## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 8: Techniques of Integration - Section 8.3 - Trigonometric Integrals - Exercises 8.3 - Page 462: 26

#### Answer

$$\int_{0}^{\pi} \sqrt{1-\cos ^{2} \theta} \mathrm{d} \theta=2$$

#### Work Step by Step

Given $$\int_{0}^{\pi} \sqrt{1-\cos ^{2} \theta} \mathrm{d} \theta$$ So, we have \begin{aligned} I&=\int_{0}^{\pi} \sqrt{1-\cos ^{2} \theta} \ d\theta\\ &=\int_{0}^{\pi}|\sin \theta| \ d \theta\\ &=\int_{0}^{\pi} \sin \theta \ d \theta\\ &=[-\cos \theta]_{0}^{\pi}\\ &=-\cos \pi+\cos0\\ &=1+1\\ &=2\\ \end{aligned}

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