University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.3 - Trigonometric Substitutions - Exercises - Page 439: 15


$-\sqrt{ 9-x^2}+C$

Work Step by Step

Let us consider $u = 9-x^2 $ and $du=-2x dx $ Now, the given integral becomes: $\int \dfrac{-1}{2}[\dfrac{-2x}{ \sqrt{9-x^2}} d x = \dfrac{-1}{2} \int \dfrac{1}{\sqrt u} du$ Now, integrate and plug in $u = 9-x^2 $: $-u^{1/2}+C=-\sqrt{ 9-x^2}+C$
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