Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 42

Answer

\[ = \sqrt 2 - 1\]

Work Step by Step

\[\begin{gathered} \int_0^{\frac{\pi }{4}} {\frac{{\sin x}}{{{{\cos }^2}x}}} \,\,\,dx \hfill \\ \hfill \\ set\,\,\,u = \cos x\,\,\,\,{\text{which implies that}}\,\,\,du = - \sin xdx \hfill \\ \hfill \\ {\text{Changing limits of integration}} \hfill \\ \hfill \\ x = 0\,\,\,\,implies\,\,u = 1 \hfill \\ x = \frac{\pi }{4}\,\,\,implies\,\,u = \frac{{\sqrt 2 }}{2} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_1^{\frac{{\sqrt 2 }}{2}} {\frac{{du}}{{{u^2}}}} \hfill \\ \hfill \\ integrate\,\, \hfill \\ \hfill \\ = \,\,\left[ {\frac{1}{u}} \right]_1^{\frac{{\sqrt 2 }}{2}} \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = \frac{1}{{\frac{{\sqrt 2 }}{2}}} - \frac{1}{1} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \sqrt 2 - 1 \hfill \\ \end{gathered} \]
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