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

Let \(\mathrm{I}=\int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \mathrm{~d} x\)
Put \(x=\sin \theta\)
\(\begin{aligned} & \Rightarrow \mathrm{d} x=\cos \theta \mathrm{d} \theta \\ & \left(1-x^2\right)^{\frac{3}{2}}=\left(1-\sin ^2 \theta\right)^{\frac{3}{2}} \\ & =\left(\cos ^2 \theta\right)^{\frac{3}{2}} \\ & =\cos ^3 \theta \\ & \therefore \quad I=\int_0^{\frac{\pi}{6}} \frac{\sin ^2 \theta \cdot \cos \theta d \theta}{\cos ^3 \theta} \\ & =\int_0^{\frac{\pi}{6}} \tan ^2 \theta d \theta \\ & =\int_0^{\frac{\pi}{6}}\left(\sec ^2 \theta-1\right) d \theta \\ & =[\tan \theta]_0^{\frac{\pi}{6}}-[\theta]_0^{\frac{\pi}{6}} \\ & =\left(\tan \frac{\pi}{6}-\tan 0\right)-\left(\frac{\pi}{6}-0\right) \\ & =\frac{1}{\sqrt{3}}-\frac{\pi}{6} \\ & =\frac{\sqrt{3}}{3}-\frac{\pi}{6} \\ & =\frac{2 \sqrt{3}-\pi}{6} \\ & \text { But, } \int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \mathrm{~d} x=\frac{\mathrm{k}}{6} \quad \ldots \text { [Given] } \\ & \therefore \quad \mathrm{k}=2 \sqrt{3}-\pi \\ & \end{aligned}\)