Answer
$$2\sqrt 2 - \ln \left( {3 + 2\sqrt 2 } \right)$$
Work Step by Step
$$\eqalign{
& y = \sec x,\,\,\,y = 2\cos x,\cr
& {\text{interval }} - \pi /4 \leqslant x \leqslant \pi /4 \cr
& 2\cos x \geqslant \sec x{\text{ for the given interval}}{\text{}} \cr
& {\text{Thus, the area is given by}} \cr
& A = \int_{ - \pi /4}^{\pi /4} {\left( {2\cos x - \sec x} \right)} dx \cr
& {\text{Integrating, we get:}} \cr
& A = \left( {2\sin x - \ln \left| {\sec x + \tan x} \right|} \right)_{ - \pi /4}^{\pi /4} \cr
& {\text{Evaluating, we get:}} \cr
& A = \left( {2\sin \left( {\frac{\pi }{4}} \right) - \ln \left| {\sec \frac{\pi }{4} + \tan \frac{\pi }{4}} \right|} \right) - \left( {2\sin \left( { - \frac{\pi }{4}} \right) - \ln \left| {\sec \left( { - \frac{\pi }{4}} \right) + \tan \left( { - \frac{\pi }{4}} \right)} \right|} \right) \cr
& A = \left( {2\left( {\frac{{\sqrt 2 }}{2}} \right) - \ln \left| {\sqrt 2 + 1} \right|} \right) - \left( {2\sin \left( { - \frac{{\sqrt 2 }}{2}} \right) - \ln \left| {\sqrt 2 - 1} \right|} \right) \cr
& A = \left( {\sqrt 2 - \ln \left| {\sqrt 2 + 1} \right|} \right) - \left( { - \sqrt 2 - \ln \left| {\sqrt 2 - 1} \right|} \right) \cr
& A = \sqrt 2 - \ln \left| {\sqrt 2 + 1} \right| + \sqrt 2 + \ln \left| {\sqrt 2 - 1} \right| \cr
& A = 2\sqrt 2 - \ln \left( {\frac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}}} \right) \cr
& A = 2\sqrt 2 - \ln \left( {3 + 2\sqrt 2 } \right) \cr} $$
