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

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