Answer
$$\cos \theta + \frac{1}{{\cos \theta }} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^3}\theta {{\cos }^{ - 2}}\theta d\theta } \cr
& {\text{split off }}\sin \theta \cr
& = \int {{{\sin }^2}\theta {{\cos }^{ - 2}}\theta } \sin \theta d\theta \cr
& {\text{pythagorean identity}} \cr
& = \int {\left( {1 - {{\cos }^2}\theta } \right){{\cos }^{ - 2}}\theta } \sin \theta d\theta \cr
& u = \cos \theta ,{\text{ }}du = - \sin \theta d\theta \cr
& = - \int {\left( {1 - {u^2}} \right){u^{ - 2}}} d\theta \cr
& = \int {\left( {{u^0} - {u^{ - 2}}} \right)} du \cr
& = \int {\left( {1 - {u^{ - 2}}} \right)} du \cr
& {\text{evaluate the integral}} \cr
& = u - \frac{{{u^{ - 1}}}}{{ - 1}} + C \cr
& = u + \frac{1}{u} + C \cr
& {\text{replace }}u = \cos \theta \cr
& = \cos \theta + \frac{1}{{\cos \theta }} + C \cr} $$
