Answer
$$\frac{{{{\tan }^3}3\theta }}{9} + \frac{{\tan 3\theta }}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^4}3\theta } d\theta \cr
& {\text{split integrand}} \cr
& = \int {{{\sec }^2}3\theta } {\sec ^2}3\theta d\theta \cr
& {\text{use trigonometric identity se}}{{\text{c}}^2}\phi = {\tan ^2}\phi + 1 \cr
& = \int {\left( {{{\tan }^2}3\theta + 1} \right)} {\sec ^2}3\theta d\theta \cr
& {\text{substitute }}u = \tan 3\theta ,{\text{ }}du = 3{\sec ^2}3\theta d\theta \cr
& = \int {\left( {{u^2} + 1} \right)} \left( {\frac{1}{3}} \right)du \cr
& = \frac{1}{3}\int {\left( {{u^2} + 1} \right)du} \cr
& {\text{find antiderivative }} \cr
& = \frac{1}{3}\left( {\frac{{{u^3}}}{3} + u} \right) + C \cr
& = \frac{{{u^3}}}{9} + \frac{u}{3} + C \cr
& {\text{write in terms of }}\theta ,{\text{ replace }}u = \tan 3\theta \cr
& = \frac{{{{\tan }^3}3\theta }}{9} + \frac{{\tan 3\theta }}{3} + C \cr} $$
