Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 275: 47

Answer

$$ \frac{1}{3} \cos (8-3 \theta)+C$$

Work Step by Step

Given $$ \int \sin (8-3 \theta) d \theta$$ Let $$u=8-3 \theta\ \ \ \Rightarrow \ \ \ du=-3d \theta $$ Then \begin{aligned} \int \sin (8-3 \theta) d \theta &=-\frac{1}{3} \int \sin u d u \\ &=-\frac{1}{3}(-\cos u+C) \\ &=\frac{1}{3} \cos u+C\\ &= \frac{1}{3} \cos (8-3 \theta)+C \end{aligned}
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