Using the substitution \(u = \tan x\), or otherwise, find \(\int \sec ^ { 2 } x \tan ^ { 2 } x \mathrm {~d} x\).
It is given that, for \(n \geqslant 0\),
$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sec ^ { n } x \tan ^ { 2 } x \mathrm {~d} x$$
Using the result that \(\frac { \mathrm { d } } { \mathrm { d } x } ( \sec x ) = \tan x \sec x\), show that, for \(n \geqslant 2\),
$$( n + 1 ) I _ { n } = ( \sqrt { } 2 ) ^ { n - 2 } + ( n - 2 ) I _ { n - 2 }$$
Hence find the mean value of \(\sec ^ { 4 } x \tan ^ { 2 } x\) with respect to \(x\) over the interval \(0 \leqslant x \leqslant \frac { 1 } { 4 } \pi\), giving your answer in exact form.