<s>[INST] Using polar coordintaes to evaluate $\int \int {\sqrt {\frac {1-x^2-y^2} {1+x^2+y^2}} }\ dx \ dy$ I want evaluate the following integral using polar coordinates.
$$ \int \int {\sqrt  {\frac {1-x^2-y^2} {1+x^2+y^2}} }\ dx \ dy $$ over the positive quadrant of the circle $$x^2+y^2=1$$
I used the substitution $x=r\cos\theta$, $y=r\sin\theta$ and reduced the integral to
$$ \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} {\sqrt {\frac {1-r^2} {1+r^2}}}r drd\theta$$
Then, I made the substitution $u=r^2$, giving me, $2rdr=du$ and got
$$\int {\sqrt{\frac{1-u}{1+u}}}du$$
I then split the integrand into partial fractions $-1 + {\frac{2}{1+u}}$
How do I now proceed? Can I reduce it to the form $\int \sqrt{x^2-a^2}\ dx$?
 [/INST] Substitute $u=1+r^2$ and $\mathrm du=2r\,\mathrm dr$, so that
$$\sqrt{\frac{1-r^2}{1+r^2}}r\,\mathrm dr=\sqrt{\frac{2-(1+r^2)}{1+r^2}}r\,\mathrm dr=\frac12\sqrt{\frac{2-u}u}\,\mathrm du$$
Then take $v=\sqrt u$ (so that $v^2=u$) and $2v\,\mathrm dv=\mathrm du$:
$$\frac12\sqrt{\frac{2-u}u}\,\mathrm du=\frac12\sqrt{\frac{2-v^2}{v^2}}(2v\,\mathrm dv)=\sqrt{2-v^2}\,\mathrm dv$$
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