Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 63

$$\int \frac{(\sin^{-1}x)^2}{\sqrt{1-x^2}}dx=\frac{1}{3}(\sin^{-1}x)^3+C$$

$$A=\int \frac{(\sin^{-1}x)^2}{\sqrt{1-x^2}}dx$$ We set $u=\sin^{-1}x$ Then $$du=\frac{1}{\sqrt{1-x^2}}dx$$ Therefore, $$A=\int u^2du=\frac{u^3}{3}+C$$ $$A=\frac{1}{3}(\sin^{-1}x)^3+C$$