 Solution

A branch point is a point around which analytic continuation changes a value; a branch cut removes curves so continuation becomes single-valued, and a branch is one such single-valued analytic choice. Defineon $\mathbb{C}\setminus ((-\infty ,-1]\cup [1,\infty ))$, choosing the square root equal to $1$ at zero. Monodromy around $\pm 1$ explains multivaluedness.

On the upper lip $0+$ of the cut convention, the branch remains positive on $(0,1)$, so ${\int }_{0+}^{1}dt/\sqrt{1-t^2}=\pi /2$. Define $\mathrm{Arccos}z=\pi /2-\mathrm{Arcsin}z$ on the same slit domain.

AlsoChanging logarithm branches changes the value by multiples of $\pi$ (already visible at $z=1$). Principal logarithms give a single branch on the plane cut from $\pm i$ outward. Differentiating both sides and checking the value at zero proveswhere the compatible principal branches are defined.