Given that
$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } t ^ { n } \sin t \mathrm {~d} t$$
show that, for \(n \geqslant 2\),
$$I _ { n } = n \left( \frac { \pi } { 2 } \right) ^ { n - 1 } - n ( n - 1 ) I _ { n - 2 } .$$
A curve \(C\) in the \(x - y\) plane is defined parametrically in terms of \(t\). It is given that
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = t ^ { 4 } ( 1 - \cos 2 t ) \quad \text { and } \quad \frac { \mathrm { d } y } { \mathrm {~d} t } = t ^ { 4 } \sin 2 t .$$
Find the length of the arc of \(C\) from the point where \(t = 0\) to the point where \(t = \frac { 1 } { 2 } \pi\).