Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 419: 18


$$\int\frac{\sin\sqrt x}{\sqrt x}dx=-2\cos\sqrt x+C$$

Work Step by Step

$$A=\int\frac{\sin\sqrt x}{\sqrt x}dx$$ Let $u=\sqrt x=x^{1/2}$. Then $du=\frac{1}{2}x^{-1/2}dx=\frac{1}{2\sqrt x}dx$, so $\frac{1}{\sqrt x}dx=2du$ Substitute into $A$, we have $$A=2\int\sin udu$$ $$A=-2\cos u+C$$ $$A=-2\cos\sqrt x+C$$
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