Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises: 35

Answer

$\frac{1}{3}sin(3θ)-θ+C$

Work Step by Step

$\int cos(3θ -1) dθ$ $=\int cos(3θ) dθ$ $-\int 1$ $dθ$ for $\int cos(3θ) dθ$, let $u=3θ$ $du=3$ $dθ$ $dθ=\frac{1}{3}du$ $\int cos(3θ) dθ$ $=\int cos (u)$ $\frac{1}{3}du$ $=\frac{1}{3}\int cos(u)$ $du$ $=\frac{1}{3}(sin(u))$ $=\frac{1}{3}sin(3θ)$ $=\int cos(3θ) dθ$ $-\int 1$ $dθ$ $=\frac{1}{3}sin(3θ)-θ+C$
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