University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 1


$$\int x\sin\frac{x}{2}dx=-2x\cos\frac{x}{2}+4\sin\frac{x}{2}+C$$

Work Step by Step

$$A=\int x\sin\frac{x}{2}dx$$ Take $u=x$ and $dv=\sin\frac{x}{2}dx$ We then have $du=dx$ and $v=-2\cos\frac{x}{2}$ Apply the formula $\int udv= uv-\int vdu$, we have $$A=x\times\Big(-2\cos\frac{x}{2}\Big)-\int-2\cos\frac{x}{2}dx$$ $$A=-2x\cos\frac{x}{2}+2\int\cos\frac{x}{2}dx$$ $$A=-2x\cos\frac{x}{2}+2\Big(2\sin\frac{x}{2}\Big)+C$$ $$A=-2x\cos\frac{x}{2}+4\sin\frac{x}{2}+C$$
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