Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 37


$$\frac{x\sin 4x}{4}+\frac{1}{16}\cos4x+c .$$

Work Step by Step

We do the integration by parts; we choose $u= x$ and $dv=\cos 4x dx$. Then, $du=dx$, $v=\frac{1}{4}\sin 4x$, hence we have $$\int x\cos 4x dx=\int udv=uv-\int vdu\\ =\frac{x\sin 4x}{4}-\frac{1}{4}\int \sin 4xdx\\ =\frac{x\sin 4x}{4}+\frac{1}{16}\cos4x+c .$$
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