# Chapter 5 - Practice Exercises - Page 343: 70

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

#### Work Step by Step

$$A=\int \frac{\sqrt{\sin^{-1}x}dx}{\sqrt{1-x^2}}$$ We set $a=\sin^{-1}x$, which means $$da=\frac{dx}{\sqrt{1-x^2}}$$ Therefore, $$A=\int \sqrt ada=\int a^{1/2}da$$ $$A=\frac{a^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{2a^{3/2}}{3}+C$$ $$A=\frac{2(\sin^{-1}x)^{3/2}}{3}+C$$

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