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