Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - Review Exercises - Page 395: 26

Answer

\[ = - \frac{{{{\cos }^9}{x^2}}}{{18}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {x\sin {x^2}{{\cos }^8}{x^2}dx} \hfill \\ \hfill \\ rewrite \hfill \\ \hfill \\ = - \frac{1}{2}\int_{}^{} {\,\left( {{{\cos }^8}{x^2}} \right)\,\left( { - \sin {x^2}} \right)\,\left( {2x} \right)dx} \hfill \\ \hfill \\ integrate\, \hfill \\ \hfill \\ use\,\,\int_{}^{} {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}} + C \hfill \\ \hfill \\ = - \frac{1}{2}\,\left( {\frac{{{{\cos }^9}{x^2}}}{9}} \right) + C \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ = - \frac{{{{\cos }^9}{x^2}}}{{18}} + C \hfill \\ \end{gathered} \]
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