University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.7 - Improper Integrals - Exercises - Page 471: 29



Work Step by Step

Here, we have $\int_{1}^{2} \dfrac{ds}{s \sqrt {s^2-1}}=\lim\limits_{a \to 1^{+}}\int_{a}^{2} \dfrac{ds}{s \sqrt {s^2-1}}$ This implies that $\lim\limits_{a \to 1^{+}}\int_{a}^{2} \dfrac{ds}{s \sqrt {s^2-1}}=\lim\limits_{a \to 1^{+}}[\sec^{-1} s]_{a}^{2}$ Thus, $\lim\limits_{a \to 1^{+}}\sec^{-1} (2)-\lim\limits_{a \to 1^{+}}\sec^{-1} (a)=\dfrac{\pi}{3}$
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