# Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 338: 42

$$\int^{\sqrt[3]2/4}_{0}\frac{ds}{\sqrt{9-4s^2}}=\frac{1}{2}\Big(\sin^{-1}\frac{\sqrt[3]2}{3}\Big)$$

#### Work Step by Step

$$A=\int^{\sqrt[3]2/4}_{0}\frac{ds}{\sqrt{9-4s^2}}=\int^{\sqrt[3]2/4}_{0}\frac{ds}{\sqrt{3^2-(2s)^2}}$$ We set $u=2s$, which means $$du=2ds$$ $$ds=\frac{1}{2}du$$ For $s=\sqrt[3]2/4$, we have $$u=\frac{\sqrt[3]2}{2}=\frac{2^{1/3}}{2}=2^{-2/3}$$ For $s=0$, we have $$u=0$$ Therefore, $$A=\frac{1}{2}\int^{2^{-2/3}}_{0}\frac{ds}{\sqrt{3^2-u^2}}$$ Since we have $$\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\Big(\frac{x}{a}\Big)+C$$ Therefore, $$A=\frac{1}{2}\sin^{-1}\Big(\frac{2s}{3}\Big)\Big]^{2^{-2/3}}_{0}$$ $$A=\frac{1}{2}\Big(\sin^{-1}\frac{2\times2^{-2/3}}{3}-\sin^{-1}0\Big)$$ $$A=\frac{1}{2}\Big(\sin^{-1}\frac{\sqrt[3]2}{3}\Big)$$

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