Answer: (i) \(I_{n+2}=\dfrac{1}{n+1}-I_n\).

(ii) \(\displaystyle \bar y=\frac{1-\frac14\pi}{\ln 2}\).

(i) Differentiate \(\cot^{n+1}x\):

\(\dfrac{\mathrm d}{\mathrm dx}\left(\cot^{n+1}x\right)=-(n+1)\cot^n x\csc^2x\).

Since \(\csc^2x=1+\cot^2x\),

\(\dfrac{\mathrm d}{\mathrm dx}\left(\cot^{n+1}x\right)=-(n+1)\cot^n x-(n+1)\cot^{n+2}x\).

Integrate from \(\dfrac\pi4\) to \(\dfrac\pi2\):

\(\left[\cot^{n+1}x\right]_{\pi/4}^{\pi/2}=-(n+1)(I_n+I_{n+2})\).

Now \(\cot(\pi/2)=0\) and \(\cot(\pi/4)=1\), so the left side is \(-1\). Hence

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

and therefore

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

(ii) The area of the region is

\(A=\int_{\pi/4}^{\pi/2}\cot x\,\mathrm dx=\left[\ln(\sin x)\right]_{\pi/4}^{\pi/2}=\ln\sqrt2=\dfrac12\ln2\).

For a region under \(y=f(x)\),

\(\displaystyle \bar y=\frac{1}{2A}\int_a^b f(x)^2\,\mathrm dx\).

Here

\(\displaystyle \bar y=\frac{1}{2A}\int_{\pi/4}^{\pi/2}\cot^2x\,\mathrm dx=\frac{I_2}{2A}\).

Using part (i) with \(n=0\),

\(I_2=1-I_0\).

Also

\(I_0=\int_{\pi/4}^{\pi/2}1\,\mathrm dx=\dfrac\pi4\),

so

\(I_2=1-\dfrac\pi4\).

Therefore

\(\displaystyle \bar y=\frac{1-\frac14\pi}{2\ln\sqrt2}=\frac{1-\frac14\pi}{\ln2}\).

Answer: The centroid is \(\left(\frac{a}{a+1},\,\frac{a}{2(a+2)}\right)\).

Let the region be under the curve \(y=x^a\) from \(x=0\) to \(x=1\), above the \(x\)-axis.

For a region under a curve \(y=f(x)\), the area is \(A=\int f(x)\,dx\), and the centroid \((\bar{x},\bar{y})\) is given by

\(\bar{x}=\frac{1}{A}\int x f(x)\,dx\), \(\bar{y}=\frac{1}{A}\int \frac{1}{2}f(x)^2\,dx\).

Here \(f(x)=x^a\).

First find the area:

\(A=\int_0^1 x^a\,dx=\left[\frac{x^{a+1}}{a+1}\right]_0^1=\frac{1}{a+1}.\)

Now find \(\bar{x}\):

\(\bar{x}=\frac{1}{A}\int_0^1 x\cdot x^a\,dx=(a+1)\int_0^1 x^{a+1}\,dx\)

\(=(a+1)\left[\frac{x^{a+2}}{a+2}\right]_0^1=\frac{a+1}{a+2}.\)

However, this is the centroid of the area under the curve measured from the \(y\)-axis, and the region extends from \(x=0\) to \(x=1\). Since the standard formula above is already correct, we keep \(\bar{x}=\frac{a+1}{a+2}\). But we must check carefully: the region enclosed by the curve, \(x=1\), and the \(x\)-axis is the area to the left of \(x=1\) under the curve, so this is the correct \(x\)-coordinate.

Now find \(\bar{y}\):

\(\bar{y}=\frac{1}{A}\int_0^1 \frac{1}{2}(x^a)^2\,dx=(a+1)\cdot \frac{1}{2}\int_0^1 x^{2a}\,dx\)

\(=(a+1)\cdot \frac{1}{2}\left[\frac{x^{2a+1}}{2a+1}\right]_0^1=\frac{a+1}{2(2a+1)}.\)

So the centroid is \(\left(\frac{a+1}{a+2},\,\frac{a+1}{2(2a+1)}\right)\).

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 \(\boxed{\frac{32}{15\pi}}\).