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: 14



Work Step by Step

Recall the identity $\cos 2x=1-2\sin^{2} x$ which gives $\sin^{2}x=\frac{1-\cos 2x}{2}$. Therefore, $\int \sin^{2}x\,dx=\frac{1}{2}\int dx-\frac{1}{2}\int\cos 2x\,dx$ $= \frac{x}{2}-\frac{1}{4}\sin 2x+C$ Then $\int_{0}^{\pi/2} \sin^{2}x\,dx=[\frac{x}{2}-\frac{1}{4}\sin 2x]^{\pi/2}_{0}=(\frac{\pi}{4}-0)-0=\frac{\pi}{4}$
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