How to find $\int_{0}^{\pi} \ln (b \cos x+c)$ without using Feynman’s integration technique? I shall find the integral by Feynman’s Technique Integration on a particular integral
$\displaystyle I(a)=\int_{0}^{\pi} \ln (a \cos x+1) d x,\tag*{} $
where $-1\leq a \leq 1.$
$\displaystyle \begin{aligned}I^{\prime}(a) &=\int_{0}^{\pi} \frac{\cos x}{a \cos x+1} d x, \\&=\frac{1}{a} \int_{0}^{\pi} \frac{(a \cos x+1)-1}{a \cos x+1} d x \\&=\frac{\pi}{a}-\frac{1}{a} \int_{0}^{\pi} \frac{d x}{a \cos x+1} \\&\stackrel{t=\tan \frac{x}{2}}{=} \frac{\pi}{a}-\frac{1}{a} \int_{0}^{\infty} \frac{1}{1+\frac{a\left(1-t^{2}\right)}{1+t^{2}}} \cdot \frac{2 d t}{1+t^{2}} \\&=\frac{\pi}{a}-\frac{2}{a} \int_{0}^{\infty} \frac{d t}{(1-a) t^{2}+(1+a)} \\&=\frac{\pi}{a}-\frac{2}{a \sqrt{1-a^{2}}} \tan^{-1}\left[\frac{\sqrt{1-a} t}{\sqrt{1+a}}\right]_{0}^{\infty} \\&=\frac{\pi}{a}-\frac{\pi}{a \sqrt{1-a^{2}}}\end{aligned}\tag*{} $
Integrating both sides w.r.t. $a$ yields
\begin{aligned}\int I^{\prime}(a) d a &=\pi\int\left(\frac{1}{a}-\frac{1}{a \sqrt{1-a^{2}}}\right) da \\& \stackrel{a=\sin \theta}{=} \pi\int\left(\frac{1}{\sin \theta}-\frac{1}{\sin \theta \cos \theta}\right) \cos \theta d \theta \\&=\pi\int \frac{\cos \theta-1}{\sin \theta} d \theta\\&I(a) =\pi \int \frac{-\sin ^{2} \theta}{\sin \theta(\cos \theta+1)} d \theta\\&=\pi \ln (1+\cos \theta) +C\end{aligned}
Putting $a=0$ gives $C=-\pi\ln 2$ and hence
$$
\boxed{\int_{0}^{\pi} \ln (a \cos x+1) d x =\pi \ln \left[1+\cos \left(\sin ^{-1} a\right)\right]= \pi \ln \left(\frac{1+\sqrt{1-a^{2}}}{2}  \right)}
$$
I now want to generalize it to
$$I(b,c)=\displaystyle \int_{0}^{\pi} \ln (b \cos x+c),\tag*{} $$
where $c\neq 0$ and $-1\leq \frac{b}{c} \leq 1.$
$$
\begin{aligned}
I(b,c)&=\int_{0}^{\pi} \ln (b \cos x+c) \\
&=\int_{0}^{\pi} \ln \left[c\left(\frac{b \cos x}{c}+1\right)\right] \\
&=\pi \ln c+\int_{0}^{\pi} \ln \left(\frac{b}{c} \cos x+1\right) d x \\
&=\pi \ln c+I\left(\frac{b}{c}\right)
\end{aligned}
$$
Putting $a=\frac{b}{c}$ yields
$$\boxed{\int_{0}^{\pi} \ln (b \cos x+c) = \pi\left[\ln c+\ln \left(1+\sqrt{1-\frac{b^{2}}{c^{2}}}\right)\right] = \pi \ln \left(\frac{c+\sqrt{c^{2}-b^{2}}}{2}\right)}
$$
For example,
$$
\int_{0}^{\pi} \ln (\cos x+1)=\pi \ln \left(\frac{1}{2}\right)=-\pi \ln 2;
$$
$$
\int_{0}^{\pi} \ln (\sqrt{3} \cos x+2) d x=\pi\ln \frac{3}{2} 
$$
Is there any  method other than Feynman’s integration technique?
 A: Concerning the antiderivative
$$I=\int \log (b \cos (x)+c)\,dx=2\int \frac 1{t^2+1}\Big[\log[ (c-b)t^2+(b+c)]-\log(t^2+1)\Big]\,dt$$ But
$$\log[ (c-b)t^2+(b+c)]=\log(c-b)+\log \left(t-\sqrt{\frac{b+c}{b-c}}\right)+\log \left(t+\sqrt{\frac{b+c}{b-c}}\right)$$
Now, using
$$\frac 1{t^2+1}=\frac 1{(t+i)(t-i)}=\frac i2  \left(\frac{1}{t+i}-\frac{1}{t-i}\right)$$ we face four integrals
$$J(\alpha,\beta)=\int \frac {\log(t+\alpha)}{t+\beta}\,dt=\text{Li}_2\left(\frac{t+\alpha }{\alpha -\beta }\right)+\log (t+\alpha ) \log
   \left(\frac{\beta +t}{\beta -\alpha }\right)$$
A: Note that, with $r= \frac cb -\sqrt{\frac{c^2}{b^2}-1}$
$$\ln (b \cos x+c)=\ln\frac{c+\sqrt{c^2-b^2}}2+\ln\left(1+2r\cos x+r^2\right)
$$
Thus, per $\int_0^\pi \ln\left(1+2r\cos x+r^2\right)dx=0$
$$\int_{0}^{\pi} \ln (b \cos x+c)dx
=\pi \ln\frac{c+\sqrt{c^2-b^2}}2
$$
A: Another option as opposed to using an infinite sequence and piecewise integration is to use a Weierstrass substitution and then a double integral.
