Answer: (i) Using integration by parts twice, for \(n\ge 2\),

\(I_n=\int_0^{\pi/2} t^n\sin t\,dt\).

Take \(u=t^n\), \(dv=\sin t\,dt\). Then \(du=nt^{n-1}dt\) and \(v=-\cos t\), so

\(I_n=\left[-t^n\cos t\right]_0^{\pi/2}+n\int_0^{\pi/2} t^{n-1}\cos t\,dt\).

Now integrate the remaining integral by parts with \(u=t^{n-1}\), \(dv=\cos t\,dt\). Then \(du=(n-1)t^{n-2}dt\) and \(v=\sin t\), giving

\(\int_0^{\pi/2} t^{n-1}\cos t\,dt=\left[t^{n-1}\sin t\right]_0^{\pi/2}-(n-1)\int_0^{\pi/2} t^{n-2}\sin t\,dt\).

Since \(\int_0^{\pi/2} t^{n-2}\sin t\,dt=I_{n-2}\), we get

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

(ii) The arc length is

\(L=\int_0^{\pi/2}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.\)

Here

\(\frac{dx}{dt}=t^4(1-\cos 2t),\qquad \frac{dy}{dt}=t^4\sin 2t.\)

So

\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=t^8\big((1-\cos 2t)^2+\sin^2 2t\big).\)

Using \((1-\cos 2t)^2+\sin^2 2t=2(1-\cos 2t)=4\sin^2 t\),

\(\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=2t^4\sin t\)

for \(0\le t\le \pi/2\). Hence

\(L=\int_0^{\pi/2}2t^4\sin t\,dt=2I_4.\)

Now use the recurrence:

\(I_4=4\left(\frac{\pi}{2}\right)^3-4\cdot 3\,I_2.\)

Also, from the recurrence with \(n=2\),

\(I_2=2\left(\frac{\pi}{2}\right)-2I_0=\pi-2\), since \(I_0=\int_0^{\pi/2}\sin t\,dt=1\).

Therefore

\(I_4=4\left(\frac{\pi}{2}\right)^3-12(\pi-2)=\frac{\pi^3}{2}-12\pi+24,\)

so

\(L=2I_4=\pi^3-24\pi+48.\)

Thus the arc length is \(\pi^3-24\pi+48\).

(i) Let \(I_n=\int_0^{\pi/2} t^n\sin t\,dt\). We integrate by parts twice.

First, choose \(u=t^n\) and \(dv=\sin t\,dt\). Then \(du=nt^{n-1}dt\) and \(v=-\cos t\). Hence

\(I_n=\left[-t^n\cos t\right]_0^{\pi/2}+n\int_0^{\pi/2} t^{n-1}\cos t\,dt.\)

Now evaluate the remaining integral by parts again, with \(u=t^{n-1}\) and \(dv=\cos t\,dt\). Then \(du=(n-1)t^{n-2}dt\) and \(v=\sin t\), so

\(\int_0^{\pi/2} t^{n-1}\cos t\,dt=\left[t^{n-1}\sin t\right]_0^{\pi/2}-(n-1)\int_0^{\pi/2} t^{n-2}\sin t\,dt.\)

Since the last integral is \(I_{n-2}\), we obtain

\(I_n=\left[-t^n\cos t\right]_0^{\pi/2}+n\left[t^{n-1}\sin t\right]_0^{\pi/2}-n(n-1)I_{n-2}.\)

At \(t=\pi/2\), \(\cos(\pi/2)=0\) and \(\sin(\pi/2)=1\), while at \(t=0\) both boundary terms are zero. Therefore

\(I_n=n\left(\frac{\pi}{2}\right)^{n-1}-n(n-1)I_{n-2},\)

for \(n\ge 2\), as required.

(ii) For a parametric curve, the arc length from \(t=0\) to \(t=\pi/2\) is

\(L=\int_0^{\pi/2}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.\)

Here

\(\frac{dx}{dt}=t^4(1-\cos 2t),\qquad \frac{dy}{dt}=t^4\sin 2t.\)

So

\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=t^8\left((1-\cos 2t)^2+\sin^2 2t\right).\)

Now simplify the bracket:

\((1-\cos 2t)^2+\sin^2 2t=1-2\cos 2t+\cos^2 2t+\sin^2 2t=2-2\cos 2t.\)

Using \(1-\cos 2t=2\sin^2 t\), this becomes

\(2-2\cos 2t=4\sin^2 t.\)

Hence, for \(0\le t\le \pi/2\),

\(\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=2t^4\sin t.\)

Therefore

\(L=\int_0^{\pi/2}2t^4\sin t\,dt=2I_4.\)

Using the recurrence from part (i),

\(I_4=4\left(\frac{\pi}{2}\right)^3-4\cdot 3\,I_2.\)

To find \(I_2\), apply the same formula with \(n=2\):

\(I_2=2\left(\frac{\pi}{2}\right)^1-2\cdot 1\,I_0.\)

Now \(I_0=\int_0^{\pi/2}\sin t\,dt=1\), so

\(I_2=\pi-2.\)

Then

\(I_4=4\left(\frac{\pi}{2}\right)^3-12(\pi-2)=\frac{\pi^3}{2}-12\pi+24.\)

So

\(L=2I_4=\pi^3-24\pi+48.\)

Hence the required arc length is \(\pi^3-24\pi+48\).