Answer
$\int sec^2(-3θ)dθ=-\frac{1}{3}tan(-3θ)+C$
Work Step by Step
$\int sec^2(-3θ)dθ$
$=-\frac{1}{3}\int-3sec^2(-3θ)dθ;$ $u=-3θ;$ $du=-3dθ$
$=-\frac{1}{3}\int sec^2(u)du$
$=-\frac{1}{3}tan(u)+C$
$=-\frac{1}{3}tan(-3θ)+C$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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