\[\displaystyle\int_0^{\frac{\pi}{2}}x\ln(\sin 2x)\text{ }dx\]

The integral above is equal to \(\dfrac{-\pi^A\ln B}{C},\) where \(A\), \(B\), and \(C\) are positive integers and \(B\) is minimized. Find \(A+B+C.\)

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