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Question Number 142886 by ZiYangLee last updated on 06/Jun/21

$$\mathrm{Evaluate}{\int }_0^{\frac{1}{2}}\frac{4x^2}{\sqrt{1-x^2}}dx$$

Answered by Ar Brandon last updated on 06/Jun/21

$$={\int }_0^{\frac{\pi }{6}}{4\sin }^2\varphi \mathrm{d}\varphi =2{\int }_0^{\frac{\pi }{6}}(1-\cos 2\varphi )\mathrm{d}\varphi =2(\frac{\pi }{6}-\frac{\sqrt{3}}{4})$$

Answered by mathmax by abdo last updated on 07/Jun/21

$$\mathrm{I}={\int }_0^{\frac{1}{2}}\frac{{4\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\Rightarrow \mathrm{I}=-4{\int }_0^{\frac{1}{2}}\frac{1-{\mathrm{x}}^2-1}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}$$$$=-4{\int }_0^{\frac{1}{2}}\sqrt{1-{\mathrm{x}}^2}\mathrm{dx}+4{\int }_0^{\frac{1}{2}}\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}$$$${\int }_0^{\frac{1}{2}}\sqrt{1-{\mathrm{x}}^2}\mathrm{dx}{=}_{\mathrm{x}=\sin \theta }{\int }_0^{\frac{\pi }{6}}{\cos }^2\theta \mathrm{d}\theta ={\int }_0^{\frac{\pi }{6}}\frac{1+\cos \left(2\theta \right)}{2}\mathrm{d}\theta$$$$=\frac{\pi }{12}+\frac{1}{4}{\left[\sin \left(2\theta \right)\right]}_0^{\frac{\pi }{6}}=\frac{\pi }{12}+\frac{1}{4}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{12}+\frac{\sqrt{3}}{8}$$$${\int }_0^{\frac{1}{2}}\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}={\left[\mathrm{arcsinx}\right]}_0^{\frac{1}{2}}=\frac{\pi }{6}\Rightarrow$$$$\mathrm{I}=\frac{2\pi }{3}-4(\frac{\pi }{12}+\frac{\sqrt{3}}{8})=\frac{2\pi }{3}-\frac{\pi }{3}-\frac{\sqrt{3}}{2}=\frac{\pi }{3}-\frac{\sqrt{3}}{2}$$

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