Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 92

Answer

$$\frac{\pi }{{12}}$$

Work Step by Step

$$\eqalign{ & \int_{\sqrt 2 }^2 {\frac{{dx}}{{x\sqrt {{x^2} - 1} }}} \cr & {\text{integrate using the rule for inverse trigonometric functions}} \cr & = \left[ {{{\sec }^{ - 1}}x} \right]_{\sqrt 2 }^2 \cr & {\text{using the fundamental theorem}} \cr & = {\sec ^{ - 1}}\left( 2 \right) - {\sec ^{ - 1}}\left( {\sqrt 2 } \right) \cr & {\text{simplify}} \cr & = \frac{\pi }{3} - \frac{\pi }{4} \cr & = \frac{\pi }{{12}} \cr} $$
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