Answer
$$\pi $$
Work Step by Step
$$\eqalign{
& \int_{ - \pi /2}^{\pi /2} {\frac{{2\cos \theta d\theta }}{{1 + {{\left( {\sin \theta } \right)}^2}}}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = \sin \theta,{\text{ so that }}du = \cos \theta d\theta \cr
& {\text{the new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}\theta = \pi /2,{\text{ then }}u = \sin \left( {\pi /2} \right) = 1 \cr
& \,\,\,\,\,\,{\text{If }}\theta = - \pi /2,{\text{ then }}u = \sin \left( { - \pi /2} \right) = - 1 \cr
& {\text{then}} \cr
& \int_{ - \pi /2}^{\pi /2} {\frac{{2\cos \theta d\theta }}{{1 + {{\left( {\sin \theta } \right)}^2}}}} = \int_{ - 1}^1 {\frac{{2du}}{{1 + {u^2}}}} \cr
& = 2\int_{ - 1}^1 {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 1 \cr
& = 2\left( {{{\tan }^{ - 1}}\left( u \right)} \right)_{ - 1}^1 + C \cr
& = 2\left( {{{\tan }^{ - 1}}\left( 1 \right) - {{\tan }^{ - 1}}\left( { - 1} \right)} \right) \cr
& = 2\left( {\frac{\pi }{4} + \frac{\pi }{4}} \right) \cr
& = \pi \cr} $$
