Expert: Paul Klarreich - 1/7/2010

Question
Hello,
I know this might not be a first-year calculus problem, but I'm so desperate that I'll ask anyway...
I can't figure out how to compute the following integral:

2*pi
{
| log((1+2*cos(x))^2) dx
}  
0

Mathematica says it's equal to 0, but I have no clue how to show it. I tried substitution (u=1+2*cos(x) and u=2*cos(x)) and various trigonometric identities (including cos^2(x)=cos(2*x)-2*sin^2(x)), but it all didn't seem to work...
Any help or ideas will be highly appreciated!
Thanks,
Philipp

Answer
Questioner:    Philipp
Private: No
Subject:  
Question:
Hello,
I know this might not be a first-year calculus problem, but I'm so desperate that I'll ask anyway...
I can't figure out how to compute the following integral:

2*pi
{
| log((1+2*cos(x))^2) dx
}  
0

Mathematica says it's equal to 0, but I have no clue how to show it. I tried substitution (u=1+2*cos(x) and u=2*cos(x)) and various trigonometric identities (including cos^2(x)=cos(2*x)-2*sin^2(x)), but it all didn't seem to work...
Any help or ideas will be highly appreciated!
Thanks,
Philipp
..............................
I am afraid there is no nice way to integrate this by your normal methods.  [Suggestion: try this site:

integrals.wolfram.com/index.jsp

which can integrate practically anything.

.......................................
But for your definite integral, it may be possible to translate it into something like:


{  pi
|     Some variation of [log((1+2*cos(x))^2)] dx
} -pi

Now if the 'Some variation of [log((1+2*cos(x))^2)]' is an odd function of x, this integral would have to be zero.

Try that.  Let me know how it goes and maybe I can help.

Unfortunately, this is an improper integral and that makes life difficult.  Do you see why?