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Math Help - Integrate ln(a + b.cosx)

  1. #1
    Member kalyanram's Avatar
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    Integrate ln(a + b.cosx)

    Hi All,
    Can anyone prove this, please...
    \int_{0}^{\pi}ln(a+b.cosx)dx=\pi.ln(\frac{a+\sqrt{  {a}^{2}+{b}^{2}}}{2})

    ~Kalyan.
    Last edited by kalyanram; June 22nd 2008 at 11:28 AM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by kalyanram View Post
    Hi All,
    Can anyone prove this, please...
    \int_{0}^{\pi}ln(a+b.cosx)dx=\pi.ln(\frac{a+\sqrt{  {a}^{2}+{b}^{2}}}{2})

    ~Kalyan.
    Maybe this would work

    \ln(a+bx)=\ln(a)+\ln\left(1+\frac{b}{a}x\right)

    Now let \alpha=\frac{b}{a}

    So we have

    I(\alpha)=\int_0^{\pi}\ln(1+\alpha{x})dx

    Now differentiate under the integral sign.

    Does anyone second this?
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  3. #3
    Super Member PaulRS's Avatar
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    <br />
J\left( \theta  \right) = \int_0^\pi  {\ln \left( {1 + \theta  \cdot \cos \left( x \right)} \right)dx}  \Rightarrow J'\left( \theta  \right) = \int_0^\pi  {\frac{{\cos \left( x \right)}}<br />
{{1 + \theta  \cdot \cos \left( x \right)}}dx} <br />

    Now find J'\left( \theta  \right) and then integrate considering that : <br />
J\left( \theta  \right) = J\left( \theta  \right) - J\left( 0 \right) = \int_0^\theta  {J'\left( t \right)dt} <br />

    Note that the original function is defined if \left| \theta  \right| \leq{ 1}<br />
    Last edited by PaulRS; June 22nd 2008 at 03:43 PM. Reason: Typo
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  4. #4
    Member kalyanram's Avatar
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    Thanks a lot Paul and Mathstud..
    I guess I will take over from here.

    ~Kalyan.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    <br />
J\left( \theta \right) = \int_0^\pi {\ln \left( {1 + \theta \cdot \cos \left( x \right)} \right)dx} \Rightarrow J'\left( \theta \right) = \int_0^\pi {\frac{{\cos \left( x \right)}}<br />
{{1 + \theta \cdot \cos \left( x \right)}}dx} <br />

    Now find J'\left( \theta \right) and then integrate considering that : <br />
J\left( \theta \right) = J\left( \theta \right) - J\left( 0 \right) = \int_0^\theta {J{\color{red}'}\left( t \right)dt} <br />
\leftarrow{\color{red}\text{here in red}}


    Note that the original function is defined if \left| \theta \right| \leq{ 1}<br />
    Is there a little mistake here?
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