Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.6 - Integration Using Tables and Computer Algebra Systems - 7.6 Exercises - Page 527: 23

Answer

$ - \ln \left( {{{\cos }^2}\theta + \sqrt {{{\cos }^4}\theta + 4} } \right) + C$

Work Step by Step

$$\eqalign{ & \int {\frac{{\sin 2\theta }}{{\sqrt {{{\cos }^4}\theta + 4} }}} d\theta \cr & {\text{Use the double angle formula sin2}}\theta = 2\sin \theta \cos \theta \cr & = \int {\frac{{2\sin \theta \cos \theta }}{{\sqrt {{{\cos }^4}\theta + 4} }}} d\theta \cr & {\text{Let }}u = {\cos ^2}\theta \Rightarrow du = - 2\cos \theta \sin \theta d\theta ,{\text{ then}} \cr & \int {\frac{{2\sin \theta \cos \theta }}{{\sqrt {{{\cos }^4}\theta + 4} }}} d\theta = - \int {\frac{{du}}{{\sqrt {{u^2} + 4} }}} \cr & {\text{Using the formula}} \cr & \int {\frac{{du}}{{\sqrt {{u^2} + {a^2}} }}} = \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C \cr & {\text{Therefore}}{\text{,}} \cr & - \int {\frac{{du}}{{\sqrt {{u^2} + 4} }}} = - \ln \left( {u + \sqrt {{u^2} + 4} } \right) + C \cr & {\text{Write in terms of }}\theta ,{\text{ }}u = {\cos ^2}\theta ,{\text{ then}} \cr & \int {\frac{{\sin 2\theta }}{{\sqrt {{{\cos }^4}\theta + 4} }}} d\theta = - \ln \left( {{{\cos }^2}\theta + \sqrt {{{\cos }^4}\theta + 4} } \right) + C \cr} $$
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