I'm looking for the value of: $$ A \exp\left(\frac{-1}{2\pi} \int_{-\pi}^{\pi} \ln(1+A+2BC \cos x) dx \right)$$ I know we could take $y=1+A+2BC \cos x$ but changing variable in this way makes the integral from $1+ A - 2BC$ to $1+ A - 2BC$ which makes the integral zero! How to solve it ?

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    $\begingroup$ $\cos$ is an even function, so the integral is $2\int_0^\pi \dotsb$. $\endgroup$ – Daniel Fischer Jul 18 '13 at 8:13

Hint:$$y=A \exp\left(\frac{-1}{2\pi} \int_{-\pi}^{\pi} \ln(1+A+2BC \cos x) \ \mathrm dx \right)\to$$ $$\ln y=\ln A+\left(\frac{-1}{2\pi} \int_{-\pi}^{\pi} \ln(1+A+2BC \cos x \mathrm dx\right)=\ln A+\left(\frac{-1}{\pi} \int_{0}^{\pi} \ln(1+A+2BC \cos x \mathrm dx \right)$$ now take $1+A+2BC \cos x=z$ then $\ \mathrm dz=-2BC\ \sin x \ \mathrm dx\to \ \mathrm dx=\pm \dfrac{-\ \mathrm dz}{2BC\sqrt{1-z^2}}$ we have

$$\ln y=\ln A-\underbrace{\frac{1}{2\pi BC} \left(\int_{0}^{\frac{\pi}{2}} \frac{-\ln z}{\sqrt{1-z^2}} \ \mathrm dz+\int_{\frac{\pi}{2}}^{\pi} \frac{\ln z}{\sqrt{1-z^2}} \ \mathrm dz\right) }_H $$ then integrate by part ...$$\large{y '=Ae^{H}\cdot H '}$$


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