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: 30


$$\int \sin (4x-9)dx =-\frac{1}{4}\cos (4x-9)+c $$

Work Step by Step

Let $ u=4x-9$, then $ du =4dx $, then we have $$\int \sin (4x-9)dx =\frac{1}{4}\int \sin u du\\ =-\frac{1}{4}\cos u+c=-\frac{1}{4}\cos (4x-9)+c $$
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