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: 31


$\int sec^2(-3θ)dθ=-\frac{1}{3}tan(-3θ)+C$

Work Step by Step

$\int sec^2(-3θ)dθ$ $=-\frac{1}{3}\int-3sec^2(-3θ)dθ;$ $u=-3θ;$ $du=-3dθ$ $=-\frac{1}{3}\int sec^2(u)du$ $=-\frac{1}{3}tan(u)+C$ $=-\frac{1}{3}tan(-3θ)+C$
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