Answer
$$ - 3\cos x - 2\tan x + C $$
Work Step by Step
$$\eqalign{
& \int {\left( {3\sin x - 2{{\sec }^2}x} \right)dx} \cr
& {\text{sum rule}} \cr
& = 3\int {\sin x} dx - 2\int {{{\sec }^2}x} dx \cr
& {\text{use integration formulas from table 4}}{\text{.2}}{\text{.1}} \cr
& = 3\left( { - \cos x} \right) - 2\left( {\tan x} \right) + C \cr
& = - 3\cos x - 2\tan x + C \cr
& \cr
& {\text{check by differentiation}} \cr
& = \frac{d}{{dx}}\left[ { - 3\cos x - 2\tan x + C} \right] \cr
& = - 3\left( { - \sin x} \right) - 2\left( {{{\sec }^2}x} \right) + 0 \cr
& = 3\sin x - 2{\sec ^2}x \cr} $$
