If the value of the integral int_{0}^{1/2} fr | vedclass

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(A) Let $I = \int_{0}^{1/2} \frac{x^{2}}{(1-x^{2})^{3/2}} dx$. We can rewrite the numerator as $x^{2} = (x^{2}-1) + 1$. Then,$I = \int_{0}^{1/2} \frac

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$2\sqrt{3}-\pi$

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(A) Let $I = \int_{0}^{1/2} \frac{x^{2}}{(1-x^{2})^{3/2}} dx$. We can rewrite the numerator as $x^{2} = (x^{2}-1) + 1$. Then,$I = \int_{0}^{1/2} \frac{x^{2}-1}{(1-x^{2})^{3/2}} dx + \int_{0}^{1/2} \frac{1}{(1-x^{2})^{3/2}} dx$. $I = -\int_{0}^{1/2} \frac{1}{(1-x^{2})^{1/2}} dx + \int_{0}^{1/2} \frac{1}{(1-x^{2})^{3/2}} dx$. For the first integral,$\int \frac{1}{\sqrt{1-x^{2}}} dx = \sin^{-1}(x)$. Evaluating from $0$ to $1/2$,we get $-[\sin^{-1}(1/2) - \sin^{-1}(0)] = -\pi/6$. For the second integral,let $x = \sin(	heta)$,then $dx = \cos(	heta) d	heta$. When $x=0, 	heta=0$; when $x=1/2, 	heta=\pi/6$. $\int_{0}^{\pi/6} \frac{\cos(	heta)}{\cos^{3}(	heta)} d	heta = \int_{0}^{\pi/6} \sec^{2}(	heta) d	heta = [	an(	heta)]_{0}^{\pi/6} = 	an(\pi/6) - 	an(0) = 1/\sqrt{3}$. Thus,$I = \frac{1}{\sqrt{3}} - \frac{\pi}{6} = \frac{\sqrt{3}}{3} - \frac{\pi}{6} = \frac{2\sqrt{3}-\pi}{6}$. Comparing this with $\frac{k}{6}$,we get $k = 2\sqrt{3}-\pi$.

If the value of the integral $\int_{0}^{1/2} \frac{x^{2}}{(1-x^{2})^{3/2}} dx$ is $\frac{k}{6}$,then $k$ is equal to

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