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.
9 (i) Using the substitution $u = \tan x$, or otherwise, find $\int \sec ^ { 2 } x \tan ^ { 2 } x \mathrm {~d} x$.\\
It is given that, for $n \geqslant 0$,
$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \sec ^ { n } x \tan ^ { 2 } x \mathrm {~d} x$$
(ii) Using the result that $\frac { \mathrm { d } } { \mathrm { d } x } ( \sec x ) = \tan x \sec x$, show that, for $n \geqslant 2$,
$$( n + 1 ) I _ { n } = ( \sqrt { } 2 ) ^ { n - 2 } + ( n - 2 ) I _ { n - 2 }$$
(iii) 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.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q9}}