Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_{ - \pi /2}^{\pi /2} {\sin \theta } d\theta \cr
& {\text{find antiderivative }} \cr
& \int_{ - \pi /2}^{\pi /2} {\sin \theta } d\theta = \left( { - \cos \theta } \right)_{ - \pi /2}^{\pi /2} \cr
& = - \left( {\cos \theta } \right)_{ - \pi /2}^{\pi /2} \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = - \left( {\cos \left( {\frac{\pi }{2}} \right) - \cos \left( { - \frac{\pi }{2}} \right)} \right) \cr
& \cos \left( { - \theta } \right) = \cos \theta \cr
& = - \left( {\cos \left( {\frac{\pi }{2}} \right) - \cos \left( {\frac{\pi }{2}} \right)} \right) \cr
& {\text{then}} \cr
& = 0 \cr} $$
