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

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Aicharya · Accepted Answer

Let $\mathrm{I}=\int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2ight)^{\frac{3}{2}}} \mathrm{~d} x$ Put $x=\sin 	heta$ $\begin{aligned} & \Rightarrow \mathrm{d} x=\cos 	heta \mathrm{d} 	heta \ & \left(1-x^2ight)^{\frac{3}{2}}=\left(1-\sin ^2 	hetaight)^{\frac{3}{2}} \ & =\left(\cos ^2 	hetaight)^{\frac{3}{2}} \ & =\cos ^3 	heta \ & 	herefore \quad I=\int_0^{\frac{\pi}{6}} \frac{\sin ^2 	heta \cdot \cos 	heta d 	heta}{\cos ^3 	heta} \ & =\int_0^{\frac{\pi}{6}} 	an ^2 	heta d 	heta \ & =\int_0^{\frac{\pi}{6}}\left(\sec ^2 	heta-1ight) d 	heta \ & =[	an 	heta]_0^{\frac{\pi}{6}}-[	heta]_0^{\frac{\pi}{6}} \ & =\left(	an \frac{\pi}{6}-	an 0ight)-\left(\frac{\pi}{6}-0ight) \ & =\frac{1}{\sqrt{3}}-\frac{\pi}{6} \ & =\frac{\sqrt{3}}{3}-\frac{\pi}{6} \ & =\frac{2 \sqrt{3}-\pi}{6} \ & 	ext { But, } \int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2ight)^{\frac{3}{2}}} \mathrm{~d} x=\frac{\mathrm{k}}{6} \quad \ldots 	ext { [Given] } \ & 	herefore \quad \mathrm{k}=2 \sqrt{3}-\pi \ & \end{aligned}$

If $\int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \mathrm{~d} x=\frac{\mathrm{k}}{6}$,…

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