University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Practice Exercises - Page 474: 29

Answer

$-\sqrt {16-y^2}+c$

Work Step by Step

Consider the integral $\int \dfrac{y dy}{\sqrt {16-y^2}}$ The given integral can be re-written as: $\int \dfrac{y dy}{\sqrt {16-y^2}}=\int (16-y^2)^{(-1/2)}y dy$ This implies that $(\dfrac{-1}{2})\int (16-y^2)^{(-1/2)}(-2y) dy=-\sqrt {16-y^2}+c$
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