University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 33



Work Step by Step

$$A=\int\frac{1}{t^2}\cos\Big(\frac{1}{t}-1\Big)dt$$ We set $u=\frac{1}{t}-1$ Then $$du=-\frac{1}{t^2}dt$$ That means $$\frac{1}{t^2}dt=-du$$ Therefore, $$A=-\int\cos udu=-\sin u+C$$ $$A=-\sin\Big(\frac{1}{t}-1\Big)+C$$
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