Answer
$$A = \frac{{3\pi }}{2}$$
Work Step by Step
$$\eqalign{
& y = \frac{{3\cos x}}{{1 + {{\sin }^2}x}} \cr
& {\text{The area of the region is given by}} \cr
& A = \int_{ - \pi /2}^{\pi /2} {\frac{{3\cos x}}{{1 + {{\sin }^2}x}}} dx \cr
& {\text{By symmetry}} \cr
& A = 2\int_0^{\pi /2} {\frac{{3\cos x}}{{1 + {{\sin }^2}x}}} dx \cr
& {\text{Let }}u = \sin x,{\text{ }}du = \cos xdx \cr
& {\text{The new limits of integration are}} \cr
& x = 0,{\text{ }}u = 0 \cr
& x = \pi /2,{\text{ }}u = 1 \cr
& {\text{Substitute}} \cr
& A = 6\int_0^1 {\frac{{\cos x}}{{1 + {{\sin }^2}x}}} dx \cr
& A = 6\int_0^1 {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{Integrating}} \cr
& A = 6\left[ {{{\tan }^{ - 1}}u} \right]_0^1 \cr
& A = 6\left( {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}0} \right) \cr
& A = 6\left( {\frac{\pi }{4} - 0} \right) \cr
& {\text{Simplify}} \cr
& A = \frac{{3\pi }}{2} \cr} $$
