Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises - Page 476: 3


$$\int x\cos 5xdx=\frac{1}{5}x\sin(5x)+\frac{1}{25}\cos(5x)+C$$

Work Step by Step

$$A=\int x\cos 5xdx$$ We would choose $u=x$ and $dv=\cos 5xdx$ We could easily see that $du=dx$. For $dv=\cos 5xdx$, we analyze as follows $$\int \cos 5xdx$$ Let $u=5x$, then $du=5dx$, so $dx=\frac{1}{5}du$ $$\int\cos 5xdx=\frac{1}{5}\int\cos udu=\frac{1}{5}\sin u+C=\frac{1}{5}\sin(5x)+C$$ Therefore, for $dv=\cos 5xdx$, $v=\frac{1}{5}\sin(5x)$ Apply Integration by Parts to A, we have $$A=uv-\int vdu$$ $$A=\frac{1}{5}x\sin(5x)-\int\frac{1}{5}\sin(5x)dx$$ $$A=\frac{1}{5}x\sin(5x)-\frac{1}{5}\int\frac{1}{5}\sin(5x)d(5x)$$ $$A=\frac{1}{5}x\sin(5x)-\frac{1}{25}(-\cos(5x)+C)$$ $$A=\frac{1}{5}x\sin(5x)+\frac{1}{25}\cos(5x)+C$$
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