Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 29



Work Step by Step

Let u=$x^{3/2}+1=>du=\frac{3}{2}x^{1/2}dx=>\frac{2}{3}du=x^{1/2}dx$ =$\int x^{1/2} sin(x^{3/2}+1)dx$ =$\int (\sin u)(\frac{2}{3}du)$ =$\frac{2}{3} \int \sin udu=\frac{2}{3}(-\cos u)+C$ =$\frac{-2}{3}cos(x^{3/2}+1)+C$
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