TITLE: Proving $\int_0^\pi \frac{\log(1+x\cos (y))}{\cos y}dy=\pi \arcsin x$
QUESTION [10 upvotes]: How would I go about proving that, $$\int_0^\pi \frac{\log(1+x\cos (y))}{\cos (y)}\,dy=\pi \arcsin (x)$$
I have tried to do this by computing the integral directly, but it appears to be too difficult. Maybe there is a better approach to this that I do not know of.

REPLY [8 votes]: Let $|x|<1$ and use the log power series (which is in this particular case uniformly convergent) to write
\begin{align} 
\int^\pi_0 \frac{\log(1+x\cos(y))} {\cos(y)}\, dy&=\int^\pi_0 \sum_{n=0}^\infty \frac{(-1)^{n}x^{n+1}\cos^{n}(y) } {n+1}\, dy\\
&= \sum_{n=0}^\infty \frac{(-1)^{n}x^{n+1} } {n+1}\int^\pi_0 \cos^{n}(y)\, dy\\
\end{align}
The integral with $\cos^{n}(y)$ is easily found using the binomial theorem and it is for $n$ even 
$$\int^\pi_0 \cos^{n}(y)\, dy = \frac{\pi} {2^{n}}\binom{n}{n/2}, $$
and zero for $n$ odd. Hence
\begin{align} \int^\pi_0 \frac{\log(1+x\cos(y))} {\cos(y)}\, dy=\pi\sum_{n=0}^\infty \frac{x^{2n+1} } {2n+1}\cdot  \frac{1} {2^{2n}}\binom{2n}{n}
\end{align}
We see the arcsine series in that. We conclude 
\begin{align} \int^\pi_0 \frac{\log(1+x\cos(y))} {\cos(y)}\, dy=\pi\arcsin(x) 
\end{align}