6 Show that $$\frac { \mathrm { d } } { \mathrm {~d} x } \left[ x ^ { n - 1 } \sqrt { } \left( 4 - x ^ { 2 } \right) \right] = \frac { 4 ( n - 1 ) x ^ { n - 2 } } { \sqrt { } \left( 4 - x ^ { 2 } \right) } - \frac { n x ^ { n } } { \sqrt { } \left( 4 - x ^ { 2 } \right) }$$ Let $$I _ { n } = \int _ { 0 } ^ { 1 } \frac { x ^ { n } } { \sqrt { } \left( 4 - x ^ { 2 } \right) } \mathrm { d } x$$ where \(n \geqslant 0\). Prove that $$n I _ { n } = 4 ( n - 1 ) I _ { n - 2 } - \sqrt { } 3$$ for \(n \geq 2\). Given that \(I _ { 0 } = \frac { 1 } { 6 } \pi\), find \(I _ { 4 }\), leaving your answer in an exact form.