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: 17



Work Step by Step

$\int^{\frac{\pi}{8}}_{0}sin2x dx$ =$[-\frac{1}{2}cos2x]^{(\frac{\pi}{8})}_0$ =$(\frac{-1}{2}cos2((\frac{\pi}{8})))-(\frac{-1}{2}cos2(0))$ =$\frac{2-\sqrt{2}}{4}$
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