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: 23


$$\int_{0}^{2 \pi} \sqrt{\frac{1-\cos x}{2}} dx =4$$

Work Step by Step

Given $$\int_{0}^{2 \pi} \sqrt{\frac{1-\cos x}{2}} dx $$ So, we have \begin{aligned} I&= \int_{0}^{2 \pi} \sqrt{\frac{1-\cos x}{2}} \ dx\\ &\text{since} \ \ \cos x= 1-2\sin^2\frac{ x}{2} \Rightarrow \sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2},\text{ we get}\\ I&=\int_{0}^{2 \pi}\left|\sin \frac{x}{2}\right| dx\\ &=2\int_{0}^{2 \pi} \frac{1}{2}\sin \frac{x}{2} \ d x\\ &=\left[-2 \cos \frac{x}{2}\right]_{0}^{2 \pi}\\ &=\left[-2 \cos\pi\right]- \left[-2 \cos0\right]\\ &=2+2\\ &=4\\ \end{aligned}
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