<s>[INST] 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
 [/INST] 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
$$I= 2\pi\ln2$$
</s>