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### Finding a Definite Integral with Natural Log and Sine

```Date: 05/14/2008 at 05:52:27
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/
```
Associated Topics:
College Calculus
High School Calculus

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