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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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Is $\alpha$ supposed to satisfy $|\alpha| \leq 1$? –  sos440 Sep 7 '12 at 15:19
    
@Ned Dabby: this question is very interesting (+1). –  Chris's sis Sep 7 '12 at 18:35
    
@Ned Dabby: where does this problem come from? –  Chris's sis Sep 7 '12 at 18:59
    
@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. :-) –  Ned Dabby Sep 11 '12 at 8:47
    
@Ned Dabby: no problem. I'm glad to see such questions around. –  Chris's sis Sep 11 '12 at 8:48

2 Answers 2

up vote 3 down vote accepted

$$I(\alpha) = \int_0^{\pi} \ln (1+ \alpha \cos(x)) dx$$ $$\dfrac{dI}{d \alpha} = \int_0^{\pi} \dfrac{\cos(x)}{1+\alpha \cos(x)} dx = \dfrac1{\alpha} \int_0^{\pi} \dfrac{\alpha \cos(x)}{1+ \alpha \cos(x)} dx = \dfrac1{\alpha} \left( \pi - \int_0^{\pi} \dfrac{dx}{1+\alpha \cos(x)}\right)$$ 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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@Thank you Marvis. Now, Iknow how to begin. I know that inyegral. –  Ned Dabby Sep 7 '12 at 15:34

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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