Proving $\int_0^\pi \frac{\log(1+x\cos (y))}{\cos y}dy=\pi \arcsin x$

HINT:

Using Feynman's Trick (differentiating under the integral), we have for $|x|<1$

$$\begin{align} \frac{d}{dx}\int_0^\pi \frac{\log(1+x\cos(y))}{\cos(y)}\,dy&=\int_0^\pi \frac{1}{1+x\cos(y)}\,dy\tag1 \end{align}$$

Use contour integration or apply the Weierstrass substitution to evaluate the integral on the right-hand side of $(1)$. Then, integrate the result.

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}