Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.6 Other Integration Strategies - 7.6 Exercises: 70

Answer

\[ = \frac{1}{{12}}{\sec ^2}4t\,\tan 4t + \frac{1}{6}\tan 4t + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {{{\sec }^4}4t\,dt} \hfill \\ \hfill \\ Inetgrate\,\,using\,\,the\,\,reduccion\,\,formula \hfill \\ \hfill \\ \int_{}^{} {{{\sec }^n}xdx} = \frac{{{{\sec }^{n - 2}}x\tan x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int_{}^{} {{{\sec }^{n - 2}}xdx} \hfill \\ \hfill \\ Let\,,\,\,\,4t = u\,\,\,\,\, \to \,\,\,\,4dy = du\, \hfill \\ then \hfill \\ \int_{}^{} {{{\sec }^4}\,\left( {4t} \right)dt} = \frac{1}{4}\int_{}^{} {{{\sec }^4}\,\left( u \right)du} \hfill \\ integrating \hfill \\ \hfill \\ \int_{}^{} {{{\sec }^4}\,\left( {4t} \right)dt} = \frac{1}{4}\int_{}^{} {{{\sec }^4}\,\left( u \right)} du \hfill \\ \hfill \\ = \frac{1}{4}\,\,\left[ {\frac{{{{\sec }^2}\tan u}}{3} + \frac{2}{3}\int_{}^{} {{{\sec }^2}udu} } \right] \hfill \\ \hfill \\ using\,\,the\,\,reduccion\,\,formula \hfill \\ \hfill \\ = \frac{1}{{12}}{\sec ^2}u\tan u + \frac{1}{6}\,\,\left[ {\frac{{{{\sec }^{2 - 2}}u\tan u}}{{2 - 1}} + \frac{{2 - 2}}{{2 - 1}}\int_{}^{} {{{\sec }^2}^{ - 2}udu} } \right] \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \int_{}^{} {{{\sec }^4}4t\,dt} = \frac{1}{{12}}{\sec ^2}u\tan u + \frac{1}{6}\tan u + C \hfill \\ \hfill \\ = \frac{1}{{12}}{\sec ^2}4t\,\tan 4t + \frac{1}{6}\tan 4t + C \hfill \\ \end{gathered} \]
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