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: 33



Work Step by Step

Let u=$\frac{1}{t}-1=t^{-1}-1=>du=-t^{-2}=>-du-\frac{1}{t^5}dt$ $\frac{1}{t^2}cos(\frac{1}{t}-1)dt$ =$\int (cos u)(-du)$ =$-\int \cos udu=2\sin u+C$ =$-sin(\frac{1}{t}-1)+C$
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