5 It is given that, for \(n \geqslant 0\),
$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \tan ^ { n } x \mathrm {~d} x$$
- By considering \(I _ { n } + I _ { n - 2 }\), or otherwise, show that, for \(n \geqslant 2\),
$$( n - 1 ) \left( I _ { n } + I _ { n - 2 } \right) = 1 .$$
- Find \(I _ { 4 }\) in terms of \(\pi\).