While solving Real Integrals, do you obtain results in Complex Numbers?

$\displaystyle \int_{\pi/4}^{\pi/3} \left(\frac{1}{\tan x\ (\ln \sin x)}+\frac{\tan x}{\ln \cos x}\right) \ dx$

$= \displaystyle \int_{\pi/4}^{\pi/3} \bigg((\ln (\ln \sin x))'-(\ln (\ln \cos x))' \bigg)\ dx$

$=\large{ \bigg(\ln(\ln(\sin x) )- \ln(\ln(\cos x)) \bigg) \Bigg|_{\pi/4}^{\pi/3} }$

$= \ln \left( \ln \left(\dfrac{\sqrt{3}}{2} \right) \right) -\ln \left( \ln \left(\dfrac{1}{\sqrt{2}} \right) \right)$

$= \displaystyle\ln\left[\frac{\ln\left(\displaystyle\frac{\sqrt{3}}{2}\right)}{\ln\left(\displaystyle\frac{1}{2}\right)} \right] \approx \boxed{-1.572}$ and thus, it's modulus is $\boxed{1.572}$ .

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Why did we suppose that the value of the integral might be a non-real (complex) entity? The answer is right below.

You can see that both

$\color{#D61F06}{\ln \left( \ln \left(\dfrac{\sqrt{3}}{2} \right) \right) \quad }$ and $\quad \color{#D61F06}{\ln \left( \ln \left(\dfrac{1}{\sqrt{2}} \right) \right) }$

are non-real (complex) numbers, but then, we have

$\left[\color{#D61F06}{\ln \left( \ln \left(\dfrac{\sqrt{3}}{2} \right) \right) -\ln \left( \ln \left(\dfrac{1}{\sqrt{2}} \right) \right)} \right] = \displaystyle \color{#3D99F6}{ \ln \left [\frac{\ln \left( \displaystyle\frac{\sqrt{3}}{2}\right)}{\ln\left(\displaystyle\frac{1}{2}\right)} \right]}$

as a real number.