evaluate $\displaystyle \int_0^\pi\ln(1+\alpha\cos x)\,dx$
$\displaystyle \int_{0}^{\pi }{\ln (1+\alpha \cos x)\,dx}=\int_{0}^{\pi }{\int_{0}^{\alpha }{\frac{\cos x}{1+y\cos x}\,dy}\,dx},$ that leads to $\displaystyle \pi \int_{0}^{\alpha }{\frac{\sqrt{1-y^{2}}-1}{y\sqrt{1-y^{2}}}\,dy},$ which is easily killed by makin' the substitution $\displaystyle u=\ln \left( 1+\sqrt{1-y^{2}} \right).$ Finally, your "nice" integral equals $\displaystyle \pi \cdot \ln \frac{1+\sqrt{1-\alpha ^{2}}}{2},$ which holds for $\displaystyle |\alpha|\le1$ and we're done.
Exactly the same method as Krizalid's just a slightly different start...it is how I solved this integral the first time I saw it... sometimes if its not apparent to you just remember that if you ever see an integral with a second variable expression in it try letting it equal to its own function.
$\displaystyle J\left(\alpha\right)=\int_0^{\pi}\ln\left(1+\alpha \cos(x)\right)dx$
So
$\displaystyle \begin{aligned}J'\left(\alpha\right)&=\frac{d}{d\a lpha}\int_0^{\pi}\ln\left(1+\alpha\cos(x)\right)dx \\
&=\int_0^{\pi}\frac{\partial}{\partial\alpha}\ln\l eft(1+\alpha\cos(x)\right)dx\\
&=\int_0^{\pi}\frac{\cos(x)}{\alpha\cos(x)+1}dx
\end{aligned}$
So the same method as Krizalid's just a different way of seeing it.