Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.3 Trigonometric Integrals - 7.3 Exercises - Page 530: 58

Answer

$$\sqrt 2 $$

Work Step by Step

$$\eqalign{ & \int_{ - \pi /4}^{\pi /4} {\sqrt {1 + \cos 4x} dx} \cr & = \int_{ - \pi /4}^{\pi /4} {\sqrt {1 + \cos 2\left( {2x} \right)} dx} \cr & {\text{Use the identity }}\cos 2\theta = 2{\cos ^2}\left( {2\theta } \right) - 1 \cr & \int_{ - \pi /4}^{\pi /4} {\sqrt {1 + \cos 2\left( {2x} \right)} dx} = \int_{ - \pi /4}^{\pi /4} {\sqrt {1 + 2{{\cos }^2}\left( {2x} \right) - 1} dx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_{ - \pi /4}^{\pi /4} {\sqrt {2{{\cos }^2}\left( {2x} \right)} dx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_{ - \pi /4}^{\pi /4} {\sqrt 2 \cos 2xdx} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 2 \int_{ - \pi /4}^{\pi /4} {\cos 2xdx} \cr & {\text{Integrate}} \cr & \sqrt 2 \int_{ - \pi /4}^{\pi /4} {\cos 2xdx} = \sqrt 2 \left[ {\frac{1}{2}\sin 2x} \right]_{ - \pi /4}^{\pi /4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt 2 }}{2}\left[ {\sin \left( {\frac{\pi }{2}} \right) - \sin \left( { - \frac{\pi }{2}} \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt 2 }}{2}\left[ {1 + 1} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 2 \cr} $$
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