Answer
$1-\frac{\pi}{4}$
Work Step by Step
$\int^{\frac{\pi}{4}}_{0}tan^2xdx$
=$\int^{\frac{\pi}{4}}_{0}(sec^2x-1)dx$
=[$tanx-x$]^{\frac{\pi}{4}}_{0}
=($tan(\frac{\pi}{4})-(\frac{\pi}{4})-(tan(0)-0)$)
$=1-(\frac{\pi}{4})$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 286: 15
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