Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - 6.10 Hyperbolic Functions - 6.10 Exercises: 57

Answer

$$ - \frac{1}{8}{\operatorname{sech} ^{ - 1}}\frac{{{x^4}}}{2} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{x\sqrt {4 - {x^8}} }}} \cr & {\text{rewriting }} \cr & = \int {\frac{{4{x^3}dx}}{{4{x^4}\sqrt {{{\left( 2 \right)}^2} - {{\left( {{x^4}} \right)}^2}} }}} \cr & = \frac{1}{4}\int {\frac{{4{x^3}dx}}{{{x^4}\sqrt {{{\left( 2 \right)}^2} - {{\left( {{x^4}} \right)}^2}} }}} \cr & {\text{find the antiderivative using the theorem 6}}{\text{.1}}2 \cr & \int {\frac{{dx}}{{u\sqrt {{a^2} - {u^2}} }}} = - \frac{1}{a}{\operatorname{sech} ^{ - 1}}\frac{u}{a} + C \cr & ,so \cr & = \frac{1}{4}\left( { - \frac{1}{2}{{\operatorname{sech} }^{ - 1}}\frac{{{x^4}}}{2}} \right) + C \cr & {\text{simplify}} \cr & = - \frac{1}{8}{\operatorname{sech} ^{ - 1}}\frac{{{x^4}}}{2} + C \cr} $$
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