# Thread: Integrate ln(a + b.cosx)

1. ## Integrate ln(a + b.cosx)

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

~Kalyan.

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

~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?

3. $\displaystyle J\left( \theta \right) = \int_0^\pi {\ln \left( {1 + \theta \cdot \cos \left( x \right)} \right)dx} \Rightarrow J'\left( \theta \right) = \int_0^\pi {\frac{{\cos \left( x \right)}} {{1 + \theta \cdot \cos \left( x \right)}}dx}$

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 $\displaystyle \left| \theta \right| \leq{ 1}$

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

~Kalyan.

5. Originally Posted by PaulRS
$\displaystyle J\left( \theta \right) = \int_0^\pi {\ln \left( {1 + \theta \cdot \cos \left( x \right)} \right)dx} \Rightarrow J'\left( \theta \right) = \int_0^\pi {\frac{{\cos \left( x \right)}} {{1 + \theta \cdot \cos \left( x \right)}}dx}$

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?