If int_0^{frac{1}{2}} frac{x^2}{left(1-x^2rig | vedclass

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$(A)$ Let $I = \int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2ight)^{\frac{3}{2}}} \,d x$. Substitute $x = \sin 	heta$, then $dx = \cos 	heta d	heta$

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$2 \sqrt{3}-\pi$

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$(A)$ Let $I = \int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2ight)^{\frac{3}{2}}} \,d x$. Substitute $x = \sin 	heta$, then $dx = \cos 	heta d	heta$. When $x = 0, 	heta = 0$. When $x = \frac{1}{2}, 	heta = \frac{\pi}{6}$. The denominator becomes $(1 - \sin^2 	heta)^{\frac{3}{2}} = (\cos^2 	heta)^{\frac{3}{2}} = \cos^3 	heta$. Thus, $I = \int_0^{\frac{\pi}{6}} \frac{\sin^2 	heta \cdot \cos 	heta}{\cos^3 	heta} d	heta = \int_0^{\frac{\pi}{6}} 	an^2 	heta d	heta$. Using the identity $	an^2 	heta = \sec^2 	heta - 1$, we get: $I = \int_0^{\frac{\pi}{6}} (\sec^2 	heta - 1) d	heta = [	an 	heta - 	heta]_0^{\frac{\pi}{6}}$. $I = (	an \frac{\pi}{6} - \frac{\pi}{6}) - (	an 0 - 0) = \frac{1}{\sqrt{3}} - \frac{\pi}{6} = \frac{\sqrt{3}}{3} - \frac{\pi}{6} = \frac{2\sqrt{3} - \pi}{6}$. Given that $I = \frac{k}{6}$, comparing the expressions gives $k = 2\sqrt{3} - \pi$.

If $\int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \,d x=\frac{k}{6}$, then the value of $k$ is

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