Definite Integral $\int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)}dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right)$ I want to prove that
$$\int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)}dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right)$$
 A: Consider $$f(z) = \frac{1}{\log(1-iz)} \frac{1}{1+z^{2}}$$ where the branch cut for $\log (1-iz)$ runs down the imaginary axis from $z=-i$.
Then integrate around a contour that consists of the line segment $[-R,R]$ and the upper half of the circle $|z|=R$, and is indented around the simple pole at the origin.
Since $ z f(z) ->0 $ uniformly as $R \to \infty$,  $\displaystyle \int f(z) \ dz$ vanishes along the upper half of $|z|=R$ as $ R \to \infty$.
So we have
$$ \begin{align} \text{PV} \int_{-\infty}^{\infty} \frac{1}{\log(1-ix)} \frac{dx}{1+x^{2}} &= \text{PV} \int_{-\infty}^{\infty} \frac{1}{\frac{1}{2}\log(1+x^{2}) - i\arctan x} \frac{dx}{1+x^{2}}  \\ &= \text{PV} \int_{-\infty}^{\infty}\frac{\frac{1}{2} \log(1+x^{2})+ i \arctan x }{\frac{1}{4} \log^{2}(1+x^{2})+ \arctan^{2} x} \frac{dx}{1+x^{2}}   \\ &= 2 \pi i \ \text{Res}[f(z), i] + \pi i \ \text{Res}[f(z),0]  \\ &= 2 \pi i \left(\frac{1}{2i \log 2} \right) + \pi i \left( -\frac{1}{i} \right) \\ &= \pi \left( \frac{1}{\log 2} - 1\right). \end{align}$$
Equating the real parts on both sides of the equation,
$$ \int_{-\infty}^{\infty} \frac{\frac{1}{2} \log(1+x^{2})}{\frac{1}{4} \log^{2}(1+x^{2}) + \arctan^{2} x} \frac{dx}{1+x^{2}} = \pi \left(\frac{1}{\log 2} -1 \right). $$
Now let $x= \tan u$.
Then
$$ \begin{align} \int_{-\pi /2}^{\pi /2} \frac{\frac{1}{2} \log(\sec^{2}u)}{\frac{1}{4} \log^{2}(\sec^{2}u) + u^{2}} du &= \int_{-\pi/2}^{\pi /2} \frac{\log (\sec u)}{\log^{2}(\sec u)+u^{2}} \ du \\ &= - \int_{-\pi/2}^{\pi/2} \frac{\log(\cos u)}{\log^{2}(\sec u) + u^{2}} \ du \\ &= \pi \left(\frac{1}{\log 2} -1 \right) \end{align}$$
which implies
$$ \int_{0}^{\pi/2} \frac{\log(\cos u)}{u^{2} + \log^{2}(\cos u)} \ du = \frac{\pi}{2} \left(1- \frac{1}{\log 2} \right).$$
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$\ds{\int_{0}^{\pi/2}{\ln\pars{\cos x} \over x^{2} + \ln^{2}\pars{\cos\pars{x}}}
     \,\dd x = {\pi \over 2}\,\bracks{1 - {1 \over \ln\pars{2}}}}$

${\sf\mbox{The integration can be performed without "leaving" the first
  quadrant !!!}}$.

We close the arc
$\ds{\braces{z = \expo{\ic\theta}\ \mid\ 0 < \theta < {\pi \over 2}}}$ with the 'segments' $\ds{\braces{z=y\ic\ \mid\ y \in \pars{0,1}}}$
and $\ds{\braces{z=x \mid\ x \in \pars{0,1}}}$. The contour is properly indented as explained below.

\begin{align}
&\color{#c00000}{%
\int_{0}^{\pi/2}{\ln\pars{\cos x} \over x^{2} + \ln^{2}\pars{\cos\pars{x}}}\,\dd x}
=\Re\int_{0}^{\pi/2}{\dd x \over \ln\pars{\cos\pars{x}} + x\ic}
\\[3mm]&=\Re
\int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi/2}}
{1 \over \ln\pars{\bracks{z^{2} + 1}/\bracks{2z}} + \ln\pars{z}}
\,{\dd z \over \ic z}
\\[3mm]&=\Im
\int_{\verts{z}\ =\ 1 \atop {\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi/2}}
{1 \over \ln\pars{\bracks{z^{2} + 1}/2}}\,{\dd z \over z}
\\[3mm]&=\Im\left\lbrack\left.%
-\int_{0}^{-\pi/2}{1 \over \ln\pars{\bracks{z^{2} + 1}/2}}\,{\dd z \over z}
\right\vert_{z\ =\ \ic\ +\ \epsilon\expo{\ic\theta}}
-\int_{1 - \epsilon}^{\epsilon}{1 \over \ln\pars{\bracks{-y^{2} + 1}/2}}
\,{\ic\,\dd y \over \ic y}
\right.
\\[3mm]&\phantom{=\Im\left\lbrack a\right.}\color{#00f}{\left.
-\int_{\pi/2}^{0}{1 \over \ln\pars{\bracks{z^{2} + 1}/2}}\,{\dd z \over z}
\right\vert_{z\ =\ \epsilon\expo{\ic\theta}}}
-\int_{\epsilon}^{1 - \epsilon}{1 \over \ln\pars{\bracks{x^{2} + 1}/2}}
\,{\dd x \over x}
\\[3mm]&\phantom{=\left\lbrack a\right.}\left.\color{#00f}{\left.-\int_{\pi}^{\pi/2}
{1 \over \ln\pars{\bracks{z^{2} + 1}/2}}\,{\dd z \over z}\right\vert_{z\ =\ 1\ +\ \epsilon\expo{\ic\theta}}}\right\rbrack
\end{align}
  The above integration is evaluated along the first quadrant, as explained above,
   and it's $\ds{"\epsilon\mbox{-indented}"}$ around
  $\ds{z = \ic, 0\ \mbox{and}\ 1}$. In the $\ds{\epsilon \to 0^{+}}$ limit, the contribution to the final result arises from the
  $\ds{\color{#00f}{\mbox{above blue terms}}}$

