Finding a Definite Integral with Natural Log and SineDate: 05/14/2008 at 05:52:27 From: Mohammad Subject: Integral LN(sin(x)) from 0 to pi I'm trying to find the following integral: - | | LN( SIN( x ) ) dx = ? | - From 0 to pi. I can evaluate integral of LN(x) but here we have LN(SIN(x)). I've tried integration-by-parts but I ended up confusing myself! Date: 05/19/2008 at 12:38:13 From: Doctor Who Subject: Re: Integral LN(sin(x)) from 0 to pi This definite integral can be evaluated by hand as follows: First we note the symmetric property of sin x in the interval [0,pi]. Therefore integral[x=0, x=pi] ln|sin x| dx = 2 integral[x=0, x=pi/2] ln|sin x| dx Further note that integral[x=0, x=pi/2] ln|sin x| dx = integral[x=0, x=pi/2] ln|sin(pi/2 - x)| dx You can verify this by using a substitution u = pi/2 - x and changing the limits. Also, sin(pi/2 - x) = cos x So we have integral[x=0, x=pi] ln|sin x| dx = integral[x=0, x=pi/2] ln|sin x| dx + integral[x=0, x=pi/2] ln|cos x| dx = integral[x=0, x=pi/2] (ln|sin x| + ln|cos x|) dx = integral[x=0, x=pi/2] ln|sin x cos x| dx = integral[x=0, x=pi/2] ln|2 sin x cos x| dx - ln(2)x [x=0, x=pi/2] = integral[x=0, x=pi/2] ln|sin 2x| dx - pi ln(2)/2 Make the substitution 2x = t and 2 dx = dt; the upper limit becomes pi and the lower limit 0. integral[x=0, x=pi] ln|sin x| dx = integral[t=0, t=pi] (1/2)ln|sin t| dt - pi ln(2)/2 Replace t by x since they are dummy variables and then subtract (1/2) integral[x=0, x=pi] ln|sin x| dx from both sides and multiply both sides by 2; we end up with integral[x=0, x=pi] ln|sin x| dx = - pi ln(2). I hope this helps. - Doctor Who, The Math Forum http://mathforum.org/dr.math/ |
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