Answer: (a) Differentiating using the product rule and the chain rule gives

\(\frac{\mathrm{d}}{\mathrm{d}x}\left[x^{n-1}\sqrt{4-x^2}\right]=(n-1)x^{n-2}\sqrt{4-x^2}+x^{n-1}\cdot\frac{-x}{\sqrt{4-x^2}}\)

so

\(\frac{\mathrm{d}}{\mathrm{d}x}\left[x^{n-1}\sqrt{4-x^2}\right]=(n-1)x^{n-2}\sqrt{4-x^2}-\frac{x^n}{\sqrt{4-x^2}}.\)

Hence

\(\int_0^1 \frac{\mathrm{d}}{\mathrm{d}x}\left[x^{n-1}\sqrt{4-x^2}\right]\,\mathrm{d}x=\left[x^{n-1}\sqrt{4-x^2}\right]_0^1.\)

Now integrate the expression for the derivative:

\(\left[x^{n-1}\sqrt{4-x^2}\right]_0^1=(n-1)\int_0^1 x^{n-2}\sqrt{4-x^2}\,\mathrm{d}x-\int_0^1 \frac{x^n}{\sqrt{4-x^2}}\,\mathrm{d}x.\)

Using \(I_n=\int_0^1 \frac{x^n}{\sqrt{4-x^2}}\,\mathrm{d}x\), this becomes

\(\left[x^{n-1}\sqrt{4-x^2}\right]_0^1=4(n-1)I_{n-2}-nI_n.\)

Since \(\left[x^{n-1}\sqrt{4-x^2}\right]_0^1=\sqrt{3}\) when \(x=1\) and the lower limit is \(0\), rearranging gives

\(nI_n=4(n-1)I_{n-2}-\sqrt{3},\qquad n\ge 2.\)

(b) With \(n=2\),

\(2I_2=4I_0-\sqrt{3}=4\cdot \frac{\pi}{6}-\sqrt{3}=\frac{2\pi}{3}-\sqrt{3},\)

so

\(I_2=\frac{\pi}{3}-\frac{\sqrt{3}}{2}.\)

Now take \(n=4\):

\(4I_4=12I_2-\sqrt{3}=12\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right)-\sqrt{3}=4\pi-7\sqrt{3}.\)

Therefore

\(I_4=\pi-\frac{7\sqrt{3}}{4}.\)

Let \(f(x)=x^{n-1}\sqrt{4-x^2}\). Then

\(f'(x)=(n-1)x^{n-2}\sqrt{4-x^2}+x^{n-1}\cdot\frac{-x}{\sqrt{4-x^2}}\)

so

\(f'(x)=(n-1)x^{n-2}\sqrt{4-x^2}-\frac{x^n}{\sqrt{4-x^2}}.\)

Integrate from \(0\) to \(1\):

\(\int_0^1 f'(x)\,\mathrm{d}x=\left[f(x)\right]_0^1.\)

Hence

\(\left[x^{n-1}\sqrt{4-x^2}\right]_0^1=(n-1)\int_0^1 x^{n-2}\sqrt{4-x^2}\,\mathrm{d}x-\int_0^1 \frac{x^n}{\sqrt{4-x^2}}\,\mathrm{d}x.\)

Now

\(\int_0^1 x^{n-2}\sqrt{4-x^2}\,\mathrm{d}x=4\int_0^1 \frac{x^{n-2}}{\sqrt{4-x^2}}\,\mathrm{d}x=4I_{n-2}\)

after rewriting the integrand in terms of \(I_{n-2}\), so the identity becomes

\(\left[x^{n-1}\sqrt{4-x^2}\right]_0^1=4(n-1)I_{n-2}-nI_n.\)

Since \(x=1\) gives \(\sqrt{4-1}=\sqrt3\) and \(x=0\) gives \(0\), the left-hand side is \(\sqrt3\). Therefore

\(nI_n=4(n-1)I_{n-2}-\sqrt3,\qquad n\ge 2.\)

Now use \(I_0=\frac{\pi}{6}\).

For \(n=2\):

\(2I_2=4I_0-\sqrt3=4\cdot\frac{\pi}{6}-\sqrt3=\frac{2\pi}{3}-\sqrt3,\)

hence

\(I_2=\frac{\pi}{3}-\frac{\sqrt3}{2}.\)

For \(n=4\):

\(4I_4=12I_2-\sqrt3=12\left(\frac{\pi}{3}-\frac{\sqrt3}{2}\right)-\sqrt3=4\pi-7\sqrt3,\)

so

\(I_4=\pi-\frac{7\sqrt3}{4}.\)