Answer
$$2\tan 2\theta - \frac{1}{2}\sec 2\theta + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2 - \sin 2\theta }}{{{{\cos }^2}2\theta }}} d\theta \cr
& = \int {\left( {\frac{2}{{{{\cos }^2}2\theta }} - \frac{{\sin 2\theta }}{{{{\cos }^2}2\theta }}} \right)} d\theta \cr
& {\text{use trigonometric identities}} \cr
& = \int {\left( {2{{\sec }^2}2\theta - \sec 2\theta \tan 2\theta } \right)} d\theta \cr
& = \int {2{{\sec }^2}2\theta } d\theta - \int {\sec 2\theta \tan 2\theta } d\theta \cr
& = \int {2{{\sec }^2}2\theta } d\theta - \frac{1}{2}\int {\sec 2\theta \tan 2\theta } \left( 2 \right)d\theta \cr
& {\text{find the antiderivative}} \cr
& = 2\tan 2\theta - \frac{1}{2}\sec 2\theta + C \cr} $$
