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$\ds{\int_{0}^{\pi/4}\ln\pars{\cos\pars{t}}\,\dd t
=-\,{\pi\ln\pars{2} \over 4} + {K \over 2}:\ {\large ?}}$ where
$\ds{K \equiv
\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{2}} \approx 0.9160}$ is the Catalan Constant.
\begin{align}
&\color{#c00000}{\int_{0}^{\pi/4}\ln\pars{\cos\pars{t}}\,\dd t}
=-\,{\pi\ln\pars{2} \over 4} + \int_{0}^{\pi/4}\ln\pars{2\cos\pars{t}}\,\dd t
\\[3mm]&=-\,{\pi\ln\pars{2} \over 4} + \half\int_{0}^{\pi/4}\ln\pars{\cot\pars{t}}\,\dd t
+\int_{0}^{\pi/4}\ln\pars{2\cos\pars{t} \over \cot^{1/2}\pars{t}}\,\dd t
\end{align}
Since ( see this link ) $\ds{K = \int_{0}^{\pi/4}\ln\pars{\cot\pars{t}}\,\dd t}$:
$$
\color{#c00000}{\int_{0}^{\pi/4}\ln\pars{\cos\pars{t}}\,\dd t}
=-\,{\pi\ln\pars{2} \over 4} + {K \over 2}
+ \half
\color{#00f}{\int_{0}^{\pi/4}\ln\pars{4\cos^{2}\pars{t} \over \cot\pars{t}}\,\dd t}
\tag{1}
$$
The problem is reduced to show that the "$\color{#00f}{\mbox{blue integral}}$"
vanishes out:
\begin{align}
&\color{#00f}{\int_{0}^{\pi/4}\ln\pars{4\cos^{2}\pars{t} \over \cot\pars{t}}\,\dd t}
=\int_{0}^{\pi/4}\ln\pars{4\sin\pars{t}\cos\pars{t}}\,\dd t
=\int_{0}^{\pi/4}\ln\pars{2\sin\pars{2t}}\,\dd t
\\[3mm]&=\half\int_{0}^{\pi/2}\ln\pars{2\sin\pars{t}}\,\dd t
={1 \over 4}\,\pi\ln\pars{2}
+ \half\,\lim_{\mu \to 0}\partiald{}{\mu}
\int_{0}^{1}t^{\mu}\pars{1 - t^{2}}^{-1/2}\,\dd t
\\[3mm]&={1 \over 4}\,\pi\ln\pars{2}
+ {1 \over 4}\,\lim_{\mu \to 0}\partiald{}{\mu}
\int_{0}^{1}t^{\pars{\mu - 1}/2}\pars{1 - t}^{-1/2}\,\dd t
\\[3mm]&={1 \over 4}\,\pi\ln\pars{2}
+ {1 \over 4}\,\lim_{\mu \to 0}\partiald{}{\mu}\bracks{%
\Gamma\pars{\mu/2 + 1/2}\Gamma\pars{1/2} \over \Gamma\pars{\mu/2 + 1}}
\\[3mm]&={1 \over 4}\,\pi\ln\pars{2}
+ {1 \over 8}\,{\Gamma\pars{1/2} \over \Gamma\pars{1}}\
\bracks{\overbrace{\Psi\pars{\half} - \Psi\pars{1}}
^{\ds{-2\ln\pars{2}}}}\,
\overbrace{\Gamma\pars{\half}}^{\ds{\root{\pi}}} = \color{#00f}{\large 0}
\quad\mbox{since}\quad\Gamma\pars{1} = 1.\tag{2}
\end{align}
$\ds{\Gamma\pars{z}}$ and $\ds{\Psi\pars{z}}$ are the Gamma and Digamma Functions, respectively.
$\pars{1}$ and $\pars{2}$ lead to:
$$
\color{#00f}{\large\int_{0}^{\pi/4}\ln\pars{\cos\pars{t}}\,\dd t
=-\,{\pi\ln\pars{2} \over 4} + {K \over 2}}
$$