\begin{align}
&\color{#c00000}{%
\int_{0}^{\pi/2}{\ln\pars{\cos x} \over x^{2} + \ln^{2}\pars{\cos\pars{x}}}\,\dd x}
\\[3mm]&=\lim_{\epsilon \to 0^{+}}\bracks{\left.%
-\pars{-\,{\pi \over 2}}\,{1 \over \ln\pars{\bracks{z^{2} + 1}/2}}
\right\vert_{z\ =\ \epsilon\expo{\ic\theta}}
-\int_{\pi}^{\pi/2}{\epsilon\expo{\ic\theta}\,\dd\theta
\over \ln\pars{1 + \epsilon\expo{\ic\theta} + \epsilon^{2}\expo{2\ic\theta}/2}}}
\\[3mm]&=-\,{\pi \over 2}\,{1 \over \ln\pars{2}} + {\pi \over 2}
\end{align}

$$\color{#66f}{\large%
\int_{0}^{\pi/2}{\ln\pars{\cos x} \over x^{2} + \ln^{2}\pars{\cos\pars{x}}}
\,\dd x = {\pi \over 2}\,\bracks{1 - {1 \over \ln\pars{2}}}}
\approx {\tt -0.6953}
$$

A: Note that for $|x| < \frac{\pi}{2}$, we have
$$ \frac{\log\cos x}{\log^2 \cos x + x^2} = \Re \frac{1}{\log\left( \frac{1+e^{2ix}}{2} \right)}. $$
Thus if $I$ denotes the given integral, we have
\begin{align*}
I
&= \frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \Re \frac{1}{\log\left( \frac{1+e^{2ix}}{2} \right)} \, dx \\
&= \frac{1}{4} \mathrm{PV}\int_{-\pi}^{\pi} \Re \frac{1}{\log\left( \frac{1+e^{ix}}{2} \right)} \, dx \\
&= \frac{1}{4} \Re \mathrm{PV}\int_{-\pi}^{\pi} \frac{1}{\log\left( \frac{1+e^{ix}}{2} \right)} \, dx.
\end{align*}
Now let $C_{\epsilon}$ be the counter-clockwised contour consisting of the circle of radius 1 centered at the origin, with two semicircular indents $\gamma_{1,\epsilon}$ around $1$ and $\gamma_{2,\epsilon}$ around $-1$ as follows:

By writing
\begin{align*}
I = \frac{1}{4} \Re \lim_{\delta\to0^{+}}\left( \int_{-\pi+\delta}^{-\delta} + \int_{\delta}^{\pi-\delta} \right) \frac{1}{\log\left( \frac{1+e^{ix}}{2} \right)} \, dx
\end{align*}
and plugging the substitution $z = e^{ix}$, we observe that
\begin{align*}
I = \frac{1}{4} \Im \lim_{\epsilon\to 0^{+}}\left(\oint_{C_{\epsilon}} - \int_{\gamma_{1,\epsilon}} - \int_{\gamma_{2,\epsilon}} \right) \frac{dz}{z \log\left(\frac{1+z}{2}\right)}
\end{align*}
Let 
\begin{align*}
f(z) = \frac{1}{z \log\left(\frac{1+z}{2}\right)}.
\end{align*}
It is plain from the logarithmic singularity that
\begin{align*}
\lim_{\epsilon \to 0^{+}} \int_{\gamma_{2,\epsilon}} f(z) \, dz = 0.
\end{align*}
Also it follows that
\begin{align*}
\lim_{\epsilon\to 0^{+}} \oint_{C_{\epsilon}} f(z) \, dz
&= 2\pi i \operatorname{Res}_{z=0} f(z)
 = -\frac{2\pi i}{\log 2}, \\
\lim_{\epsilon\to 0^{+}} \int_{\gamma_{1,\epsilon}} f(z) \, dz
&= -\pi i \operatorname{Res}_{z=1} f(z)
 = -2\pi i.
\end{align*}
Therefore we have
\begin{align*}
I
&= \frac{1}{4} \Im \left( 2\pi i - \frac{2\pi i}{\log 2} \right)
 = \frac{\pi}{2} \left( 1 - \frac{1}{\log 2} \right).
\end{align*}
