Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 252: 33



Work Step by Step

$\int25sec^2(3z)dz$ $=\frac{25}{3}\int3sec^2(3z)dz;$ $u=3z;$ $du=3dz$ $=\frac{25}{3}\int sec^2(u)du$ $=\frac{25}{3}tan(u)+C$ $=\frac{25}{3}tan(3z)+C$
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