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: 46

Answer

\[ = \sqrt 2 \]

Work Step by Step

\[\begin{gathered} \int\limits_0^{\frac{\pi }{2}} {\sqrt {1 + \cos 2x} \,\,dx} \hfill \\ \hfill \\ using\,\,the\,\,identity\,\,\,1 = {\sin ^2}x + {\cos ^2}x\,\,and\,\,\cos \,2x = {\cos ^2}x - {\sin ^2}x \hfill \\ integral\,becomes \hfill \\ \hfill \\ \int\limits_0^{\frac{\pi }{2}} {\sqrt {1 + \cos 2x} \,\,dx} = \int\limits_0^{\frac{\pi }{2}} {\sqrt {{{\cos }^2}x + {{\sin }^2}x + {{\cos }^2}x - {{\sin }^2}x} \,dx} \hfill \\ \hfill \\ simplify\,\,inside\,\,the\,\,\,radical \hfill \\ \hfill \\ = \sqrt 2 \int\limits_0^{\frac{\pi }{2}} {\cos xdx} \hfill \\ \hfill \\ integrate\,and\,use\,\,the\,\,ftc \hfill \\ \hfill \\ = \sqrt 2 \,\,\left[ {\sin x} \right]_0^{\frac{\pi }{2}} \hfill \\ \hfill \\ = \sqrt 2 \,\left( {1 - 0} \right) = \sqrt 2 \hfill \\ \end{gathered} \]
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