Using $$\int_0^{\pi/2} \log\sin x\,\mathrm dx= -\dfrac{\pi}{2} \log 2$$ how to find
$$I=\int_0^{\infty} \log{(x+1/x)}\,\frac{dx}{1+x^2}. $$
Putting $x=\tan z$,
$I=\int_0^{\pi/2} (\log 2-\log(\sin(2z)))dz=\frac{\pi}{2}\log 2-1/2\int_0^{\pi} \log(\sin(u))du$ for $2z=u$
what to do next?
Answer
Split the second integral:
$$\int_0^{\pi}\log(\sin(u)) du = \int_0^{\pi/2}\log(\sin(u)) du + \int_{\pi/2}^{\pi}\log(\sin(u)) du$$
And use a change of variable $u=\pi-x$ for the second part.
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