Answer
$-\frac{\pi}{4}+1$
Work Step by Step
$\int_{0}^{\pi/4} (tan^2x) dx=\int_{0}^{\pi/4} (-1+sec^2x)=[-x+tanx]_{0}^{\pi/4}=[-\frac{\pi}{4}+tan(\frac{\pi}{4})]-[-0+tan0]=-\frac{\pi}{4}+1$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 15
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