Thomas' Calculus 13th Edition

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

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} =($tan(\frac{\pi}{4})-(\frac{\pi}{4})-(tan(0)-0)$) $=1-(\frac{\pi}{4})$
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