Question

integration of (2log sinx - log sin2x) dx from 0-π/2.

Answer 1

I = integral  of  Log sin^2 x  - Log (2 sinx cosx)  dx  from x = 0 to pi/2

f(x) = Log [ sin^2 x / (2 sinx cosx ) ] = Log tanx  - Log 2

$I1= \int\limits_{x=0}^{\pi/2} {Log\ tanx} \, dx\\\\y=\pi/2 - x, dy=-dx,tanx=coty\\\\I1= \int\limits_{y=0}^{\pi/2} {Log\ cot y} \, dy\\\\=-\int\limits_{y=0}^{\pi/2} {Log\ tan y} \, dy\\\\=-I1\\\\So\ I1=0\\\\I=\int \limits_0^{\pi/2} {-Log 2} \, dx=-\pi/2 * Log2$

ans:   -π/2 * Log 2

Answer 2

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