Answer: (i) \(I_{2}=1\).

(ii) For \(n\gt 2\), \((n-1)I_n=2^{\frac{n}{2}-1}+(n-2)I_{n-2}\).

(iii) The volume of revolution is \(\frac{28\pi}{15}\).

(i) \(I_2=\int_0^{\frac{\pi}{4}} \sec^2 x\,\mathrm{d}x=[\tan x]_0^{\frac{\pi}{4}}=1-0=1.\)

(ii) Start from \(I_n=\int_0^{\frac{\pi}{4}} \sec^{n-2}x\,\sec^2x\,\mathrm{d}x\). Integrate by parts with \(u=\sec^{n-2}x\) and \(\mathrm{d}v=\sec^2x\,\mathrm{d}x\). Then \(\mathrm{d}u=(n-2)\sec^{n-3}x\sec x\tan x\,\mathrm{d}x\) and \(v=\tan x\).

So

\(I_n=\left[\sec^{n-2}x\tan x\right]_0^{\frac{\pi}{4}}-(n-2)\int_0^{\frac{\pi}{4}} \sec^{n-3}x\,(\sec x\tan x)\tan x\,\mathrm{d}x.\)

Since \(\sec x\tan x\tan x=\sec x\,\tan^2x\), this becomes

\(I_n=\left[\sec^{n-2}x\tan x\right]_0^{\frac{\pi}{4}}-(n-2)\int_0^{\frac{\pi}{4}} \sec^{n-2}x\,\tan^2x\,\mathrm{d}x.\)

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

\(I_n=\left[\sec^{n-2}x\tan x\right]_0^{\frac{\pi}{4}}-(n-2)\int_0^{\frac{\pi}{4}} \sec^{n-2}x(\sec^2x-1)\,\mathrm{d}x.\)

The boundary term is

\(\left[\sec^{n-2}x\tan x\right]_0^{\frac{\pi}{4}}=(\sqrt{2})^{n-2}\cdot 1-1\cdot 0=2^{\frac{n}{2}-1}.\)

Also,

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

Hence

\(I_n=2^{\frac{n}{2}-1}-(n-2)(I_n-I_{n-2}).\)

Rearranging gives

\((n-1)I_n=2^{\frac{n}{2}-1}+(n-2)I_{n-2}.\)

(iii) Rotating the region about the \(x\)-axis gives volume

\(V=\pi\int_0^{\frac{\pi}{4}} y^2\,\mathrm{d}x=\pi\int_0^{\frac{\pi}{4}} \sec^6x\,\mathrm{d}x=\pi I_6.\)

Use the recurrence for \(n=4\):

\(3I_4=2^{1}+2I_2=2+2(1)=4,\)

so \(I_4=\frac{4}{3}.\)

Then for \(n=6\):

\(5I_6=2^{2}+4I_4=4+4\cdot\frac{4}{3}=\frac{28}{3},\)

so \(I_6=\frac{28}{15}.\)

Therefore

\(V=\pi I_6=\frac{28\pi}{15}.\)