Answer
$$\frac{1}{3}{\sec ^3}\theta + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^3}\theta \tan \theta } d\theta \cr
& = \int {\frac{1}{{{{\cos }^3}\theta }}\left( {\frac{{\sin \theta }}{{\cos \theta }}} \right)} d\theta \cr
& = \int {\frac{{\sin \theta }}{{{{\cos }^4}\theta }}} d\theta \cr
& \cr
& {\text{Let }}u = \cos \theta ,\,\,\,du = - \sin \theta d\theta \cr
& \int {\frac{{-\sin \theta d\theta }}{{{{\cos }^4}\theta }}} = \int {\frac{{ - du}}{{{u^4}}}} \cr
& = - \int {{u^{ - 4}}du} \cr
& = - \frac{{{u^{ - 3}}}}{{ - 3}} + C \cr
& = \frac{1}{{3{u^3}}} + C \cr
& {\text{Substitute back }}u = \cos \theta \cr
& = \frac{1}{{3{{\cos }^3}\theta }} + C \cr
& = \frac{1}{3}{\sec ^3}\theta + C \cr
& \cr
& {\text{Let }}u = \sec \theta ,\,\,\,du = \sec \theta \tan \theta d\theta \cr
& \int {{{\sec }^3}\theta \tan \theta } d\theta = \int {{{\sec }^2}\theta \tan \theta \sec \theta } d\theta \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int {{u^2}du} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{u^3}}}{3} + C \cr
& {\text{Substitute back }}u = \sec \theta \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{3}{\sec ^3}\theta + C \cr} $$
