Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 7 - Principles Of Integral Evaluation - 7.2 Integration By Parts - Exercises Set 7.2 - Page 498: 44

Answer

\[2\sqrt x \sin \sqrt x + 2\cos \sqrt x + C\]

Work Step by Step

\[\begin{gathered} \int {\cos \sqrt x } dx \hfill \\ {\text{Let }}z = \sqrt x ,{\text{ }}dz = \frac{1}{{2\sqrt x }}dx = \frac{1}{{2z}}dx{\text{ }} \Rightarrow dx = 2zdz,{\text{ then}} \hfill \\ \int {\cos \sqrt x } dx = \int {2z\cos z} dz \hfill \\ {\text{Integrate by parts, }} \hfill \\ {\text{let }}u = 2z{\text{ }}du = 2dz \hfill \\ dv = \cos zdz{\text{ }}v = \sin z \hfill \\ {\text{Therefore,}} \hfill \\ \int {2z\cos z} dz = 2z\sin z - \int {\sin z\left( 2 \right)} dz \hfill \\ {\text{ }} = 2z\sin z - 2\int {\sin z} dz \hfill \\ {\text{ }} = 2z\sin z + 2\cos z + C \hfill \\ {\text{Back - substitute }}z = \sqrt x \hfill \\ {\text{ }} = 2\sqrt x \sin \sqrt x + 2\cos \sqrt x + C \hfill \\ \end{gathered} \]
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