Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.7 Derivatives And Integrals Involving Inverse Trigonometric Functions - Exercises Set 6.7 - Page 470: 40

Answer

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

Work Step by Step

$$\eqalign{ & \int_{ - 2}^{ - 2/\sqrt 3 } {\frac{{dx}}{{x\sqrt {{x^2} - 1} }}} \cr & {\text{integrate using the formula }}\int {\frac{1}{{\left| u \right|\sqrt {{u^2} - 1} }} = {{\sec }^{ - 1}}u + C,{\text{ }}} \left( {{\text{see page 468}}} \right) \cr & {\text{let }}a = 1 \cr & \int_{ - \sqrt 2 }^{ - 2/\sqrt 3 } {\frac{{dx}}{{x\sqrt {{x^2} - 1} }}} = \left. {\left( {{{\sec }^{ - 1}}u} \right)} \right|_{ - \sqrt 2 }^{ - 2/\sqrt 3 } \cr & {\text{part 1 of fundamental theorem of calculus}} \cr & = {\sec ^{ - 1}}\left( { - \frac{2}{{\sqrt 2 }}} \right) - {\sec ^{ - 1}}\left( { - \sqrt 2 } \right) \cr & or \cr & = co{s^{ - 1}}\left( { - \frac{{\sqrt 3 }}{2}} \right) - {\cos ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right) \cr & {\text{simplify}} \cr & = \frac{{5\pi }}{6} - \frac{{3\pi }}{4} \cr & = \frac{{10\pi - 9\pi }}{{12}} \cr & = - \frac{\pi }{{12}} \cr} $$
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