Results 1 to 5 of 5

Thread: Integrate ln(a + b.cosx)

  1. #1
    Member kalyanram's Avatar
    Joined
    Jun 2008
    From
    Uppsala, Sweden
    Posts
    149
    Thanks
    14

    Integrate ln(a + b.cosx)

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

    ~Kalyan.
    Last edited by kalyanram; Jun 22nd 2008 at 10:28 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by kalyanram View Post
    Hi All,
    Can anyone prove this, please...
    $\displaystyle \int_{0}^{\pi}ln(a+b.cosx)dx=\pi.ln(\frac{a+\sqrt{ {a}^{2}+{b}^{2}}}{2})$

    ~Kalyan.
    Maybe this would work

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

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

    So we have

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

    Now differentiate under the integral sign.

    Does anyone second this?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    $\displaystyle
    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)}}
    {{1 + \theta \cdot \cos \left( x \right)}}dx}
    $

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

    Note that the original function is defined if $\displaystyle \left| \theta \right| \leq{ 1}
    $
    Last edited by PaulRS; Jun 22nd 2008 at 02:43 PM. Reason: Typo
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member kalyanram's Avatar
    Joined
    Jun 2008
    From
    Uppsala, Sweden
    Posts
    149
    Thanks
    14
    Thanks a lot Paul and Mathstud..
    I guess I will take over from here.

    ~Kalyan.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by PaulRS View Post
    $\displaystyle
    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)}}
    {{1 + \theta \cdot \cos \left( x \right)}}dx}
    $

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


    Note that the original function is defined if $\displaystyle \left| \theta \right| \leq{ 1}
    $
    Is there a little mistake here?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. (cosx^2)/x=cosx true or false. Explain.
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Aug 23rd 2011, 05:40 PM
  2. sin(cosx)=0
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Mar 21st 2010, 05:53 PM
  3. Prove that (cscx - cotx)^2 = (1-cosx)/(1+cosx)
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Feb 10th 2009, 08:02 PM
  4. Verify that √((1-cosx)/(1+cosx)) = (1-cosx)/|sinx|
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Dec 10th 2008, 08:39 PM
  5. (x^2)cosx
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 4th 2006, 11:11 AM

Search Tags


/mathhelpforum @mathhelpforum