Answer
$$A = 1$$
Work Step by Step
$$\eqalign{
& y = {\cos ^2}x,{\text{ }}y = {\sin ^2}x,{\text{ }}x = - \frac{\pi }{4},{\text{ }}x = \frac{\pi }{4} \cr
& {\cos ^2}x \geqslant {\sin ^2}x{\text{ on the interval }}\left( { - \frac{\pi }{4},\frac{\pi }{4}} \right),{\text{ then}} \cr
& {\text{The area of the region is given by }} \cr
& A = \int_{ - \pi /4}^{\pi /4} {\left( {{{\cos }^2}x - {{\sin }^2}x} \right)} dx \cr
& {\text{By symmetry}} \cr
& A = 2\int_0^{\pi /4} {\left( {{{\cos }^2}x - {{\sin }^2}x} \right)} dx \cr
& {\text{Use the trigonometric identity co}}{{\text{s}}^2}x - {\sin ^2}x = \cos 2x \cr
& A = 2\int_0^{\pi /4} {\cos 2x} dx \cr
& A = \int_0^{\pi /4} {2\cos 2x} dx \cr
& {\text{Integrate}} \cr
& A = \left[ {\sin 2x} \right]_0^{\pi /4} \cr
& A = \sin \left[ {2\left( {\frac{\pi }{4}} \right)} \right] - \sin \left[ 0 \right] \cr
& A = 1 - 0 \cr
& A = 1 \cr} $$
