Let $y=f(x)$ be the solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{x y}{x^2-1}=\frac{x^6+4 x}{\sqrt{1-x^2}},-1< x<1$ such that $f(0)=0$. If $6 \int_{-1 / 2}^{1 / 2} f(x) \mathrm{d} x=2 \pi-\alpha$ then $\alpha^2$ is equal to _________ .

<p>I.F. $\mathrm{e}^{-\frac{1}{2} \int \frac{2 \mathrm{x}}{1-\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{-\frac{1}{2} \ln \left(1-\mathrm{x}^2\right)}=\sqrt{1-\mathrm{x}^2}$</p> <p>$y \times \sqrt{1-x^2}=\int\left(x^6+4 x\right) d x=\frac{x^7}{7}+2 x^2+c$</p> <p>Given $y(0)=0 \Rightarrow c=0$</p> <p>$y=\frac{\frac{x^7}{7}+2 x^2}{\sqrt{1-x^2}}$</p> <p>$$\begin{aligned} &\text { Now, } 6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\frac{x^7}{7}+2 x^2}{\sqrt{1-x^2}} d x=6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{2 x^2}{\sqrt{1-x^2}} d x\\ &=24 \int_0^{\frac{1}{2}} \frac{\mathrm{x}^2}{\sqrt{1-\mathrm{x}^2}} \mathrm{dx} \end{aligned}$$</p> <p>$$\begin{aligned} & \text { Put } x=\sin \theta \\ & d x=\cos \theta d \theta \\ & =24 \int_0^{\frac{\pi}{6}} \frac{\sin ^2 \theta}{\cos \theta} \cos \theta d \theta \\ & =24 \int_0^{\frac{\pi}{6}}\left(\frac{1-\cos 2 \theta}{2}\right) d \theta=12\left[\theta-\frac{\sin 2 \theta}{2}\right]_0^{\frac{\pi}{6}} \\ & =12\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) \\ & =2 \pi-3 \sqrt{3} \\ & \alpha^2=(3 \sqrt{3})^2=27 \end{aligned}$$</p>