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integrate [log(sin x) - log(2cos x)] dx from 0 to pi/2

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Prove that: `int_(0)^(pi//2) log (sin x) dx =int_(0)^(pi//2) log (cos x) dx =(-pi)/(2) log 2`

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integration 0 to 'pi/2' ln(cosx)dx || integration 0 to 'pi/2' log(sinx)dx || 0 to pi/2 ln(cosx)dx

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`int_(pi//4)^(pi//2) cos 2 x log sin x " " dx=`

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Prove that integral of 0 to pi/2 sin^2x/sinx+cosx dx= log(√2+1)/√2|int_0^(pi/2) sin^2x/(sinx+cosx)dx

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