The 2 seems out of place

Calculus Level 5

0π2xln(sin2x) dx\displaystyle\int_0^{\frac{\pi}{2}}x\ln(\sin 2x)\text{ }dx

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

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