Answer
$$ - \frac{1}{7}{\left( {\frac{{\sqrt {1 - {r^2}} }}{r}} \right)^7} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\left( {1 - {r^2}} \right)}^{5/2}}}}{{{r^8}}}} dr \cr
& {\text{Use the trigonometric substitutions:}} \cr
& {\text{ }}r = \sin \theta,{\text{ }}dr = \cos \theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int {\frac{{{{\left( {1 - {r^2}} \right)}^{5/2}}}}{{{r^8}}}} dr = \int {\frac{{{{\left( {1 - {{\sin }^2}\theta } \right)}^{5/2}}}}{{{{\sin }^8}\theta }}} \left( {\cos \theta d\theta } \right) \cr
& {\text{Use the fundamental identity }}\cr
& {\cos ^2}\theta + {\sin ^2}\theta = 1 \cr
& = \int {\frac{{{{\left( {{{\cos }^2}\theta } \right)}^{5/2}}}}{{{{\sin }^8}\theta }}} \left( {\cos \theta d\theta } \right) \cr
& {\text{Simplifying, we get:}} \cr
& = \int {\frac{{{{\cos }^6}\theta }}{{{{\sin }^8}\theta }}} d\theta \cr
& = \int {\left( {\frac{{{{\cos }^6}\theta }}{{{{\sin }^6}\theta }}} \right)} \left( {\frac{1}{{{{\sin }^2}\theta }}} \right)d\theta \cr
& = \int {{{\cot }^6}\theta } cs{c^2}\theta d\theta \cr
& {\text{Integrate}} \cr
& = - \frac{1}{7}{\cot ^7}\theta + C \cr
& {\text{Write in terms of }}x,{\text{ }}\cot \theta = \frac{{\sqrt {1 - {r^2}} }}{r} \cr
& = - \frac{1}{7}{\left( {\frac{{\sqrt {1 - {r^2}} }}{r}} \right)^7} + C \cr} $$
