Answer
$$\frac{2}{3}{\left( {1 + \sec \theta } \right)^{3/2}} + C $$
Work Step by Step
$$\eqalign{
& \int {\left( {\sec \theta \tan \theta } \right)} \sqrt {1 + \sec \theta } d\theta \cr
& {\text{use the substitution method }} \cr
& u = 1 + \sec \theta,\,\,\,\,du = \sec \theta \tan \theta d\theta,\,\,\,\,\,d\theta = \frac{{du}}{{\sec \theta \tan \theta }} \cr
& {\text{Then}}{\text{,}} \cr
& \int {\left( {\sec \theta \tan \theta } \right)} \sqrt {1 + \sec \theta } d\theta = \int {\left( {\sec \theta \tan \theta } \right)} \sqrt u \left( {\frac{{du}}{{\sec \theta \tan \theta }}} \right) \cr
& = \int {\sqrt u } du \cr
& = \int {{u^{1/2}}} du \cr
& {\text{integrate}} \cr
& = \frac{{{u^{3/2}}}}{{3/2}} + C \cr
& = \frac{2}{3}{u^{3/2}} + C \cr
& {\text{write in terms of }}\theta;{\text{ replace }}u = 1 + \sec \theta \cr
& = \frac{2}{3}{\left( {1 + \sec \theta } \right)^{3/2}} + C \cr} $$
