Answer
$25sec^2(3z)dz=\frac{25}{3}tan(3z)+C$
Work Step by Step
$\int25sec^2(3z)dz$
$=\frac{25}{3}\int3sec^2(3z)dz;$ $u=3z;$ $du=3dz$
$=\frac{25}{3}\int sec^2(u)du$
$=\frac{25}{3}tan(u)+C$
$=\frac{25}{3}tan(3z)+C$
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: 33
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