Answer is: 27
\(I.F. \mathrm{e}^{-\frac{1}{2} \int \frac{2 \mathrm{x}}{1-x^{2}} \mathrm{dx}}=\mathrm{e}^{-\frac{1}{2} \ell n \left(1-\mathrm{x}^{2}\right)}=\sqrt{1-\mathrm{x}^{2}}\)
\(y \times \sqrt{1-x^{2}}=\int\left(x^{6}+4 x\right) d x=\frac{x^{7}}{7}+2 x^{2}+c\)
Given \(\mathrm{y}(0)=0 \Rightarrow \mathrm{c}=0\)
\(y=\frac{\frac{x^{7}}{7}+2 x^{2}}{\sqrt{1-x^{2}}}\)
Now, \(6 \int^{\frac{1}{2}}_{-\frac{1}{2}} \frac{\frac{x^7}{7}+2x^2}{\sqrt{1-x^2}}dx=6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{2 x^{2}}{\sqrt{1-x^{2}}} d x\)
\(=24 \int_{0}^{\frac{1}{2}} \frac{\mathrm{x}^{2}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}\)
Put \(\mathrm{x}=\sin \theta\)
\(\mathrm{dx}=\cos \theta \mathrm{d} \theta\)
\(=24 \int_{0}^{\frac{\pi}{6}} \frac{\sin ^{2} \theta}{\cos \theta} \cos \theta \mathrm{~d} \theta\)
\(=24 \int_{0}^{\frac{\pi}{6}}\left(\frac{1-\cos 2 \theta}{2}\right) \mathrm{d} \theta=12\left[\theta-\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{6}}\)
\(=12\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)\)
\(=2 \pi-3 \sqrt{3}\)
\(\alpha^{2}=(3 \sqrt{3})^{2}=27\)
