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FInd this integral

  1. Sep 23, 2014 #1


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    1. The problem statement, all variables and given/known data
    Evaluate [itex]\displaystyle \int_0^{\pi} \log (1+a\cos x) dx[/itex]

    2. Relevant equations

    3. The attempt at a solution
    Using Leibnitz's Rule,
    F'(a)=[itex]\displaystyle \int_0^{\pi} \dfrac{\cos x}{1+a \cos x} dx [/itex]

    Now, If I assume sinx=t, then the above integral changes to
    [itex]\displaystyle \int_0^{0} \dfrac{dt}{1+a \sqrt{1-t^2}} [/itex]

    Since both the limits are zero now, shouldn't the value of integral be 0! :confused:
  2. jcsd
  3. Sep 23, 2014 #2


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    No. For one thing, the use of the square root function hides the fact that cos(t) will change sign over the range. Split it into two integrals to be safe.
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