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: 53

Answer

\[ = \pi \]

Work Step by Step

\[\begin{gathered} \int_{ - \pi }^\pi {{{\cos }^2}xdx} \hfill \\ \hfill \\ where\,\,{\cos ^2}x = \frac{{1 + \cos 2x}}{2} \hfill \\ \hfill \\ = \int_{ - \pi }^\pi {\,\left( {\frac{{1 + \cos 2x}}{2}} \right)} \,dx \hfill \\ \hfill \\ integrate\,\,and\,\,use\,\,the\,\,ftc \hfill \\ \hfill \\ = \,\left( {\frac{\pi }{2} + \frac{{\sin \,\left( {2\pi } \right)}}{4}} \right) - \,\left( { - \frac{\pi }{2} + \frac{{\sin \,\left( { - 2\pi } \right)}}{4}} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{\pi }{2} + \frac{\pi }{2} \hfill \\ \hfill \\ = \pi \hfill \\ \end{gathered} \]
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