Integrate ln(a + b.cosx)

**kalyanram** · June 22nd 2008, 09:56 AM

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.

**Mathstud28** · June 22nd 2008, 11:13 AM

Originally Posted by kalyanram

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?

**PaulRS** · June 22nd 2008, 11:28 AM

$ 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 $J'\left( \theta \right)$ and then integrate considering that : $ 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 $\left| \theta \right| \leq{ 1} $

**kalyanram** · June 22nd 2008, 12:17 PM

Thanks a lot Paul and Mathstud..
I guess I will take over from here.

~Kalyan.

**Mathstud28** · June 22nd 2008, 12:27 PM

Originally Posted by PaulRS

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

Note that the original function is defined if $\left| \theta \right| \leq{ 1} $

Is there a little mistake here?

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