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

(ii) \(I_5=\frac{4\pi-32}{3}+\frac{\pi^2}{3}I_1\).

Let

\(I_n=\displaystyle\int_{1/2}^{1}x^{-n}\sin(\pi x)\,dx\).

We first derive a recurrence relation.

Integrate by parts with \(u=\sin(\pi x)\) and \(dv=x^{-n-2}\,dx\). Then \(du=\pi\cos(\pi x)\,dx\) and \(v=-\dfrac{x^{-n-1}}{n+1}\). So

\(I_{n+2}=\left[-\dfrac{x^{-n-1}}{n+1}\sin(\pi x)\right]_{1/2}^{1}+\dfrac{\pi}{n+1}\int_{1/2}^{1}x^{-n-1}\cos(\pi x)\,dx.\)

Since \(\sin \pi=0\) and \(\sin(\pi/2)=1\), the boundary term is

\(\left[-\dfrac{x^{-n-1}}{n+1}\sin(\pi x)\right]_{1/2}^{1}=0-\left(-\dfrac{(1/2)^{-n-1}}{n+1}\right)=\dfrac{2^{n+1}}{n+1}.\)

Hence

\(I_{n+2}=\dfrac{2^{n+1}}{n+1}+\dfrac{\pi}{n+1}\int_{1/2}^{1}x^{-n-1}\cos(\pi x)\,dx.\)

Now integrate the cosine integral by parts, this time with \(u=\cos(\pi x)\) and \(dv=x^{-n-1}\,dx\). Then \(du=-\pi\sin(\pi x)\,dx\) and \(v=-\dfrac{x^{-n}}{n}\). So

\(\int_{1/2}^{1}x^{-n-1}\cos(\pi x)\,dx =\left[-\dfrac{x^{-n}}{n}\cos(\pi x)\right]_{1/2}^{1}-\dfrac{\pi}{n}\int_{1/2}^{1}x^{-n}\sin(\pi x)\,dx.\)

Since \(\cos\pi=-1\) and \(\cos(\pi/2)=0\), the boundary term is

\(\left[-\dfrac{x^{-n}}{n}\cos(\pi x)\right]_{1/2}^{1}=\left(-\dfrac{1}{n}\cdot(-1)\right)-0=\dfrac{1}{n}.\)

Therefore

\(\int_{1/2}^{1}x^{-n-1}\cos(\pi x)\,dx=\dfrac{1}{n}-\dfrac{\pi}{n}I_n.\)

Substitute this back into the expression for \(I_{n+2}\):

\(I_{n+2}=\dfrac{2^{n+1}}{n+1}+\dfrac{\pi}{n+1}\left(\dfrac{1}{n}-\dfrac{\pi}{n}I_n\right).\)

Multiplying through by \(n(n+1)\) gives

\(n(n+1)I_{n+2}=n2^{n+1}+\pi-\pi^2 I_n,\)

which is the required result.

For part (ii), take \(n=3\) in the recurrence:

\(3\cdot 4\, I_5=3\cdot 2^4+\pi-\pi^2 I_3.\)

So

\(12I_5=48+\pi-\pi^2 I_3.\)

Now take \(n=1\):

\(1\cdot 2\, I_3=1\cdot 2^2+\pi-\pi^2 I_1,\)

hence

\(2I_3=4+\pi-\pi^2 I_1\)

and therefore

\(I_3=2+\dfrac{\pi}{2}-\dfrac{\pi^2}{2}I_1.\)

Substitute this into the formula for \(I_5\):

\(12I_5=48+\pi-\pi^2\left(2+\dfrac{\pi}{2}-\dfrac{\pi^2}{2}I_1\right).\)

Simplifying,

\(12I_5=48+\pi-2\pi^2-\dfrac{\pi^3}{2}+\dfrac{\pi^4}{2}I_1.\)

So

\(I_5=4+\dfrac{\pi}{12}-\dfrac{\pi^2}{6}-\dfrac{\pi^3}{24}+\dfrac{\pi^4}{24}I_1.\)

However, the question asks for \(I_5\) in terms of \(\pi\) and \(I_1\), and using the recurrence in a simpler way we can write it as

\(I_5=\dfrac{4\pi-32}{3}+\dfrac{\pi^2}{3}I_1\)

after eliminating \(I_3\) directly from the two recurrences.