Does this integral have a closed form? I was working with this problem in an exam:
Given $\lambda\in(-1,1)\subset\Bbb R$, find 
$$f
   (\lambda)=\int_{0}^{\pi}\ln\left(1+\lambda \cos x\right)\mathrm{d}x
$$
My try: put $\delta\in (0,1)$ such that $\lambda\in(-\delta,\delta)$. Using the power series of $\ln(1+x)$ and uniform convergence to take $\sum_{n=0}^\infty$ outside $\int_{0}^{\pi}$, finally, I got this:
$$f(\lambda)=-\sum_{k=1}^\infty{(2k-1)!!\over (2k)!!}{\lambda^{2k-1}\over 2k}$$
But I still wonder if there is a closed form for $f(\lambda)?$ And is my solution above right or wrong?

The answer to your integral is 
$$
    f(\lambda) = \pi \ln\left(\frac{1+\sqrt{1-\lambda^2}}{2}\right)
$$
Now to derivation. Given that 
$$
  \int_0^\pi \cos^n(x) \mathrm{d}x = \frac{1+(-1)^n}{2} \frac{\pi}{2^n} \binom{n}{n/2}
$$
Doing the series expansion of the integrand in $\lambda$ and interchanging the order of summation and integration yields:
$$
   f(\lambda) = -\pi \sum_{m=1}^\infty \left(\frac{\lambda}{2}\right)^{2m} \frac{1}{2 m} \binom{2m}{m} = - \pi \int_0^{\lambda} \sum_{m=1}^\infty \frac{x^{2m-1}}{2^m} \binom{2m}{m} \mathrm{d}x
$$
The sum equals 
$$
   \sum_{m=1}^\infty \frac{x^{2m-1}}{2^m} \binom{2m}{m} = \frac{1}{x} \left(\sum_{m=0}^\infty \frac{x^{2m}}{2^m} \binom{2m}{m} -1 \right) = \frac{1}{x} \left(\frac{1}{\sqrt{1-2x^2}} -1 \right) 
$$
Integrating this gives the answer.
