\[ I\da \int_{-1}^1 \sqrt{1-x^2} \dx = {\pi \over 2} .\]
Solution: Dogbone and residue at $\infty$
Take a branch cut \([-1, 1]\) and \(\Gamma\) the standard dogbone contour:
Orient \(\Gamma\) positively about infinity, i.e. counterclockwise.
Contribution from \(\gamma_1 \da\ts{t+i\eps \st t\in [-1, 1]}\), the upper horizontal piece: \[ \int_{\gamma_1}f(z)\dz \to \int_{-1}^1 \sqrt{1-t^2} \dt = I .\]
Contribution from \(\gamma_2\da\ts{t-i\eps \st t\in [-1, 1]}\), the lower horizontal piece: note that following \(e^{2\pi i t}z\) to \(t=1\) sends \(\sqrt{z}\) to \(-\sqrt{z}\), so \[ \int_{\gamma_2}f(z)\dz \to \int_{1}^{-1} - \sqrt{1-t^2} \dt = I .\]
Contributions from the circles: use that \(f(z) \da \sqrt{z^2-1}\) is a continuous function and these arcs are compact, so they are uniformly bounded. Thus \(\int f\to 0\) by the ML estimate on these arcs.
The total contribution: \[ \qty{\int_{\gamma_1} + \int_{\gamma_2}} f = 2I .\]
The residue at infinity: \[ \Res_{z=\infty}f(z) &= \Res_{z=0} - {1\over z^2}\sqrt{1 - {1\over z^2}} \\ &= \Res_{z=0} - {1\over z^2}\sqrt{z^2-1 \over z^2} \\ &= \Res_{z=0} - {i\over z^3}\sqrt{1-z^2} \\ &= \Res_{z=0} - {i\over z^3}\sum_{k\geq 0} {1/2\choose k}z^{2k} \\ &= \Res_{z=0} - {i\over z^3} \qty{1 - {1\over 2}z^2 - \bigo(z^3) } \\ &= {i\over 2} ,\] thus \[ 2\pi i \cdot -{i\over 2} = 2I \implies I = {\pi \over 2} .\]