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}

Leibniz integral rule

This states that: $$\frac{d}{dx}\int_a^b f(x,y)dy=\int_a^b\frac{\partial}{\partial_x}[f(x,y)]dy$$ so we can apply this: $$I(x)=\int_0^\pi\frac{\ln(1+x\cos y)}{\cos y}dy$$ $$I'(x)=\int_0^\pi\frac{\cos y}{(1+x\cos y)\cos y}dy=\int_0^\pi\frac{1}{1+x\cos y}dy$$ $$I'(x)=2\int_0^\infty\frac{1}{1+x\frac{1-t^2}{1+t^2}}\frac{1}{1+t^2}dt=2\int_0^\infty\frac{1}{(1+t^2)+x(1-t^2)}dt$$$$=2\int_0^\infty\frac{1}{(1+x)+(1-x)t^2}dt$$$$=\frac{2}{1+x}\int_0^\infty\frac{1}{1+\frac{1-x}{1+x}t^2}dt$$ now by using $$\sqrt{\frac{1-x}{1+x}}t=\tan(u)$$ we can obtain: $$I'(x)=\frac{2}{\sqrt{(1+x)(1-x)}}\int_0^{\pi/2}du=\frac{\pi}{\sqrt{1-x^2}}$$ now finally we get: $$I(x)=\int\frac{\pi}{\sqrt{1-x^2}}dx=\pi\arcsin(x)+C$$ and we can prove by evaluating $I(0)$ that $C=0$