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: 56

Answer

\[ = \frac{\pi }{8}\]

Work Step by Step

\[\begin{gathered} \int_0^{\frac{\pi }{4}} {{{\cos }^2}8\theta d\theta } \hfill \\ \hfill \\ use\,\,the\,identity\,\,{\cos ^2}x = \frac{{1 + \cos 2x}}{2} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \int_0^{\frac{\pi }{4}} {\frac{{1 + \cos 16\theta }}{2}} d\theta \hfill \\ \hfill \\ = \int_0^{\frac{\pi }{4}} {\frac{1}{2}d\theta + \frac{1}{{32}}\int_0^{\frac{\pi }{4}} {\cos 16\theta \,\left( {16} \right)d\theta } } \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \,\,\left[ {\frac{\theta }{2}} \right]_0^{\frac{\pi }{4}} + \frac{1}{{32}}\,\,\left[ {\sin 16\theta } \right]_0^{\frac{\pi }{4}} \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = \frac{\pi }{8} - 0 + \frac{1}{{32}}\,\,\left[ {\sin \,\left( {\frac{{16\pi }}{4}} \right) - \sin \theta } \right] \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ = \frac{\pi }{8} \hfill \\ \hfill \\ \hfill \\ \end{gathered} \]
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