Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 62

$$\int \frac{e^{\cos^{-1}x}dx}{\sqrt{1-x^2}}=-e^{\cos^{-1}x}+C$$

$$A=\int \frac{e^{\cos^{-1}x}dx}{\sqrt{1-x^2}}$$ We set $u=\cos^{-1}x$ Then $$du=-\frac{1}{\sqrt{1-x^2}}dx$$ $$\frac{1}{\sqrt{1-x^2}}dx=-du$$ Therefore, $$A=-\int e^udu=-e^u+C$$ $$A=-e^{\cos^{-1}x}+C$$