Calculus, 10th Edition (Anton)

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

Chapter 4 - Integration - 4.3 Integration By Substitution - Exercises Set 4.3 - Page 286: 2

Answer

$${\text{part a) }} - 2\cos \sqrt x + C{\text{ part b) }}\frac{3}{4}\sqrt {4{x^2} + 5} + C$$

Work Step by Step

$$\eqalign{ & {\text{part a}} \cr & \cr & \int {\frac{1}{{\sqrt x }}\sin \sqrt x } dx \cr & {\text{substitute }}u = \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx \cr & = \int {\frac{1}{{\sqrt x }}\sin \sqrt x } dx = \int {2\sin u} du \cr & {\text{find antiderivative}} \cr & = - 2\cos u + C \cr & {\text{write in terms of }}x \cr & = - 2\cos \sqrt x + C \cr & \cr & {\text{part b}} \cr & \cr & \int {\frac{{3xdx}}{{\sqrt {4{x^2} + 5} }}} \cr & {\text{substitute }}u = 4{x^2} + 5,{\text{ }}du = 8xdx \cr & = \int {\frac{{3xdx}}{{\sqrt {4{x^2} + 5} }}} = 3\int {\frac{{1/8du}}{{\sqrt u }}} \cr & = \frac{3}{8}\int {{u^{ - 1/2}}} du \cr & {\text{find antiderivative}} \cr & = \frac{3}{8}\frac{{{u^{1/2}}}}{{1/2}} + C \cr & = \frac{3}{4}{u^{1/2}} + C \cr & {\text{write in terms of }}x \cr & = \frac{3}{4}\sqrt {4{x^2} + 5} + C \cr} $$
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