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.

• Hint: make ur substitution and use $1-\cos(x)=2\sin(x/2)^2$ in the denominator to kill the noninteger exponent – tired Mar 14 '16 at 19:23
• Did you find a simple integration after apply $1-cosx= 2 sin ^2 (x/2)$? – Khosrotash Mar 14 '16 at 19:27
• it is not nice but rather straightforward – tired Mar 14 '16 at 19:32

$$\int\frac{1}{\left(1-\sqrt{1-\left(\frac{r}{\text{R}}\right)^2}\right)^{\frac{3}{2}}}\space\text{d}r=\int\frac{1}{\left(1-\sqrt{1-\frac{r^2}{\text{R}^2}}\right)^{\frac{3}{2}}}\space\text{d}r=$$ $$\int\frac{1}{\left(1-\sqrt{\frac{\text{R}^2-r^2}{\text{R}^2}}\right)^{\frac{3}{2}}}\space\text{d}r=\int\frac{1}{\left(1-\frac{\sqrt{\text{R}^2-r^2}}{\text{R}}\right)^{\frac{3}{2}}}\space\text{d}r=\int\frac{1}{\left(\frac{\text{R}-\sqrt{\text{R}^2-r^2}}{\text{R}}\right)^{\frac{3}{2}}}\space\text{d}r=$$ $$\int\frac{1}{\frac{\left(\text{R}-\sqrt{\text{R}^2-r^2}\right)^{\frac{3}{2}}}{\text{R}^{\frac{3}{2}}}}\space\text{d}r=\text{R}^{\frac{3}{2}}\int\frac{1}{\left(\text{R}-\sqrt{\text{R}^2-r^2}\right)^{\frac{3}{2}}}\space\text{d}r=$$
Substitute $u=\text{R}-\sqrt{\text{R}^2-r^2}$ and $\text{d}u=\frac{r}{\sqrt{\text{R}^2-r^2}}\space\text{d}r$:
$$\text{R}^{\frac{3}{2}}\int\frac{\text{R}-u}{u^2\sqrt{2\text{R}-u}}\space\text{d}u=\text{R}^{\frac{3}{2}}\int\left[\frac{\text{R}}{u^2\sqrt{2\text{R}-u}}-\frac{u}{u^2\sqrt{2\text{R}-u}}\right]\space\text{d}u=$$ $$\text{R}^{\frac{3}{2}}\left[\int\frac{\text{R}}{u^2\sqrt{2\text{R}-u}}\space\text{d}u-\int\frac{u}{u^2\sqrt{2\text{R}-u}}\space\text{d}u\right]=$$ $$\text{R}^{\frac{3}{2}}\left[\text{R}\int\frac{1}{u^2\sqrt{2\text{R}-u}}\space\text{d}u-\int\frac{1}{u\sqrt{2\text{R}-u}}\space\text{d}u\right]$$