CalculusChallengeNon-Elementary Integration ShareNon-Elementary Integration · Challenge problemGiven ∫0π/2ln(sinx) dx=−π2ln2\displaystyle\int_0^{\pi/2}\ln(\sin x)\,dx=-\dfrac\pi2\ln2∫0π/2ln(sinx)dx=−2πln2, evaluate ∫0π/2x2sin2x dx\displaystyle\int_0^{\pi/2}\dfrac{x^2}{\sin^2x}\,dx∫0π/2sin2xx2dx.Check Show answer Walk through it step by step Practice more like this Review this concept