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: 45

Answer

\[ = \sqrt 2 - 1\]

Work Step by Step

\[\begin{gathered} \int_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{\cos x}}{{{{\sin }^2}x}}\,dx} \hfill \\ \hfill \\ set\,\,u = \sin x\,\,\,then\,\,\,du = \cos xdx \hfill \\ \hfill \\ x = \frac{\pi }{4}\,\,\,\,implies\,\,u = \sin \,\left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2} \hfill \\ \hfill \\ x = \frac{\pi }{2}\,\,\,\,\,\,implies\,\,u = \sin \,\left( {\frac{\pi }{2}} \right) = 1 \hfill \\ \hfill \\ apply\,\,the\,\,\,substitution \hfill \\ \hfill \\ \int_{\frac{{\sqrt 2 }}{2}}^1 {\frac{{du}}{{{u^2}}}} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \,\,\left[ { - \frac{1}{u}} \right]_{\frac{{\sqrt 2 }}{2}}^1\, = \,\,\left[ {\frac{1}{u}} \right]_1^{\frac{{\sqrt 2 }}{2}} \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = \frac{1}{{\frac{{\sqrt 2 }}{2}}} - \frac{1}{1} \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ = \sqrt 2 - 1 \hfill \\ \hfill \\ \end{gathered} \]
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