Compute $\int\limits_Q^1\sqrt{(1-x^2)(1-\frac{Q^2}{x^2})}\mathrm{d}x$ I am trying to compute the integral
$$\int\limits_Q^1\sqrt{(1-x^2)(1-\frac{Q^2}{x^2})}\mathrm{d}x$$
where $0\leq Q <1$ is a real number.
I tried to substitute $x=\cos y,$ but this didn't bring much.
My other idea was to use complex integration along the lines of
https://en.wikipedia.org/wiki/Methods_of_contour_integration#Example_.28VI.29_.E2.80.93_logarithms_and_the_residue_at_infinity.
However I am not completely sure how to proceed and to take my branch cuts.
Hint: Maple yields to solution
$$\frac{\pi}{4}Q^2-\frac{\pi}{2}Q+\frac{\pi}{4}.$$
 A: $$\mathrm{
I=\int_Q^1{\sqrt{(1-x^2)(x^2-Q^2)}\over x}\,dx
}$$
Use the substitution
$$\mathrm{
(1-x^2)=z^2(x^2-Q^2)\\
\implies -2xdx=2z(x^2-Q^2)dz+2z^2xdx\\
\implies -x(1+z^2)dx=z(x^2-Q^2)dz\quad\cdots(1)
}$$
The definition of $\mathrm z$ implies
$$\mathrm{
-x^2(1+z^2)=-(1+z^2Q^2)\;\cdots(2)
}$$
Combining $(1),(2)$ we get
$$\mathrm{
-{(1+z^2Q^2)\over x}dx=z(x^2-Q^2)dz=z\left({1+z^2Q^2\over1+z^2}-Q^2\right)dz=z{1-Q^2\over1+z^2}dz
}$$
And thus our integral turns into
$$\mathrm{
I=(1-Q^2)^2\int_0^\infty{z^2\over(1+z^2)^2(1+z^2Q^2)}dz
}$$
The integrand has poles of order $2$ at $\mathrm{z=\pm i}$ and simple poles at $\mathrm{\pm{i\over Q}}$. If we use a semi-circular contour in the upper half-plane, we only enclose the poles $\mathrm{ i, {i\over Q}}$ at which residues are $-\mathrm{i(1+Q^2)\over4(1-Q^2)^2}$ and $\mathrm{iQ\over2(1-Q^2)^2}$ respectively and therefore (omitting some details) the required integral is
$$\mathrm{
(1-Q^2)^2{1\over2}\times2\pi i\left[\mathrm{i(1+Q^2)\over4(1-Q^2)^2}-\mathrm{iQ\over2(1-Q^2)^2}\right]\\
=\frac{\pi}{4}Q^2-\frac{\pi}{2}Q+\frac{\pi}{4}
}$$
A: You can check that (since there is a selected answer, I don't have the energy/motivation to give all details, but start with $x\mapsto x^2$) there is a primitive
$$
\frac{1}{2}\sqrt{(x^2-Q^2)(1-x^2)}+\frac{1}{2}(1+Q^2)\arctan\sqrt{\frac{Q^2-x^2}{1-x^2}}
+Q\arctan\biggl(Q\sqrt{\frac{1-x^2}{x^2-Q^2}}\biggr)
$$
Insert limits (you have to use limits) to find that
$$
\int_Q^1\sqrt{(1-x^2)(1-Q^2/x^2)}\,dx=\frac{\pi(1-Q)^2}{4}.
$$
