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:
$G(v) = 2 sinv - 3arcsinv + C$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 355: 18
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