Zeroes of a non elementary integral

Consider the integral as a function of $x$ and then differentiate with respect to $x$. This yields $$f'(x)=\int_0^{2\pi} \frac{1}{x+\sin t}\,dt$$ which can be easily solved by a tangent half angle substitution to yield $$f'(x)=\frac{2\pi}{\sqrt{x^2-1}}.$$ Integrating with respect to $x$ yields $f(x)=2\pi\cosh^{-1}(x)+C.$

To solve for $C,$ consider $f(1).$ The integral is $$\int_0^{2\pi}\ln (1+\sin x)\,dx=\int_0^{2\pi}\ln (1+\cos x)\,dx=2\pi\ln(2)+\int_0^{2\pi}\ln (\cos^2 (x/2))\,dx\\=2\pi\ln(2)+\int_0^{2\pi}\ln (\sin^2 (x/2))\,dx=2\pi\ln(2)+4\int_0^{\pi}\ln (\sin (x))\,dx,$$ which upon combining with the famous result $\int_0^{\pi}\ln(\sin(x))=-\pi\ln(2),$ and $f(1)=C,$ yields $C=-2\pi\ln2$. And lo and behold, $\ln 2= \cosh^{-1}(5/4)!!$ This immediately yields $x=\frac54.$