Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.5 - Strategy for Integration - 7.5 Exercises - Page 522: 12

Answer

${e^{\arcsin x}} + C$

Work Step by Step

$$\eqalign{ & \int {\frac{{{e^{\arcsin x}}}}{{\sqrt {1 - {x^2}} }}} dx \cr & \int {\frac{{{e^{\arcsin x}}}}{{\sqrt {1 - {x^2}} }}} dx = \int {{e^{\arcsin x}}\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)} dx \cr & {\text{Use the substitution method}} \cr & {\text{Let }}u = \arcsin x,{\text{ }}du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr & {\text{Substituting}} \cr & \int {{e^{\arcsin x}}\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)} dx = \int {{e^u}du} \cr & {\text{Integrating}} \cr & = {e^u} + C \cr & {\text{Write in terms of }}x,{\text{ substitute }}\arcsin x{\text{ for }}u \cr & = {e^{\arcsin x}} + C \cr} $$
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