Calculus: Early Transcendentals 8th Edition

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

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises: 34

Answer

$$\int\frac{\cos(\pi/x)}{x^2}dt=-\frac{\sin(\pi/x)}{\pi}+C$$

Work Step by Step

$$A=\int\frac{\cos(\pi/x)}{x^2}dt$$ Let $u=\frac{\pi}{x}$ Then we have $du=\frac{(\pi)'x-\pi(x)'}{x^2}dt=\frac{-\pi}{x^2}dt$. That means $\frac{1}{x^2}dt=-\frac{1}{\pi}du$ Substitute into $A$: $$A=-\frac{1}{\pi}\int\cos udu$$ $$A=-\frac{1}{\pi}\sin u+C$$ $$A=-\frac{\sin(\pi/x)}{\pi}+C$$
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