I.F. $\mathrm{e}^{-\frac{1}{2} \int \frac{2 \mathrm{x}}{1-\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{-\frac{1}{2} \ln \left(1-\mathrm{x}^2\right)}=\sqrt{1-\mathrm{x}^2}$

$$y \times \sqrt{1-x^2}=\int\left(x^6+4 x\right) d x=\frac{x^7}{7}+2 x^2+c$$

Given $y(0)=0 \Rightarrow c=0$

$$y=\frac{\frac{x^7}{7}+2 x^2}{\sqrt{1-x^2}}$$

$$\begin{aligned} &\text { Now, } 6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\frac{x^7}{7}+2 x^2}{\sqrt{1-x^2}} d x=6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{2 x^2}{\sqrt{1-x^2}} d x\\ &=24 \int_0^{\frac{1}{2}} \frac{\mathrm{x}^2}{\sqrt{1-\mathrm{x}^2}} \mathrm{dx} \end{aligned}$$

$$\begin{aligned} & \text { Put } x=\sin \theta \\ & d x=\cos \theta d \theta \\ & =24 \int_0^{\frac{\pi}{6}} \frac{\sin ^2 \theta}{\cos \theta} \cos \theta d \theta \\ & =24 \int_0^{\frac{\pi}{6}}\left(\frac{1-\cos 2 \theta}{2}\right) d \theta=12\left[\theta-\frac{\sin 2 \theta}{2}\right]_0^{\frac{\pi}{6}} \\ & =12\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) \\ & =2 \pi-3 \sqrt{3} \\ & \alpha^2=(3 \sqrt{3})^2=27 \end{aligned}$$