Evaluation of $ \int_{0}^{\pi}\ln(5-4\cos x)\,dx$ 
Evaluation of $\displaystyle \int_{0}^{\pi}\ln(5-4\cos x)dx = \int_{0}^{\pi}\ln(5+4\cos x)dx$

$\bf{My\; Try::}$ Let $\displaystyle I(a,b) = \int_{0}^{\pi}\ln(a+b\cos x)dx$
Then $$\frac{d}{db}(a,b) = \frac{d}{db}\left[\int_{0}^{\pi}\ln(a+b\cos x)dx\right]db$$
So $$I'(a,b) = \int_{0}^{\pi}\frac{\cos x}{a+b\cos x}dx = \frac{1}{b}\int_{0}^{\pi}\frac{(a+b\cos x)-a}{a+b\cos x}dx$$
So we get $$I'(a,b) = \frac{\pi}{b}-\frac{a}{b}\int_{0}^{\pi}\frac{1}{a+\cos x}dx$$
Using half angle formula $\displaystyle \tan x = \frac{1-\tan^2 x/2}{1+\tan^2 x/2}$
so we get $$I'(a,b) = \frac{\pi}{b}-\frac{a}{b}\int_{0}^{\pi}\frac{\sec^2 x/2}{(a+b)+(a-b)\tan^2 x/2}dx$$
Now Put $\tan x/2= t\; $ Then $\displaystyle \sec^2 \frac{x}{2}dx = 2dt$
So we get $$I'(a,b) = \frac{\pi}{b}-\int_{0}^{\infty}\frac{1}{(a+b)+(a-b)t^2}=\frac{\pi}{b}-\frac{a\pi}{2b\sqrt{a^2-b^2}}$$
So we get $$I(a,b) = \pi\ln|b|-\frac{\pi a}{2}\left[-\frac{\ln|b^2-a^2|+2\ln |b|}{2a^2}\right]$$
So we put $a = 5$ and $b=4$
We get $$I(5,4) = \pi\cdot \ln (5)-\frac{\pi}{2}\left[\frac{-\ln(9)+2\ln(5)}{2\cdot 5}\right]$$
So We get $$I(5,4) = \left[\frac{18\ln(5)+\ln(9)}{20}\right]\cdot \pi$$
I did not understand where i have done mistake, Help me
Thanks
 A: This approach seems really good, but the first mistake was in
$$\int_0^{\pi}\frac1{a+b\cos x}dx=\frac1a\int_0^{\pi}\frac1{1+e\cos x}dx$$
Where $e=\frac ba$. If you used the eccentric anomaly at this point,
$$\sin E=\frac{\sqrt{1-e^2}\sin x}{1+e\cos x}$$
So that $$dE=\frac{\sqrt{1-e^2}}{1+e\cos x}dx$$
You would get
$$\int_0^{\pi}\frac1{a+b\cos x}dx=\frac1a\int_0^{\pi}\frac{dE}{\sqrt{1-e^2}}=\frac{\pi}{\sqrt{a^2-b^2}}$$
So you were off by a factor of $2$ at this point. Next, if we let $b=a\sin\theta$, then
$$\begin{align}\int\frac{db}{b\sqrt{a^2-b^2}}&=\frac1a\int\csc\theta\,d\theta=-\frac1a\ln\left|\csc\theta+\cot\theta\right|+C_1(a)\\
&=-\frac1a\ln\left(\frac ab+\frac{\sqrt{a^2-b^2}}b\right)+C_1(a)\\
&=-\frac1a\ln\left(a+\sqrt{a^2-b^2}\right)+\frac1a\ln b+C_1(a)\end{align}$$
So you would arrive at
$$I(a,b)=\pi\ln\left(a+\sqrt{a^2-b^2}\right)+C(a)$$
At $b=0$ we get $\pi\ln a=\pi\ln2a+C(a)$ so $C(a)=-\pi\ln2$. Then
$$I(5,4)=\pi\ln8-\pi\ln2=\pi\ln4$$
Numerical quadrature confirms this result.
A: Here is an alternative, perhaps simpler, procedure
\begin{align}
I=\int_0^\pi \ln(5-4\cos x) \overset{x\to 2x}{dx}
=2\pi\ln2+2\int_0^{\pi/2} \ln\left(\frac54-\cos 2x\right)dx
\end{align}
where, with $\cos 2x =\frac{1-\tan^2x}{1+\tan^2x}$
\begin{align}
J=&\int_0^{\pi/2} \ln\left(\frac54-\cos 2x\right)dx
 =\int_0^{\pi/2}\ln \frac{1+9\tan^2x}{4(1+\tan^2x)}dx\\
 =& \int_0^{\pi/2} \int_0^{1/2}\frac{2(1+t)\tan^2x-2(1-t)}{(1+t)^2\tan^2x +(1-t)^2}dt\ dx\>\>\>\>\> r=\frac{1+t}{1-t}\\
 =& \int_0^{1/2} \frac2{1-t}\int_0^{\pi/2}\frac{r\tan^2x-1}{r^2\tan^2x +1}dx\ dt\>\>\>\>\> r\tan x = \cot y\\
 =& \int_0^{1/2} \frac2{1-t}\int_0^{\pi/2}\frac{1-r\tan^2y}{r^2\tan^2y +1}dy\ dt=-J=0
\end{align}
Thus
$$I= 2\pi\ln2$$
A: I get, assuming there's a $\mathrm{d}t$,  $$
  \int_0^\infty \frac{1}{(a+b)(a-b)t^2} \,\mathrm{d}t = \frac{\pi}{2 \sqrt{a^2-b^2}}  \text{.}
$$
Integrating that with respect to $b$, I get $\displaystyle \frac{-\pi}{2}\arctan \frac{b\sqrt{a^2 - b^2}}{b^2 - a^2} $, which is not equivalent to your expression containing logarithms.
I haven't checked for further errors.
Edit:  @user5713492 has (perhaps unintentionally) pointed out another error.  Your second $\int_0^\infty$ has no $\frac{a}{b}$ prefactor.  Assuming this was intended, I get $$
  \frac{\partial I(a,b)}{\partial b} = \frac{\pi}{b} - \frac{a}{b} \int_0^\infty \frac{1}{(a+b)+(a-b)t^2} \,\mathrm{d}t = \frac{\pi}{b} - \frac{a \pi}{2b\sqrt{a^2 - b^2}}  \text{,}
$$ which is what you get, so you probably dropped the prefactor.
Integrating that with respect to $b$, I get  $$
  \frac{\pi}{2}\ln b + \ln(a + \sqrt{a^2 - b^2}) + C(a)  \text{,}
$$  where $C$ is an arbitrary function of $a$.  This is not particularly close to your expression with logs and I see no path to equivalence, even ignoring the arbitrary function of integration.
