Answer
$A = \frac{{3\pi }}{8} + \frac{1}{2}$
Work Step by Step
$$\eqalign{
& y = {\cos ^2}x,{\text{ }}y = \sin x\cos x,{\text{ }}x = - \frac{\pi }{2},{\text{ }}x = \frac{\pi }{4} \cr
& {\cos ^2}x \geqslant \sin x\cos x{\text{ on the interval }}\left( { - \frac{\pi }{2},\frac{\pi }{4}} \right),{\text{ then}} \cr
& {\text{The area of the region is given by }} \cr
& A = \int_{ - \pi /2}^{\pi /4} {\left( {{{\cos }^2}x - \sin x\cos x} \right)} dx \cr
& {\text{Use the identity }}{\cos ^2}x = \frac{{1 + \cos 2x}}{2} \cr
& A = \int_{ - \pi /2}^{\pi /4} {\left( {\frac{{1 + \cos 2x}}{2} - \sin x\cos x} \right)} dx \cr
& A = \int_{ - \pi /2}^{\pi /4} {\left( {\frac{1}{2} + \frac{{\cos 2x}}{2} - \sin x\cos x} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ {\frac{1}{2}x + \frac{1}{4}\sin 2x - \frac{{{{\sin }^2}x}}{2}} \right]_{ - \pi /2}^{\pi /4} \cr
& A = \left[ {\frac{1}{2}\left( {\frac{\pi }{4}} \right) + \frac{1}{4}\sin 2\left( {\frac{\pi }{4}} \right) - \frac{1}{2}{{\sin }^2}\left( {\frac{\pi }{4}} \right)} \right] \cr
& {\text{ }} - \left[ {\frac{1}{2}\left( { - \frac{\pi }{2}} \right) + \frac{1}{4}\sin 2\left( { - \frac{\pi }{2}} \right) - \frac{1}{2}{{\sin }^2}\left( { - \frac{\pi }{2}} \right)} \right] \cr
& {\text{Simplifying}} \cr
& A = \left[ {\frac{\pi }{8} + \frac{1}{4}\sin \left( {\frac{\pi }{2}} \right) - \frac{1}{2}{{\sin }^2}\left( {\frac{\pi }{4}} \right)} \right] \cr
& {\text{ }} - \left[ { - \frac{\pi }{4} + \frac{1}{4}\sin \left( { - \pi } \right) - \frac{1}{2}{{\sin }^2}\left( { - \frac{\pi }{2}} \right)} \right] \cr
& A = \frac{\pi }{8} + \frac{1}{4} - \frac{1}{4} + \frac{\pi }{4} + 0 + \frac{1}{2} \cr
& A = \frac{{3\pi }}{8} + \frac{1}{2} \approx 1.67809 \cr} $$
