Is there any method other than Feynman’s Integration Technique to find $ \int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x?$ We are going to find the formula, by Feynman’s Integration Technique, for
$$\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x,$$
where $a+c$ $\textrm{ and }$ $b+c$ are positive real numbers.
First of all, let’s ‘kill’ the term $\sin x$ and the square of $\cos x$ by identity and double angle formula.
$\displaystyle \begin{aligned}\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x =& \int_{0}^{\frac{\pi}{2}} \ln \left[(a-b) \cos ^{2} x+(b+c)\right] d x \\=& \int_{0}^{\frac{\pi}{2}} \ln \left[(a-b)\left(\frac{1+\cos 2 x}{2}\right)+(b+c)\right] d x \\=& \int_{0}^{\frac{\pi}{2}} \ln \left[\frac{a-b}{2} \cos 2 x+\left(b+c+\frac{a-b}{2}\right)\right] d x \\\stackrel{2x\mapsto x}{=}  & \frac{1}{2} \int_{0}^{\pi} \ln \left[\frac{a-b}{2} \cos x+\left(\frac{a+b+2 c}{2}\right)\right] d x\end{aligned}\tag*{} $
By my post, I found, by Feynman’s Integration Technique, that
$\displaystyle \int_{0}^{\pi} \ln (b \cos x+c)=\pi \ln \left(\frac{c+\sqrt{c^{2}-b^{2}}}{2}\right),\tag*{} $
where $\left|\frac{c}{b}\right|\geq 1$.
$\displaystyle I=\frac{\pi}{2} \ln \left[\frac{\frac{a+b+2 c}{2}+\sqrt{\left(\frac{a+b+2 c}{2}\right)^{2}-\left(\frac{a-b}{2}\right)^{2}}}{2}\right]\tag*{} $
Simplifying gives the result
$\displaystyle \begin{aligned}I&=\frac{\pi}{2} \ln \left[\frac{a+b+2 c+2 \sqrt{(a+c)(b+c)}]}{4}\right] \\&=\frac{\pi}{2} \ln \left(\frac{\sqrt{a+c}+\sqrt{b+c}}{2}\right)^{2} \\&=\pi \ln \left(\frac{\sqrt{a+c}+\sqrt{b+c}}{2}\right)\end{aligned}\tag*{} $

Use the short-hands $p^2=a+c$,  $q^2=b+c$ and $r=\frac{p-q}{p+q}$ to rewrite the integral as
\begin{align}
I=&\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x\\
=& \int_{0}^{\frac{\pi}{2}} \ln \left(p^2 \cos ^{2} x+q^2 \sin ^{2} x\right) d x\\
= & \int_{0}^{\frac{\pi}{2}}\bigg(\ln\frac{(p+q)^2}4+ \ln \left(1+2 r\cos 2 x+r^2 \right) \bigg)\ d x\\
\end{align}
and note that
\begin{align} 
&\int_0^{\frac\pi2}\ln( 1 + 2r\cos 2x + r^2) dx
 = 2\sum_{k=1}^\infty \frac{(-r)^{k+1}}k  \int_0^{\frac\pi2}\cos 2kx\ dx =0
\end{align}
Thus
\begin{align}
I=\int_{0}^{\frac{\pi}{2}}\ln\frac{(p+q)^2}4\ d x=\pi\ln \frac{p+q}2\
\end{align}
