CalculusHardNon-Elementary Integration ShareNon-Elementary Integration · Hard problemGiven ∫0π/2ln(sinx) dx=−π2ln2\displaystyle\int_0^{\pi/2}\ln(\sin x)\,dx=-\dfrac{\pi}{2}\ln 2∫0π/2ln(sinx)dx=−2πln2, evaluate ∫0πln(1−cosx) dx\displaystyle\int_0^{\pi}\ln(1-\cos x)\,dx∫0πln(1−cosx)dx.Check Show answer Walk through it step by step Practice more like this Review this concept