}. $$

Although non-trivial, this is a standard result in complex analysis. So we omit the proof.

Lemma 3. Let $f(s) = (1 - e^{-2s})(s\coth s - 1)$. Then
  
  
*
  
*$f(s) = (s-1) + (s+1)e^{-2s}$ and hence $f''(s) = 4s e^{-2s}$. 
  
*$f(s)/s^{2}$ and $f'(s)/s$ converges to $0$ as $s \to 0$ and $s \to +\infty$.
  

Proof. The first assertion is just a simple calculation. To prove the second assertion, it suffices to look into the McLaurin series expansion $f(s) = \frac{2}{3}s^{3} - \frac{2}{3}s^{4} + \cdots$. ////
2. Calculation
Now we are ready to calculate the integral. Let $I$ denote the integral. Then with the substitution $\sin^{2}\theta = e^{-t}$ (so that $d\theta/\tan\theta = -dt/2t$), we have
\begin{align*}
I
&= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \log \left( \frac{\log^{2} \sin^{2}\theta}{4\pi^{2} + \log^{2}\log^{2}\theta} \right) \frac{\log^{2}\cos^{2}\theta}{\tan\theta} \, d\theta \\
&= \frac{1}{4} \int_{0}^{\infty} \log(1 - e^{-t}) \log\left( \frac{t^{2}}{4\pi^{2} + t^{2}} \right) \, dt\\
&= \frac{1}{4} \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{e^{-nt}}{n} \log \left(1 + \frac{4\pi^{2}}{t^{2}} \right) \, dt.
\end{align*}
Now we utilize the Tonelli's theorem to interchange the summation and integral. Then
\begin{align*}
I
&= \frac{1}{4} \sum_{n=1}^{\infty} \int_{0}^{\infty} \frac{e^{-nt}}{n} \log \left(1 + \frac{4\pi^{2}}{t^{2}} \right) \, dt \\
&= \frac{1}{4} \sum_{n=1}^{\infty} \int_{0}^{\infty} \frac{e^{-u}}{n^{2}} \log \left(1 + \frac{4\pi^{2}n^{2}}{u^{2}} \right) \, du, \quad (u = nt) \\
&= \frac{\pi^{2}}{4} \sum_{n=1}^{\infty} \int_{0}^{\infty} e^{-u} \left( \int_{u/2}^{\infty} \frac{2}{s^{2} + n^{2}\pi^{2}} \, \frac{ds}{s} \right) \, du,
\end{align*}
where the last equality follows from Lemma 1. Applying the Tonelli's theorem again, Lemma 2 shows that
\begin{align*}
I
&= \frac{\pi^{2}}{4} \int_{0}^{\infty} e^{-u} \left( \int_{u/2}^{\infty} \frac{s \coth s - 1}{s^{2}} \, \frac{ds}{s} \right) \, du \\
&= \frac{\pi^{2}}{4} \int_{0}^{\infty} \left( \int_{0}^{2s} e^{-u} \, du \right) \frac{s \coth s - 1}{s^{3}} \, ds \\
&= \frac{\pi^{2}}{4} \int_{0}^{\infty} \frac{f(s)}{s^{3}} \, ds,
\end{align*}
where we applied Tonelli's theorem again in the second line, and $f(s)$ denotes the function in Lemma 3. So it suffices to prove that the last integral, without the constant $\pi^{2}/4$, equals 1. Indeed, Lemma 3 shows that
\begin{align*}
\int_{0}^{\infty} \frac{f(s)}{s^{3}} \, ds
&= \left[ -\frac{f(s)}{2s^{2}} \right]_{0}^{\infty} + \frac{1}{2} \int_{0}^{\infty} \frac{f'(s)}{s^{2}} \, ds \\
&= \left[ -\frac{\smash{f'}(s)}{2s} \right]_{0}^{\infty} + \frac{1}{2} \int_{0}^{\infty} \frac{f''(s)}{s} \, ds \\
&= \int_{0}^{\infty} 2e^{-2s} \, ds
 = 1
\end{align*}
and therefore we get $I = \pi^{2}/4$ as desired.
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