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 - 6.8 Hyperbolic Functions And Hanging Cables - Exercises Set 6.8 - Page 481: 44

Answer

$$\ln \left( {2 + \sqrt 3 } \right)$$

Work Step by Step

$$\eqalign{ & \int_0^{\sqrt 3 } {\frac{{dt}}{{\sqrt {{t^2} + 1} }}} \cr & {\text{find the antiderivarive using the theorem 6}}{\text{.8}}{\text{.6 }}\left( {{\text{see page 480}}} \right) \cr & \int {\frac{{du}}{{\sqrt {{a^2} + {u^2}} }}} = {\sinh ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr & \int {\frac{{du}}{{\sqrt {{a^2} + {u^2}} }}} = \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C \cr & \cr & \int_0^{\sqrt 3 } {\frac{{dt}}{{\sqrt {{t^2} + 1} }}} = \left. {\left( {\ln \left( {t + \sqrt {{t^2} + 1} } \right)} \right)} \right|_0^{\sqrt 3 } \cr & {\text{part 1 of fundamental theorem of calculus}} \cr & = \ln \left( {\sqrt 3 + \sqrt {{{\left( {\sqrt 3 } \right)}^2} + 1} } \right) - \ln \left( {0 + \sqrt {{{\left( 0 \right)}^2} + 1} } \right) \cr & {\text{simplifiyng}} \cr & = \ln \left( {\sqrt 3 + \sqrt 4 } \right) - \ln \left( {\sqrt 1 } \right) \cr & = \ln \left( {2 + \sqrt 3 } \right) \cr} $$
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