Answer
$$\frac{1}{3}{\left( {\tan 3x} \right)^3} + \tan 3x + C $$
Work Step by Step
$$\eqalign{
& \int {3{{\sec }^4}3x} dx \cr
& {\text{We write }}{\sec ^4}3x{\text{ as }}{\sec ^2}3x{\sec ^2}3x{\text{ }} \cr
& = \int {3{{\sec }^2}3x} {\sec ^2}3xdx \cr
& {\text{use the fundamental identiy ta}}{{\text{n}}^2}\theta = {\sec ^2}\theta - 1 \cr
& = \int {3\left( {{{\tan }^2}3x + 1} \right)} {\sec ^2}3xdx \cr
& {\text{Integrate by substitution method}} \cr
& {\text{Let }}u = {\tan ^2}3x,\,\,\,\,du = 3{\sec ^2}3xdx,\,\,\,\,dx = \frac{{du}}{{3{{\sec }^2}3x}} \cr
& {\text{Then}}{\text{,}} \cr
& = \int {3\left( {{u^2} + 1} \right)} {\sec ^2}3x\left( {\frac{{du}}{{3{{\sec }^2}3x}}} \right) \cr
& {\text{Cancel common factors}} \cr
& = \int {\left( {{u^2} + 1} \right)} du \cr
& {\text{integrating}} \cr
& = \frac{1}{3}{u^3} + u + C \cr
& {\text{write in terms of }}x,\,\,\,u = \tan 3x \cr
& = \frac{1}{3}{\left( {\tan 3x} \right)^3} + \tan 3x + C \cr} $$
