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*{} $

This isn't really an answer as such, but as a general comment, the next trick you should know is how to "peel off" the logarithm. Let's guess that for some functions $g$ and $h$:
$$\int \log f(x)\,dx = h(x) \log f(x) + g(x)$$
Taking the derivative of both sides:
$$\log f(x) = h'(x) \log f(x) + h(x) \frac{f'(x)}{f(x)} + g'(x)$$
This is clearly going to have a solution if $h'(x) = 1$, that is, $h(x) = x$. So:
$$\int \log f(x)\,dx = x \log f(x) + g(x)$$
where:
$$g(x) = - \int x \frac{f'(x)}{f(x)}\,dx$$
In your case:
$$\int \log \left(b \cos x + c\right)\,dx = x \log \left(b \cos x + c\right) + g(x)$$
where:
$$g(x) = \int \frac{x \sin x}{\cos x + c/b}\,dx$$
This works for any elementary field extension, not just the logarithm function. You can think of this as integration by parts, but I find the Risch algorithm interpretation easier to apply.
