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: 29


$$-\frac{1}{12} \sin ^{3}(3 x) \cos (3 x)-\frac{1}{8} \sin (3 x) \cos (3 x)+\frac{3}{8} x+c$$

Work Step by Step

Given $$\int \sin ^{4}(3 x) d x$$ Let $$u=3x\ \ \Rightarrow \ du= 3dx $$ Then by using $$\int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x$$ \begin{align*} \int \sin ^{4}(3 x) d x&=\frac{1}{3} \int \sin ^{4} u d u\\ &=-\frac{1}{12} \sin ^{3} u \cos u+\frac{1}{4} \int \sin ^{2} u d u \quad(n=4)\\ &=-\frac{1}{12} \sin ^{3} u \cos u+\frac{1}{4}\left(-\frac{1}{2} \sin u \cos u+\frac{1}{2} \int d u\right) \quad(n=2)\\ &=-\frac{1}{12} \sin ^{3} u \cos u-\frac{1}{8} \sin u \cos u+\frac{1}{8} u+c\\ &=-\frac{1}{12} \sin ^{3}(3 x) \cos (3 x)-\frac{1}{8} \sin (3 x) \cos (3 x)+\frac{3}{8} x+c \end{align*}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.