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\\&=-\pi\int \frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} d \theta\\&=2\pi \ln \left(\cos \frac{\theta}{2}\right)+C\end{aligned} Hence
$$ \boxed{\int_{0}^{\pi} \ln (a \cos x+1) d x =2\pi \ln \left[\cos \left(\frac{\sin ^{-1} a}{2}\right)\right]} $$
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 $$\boxed{\int_{0}^{\pi} \ln (b \cos x+c) =\pi\left\{\ln c+2 \ln \left[\cos \left(\frac{\sin ^{-1}\left(\frac{b}{c}\right)}{2}\right)\right]\right\}} $$
For example,
$$ \int_{0}^{\pi} \ln (\cos x+1)=2 \pi \ln \left(\frac{1}{\sqrt{2}}\right)=-\pi \ln 2; $$
$$ \int_{0}^{\pi} \ln (\sqrt{3} \cos x+2) d x=\pi\ln \frac{3}{2} $$
Is there any method other than Feynman’s integration technique?
