Quick question on complex integral

Question

For $f(z) = (1-z^2)^{\frac{1}{2}}$, how do I show that the integral of $f(z)$ from $0$ to $\pi$ is $O(R^{-2})$?

$$\int f(z) dz = \int \frac{1}{z(1-z^2)^{\frac{1}{2}}} dz $$

Answer 1

$$\left | \int_0^\pi f(Re^{i \theta}) d ( R e^{i \theta}) \right | = \left | \int_0^\pi \frac{ i d \theta}{ \sqrt{ 1- R^2 e^{2 i \theta}}} \right | \leq \int_0^ \pi \frac{ d \theta}{ \sqrt{1 + R^2} } = \frac{ \pi }{ \sqrt{ 1 + R^2}} \to 0$$

as $R \to \infty$. Notice that it's just $O(R^{-1})$