## Calculus 8th Edition

Published by Cengage

# Chapter 7 - Techniques of Integration - 7.3 Trigonometric Substitution - 7.3 Exercises - Page 531: 10

#### Answer

$$\displaystyle{\int_0^{\frac{2}{3}}\sqrt{4-9{x}^2}dx=\frac{\pi }{3}}\\$$

#### Work Step by Step

$\displaystyle{I=\int_0^{\frac{2}{3}}\sqrt{4-9x^2}dx}\\ \displaystyle{I=\int_0^{\frac{2}{3}}\sqrt{4-{\left(3x\right)}^2}dx}\\$ $\displaystyle \left[\begin{array}{ll} 3x=2\sin\theta & 9 x^2=4\sin^2\theta \\ & \\ \frac{dx}{d\theta}=\frac{2}{3}\cos\theta & dx=\frac{2}{3}\cos\theta\ d\theta \end{array}\right]$ Integration by substitution $\displaystyle{I=\int_0^{\frac{\pi}{2}}\sqrt{4-4\sin^2\theta}\times\frac{2}{3}\cos\theta\ d\theta}\\ \displaystyle{I=\int_0^{\frac{\pi}{2}}\sqrt{4\left(1-\sin^2\theta\right)}\times\frac{2}{3}\cos\theta\ d\theta}\\ \displaystyle{I=\int_0^{\frac{\pi}{2}}2\cos\theta\times\frac{2}{3}\cos\theta\ d\theta}\\ \displaystyle{I=\frac{4}{3}\int_0^{\frac{\pi}{2}}\cos^2\theta\ d\theta}\\ \displaystyle{I=\frac{2}{3}\int_0^{\frac{\pi}{2}}\cos2\theta+1\ d\theta}\\ \displaystyle{I=\frac{2}{3}\left[\frac{1}{2}\sin2\theta+ x\right]_0^{\frac{\pi}{2}}}\\ \displaystyle{I=\frac{1}{3}\sin\left(2\pi\times\frac{1}{2}\right)\ +\frac{\pi }{3}-0-0}\\ \displaystyle{I=\frac{1}{3}\sin\left(\pi\right)\ +\frac{\pi }{3}}\\ \displaystyle{I=\frac{\pi }{3}}\\$

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