Integrate the semicircle radical in two ways

Take $-\pi/2<\theta<\pi/2$, so $dx=\cos\theta\,d\theta$ and $\sqrt{1-\sin^2\theta}=\cos\theta$. Then the integral is $\int\cos^2\theta\,d\theta=\frac12\int(1+\cos2\theta)\,d\theta=\frac12\theta+\frac14\sin2\theta+C=\frac12\theta+\frac12\sin\theta\cos\theta+C$. Thus $F_s(x)=\frac12x\sqrt{1-x^2}+\frac12\sin^{-1}x+C$ on $-1<x<1$.

The source also gives $F_c(x)=\frac12x\sqrt{1-x^2}-\frac12\cos^{-1}x+C$. Take $x=\cos\theta$, $0<\theta<\pi$, so $dx=-\sin\theta\,d\theta$ and the integral becomes $-\int\sin^2\theta\,d\theta=-\theta/2+\sin2\theta/4+C$. Using $\sin^{-1}x+\cos^{-1}x=\pi/2$, the displayed nonconstant forms differ by $F_c-F_s=-\pi/4$. This is absorbed by the arbitrary constant.