Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.1 - Using Basic Integration Formulas - Exercises 8.1 - Page 449: 41

Answer

$$2\sqrt 2 - \ln \left( {3 + 2\sqrt 2 } \right)$$

Work Step by Step

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