Answer:

r For the first part, write r \(I_n=\int_0^{\frac{\pi}{2}}\cos^{n-1}x\,\cos x\,dx\). Integrating by parts with \(u=\cos^{n-1}x\) and \(dv=\cos x\,dx\), we get \(du=-(n-1)\cos^{n-2}x\sin x\,dx\) and \(v=\sin x\).

So \(I_n=\left[\cos^{n-1}x\sin x\right]_0^{\frac{\pi}{2}}+(n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}x\sin^2x\,dx\). The bracketed term is zero at both limits, hence \(I_n=(n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}x(1-\cos^2x)\,dx\).

This gives \(I_n=(n-1)\left(I_{n-2}-I_n\right)\), so \(nI_n=(n-1)I_{n-2}\). Therefore \(I_n=\frac{n-1}{n}I_{n-2}\) for \(n\ge 2\).

For the curve, the mean value of \(y\) over \(0\le x\le a\) is \(m=\frac{1}{a}\int_0^a y\,dx\).

Using the parametric equations \(x=a\sin^3t\), \(y=a\cos^3t\) with \(0\le t\le \frac{\pi}{2}\), we have \(\frac{dx}{dt}=3a\sin^2t\cos t\). Hence \(m=\frac{1}{a}\int_0^{\frac{\pi}{2}} a\cos^3t\,(3a\sin^2t\cos t)\,dt=3a\int_0^{\frac{\pi}{2}}\cos^4t\sin^2t\,dt\).

Now \(\sin^2t=1-\cos^2t\), so \(m=3a\int_0^{\frac{\pi}{2}}\left(\cos^4t-\cos^6t\right)dt\), as required.

To find the exact value, use the reduction formula.

Let \(I_n=\int_0^{\frac{\pi}{2}}\cos^n x\,dx\). Then \(I_0=\frac{\pi}{2}\), and \(I_2=\frac{1}{2}I_0=\frac{\pi}{4}\), \(I_4=\frac{3}{4}I_2=\frac{3\pi}{16}\), \(I_6=\frac{5}{6}I_4=\frac{5\pi}{32}\).

Therefore \(m=3a(I_4-I_6)=3a\left(\frac{3\pi}{16}-\frac{5\pi}{32}\right)=3a\cdot\frac{\pi}{32}=\frac{3\pi a}{32}\).

For \(I_n=\int_0^{\frac{\pi}{2}}\cos^n x\,dx\), write \(\cos^n x=\cos^{n-1}x\cdot \cos x\) and integrate by parts with \(u=\cos^{n-1}x\), \(dv=\cos x\,dx\).

Then \(du=-(n-1)\cos^{n-2}x\sin x\,dx\) and \(v=\sin x\), so

\(I_n=\left[\cos^{n-1}x\sin x\right]_0^{\frac{\pi}{2}}+(n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}x\sin^2x\,dx\).

The boundary term is zero because \(\sin 0=0\) and \(\cos\left(\frac{\pi}{2}\right)=0\). Hence

\(I_n=(n-1)\int_0^{\frac{\pi}{2}}\cos^{n-2}x\sin^2x\,dx\).

Using \(\sin^2x=1-\cos^2x\),

\(I_n=(n-1)\int_0^{\frac{\pi}{2}}\left(\cos^{n-2}x-\cos^n x\right)dx=(n-1)(I_{n-2}-I_n)\).

So \(nI_n=(n-1)I_{n-2}\), and therefore

\(I_n=\frac{n-1}{n}I_{n-2}\), for \(n\ge 2\).

For the curve, the mean value of \(y\) over \(0\le x\le a\) is

\(m=\frac{1}{a}\int_0^a y\,dx\).

Now \(x=a\sin^3t\) and \(y=a\cos^3t\), with \(0\le t\le \frac{\pi}{2}\). Also

\(\frac{dx}{dt}=3a\sin^2t\cos t\).

So

\(m=\frac{1}{a}\int_0^{\frac{\pi}{2}} a\cos^3t\cdot 3a\sin^2t\cos t\,dt=3a\int_0^{\frac{\pi}{2}}\cos^4t\sin^2t\,dt\).

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

\(m=3a\int_0^{\frac{\pi}{2}}\left(\cos^4t-\cos^6t\right)dt\).

To evaluate it exactly, use the reduction formula for \(I_n=\int_0^{\frac{\pi}{2}}\cos^n x\,dx\).

First, \(I_0=\frac{\pi}{2}\). Then

\(I_2=\frac{1}{2}I_0=\frac{\pi}{4}\),

\(I_4=\frac{3}{4}I_2=\frac{3\pi}{16}\),

\(I_6=\frac{5}{6}I_4=\frac{5\pi}{32}\).

Hence

\(m=3a(I_4-I_6)=3a\left(\frac{3\pi}{16}-\frac{5\pi}{32}\right)=3a\cdot\frac{\pi}{32}=\frac{3\pi a}{32}\).