Integrate via substitution and derivation rule I have to solve 
this integral
$$\int_{-r}^{+r}\int_{-\sqrt{r^2-x^2}}^{+\sqrt{r^2-x^2}} \sqrt{1-\frac{x^2+y^2}{x^2+y^2-r^2}} \operatorname d y \operatorname d x$$
with substitution and then the trick that $\dfrac 1 {\sqrt{1-x^2}} = \dfrac{\mathsf d\;\arcsin(x)}{\mathsf d\;x\qquad\quad\;\,} $
can someone give me a tip on what I should substitute in order to continue?
 A: I would say
$\displaystyle \int_{-r}^{+r}\int_{-\sqrt{r^2-x^2}}^{+\sqrt{r^2-x^2}} \sqrt{1-\frac{x^2+y^2}{x^2+y^2-r^2}} \operatorname d y \operatorname d x=4\int_{0}^{r}\int_{0}^{\sqrt{r^2-x^2}} \sqrt{\frac{r^2}{r^2-x^2-y^2}} \operatorname d y \operatorname d x=$
$\displaystyle=4\int_{0}^{r}\int_{0}^{\sqrt{r^2-x^2}} \frac{r}{\sqrt{r^2-x^2}}\frac{1}{\sqrt{1-\left(\frac{y}{\sqrt{r^2-x^2}}\right)^2}} \operatorname d y \operatorname d x=$
$\displaystyle =4r\int_{0}^{r}\left[\arcsin \frac{y}{\sqrt{r^2-x^2}}\right]_{0}^{\sqrt{r^2-x^2}}\operatorname dx=\cdots $
A: First use $x=r \sin(u)$ so, $\mathrm d x = r \cos u\, \mathrm d u$
$$\begin{align}
& \int_{-r}^{+r}\int_{-\sqrt{r^2-x^2}}^{+\sqrt{r^2-x^2}} \sqrt{1-\frac{x^2+y^2}{x^2+y^2-r^2}}\; \operatorname d y \operatorname d x
\\
& = \int_{-r}^{+r}\int_{-\sqrt{r^2-x^2}}^{+\sqrt{r^2-x^2}} \sqrt{\frac{r^2}{r^2-x^2-y^2}} \;\operatorname d y \operatorname d x
\\
& = \int_{-\pi/2}^{+\pi/2}r|r| \cos u \int_{-r \cos u}^{+r\cos u} \sqrt{\frac{1}{r^2\cos^2 u - y^2}} \;\operatorname d y \operatorname d u
\end{align}$$
Now can you see how to use the identity?
