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: 25

Answer

\[ = \ln \left| {x + \sqrt {{x^2} - 81} } \right| + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{dx}}{{\sqrt {{x^2} - 81} }}\,\,,\,\,\,\,x > 9} \hfill \\ \hfill \\ the\,\,integral\,\,contains\,\,{a^2} - {x^2}\,\,so \hfill \\ \hfill \\ x = 9\sec \theta \,\,\,\,then\,\,\,\,dx = 9\sec \theta \tan \theta d\theta \hfill \\ and\,\,\sqrt {{x^2} - 81} = 9\tan \theta \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{\sqrt {{x^2} - 81} }}} = \int_{}^{} {\frac{{9\sec \theta \tan \theta d\theta }}{{9\tan \theta }}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \int_{}^{} {\sec \theta d\theta } \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \ln \left| {\sec \theta + \tan \theta } \right| + C \hfill \\ \hfill \\ substitute\,\,back \hfill \\ \hfill \\ = \ln \left| {\frac{x}{9} + \frac{{\sqrt {{x^2} - 81} }}{9}} \right| + C \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \ln \left| {x + \sqrt {{x^2} - 81} } \right| + C \hfill \\ \end{gathered} \]
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