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: 5

Answer

$\int(2cos(x) - 9sin(x)) dx = 2sin(x) +9cos(x) + C$

Work Step by Step

Rules to remember: $\int sin(x)dx = -cos(x)$, $\int cos(x)dx = sin(x)$ $sin(x)' = cos(x)$, $cos(x)' = -sin(x)$ Anti-differentiation: $\int(2cos(x) - 9sin(x)) dx = 2sin(x) -(-9cos(x)) + C= 2sin(x) +9cos(x) + C$ Checking with differentiation: $(2sin(x) +9cos(x) + C)'= 2cos(x) -9sin(x)$
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