Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.3 Trigonometric Integrals - 7.3 Exercises: 44

Answer

\[ = \frac{1}{4}\]

Work Step by Step

\[\begin{gathered} \int_0^{\frac{\pi }{4}} {{{\tan }^3}\theta \,{{\sec }^2}\theta \,d\theta } \hfill \\ \hfill \\ {u^3} = {\tan ^3}\theta \,\,\,then\,\,\,\,du = {\sec ^2}\theta d\theta \hfill \\ \hfill \\ \int_{}^{} {{{\tan }^3}\theta \,{{\sec }^2}\theta \,d\theta } = \int_{}^{} {{u^3}du} \hfill \\ \hfill \\ Use\,\,\,\int_{}^{} {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}} + C\, \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ \int_{}^{} {{u^3}du} = \frac{{{u^4}}}{4} + C \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_0^{\frac{\pi }{4}} {{{\tan }^3}\theta \,{{\sec }^2}\theta \,d\theta } = \,\,\left[ {\frac{{{{\tan }^4}\theta }}{4}} \right]_0^{\frac{\pi }{4}} \hfill \\ \hfill \\ Evaluate\,\,\,the\,\,limits \hfill \\ \hfill \\ = \frac{1}{4}{\tan ^4}\,\left( {\frac{\pi }{4}} \right) - \frac{1}{4}{\tan ^4}\,\left( 0 \right) \hfill \\ \hfill \\ solution \hfill \\ \hfill \\ = \frac{1}{4} \hfill \\ \, \hfill \\ \end{gathered} \]
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