Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 25


$$\frac{1}{12} \cos ^{3}(3 x+2) \sin (3 x+2)+\frac{1}{8}(3 x+2)+\frac{1}{16} \sin (6 x+4)+c$$

Work Step by Step

Given $$\int \cos ^{4}(3 x+2) d x $$ Let $$ u=3x+2 \ \ \ \Rightarrow \ \ \ du=3dx $$ Then $$\int \cos ^{4}(3 x+2) d x=\frac{1}{3} \int \cos ^{4} u d u$$ Hence by using $$ \int \cos ^{n} x d x=\frac{\cos ^{n-1} x \sin x}{n}+\frac{n-1}{n} \int \cos ^{n-2} x d x$$ We get \begin{align*} \int \cos ^{4}(3 x+2) d x&=\frac{1}{3} \int \cos ^{4} u d u\\ &= \frac{1}{3}\left(\frac{1}{4} \cos ^{3} u \sin u+\frac{3}{4} \int \cos ^{2} u d u\right)\\ &=\frac{1}{12} \cos ^{3} u \sin u+\frac{1}{4}\left(\frac{\sin u \cos u}{2}+\frac{1}{2} \int \cos ^{\theta} u d u\right) \\ &=\frac{1}{12} \cos ^{3} u \sin u+\frac{1}{4}\left(\frac{\sin 2 u}{4}+\frac{u}{2}\right)+c\\ &=\frac{1}{12} \cos ^{3}(3 x+2) \sin (3 x+2)+\frac{1}{8}(3 x+2)+\frac{1}{16} \sin (6 x+4)+c \end{align*}
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