Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises: 18

Answer

$G(v) = 2 sinv - 3arcsinv + C$

Work Step by Step

From definition: $f(x) = -sinx$ the derivative is $f'(x) = \frac{1}{\sqrt {1-x^{2}}}$. $f(x) = sin x$ the derivative is $f'(x) = cosx$. So using this: $G(v) = 2 sinv - 3arcsinv + C$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.