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: 15

Answer

\[ = - 3\ln \left| {\frac{{3 + \sqrt {9 - {x^2}} }}{x}} \right| + \sqrt {9 - {x^2}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{\sqrt {9 - {x^2}} }}{x}} \,\,dx \hfill \\ \hfill \\ the\,\,integral\,\,has\,\,the\,\,form\,\,{a^2} - {x^2}\,\, \hfill \\ \hfill \\ x = 3\sin \theta \,\,\,\,\,\,then\,\,\,\,\,dx = 3\cos \theta d\theta \hfill \\ and\,\,\sqrt {9 - {x^2}} = 3\cos \theta \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \int_{}^{} {\frac{{\sqrt {9 - {x^2}} }}{x}dx} \hfill \\ \hfill \\ = \int_{}^{} {\frac{{\,\left( {3\cos \theta } \right)}}{{\,\left( {3\sin \theta } \right)}}\,\left( {3\cos \theta } \right)d\theta } \hfill \\ \hfill \\ = 3\int_{}^{} {\frac{{{{\cos }^2}\theta }}{{\sin \theta }}d\theta \,} \hfill \\ \hfill \\ = 3\int_{}^{} {\frac{{\,\left( {1 - {{\sin }^2}\theta } \right)}}{{\sin \theta }}d\theta } \hfill \\ \hfill \\ distribute \hfill \\ \hfill \\ = 3\int_{}^{} {\csc d\theta } - 3\int_{}^{} {\sin \theta d\theta } \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = - \ln \left| {\csc \,\,\,\theta + \cot \,\,\theta } \right| + 3\cos \,\,\theta + C \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = - 3\ln \left| {\frac{{3 + \sqrt {9 - {x^2}} }}{x}} \right| + \sqrt {9 - {x^2}} + C \hfill \\ \hfill \\ \end{gathered} \]
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