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Help me with this logarithmic integral

  1. Oct 18, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    [itex] \displaystyle \int_0^{2 \pi} x \ln \dfrac{3+ \cos x}{3- \cos x} dx [/itex]


    2. Relevant equations

    3. The attempt at a solution

    Using property of definite integral
    2I = [itex]\displaystyle \int_0^{2 \pi} 2 \pi \ln \dfrac{3+ \cos x}{3- \cos x} dx [/itex]
     
  2. jcsd
  3. Oct 18, 2013 #2

    Dick

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    I'm not sure what "property of definite integral" would give you that. You'll have to spell it out. Here's a hint. Change the variable to u=2pi-x and see what happens.
     
  4. Oct 18, 2013 #3

    SteamKing

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    You can also rewrite the log of a quotient into something simpler.
     
  5. Oct 19, 2013 #4

    utkarshakash

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    [itex]\int_0^a f(x) = \int_0^a f(a-x)[/itex]. This is what I've used.

    Then I added both integrals to get rid of x outside log.
     
  6. Oct 19, 2013 #5

    utkarshakash

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    Can you please elaborate? I didn't get you.
     
  7. Oct 19, 2013 #6
    Integrating from one point to another finds the area under the curve between those two points. Think about the graphs of f(x) and f(a-x). f(a-x) is going to be flipped over the x-axis and shifted to the left by "a" units.. Most of the time, the value of the integral wouldn't be the same for both functions. Sorry to tell ya, but I think the property of definite integral may only apply to periodic functions. I don't think this function is periodic

    [itex]log(\frac{a}{b})=log(a)-log(b)[/itex]
    [itex]log(a*b)=log(a)+log(b)[/itex]
     
  8. Oct 19, 2013 #7
    Okay. Now observe that
    $$\int_0^{2 \pi} \ln \dfrac{3+ \cos x}{3- \cos x} dx=2\int_0^{\pi} \ln \dfrac{3+ \cos x}{3- \cos x} dx$$

    Use the same property you used before.
     
  9. Oct 19, 2013 #8

    utkarshakash

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    Well done.
     
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