Answer
$$\frac{1}{3}\tan 3x - x + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^2}3x} dx \cr
& {\text{Use the pythagorean identity ta}}{{\text{n}}^2}\theta + 1 = {\sec ^2}\theta \cr
& = \int {\left( {{{\sec }^2}3x - 1} \right)} dx \cr
& = \int {{{\sec }^2}3x} dx - \int {dx} \cr
& = \frac{1}{3}\int {{{\sec }^2}3x} \left( 3 \right)dx - \int {dx} \cr
& {\text{Integrating}} \cr
& = \frac{1}{3}\tan 3x - x + C \cr} $$
