Answer
$\arcsin(\frac{x}{3})+C$
Work Step by Step
$\int \frac{1}{\sqrt{9-x^2}}dx$
let u= $\frac{x}{3}$
$du=\frac{1}{3}dx$
$dx=3du$
=$\int \frac{1}{\sqrt{1-u^2}}du$
=$\arcsin(u)+C$
$=\arcsin(\frac{x}{3})+C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.7 Exercises - Page 380: 1
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