University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Practice Exercises - Page 278: 108


$\dfrac{-1}{\pi} \cot (\pi s)+c$

Work Step by Step

Calculate the anti-derivative . Since, we know $\int \csc^2 x =-\cot t+C$ Substitute $k=\pi s \implies du=\pi ds$ $\int \csc^2 k (1/\pi) dk=\dfrac{1}{\pi} \int \csc^2 k dk$ or, $=\dfrac{-1}{\pi} \cot k+c$ Back substitution: $k=\pi s$ $=\dfrac{-1}{\pi} \cot (\pi s)+c$
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