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I have to verify that

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

with $|\alpha|<1$. It is my homework and don't know where to begin.

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  • $\begingroup$ Is $\alpha$ supposed to satisfy $|\alpha| \leq 1$? $\endgroup$ – Sangchul Lee Sep 7 '12 at 15:19
  • $\begingroup$ @Ned Dabby: this question is very interesting (+1). $\endgroup$ – user 1357113 Sep 7 '12 at 18:35
  • $\begingroup$ @Ned Dabby: where does this problem come from? $\endgroup$ – user 1357113 Sep 7 '12 at 18:59
  • $\begingroup$ @Chris'ssister: Oh sorry for the delay. This was one of the problem I had to solve. It comes from ADVANCED CALCULUS by Schaum. Thank you for +1. :-) $\endgroup$ – Ned Dabby Sep 11 '12 at 8:47
  • $\begingroup$ @Ned Dabby: no problem. I'm glad to see such questions around. $\endgroup$ – user 1357113 Sep 11 '12 at 8:48
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$$I(\alpha) = \int_0^{\pi} \ln (1+ \alpha \cos(x)) dx$$ $$\dfrac{dI}{d \alpha} = \int_0^{\pi} \dfrac{\cos(x)}{1+\alpha \cos(x)} dx = \dfrac1{\alpha} \int_0^{\pi} \dfrac{\alpha \cos(x)}{1+ \alpha \cos(x)} dx = \dfrac1{\alpha} \left( \pi - \int_0^{\pi} \dfrac{dx}{1+\alpha \cos(x)}\right)$$ And integral $$\int_0^{\pi} \dfrac{dx}{1+\alpha \cos(x)}$$ can be evaluated using the standard complex analytic technique by using the transformation $z = \exp(ix)$.

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  • $\begingroup$ @Thank you Marvis. Now, Iknow how to begin. I know that inyegral. $\endgroup$ – Ned Dabby Sep 7 '12 at 15:34
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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 $\alpha=0$.

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