Verification the integral identity $\int_0^\pi \ln(1+\alpha\cos(x))dx=\pi\ln\left(\frac{1+\sqrt{1-\alpha^2}}{2}\right)$.

$$I(\alpha) = \int_0^{\pi} \ln (1+ \alpha \cos(x)) dx$$ $$\dfrac{dI}{d \alpha} = \int_0^{\pi} \dfrac{\cos(x)}{1+\alpha \cos(x)} dx = \dfrac1{\alpha} \int_0^{\pi} \dfrac{\alpha \cos(x)}{1+ \alpha \cos(x)} dx = \dfrac1{\alpha} \left( \pi - \int_0^{\pi} \dfrac{dx}{1+\alpha \cos(x)}\right)$$ And integral $$\int_0^{\pi} \dfrac{dx}{1+\alpha \cos(x)}$$ can be evaluated using the standard complex analytic technique by using the transformation $z = \exp(ix)$.

Hint: Differentiate the left-hand sign with respect to $\alpha$. The details are probably in your book, but if not, there is a useful Wikipedia article. You will get something you can integrate explicitly with respect to $x$.

Compare with the derivative of the right-hand side with respect to $\alpha$. Finally, observe that the left-hand side and the right-hand side agree at $\alpha=0$.