# Chapter 7: Transcendental Functions - Section 7.6 - Inverse Trigonometric Functions - Exercises 7.6 - Page 421: 81

$${e^{{{\sin }^{ - 1}}x}} + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{{e^{{{\sin }^{ - 1}}x}}dx}}{{\sqrt {1 - {x^2}} }}} \cr & {\text{use the substitution method}}{\text{.}} \cr & u = {\sin ^{ - 1}}x,{\text{ so that }}du = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr & {\text{then}} \cr & \int {{e^{{{\sin }^{ - 1}}x}}\frac{1}{{\sqrt {1 - {x^2}} }}dx} = \int {{e^u}} du \cr & {\text{Integrating}} \cr & = {e^u} + C \cr & {\text{write in terms of }}x;{\text{ replace }}{\sin ^{ - 1}}x{\text{ for }}u \cr & = {e^{{{\sin }^{ - 1}}x}} + C \cr}

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