Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.4 Trigonometric Substitutions - 7.4 Exercises: 7

Answer

\[ = \frac{\pi }{6}\]

Work Step by Step

\[\begin{gathered} \int_0^{5/2} {\frac{{dx}}{{\sqrt {25 - {x^2}} }}} \hfill \\ \hfill \\ the\,\,integral\,\,has\,\,the\,\,form\,\,{a^2} - {x^2}\,\, \hfill \\ \hfill \\ x = 5\sin \theta \,\,\,\,\,then\,\,\,\,dx = 5\cos \theta d\theta \hfill \\ \hfill \\ and\,\,\sqrt {25 - {x^2}} = 5\cos \theta \hfill \\ \hfill \\ \int_{}^{} {\frac{{5\cos \theta d\theta }}{{5\cos \theta }}} = \int_{}^{} {d\theta } \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \theta + C \hfill \\ \hfill \\ where\,\,\sin \theta = \frac{x}{5} \to \theta = {\sin ^{ - 1}}\,\left( {\frac{x}{5}} \right) \hfill \\ \hfill \\ = \,\,\left[ {{{\sin }^{ - 1}}\,\left( {\frac{x}{5}} \right)} \right]_0^{5/2} \hfill \\ \hfill \\ evaluate\,\,the\,limits \hfill \\ \hfill \\ = {\sin ^{ - 1}}\,\left( {\frac{{5/2}}{5}} \right) - {\sin ^{ - 1}}\,\left( {\frac{0}{5}} \right) \hfill \\ \hfill \\ = \frac{\pi }{6} \hfill \\ \end{gathered} \]
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