Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.5 The Fundamental Theorem of Calculus, Part II - Exercises - Page 263: 30


$-\frac{ \cos ^{3} (1/x) }{x^2}.$

Work Step by Step

By making use of Theorem 1 (FTC II), we have $$ \frac{d}{d x} \int_{1}^{1 / x} \cos ^{3} t d t=\left(\frac{d}{d u} \int_{0}^{u} \cos ^{3} t d t\right)\frac{du}{dx}= \cos ^{3} (1/x) \ (-\frac{1}{x^2})\\ =-\frac{ \cos ^{3} (1/x) }{x^2}. $$
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