University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises: 15

Answer

$-\frac{\pi}{4}+1$

Work Step by Step

$\int_{0}^{\pi/4} (tan^2x) dx=\int_{0}^{\pi/4} (-1+sec^2x)=[-x+tanx]_{0}^{\pi/4}=[-\frac{\pi}{4}+tan(\frac{\pi}{4})]-[-0+tan0]=-\frac{\pi}{4}+1$
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