What's a good substitution to solve this integral? Is there a good substitution (or other simple method) to more easily solve this integral?
$$\int \frac{1}{\left(1-\sqrt{1-(\frac{r}{R})^2}\right)^{\frac{3}{2}}}dr$$
Honestly, I was trying  $\frac{r}{R}=\sin(\alpha)$, but the result was complicated...
I appreciate anyone's hints or good ideas.
 A: HINT:
$$\int\frac{1}{\left(1-\sqrt{1-\left(\frac{r}{\text{R}}\right)^2}\right)^{\frac{3}{2}}}\space\text{d}r=\int\frac{1}{\left(1-\sqrt{1-\frac{r^2}{\text{R}^2}}\right)^{\frac{3}{2}}}\space\text{d}r=$$
$$\int\frac{1}{\left(1-\sqrt{\frac{\text{R}^2-r^2}{\text{R}^2}}\right)^{\frac{3}{2}}}\space\text{d}r=\int\frac{1}{\left(1-\frac{\sqrt{\text{R}^2-r^2}}{\text{R}}\right)^{\frac{3}{2}}}\space\text{d}r=\int\frac{1}{\left(\frac{\text{R}-\sqrt{\text{R}^2-r^2}}{\text{R}}\right)^{\frac{3}{2}}}\space\text{d}r=$$
$$\int\frac{1}{\frac{\left(\text{R}-\sqrt{\text{R}^2-r^2}\right)^{\frac{3}{2}}}{\text{R}^{\frac{3}{2}}}}\space\text{d}r=\text{R}^{\frac{3}{2}}\int\frac{1}{\left(\text{R}-\sqrt{\text{R}^2-r^2}\right)^{\frac{3}{2}}}\space\text{d}r=$$

Substitute $u=\text{R}-\sqrt{\text{R}^2-r^2}$ and $\text{d}u=\frac{r}{\sqrt{\text{R}^2-r^2}}\space\text{d}r$:

$$\text{R}^{\frac{3}{2}}\int\frac{\text{R}-u}{u^2\sqrt{2\text{R}-u}}\space\text{d}u=\text{R}^{\frac{3}{2}}\int\left[\frac{\text{R}}{u^2\sqrt{2\text{R}-u}}-\frac{u}{u^2\sqrt{2\text{R}-u}}\right]\space\text{d}u=$$
$$\text{R}^{\frac{3}{2}}\left[\int\frac{\text{R}}{u^2\sqrt{2\text{R}-u}}\space\text{d}u-\int\frac{u}{u^2\sqrt{2\text{R}-u}}\space\text{d}u\right]=$$
$$\text{R}^{\frac{3}{2}}\left[\text{R}\int\frac{1}{u^2\sqrt{2\text{R}-u}}\space\text{d}u-\int\frac{1}{u\sqrt{2\text{R}-u}}\space\text{d}u\right]$$
