11 Show that \(\int x \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x = - \frac { 1 } { 3 } \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } + c\), where \(c\) is a constant. Given that \(I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x\), prove that, for \(n \geqslant 2\), $$( n + 2 ) I _ { n } = ( n - 1 ) I _ { n - 2 }$$ Use the substitution \(x = \sin u\) to show that $$\int _ { 0 } ^ { 1 } \left( 1 - x ^ { 2 } \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x = \frac { 1 } { 4 } \pi$$ Find \(I _ { 4 }\).