Procedure for evaluating $\int_{x=\ -1}^1\int_{y=\ -\sqrt{1-x^2}}^{\sqrt{1-x^2}}\frac{x^2+y^2}{\sqrt{{1-x^2-y^2}}}\,dy\,dx$ While solving another problem I have come across this integral which I am unable to evaluate. Can someone please evaluate the following integral? Thank you.
$$\int_{x=\ -1}^1\int_{\large y=\ -\sqrt{1-x^2}}^{\large\sqrt{1-x^2}}\frac{x^2+y^2}{\sqrt{{1-x^2-y^2}}}\,dy\,dx.$$
I know the answer is $\dfrac{4\pi}3$, but I am more interested in the procedure followed to get to this answer.

Based on the limit of integral $-\sqrt{1-x^2} < y < \sqrt{1-x^2}$ and $-1 < x < 1$, the region of integration is a unit circle in the Cartesian coordinate. See this plot to visualize the region of integration. Using polar coordinate, we have $x^2+y^2=r^2$ and the region of integration will be $0<r<1$ and $0<\theta<2\pi$. Therefore
\begin{align}
\int_{x=\ -1}^1\int_{\large y=\ -\sqrt{1-x^2}}^{\large\sqrt{1-x^2}}\frac{x^2+y^2}{\sqrt{{1-x^2-y^2}}}\,dy\,dx&=\int_{\theta=0}^{2\pi}\int_{r=0}^{1}\frac{r^2}{\sqrt{{1-r^2}}}\,r\ dr\,d\theta\\
&=\int_{\theta=0}^{2\pi}\,d\theta\ \int_{r=0}^{1}\frac{r^3}{\sqrt{{1-r^2}}}\ dr.
\end{align}
Let $r=\sin t\;\Rightarrow\;dr=\cos t\ dt$ and the corresponding region is $0<t<\dfrac\pi2$, then
\begin{align}
\int_{x=\ -1}^1\int_{\large y=\ -\sqrt{1-x^2}}^{\large\sqrt{1-x^2}}\frac{x^2+y^2}{\sqrt{{1-x^2-y^2}}}\,dy\,dx&=\int_{\theta=0}^{2\pi}\,d\theta\ \int_{t=0}^{\Large\frac\pi2}\frac{\sin^3t}{\sqrt{{1-\sin^2t}}}\cdot \cos t\ dt\\
&=2\pi\int_{t=0}^{\Large\frac\pi2} \sin^3t\ dt\\
&=2\pi\int_{t=0}^{\Large\frac\pi2} \sin^2t\ \sin t\ dt\\
&=2\pi\int_{t=0}^{\Large\frac\pi2} (\cos^2t-1)\ d(\cos t).
\end{align}
Set $\theta=\cos t$ and you can take it from here.
