Answer
$$\int \sin (4x-9)dx =-\frac{1}{4}\cos (4x-9)+c $$
Work Step by Step
Let $ u=4x-9$, then $ du =4dx $, then we have
$$\int \sin (4x-9)dx =\frac{1}{4}\int \sin u du\\
=-\frac{1}{4}\cos u+c=-\frac{1}{4}\cos (4x-9)+c $$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 30
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