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\[ {\int_0^1\frac{\ln\left(x^4-2x^2+5\right)-\ln\left(5x^4-2x^2+1\right)}{1-x^2}\, dx} \]

Given that the integral above is equal to ...

\[\displaystyle \int _{ 0 }^{ \pi /4 }{ \frac { \ln { \left( \sin { (x) } \right) } \ln { \left( \cos { (x) } \right) } }{ \sin { (x) } \cos { (x) } } dx } \]

\[ \large \int_0^{\pi /2} \ln \left( \left | \cos(\tan x ) \right | \right) \, dx \]

The value of the integral above is equal to ...

Hi Brilliant! Due to many problems in part 1 it had become slow to load. So this is a sequel of Brilliant Integration - Season 2(Part 1)

The aims of ...

Here's another proof of Euler reflection formula :

\(Lemma1\quad :\)

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