Calculus, 10th Edition (Anton)

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

Chapter 7 - Principles Of Integral Evaluation - 7.1 An Overview Of Integration Methods - Exercises Set 7.1 - Page 491: 28

Answer

$${\sin ^{ - 1}}\left( {\frac{{{e^x}}}{2}} \right) + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{e^x}}}{{\sqrt {4 - {e^{2x}}} }}dx} \cr & {\text{substitute }}u = {e^x},{\text{ }} \cr & du = {e^x}dx \cr & \int {\frac{{{e^x}}}{{\sqrt {4 - {e^{2x}}} }}dx} = \int {\frac{{{e^x}}}{{\sqrt {{2^2} - {{\left( {{e^x}} \right)}^2}} }}} dx \cr & = \int {\frac{{du}}{{\sqrt {{2^2} - {u^2}} }}} \cr & {\text{find the antiderivative }}\left( {{\text{formula 21 page 489}}} \right) \cr & = \int {\frac{{du}}{{\sqrt {{2^2} - {u^2}} }}} = {\sin ^{ - 1}}\left( {\frac{u}{2}} \right) + C \cr & {\text{write in terms of }}x,{\text{ replace }}u = {e^x} \cr & = {\sin ^{ - 1}}\left( {\frac{{{e^x}}}{2}} \right) + C \cr} $$
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