Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 66

$$\int \frac{dy}{(\sin^{-1}y)\sqrt{1-y^2}}=\ln|\sin^{-1}y|+C$$

$$A=\int \frac{dy}{(\sin^{-1}y)\sqrt{1-y^2}}$$ We set $u=\sin^{-1}y$ Then $$du=\frac{dy}{\sqrt{1-y^2}}$$ Therefore, $$A=\int \frac{1}{u}du=\ln|u|+C$$ $$A=\ln|\sin^{-1}y|+C$$