Calculus (3rd Edition)

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

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 19


$-\cos (\theta-8) +c $

Work Step by Step

Let $ u=\theta-8$, then $ du=d\theta $ and hence $$\int \sin(\theta-8) d\theta=\int \sin u du=-\cos u +c= -\cos (\theta-8) +c $$ Where we used the fact that the integral of sine is negative cosine.
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