Answer
$$\frac{{ - {{\cot }^7}\theta }}{7} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\left( {1 - {u^2}} \right)}^{5/2}}}}{{{u^8}}}} du \cr
& {\text{The integrand contains the form }}{a^2} - {x^2} \cr
& 1 - {u^2} \to a = 1 \cr
& {\text{Use the change of variable }}x = a\sin \theta \cr
& u = \sin \theta ,\,\,\,du = \cos \theta d\theta \cr
& \int {\frac{{{{\left( {1 - {u^2}} \right)}^{5/2}}}}{{{u^8}}}} du = \int {\frac{{{{\left( {1 - {{\sin }^2}\theta } \right)}^{5/2}}}}{{{{\sin }^8}\theta }}} \cos \theta d\theta \cr
& = \int {\frac{{{{\left( {{{\cos }^2}\theta } \right)}^{5/2}}}}{{{{\sin }^8}\theta }}} \cos \theta d\theta \cr
& = \int {\frac{{{{\cos }^6}\theta }}{{{{\sin }^8}\theta }}} d\theta = \int {\left( {\frac{{{{\cos }^6}\theta }}{{{{\sin }^6}\theta }}} \right)\left( {\frac{1}{{{{\sin }^2}\theta }}} \right)} d\theta \cr
& = \int {{{\cot }^6}\theta {{\csc }^2}\theta } d\theta \cr
& = - \int {{{\cot }^6}\theta \left( { - {{\csc }^2}\theta } \right)} d\theta \cr
& {\text{Integrate by using the power rule}} \cr
& = \frac{{ - {{\cot }^7}\theta }}{7} + C \cr} $$
