9 Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x \sin x \mathrm {~d} x\). Given that \(I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x ^ { n } \sin x \mathrm {~d} x\), prove that, for \(n > 1\), $$I _ { n } = n \left( \frac { 1 } { 2 } \pi \right) ^ { n - 1 } - n ( n - 1 ) I _ { n - 2 }$$ By first using the substitution \(x = \cos ^ { - 1 } u\), find the value of $$\int _ { 0 } ^ { 1 } \left( \cos ^ { - 1 } u \right) ^ { 3 } \mathrm {~d} u$$ giving your answer in an exact form.