Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.3 - Trigonometric Integrals - Exercises 8.3 - Page 462: 13



Work Step by Step

Recall: $\cos 2x= 2\cos^{2}x-1$ or $\cos^{2}x= \frac{1+\cos2x}{2}$ We have, $\int \cos^{2}xdx= \frac{1}{2}\int (1+\cos2x)dx$ $=\frac{1}{2}\int dx+\frac{1}{2}\int \cos 2xdx$ $=\frac{x}{2}+\frac{\sin2x}{4}+C$
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