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\[\large \int_0^{\pi/2} \left(\dfrac{\theta}{\sin \theta}\right)^2 \; d\theta = A\pi \ln B \]

If the equation above holds true for positive integers \(A\) and \(B\), find ...

\[\sum _{n=2}^{\infty } \frac{(\zeta (n)-1) \cos \left(\frac{\pi n}{3}\right)}{n}=\frac{1}{2} \left(\log \left(\pi \text{sech}\left(\frac{\pi \sqrt{a}}{2}\right)\right)-\gamma b+c\right)\]

\[\large \sum _{n=2}^{\infty } \frac{\zeta (n) \cos \left(\frac{\pi n}{3}\right)}{n}=\frac{1}{2} \left(\log \left(\pi \text{sech}\left(\frac{\pi \sqrt{a}}{2}\right)\right)-\gamma b\right)\]

Compute

\[\int_0^1 \left(\sum _{n=2}^{\infty } \frac{\cos (\pi n x)}{n}\right) \, dx\]

\[\displaystyle \frac{\displaystyle\int_{0}^{\frac{\pi}{2}}\ln(\sin{x})\ln(\cos{x})dx}{\displaystyle \int_{0}^{\frac{\pi}{2}}(\ln\tan{x})^2dx}=\frac{A\ln^2{2}}{\pi^2}-\frac{1}{B}\]

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