Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises: 8

Answer

$$\frac{1}{9}\cos(4-9x)+C$$

Work Step by Step

Finding the general antiderivative of $f$: $\int{f(x)}dx$ $=\int\sin(4-9x)dx$ Given $$\int\sin({kx})dx=-\frac{1}{k}\cos({kx})+C$$ $\int\sin(4-9x)dx$ $=-\frac{1}{-9}\cos(4-9x)$ $=\frac{1}{9}\cos(4-9x)+C$
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