Answer:

Let \(I_n=\int_0^2 x^n(4-x^2)^{1/2}\,dx\).

Differentiate \(x^n(4-x^2)^{3/2}\):

\(\dfrac{d}{dx}\{x^n(4-x^2)^{3/2}\}=nx^{n-1}(4-x^2)^{3/2}-3x^{n+1}(4-x^2)^{1/2}.\)

Integrate from \(0\) to \(2\). Since \((4-x^2)^{3/2}=0\) at both endpoints,

\(0=\int_0^2 \left(nx^{n-1}(4-x^2)^{3/2}-3x^{n+1}(4-x^2)^{1/2}\right)dx.\)

Now write \((4-x^2)^{3/2}=(4-x^2)(4-x^2)^{1/2}\):

\(0=n\int_0^2 x^{n-1}(4-x^2)^{1/2}(4-x^2)\,dx-3I_{n+1}.\)

So

\(0=n\left(4I_{n-1}-I_{n+1}\right)-3I_{n+1}=4nI_{n-1}-(n+3)I_{n+1},\)

hence \((n+3)I_{n+1}=4nI_{n-1}\) for \(n\ge 2\).

To find \(I_1\), note that

\(\dfrac{d}{dx}(4-x^2)^{3/2}=-3x(4-x^2)^{1/2}.\)

Therefore

\(I_1=\int_0^2 x(4-x^2)^{1/2}\,dx=-\dfrac13\int_0^2 \dfrac{d}{dx}(4-x^2)^{3/2}\,dx\)

\(\qquad =-\dfrac13\left[(4-x^2)^{3/2}\right]_0^2=-\dfrac13(0-8)=\dfrac83.\)

Now use the recurrence with \(n=2\):

\(5I_3=8I_1=8\cdot \dfrac83=\dfrac{64}{3},\)

so

\(I_3=\dfrac{64}{15}.\)