<s>[INST] 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
$$\int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)}dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right)$$
 [/INST] 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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