Take $x\mapsto\tan\left(\tfrac x2\right)$ such that $\cos x\mapsto(1-x^2)/(1+x^2)$. The integral then becomes
$$\begin{align*}I & \equiv\int\limits_0^\pi\mathrm dx\,\log(1+a\cos x)\\ & =2\int\limits_0^{+\infty}\frac {\mathrm dx}{1+x^2}\log\left(1+a\frac {1-x^2}{1+x^2}\right)\\ & =2\int\limits_0^{+\infty}\frac {\mathrm dx}{1+x^2}\left[\log(1+a)+\log\left(1+\frac {1-a}{1+a}x^2\right)-\log(1+x^2)\right]\end{align*}$$
The first integral is trivial
$$\log(1+a)\int\limits_0^{+\infty}\frac {\mathrm dx}{1+x^2}=\frac \pi2\log(1+a)$$
The last two integrals can be evaluated using the general case
$$J\equiv\int\limits_0^{+\infty}\mathrm dx\,\frac {\log(1+z^2 x^2)}{1+x^2}$$
With the middle integral being the special case $z^2=(1-a)/(1+a)$ and the last integral being the case $z=1$. To evaluate $J$, we make use of a double integral, namely
$$\log(1+z^2)=\int\limits_0^1\mathrm dy\,\frac {2yz^2}{1+y^2z^2}$$
Substituting the identity above into $J$ and switching the order of integration gives
$$\begin{align*}J & =\int\limits_0^1\mathrm dy\,\int\limits_0^{+\infty}\mathrm dx\,\frac {2x^2yz^2}{(1+x^2)(1+x^2y^2z^2)}\\ & =\int\limits_0^1\mathrm dy\,\frac {2yz^2}{y^2z^2-1}\int\limits_0^{+\infty}\mathrm dx\,\left(\frac 1{1+x^2}-\frac 1{1+x^2y^2z^2}\right)\\ & =\pi z^2\int\limits_0^1\mathrm dy\,\frac y{y^2z^2-1}\left(1-\frac 1{yz}\right)\\ & =\pi z\int\limits_0^1\frac {\mathrm dy}{1+yz}\\ & =\pi\log(1+z)\end{align*}$$
Hence
$$\int\limits_0^{+\infty}\frac {\mathrm dx}{1+x^2}\log\left(1+\frac {1-a}{1+a}x^2\right)=\pi\log\left(1+\sqrt{\frac {1-a}{1+a}}\right)$$
And
$$\int\limits_0^{+\infty}\frac {\mathrm dx}{1+x^2}\log(1+x^2)=\pi\log 2$$
Putting everything together, then
$$\int\limits_0^\pi\mathrm dx\,\log(1+a\cos x)\color{blue}{=\pi\log\left(\frac {1+a}4\right)+2\pi\log\left(1+\sqrt{\frac {1-a}{1+a}}\right)}$$
Let $a=1$ and we get the integral stated in the question
$$\int\limits_0^\pi\mathrm dx\,\log(1+\cos x)=\pi\log\left(\frac 12\right)=-\pi\log 2$$

The more general case considering $\log(a+b\cos x)$ can be evaluated by transforming the integrand such that the constant term becomes one. Factoring out an $a$ gives
$$K\equiv\int\limits_0^\pi\mathrm dx\,\log(a+b\cos x)=\pi\log a+\int\limits_0^\pi\mathrm dx\,\log\left(1+\frac ba\cos x\right)$$
Using the formula we derived above and after some simplification, then
$$\int\limits_0^\pi\mathrm dx\,\log(a+b\cos x)\color{blue}{=\pi\log\left(\frac {a+b}4\right)+2\pi\log\left(1+\sqrt{\frac {a-b}{a+b}}\right)}$$
Substituting $a=2$ and $b=\sqrt3$ gives
$$\int\limits_0^\pi\mathrm dx\,\log(2+\sqrt3\cos x)=\pi\log\left(\frac {2+\sqrt3}4\right)+2\pi\log\left(1+\sqrt{\frac {2-\sqrt3}{2+\sqrt3}}\right)=\pi\log\left(\frac 32\right)$$
A: You can use Chebyshev polynomials of the second kind. The identity
$$ U_n(x) = 2^n \prod_{k=1}^{n}\left(x-\cos\frac{\pi k}{n+1}\right) $$
leads to
$$\log U_n(A) = n\log 2+\sum_{k=1}^{n}\log\left(A-\cos\frac{\pi k}{n+1}\right) $$
then to
$$ \int_{0}^{\pi}\log(A-\cos\theta)\,d\theta = -\pi\log 2+\lim_{n\to +\infty}\frac{\pi \log U_n(A)}{n+1} $$
for any $A>1$.
