Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises: 52

Answer

\[ = \frac{1}{{\sqrt 2 }}\,\left( {1 - \frac{1}{{\sqrt 2 }}} \right)\]

Work Step by Step

\[\begin{gathered} \int\limits_0^{\frac{\pi }{8}} {\sqrt {1 - \cos 4x} \,dx} \hfill \\ \hfill \\ use\,\,the\;\;identities \hfill \\ \hfill \\ 1 = {\sin ^2}2x + {\cos ^2}2x\,\,\,\,and\,\,\,\,\cos \,4x = {\cos ^2}2x - {\sin ^2}2x \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \int\limits_0^{\frac{\pi }{8}} {\sqrt {{{\sin }^2}2x + {{\cos }^2}2x - {{\cos }^2}2x + {{\sin }^2}2x} \,\,dx} \hfill \\ \hfill \\ simplify\,\,inside\,\,the\,\,radical \hfill \\ \hfill \\ = \sqrt 2 \int\limits_0^{\frac{\pi }{8}} {\sin 2x\,dx} \hfill \\ \hfill \\ set\,\,\,\,\,2x = t\,\,\, \to \,\,2dx = dt\,\,and\,\,dx = \frac{{dt}}{2} \hfill \\ \hfill \\ = \frac{{\sqrt 2 }}{2}\,\int\limits_0^{\frac{\pi }{4}} {\sin t\,dt} \hfill \\ \hfill \\ \,integrate\,and\,use\,\,the\,\,ftc \hfill \\ \hfill \\ = \frac{1}{{\sqrt 2 }}\,\,\left[ { - \cos t} \right]_0^{\frac{\pi }{4}} \hfill \\ \hfill \\ = \frac{{ - 1}}{{\sqrt 2 }}\,\left( {\frac{1}{{\sqrt 2 }} - 1} \right) \hfill \\ \hfill \\ = \frac{1}{{\sqrt 2 }}\,\left( {1 - \frac{1}{{\sqrt 2 }}} \right) \hfill \\ \hfill \\ \end{gathered} \]
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