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



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} =($tan(\frac{\pi}{4})-(\frac{\pi}{4})-(tan(0)-0)$) $=1-(\frac{\pi}{4})$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.