I've been stuck with the following integral, I know I have to use substitution, but I don't know how.
Let $S=\{(x,y):\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1\}$. Show that $$\int \int_{S} \left( \dfrac{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}{1+\frac{x^2}{a^2}+\frac{y^2}{b^2}}\right)^{\frac{1}{2}}=\frac{\pi}{4}\left(\frac{\pi}{2}-1\right)ab$$
If someone can give me a hint I would be very grateful :)
Hint:
The 'normal' parameterization works here: $x = ar\cos\theta$, $y = br\sin\theta$ with $0 < r \leq 1$ and $0 \leq \theta < 2\pi$. The Jacobian $\displaystyle\frac{\partial(r,\theta)}{\partial(x,y)} = abr$. Hence we can write the integral as
$$\int_0^{2\pi}\int_0^1 \left( \frac{1-r^2}{1+r^2} \right)^{1/2} . abr \ dr \ d\theta$$
This integral separates
$$ab \left( \int_0^{2\pi} d\theta \right) \left( \int_0^1 \left( \frac{1-r^2}{1+r^2} \right)^{1/2} r \ dr \right) = 2\pi ab \left( \frac{1}{2} \int_0^1 \left( \frac{1-u}{1+u} \right)^{1/2} \ du \right)$$
where $u = r^2$. Evaluating this last integral takes some work, but from the answer you know what you're shooting for.