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: 30

Answer

\[ = \frac{1}{{1458}}{\tan ^{ - 1}}\,\left( {\frac{x}{9}} \right) + \frac{x}{{162\,\left( {81 + {x^2}} \right)}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{dx}}{{\,{{\left( {81 + {x^2}} \right)}^2}}}} \hfill \\ \hfill \\ rewrite,\,\,use\,\,\sqrt[n]{{{a^m}}} = {a^{m/n}} \hfill \\ \hfill \\ = \int_{}^{} {\frac{{dx}}{{\,{{\left( {\sqrt {{x^2} + {9^2}} } \right)}^4}}}} \hfill \\ \hfill \\ the\,\,\,integral\,contains\,\,the\,\,form\,\,{x^2} - {a^2}\,\, \hfill \\ \hfill \\ x = 9\tan \theta \,\,\,\,\,\, \to \,\,\,\,\,dx = 9{\sec ^2}\theta d\theta \hfill \\ and\,\,\sqrt {{x^2} + 81} = 9\sec \theta \hfill \\ \hfill \\ apply\,\,the\,\,substitution \hfill \\ \hfill \\ = \int_{}^{} {\frac{{dx}}{{\,\left( {\sqrt {{x^2} + {9^2}} } \right)}}} = \int_{}^{} {\frac{{9{{\sec }^2}\theta d\theta }}{{{9^4}{{\sec }^4}\theta }}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{1}{{{9^3}}}\int_{}^{} {\frac{1}{{\sec \theta }}} d\theta \hfill \\ \hfill \\ = \frac{1}{{729}}\,\int_{}^{} {{{\cos }^2}\theta d\theta } = \frac{1}{{729}}\int_{}^{} {\,\left( {\frac{{1 + \cos 2\theta }}{2}} \right)d\theta } \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \frac{1}{{729}}\,\left( {\frac{\theta }{2} + \frac{{\sin 2\theta }}{4}} \right) + C \hfill \\ \hfill \\ = \frac{1}{{729}}\,\left( {\frac{\theta }{2} + \frac{{\sin \theta \cos \theta }}{2}} \right) + C \hfill \\ \hfill \\ substitute\,\,for\,\,\sin \theta {\text{ and cos}}\theta \hfill \\ \hfill \\ = \frac{1}{{1458}}{\tan ^{ - 1}}\,\left( {\frac{x}{9}} \right) + \frac{1}{{1458}}\,\left( {\frac{x}{{\sqrt {81 + 2} }}} \right)\,\left( {\frac{9}{{\sqrt {81 + {x^2}} }}} \right) + C \hfill \\ \hfill \\ = \frac{1}{{1458}}{\tan ^{ - 1}}\,\left( {\frac{x}{9}} \right) + \frac{x}{{162\,\left( {81 + {x^2}} \right)}} + C \hfill \\ \end{gathered} \]
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