## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 428: 64

#### Answer

See below for detailed work.

#### Work Step by Step

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

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