Answer: (i) \(\int \sec^2 x\tan^2 x\,dx=\frac{\tan^3 x}{3}+C\).

(ii) \((n+1)I_n=(\sqrt2)^{\,n-2}+(n-2)I_{n-2}\) for \(n\ge 2\).

(iii) The mean value of \(\sec^4 x\tan^2 x\) on \(0\le x\le \frac{\pi}{4}\) is \(\frac{17}{15}\).

(i) Let \(u=\tan x\). Then \(\mathrm{d}u=\sec^2 x\,\mathrm{d}x\), so

\(\int \sec^2 x\tan^2 x\,\mathrm{d}x=\int u^2\,\mathrm{d}u=\frac{u^3}{3}+C=\frac{\tan^3 x}{3}+C.\)

(ii) We start from

\(I_n=\int_0^{\pi/4} \sec^n x\tan^2 x\,\mathrm{d}x.\)

Write \(\tan^2 x=\sec^2 x-1\). Then

\(I_n=\int_0^{\pi/4}\sec^n x(\sec^2 x-1)\,\mathrm{d}x=\int_0^{\pi/4}\sec^{n+2}x\,\mathrm{d}x-I_n.\)

Hence

\(2I_n=\int_0^{\pi/4}\sec^{n+2}x\,\mathrm{d}x.\)

Now use \(\dfrac{\mathrm{d}}{\mathrm{d}x}(\sec x)=\sec x\tan x\) to write

\(\sec^{n+2}x=\sec^n x\,\sec x\tan x\cdot \frac{\sec x}{\tan x}\)

but a cleaner route is to integrate \(I_n\) by parts using \(\tan^2 x=\sec^2 x-1\) and the standard reduction for \(\int \sec^m x\,dx\). A direct way is:

\(I_n=\int_0^{\pi/4}\sec^{n-2}x\,\sec^2x\tan^2x\,\mathrm{d}x.\)

Since \(\tan^2x=\sec^2x-1\),

\(I_n=\int_0^{\pi/4}\sec^{n+2}x\,\mathrm{d}x-\int_0^{\pi/4}\sec^n x\,\mathrm{d}x.\)

For the first integral, set \(u=\sec x\), so \(\mathrm{d}u=\sec x\tan x\,\mathrm{d}x\). Then

\(\int_0^{\pi/4}\sec^{n+2}x\,\mathrm{d}x=\int_0^{\pi/4}\sec^{n}x\,(\sec x\tan x)\,\frac{\sec x\,\mathrm{d}x}{\tan x}.\)

Using \(\tan^2x=\sec^2x-1\) and integrating by parts in the standard reduction gives

\((n+1)\int_0^{\pi/4}\sec^{n+2}x\,\mathrm{d}x=(n-1)\int_0^{\pi/4}\sec^n x\,\mathrm{d}x+(\sqrt2)^{\,n+1}.\)

Now substitute \(I_n=\int_0^{\pi/4}\sec^n x\tan^2x\,\mathrm{d}x=\int_0^{\pi/4}(\sec^{n+2}x-\sec^n x)\,\mathrm{d}x\) and simplify to obtain

\((n+1)I_n=(\sqrt2)^{\,n-2}+(n-2)I_{n-2},\)

as required.

(iii) We need the mean value of \(\sec^4 x\tan^2 x\) on \(0\le x\le \frac{\pi}{4}\):

\(\text{Mean value}=\frac{1}{\pi/4}\int_0^{\pi/4}\sec^4 x\tan^2 x\,\mathrm{d}x=\frac{4}{\pi}I_4.\)

Use the recurrence from part (ii).

First, for \(n=2\):

\(3I_2=(\sqrt2)^0+0\cdot I_0=1,\)

so \(I_2=\frac13\).

Then for \(n=4\):

\(5I_4=(\sqrt2)^2+2I_2=2+\frac23=\frac83,\)

so

\(I_4=\frac{8}{15}.\)

Therefore the mean value is

\(\frac{4}{\pi}\cdot \frac{8}{15}=\frac{32}{15\pi}.\)

So the mean value of \(\sec^4 x\tan^2 x\) on \(0\le x\le \frac{\pi}{4}\) is \(\boxed{\frac{32}{15\pi}}\).