Better substitution calculating integral?

Question

I'm calculating $$ \iint\limits_S \, \left(\frac{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}{1+\frac{x^2}{a^2}+\frac{y^2}{b^2}} \right)^\frac{1}{2} \, dA$$ with $$S =\left\{ (x, \, y) \in \mathbb{R}^2 : \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1\right\}.$$

I take $$x = ar\cos \theta$$ $$y= br\sin \theta$$ and the integral becomes $$ ab\int_0^{2\pi}d\theta\int_0^1\, \left(\frac{1-r^2}{1+r^2} \right)^\frac{1}{2} r \, dr$$ What is better substitution to calculate inner integral? I tried with $r= \sin \vartheta$, a friend mine told me $u=1+r^2$. Thanks for any suggestions and helping ideas.

Answer 1

Sub $u = r^2$, then $u=\cos{\theta}$. If you get things right, you get the following integral

$$2 \int_0^{\pi/2} d\theta \, \sin^2{\frac{\theta}{2}} $$

which I imagine you can do.

Answer 2

One way to do this is to substitute $u^2 = 1 + r^2$, so $2u \, du = 2r\, dr$, or $r\, dr = u\, du$

$$\int_0^1 \sqrt{\frac{1-r^2}{1+r^2}}\,r\,dr = \int_0^1 \sqrt{\frac{2-(1+r^2)}{1+r^2}}\,r\,dr\\ = \int_1^\sqrt{2} \sqrt{\frac{2-u^2}{u^2}}\,u\,du = \int_1^\sqrt{2} \sqrt{2- u^2} \, du $$

Now substitute $u = \sqrt{2} \sin \phi$