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Formula 4.224 (9) from Gradshteyn and Ryzhik (7ed, 2007) states that $$\int_0^\pi \ln (a + b\,\cos x)\,dx = π\,\ln \left(\frac{a +\sqrt{a^2-b^2}}{2}\right), \quad [a ≥ |b| > 0]$$ Could someone help me to understand how they got to the right-hand side expression? Thanks!!

Ps: I've found this similar topic that may be helpful: A question in Complex Analysis $\int_0^{2\pi}\log(1-2r\cos x +r^2)\,dx$

MJ. Karling
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1 Answers1

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Hint: If $f(a) = \displaystyle\int_0^\pi \ln(a + b \cos(x)) d x$, then \begin{equation} f'(a) = \int_0^\pi \frac{1}{a + b \cos x} dx = \frac{\pi}{\sqrt{a^2-b^2}} \end{equation} according to Integrating $\dfrac{1}{a+b\cos(x)}$ using $e^{ix}$

Gribouillis
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