# Integrate ln(a + b.cosx)

• Jun 22nd 2008, 10:56 AM
kalyanram
Integrate ln(a + b.cosx)
Hi All,
$\int_{0}^{\pi}ln(a+b.cosx)dx=\pi.ln(\frac{a+\sqrt{ {a}^{2}+{b}^{2}}}{2})$

~Kalyan.
• Jun 22nd 2008, 12:13 PM
Mathstud28
Quote:

Originally Posted by kalyanram
Hi All,
$\int_{0}^{\pi}ln(a+b.cosx)dx=\pi.ln(\frac{a+\sqrt{ {a}^{2}+{b}^{2}}}{2})$

~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?
• Jun 22nd 2008, 12:28 PM
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'\left( t \right)dt}
$

Note that the original function is defined if $\left| \theta \right| \leq{ 1}
$
• Jun 22nd 2008, 01:17 PM
kalyanram
Thanks a lot Paul and Mathstud..
I guess I will take over from here.

~Kalyan.
• Jun 22nd 2008, 01:27 PM
Mathstud28
Quote:

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?