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 $\_\_\_\_$