Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 57

Answer

\[ = \frac{\pi }{4}\]

Work Step by Step

\[\begin{gathered} \int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {{{\sin }^2}2\theta d\theta } \hfill \\ \hfill \\ use\,\,the\,\,idenity\,\,{\sin ^2}x\, = \frac{{1 - \cos 2x}}{2} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\frac{{1 - \cos 4\theta }}{2}} \,d\theta \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \,\,\left[ {\frac{\theta }{2} - \frac{{\sin 4\theta }}{8}} \right]_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = \,\left( {\frac{\pi }{8} - \frac{{\sin \,\left( \pi \right)}}{8}} \right) - \,\left( {\frac{\pi }{8} - \frac{{\sin \,\left( { - \pi } \right)}}{8}} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{\pi }{8} + \frac{\pi }{8} \hfill \\ \hfill \\ solution \hfill \\ \hfill \\ = \frac{\pi }{4} \hfill \\ \end{gathered} \]
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