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: 58

Answer

\[\frac{{{x^2}}}{4} + \frac{1}{8}\sin 2{x^2} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {x{{\cos }^2}\,\left( {{x^2}} \right)dx} = \frac{1}{2}\int_{}^{} {{{\cos }^2}\,\left( {{x^2}} \right)\,\left( {2x} \right)dx} \hfill \\ \hfill \\ or \hfill \\ \hfill \\ = \frac{1}{2}\int_{}^{} {{{\cos }^2}udu} \hfill \\ \hfill \\ where\,\,\,{\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ = \frac{1}{4}\int_{}^{} {\,\left( {1 + \cos 2u} \right)} du \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \frac{u}{4} + \frac{1}{8}\sin 2u + C \hfill \\ \hfill \\ substitute\,\,u = {x^2} \hfill \\ \hfill \\ \frac{{{x^2}}}{4} + \frac{1}{8}\sin 2{x^2} + C \hfill \\ \end{gathered} \]
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