Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 513: 63


$\frac{\pi}{18} \approx$ 0.175

Work Step by Step

$\int_0^\frac{2}{\sqrt 3}\frac{1}{{4+9x^{2}}}dx$ $=\frac{1}{3}\int_0^\frac{2}{\sqrt 3}\frac{1}{{2^{2}+(3x)^{2}}}dx$ Use arctan identity: where a = 2, u = 3x, and du = 3dx $\int{\frac{du}{a^{2}+u^{2}}} = \frac{1}{a}arctan(\frac{u}{a}) + C$ $=[\frac{1}{6}arctan(\frac{3x}{2})]^{\frac{2}{\sqrt 3}}_{0}$ $=[\frac{1}{6}arctan(\frac{6}{2\sqrt 3})]-[\frac{1}{6}arctan(0)]$ $=[(\frac{1}{6})(\frac{\pi}{3})]-0$ $=\frac{\pi}{18}$
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