Answer
$\frac{1}{2}{\sec ^2}\theta + \frac{1}{3}{\sec ^3}\theta + C$
Work Step by Step
$$\eqalign{
&\text{Let }I=\int {\frac{{\sin \theta + \tan \theta }}{{{{\cos }^3}\theta }}} d\theta \cr
& {\text{Use the basic identity }}\tan \theta = \frac{{\sin \theta }}{{\cos \theta }} \cr
& I = \int {\frac{{\sin \theta + \frac{{\sin \theta }}{{\cos \theta }}}}{{{{\cos }^3}\theta }}} d\theta \cr
& = \int {\frac{{\sin \theta \cos \theta + \sin \theta }}{{{{\cos }^4}\theta }}} d\theta \cr
& {\text{Distriubute the numerator}} \cr
& I= \int {\left( {\frac{{\sin \theta }}{{{{\cos }^3}\theta }} + \frac{{\sin \theta }}{{{{\cos }^4}\theta }}} \right)} d\theta \cr
& = \int {\left( {\frac{1}{{{{\cos }^3}\theta }} + \frac{1}{{{{\cos }^4}\theta }}} \right)\sin \theta } d\theta \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}u = \cos \theta ,{\text{ }}du = - \sin \theta d\theta \cr
& I= \int {\left( {\frac{1}{{{u^3}}} + \frac{1}{{{u^4}}}} \right)\left( { - 1} \right)} du \cr
& = - \int {\left( {{u^{ - 3}} + {u^{ - 4}}} \right)} du \cr
& {\text{Use the power rule }}\int {{u^n}} du = \frac{1}{{n + 1}}{u^{n + 1}} + C \cr
& I= - \frac{{{u^{ - 2}}}}{{ - 2}} - \frac{{{u^{ - 3}}}}{{ - 3}} + C \cr
& = \frac{1}{{2{u^2}}} + \frac{1}{{3{u^3}}} + C \cr
& {\text{Write in terms of }}\theta ,{\text{ substitute }}\cos \theta {\text{ for }}u \cr
& I= \frac{1}{{2{{\cos }^2}\theta }} + \frac{1}{{3{{\cos }^3}\theta }} + C \cr
& = \frac{1}{2}{\sec ^2}\theta + \frac{1}{3}{\sec ^3}\theta + C \cr} $$
