Integral $\int_{-1}^{1}\frac{\sqrt{(1-x^{2})}}{1+x^{2}}dx$ Consider $$\int_{-1}^{1}\frac{\sqrt{(1-x^{2})}}{1+x^{2}}dx$$ I have a problem with this integral; the method I know consists in calculating the complex integral of $$f(z) = \left( \frac{z-1}{z+1} \right)^{\frac{1}{2}} \frac{1+z}{1+z^{2}}$$ along the curve formed by the " shrinking dogbone contour" withe centres $\\{-1,1\\}$ and the circumference of radius $R \rightarrow \infty$. But I obtain $0$, impossible. What's wrong with this method ? In particular, what are the residues of $f(z) $ in $\\{-i,i\\} ? $

**Edited answer**

If you need to use the residue method, you should try the following change of variable : $x=\frac{1-t^2}{1+t^2}$ which will allow you to write $I$ as an integral of a rational function from $-\infty$ to $\infty$. You will be able to use the method you know by using a contour a circumference of radius $R\to\infty$.

_Note_ : In my previous answer, I doubted the pertinence of residues, a statement that has been denied by @RonGordon.