Answer
$$4\sqrt 3 - 3$$
Work Step by Step
$$\eqalign{
& \int_{ - \pi /3}^{ - \pi /4} {\left( {4{{\sec }^2}t + \frac{\pi }{{{t^2}}}} \right)} dt \cr
& {\text{Separate}} \cr
& = \int_{ - \pi /3}^{ - \pi /4} {\left( {4{{\sec }^2}t} \right)} dt + \int_{ - \pi /3}^{ - \pi /4} {\left( {\pi {t^{ - 2}}} \right)} dt \cr
& {\text{Integrate}} \cr
& = \left( {4\tan t} \right)_{ - \pi /3}^{ - \pi /4} + \left( {\frac{{\pi {t^{ - 1}}}}{{ - 1}}} \right)_{ - \pi /3}^{ - \pi /4} \cr
& = \left( {4\tan t - \frac{\pi }{t}} \right)_{ - \pi /3}^{ - \pi /4} \cr
& {\text{Evaluating, we get:}} \cr
& = \left( {4\tan \left( { - \frac{\pi }{4}} \right) - \frac{\pi }{{\left( { - \pi /4} \right)}}} \right) - \left( {4\tan \left( { - \frac{\pi }{3}} \right) - \frac{\pi }{{\left( { - \pi /3} \right)}}} \right) \cr
& = \left( { - 4 + 4} \right) - \left( { - 4\sqrt 3 + 3} \right) \cr
& = 4\sqrt 3 - 3 \cr} $$
