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 428: 63

Answer

See below for detailed work.

Work Step by Step

$$A=\int x^n\cos xdx$$ We set $u= x^n$ and $dv=\cos xdx$ That makes $du=nx^{n-1}dx$ and $v=\sin x$ Applying integration by parts $\int udv=uv-\int vdu$, we have $$A=x^n\sin x-\int nx^{n-1}\sin xdx$$ $$A=x^n\sin x-n\int x^{n-1}\sin xdx$$ The reduction formula has been established.
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