Calculus: Early Transcendentals 8th Edition

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