Answer: The reduction formula is \((2n+3)I_n=2nI_{n-1}\), for \(n\geqslant 1\).

The exact value is \(I_3=\dfrac{32}{315}\).

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

Integrate by parts with \(u=x^n\) and \(dv=(1-x)^{1/2}\,dx\). Then \(du=nx^{n-1}\,dx\) and \(v=-\dfrac{2}{3}(1-x)^{3/2}\).

So

\(I_n=\left[-\dfrac{2}{3}x^n(1-x)^{3/2}\right]_0^1+\dfrac{2}{3}\int_0^1 nx^{n-1}(1-x)(1-x)^{1/2}\,dx\).

The boundary term is zero at both limits. Hence

\(I_n=\dfrac{2n}{3}\int_0^1 x^{n-1}(1-x)^{1/2}\,dx-\dfrac{2n}{3}\int_0^1 x^n(1-x)^{1/2}\,dx\).

Recognising the integrals gives

\(I_n=\dfrac{2n}{3}I_{n-1}-\dfrac{2n}{3}I_n\).

Therefore

\(3I_n=2nI_{n-1}-2nI_n\),

so

\((2n+3)I_n=2nI_{n-1}\).

Now

\(I_0=\int_0^1(1-x)^{1/2}\,dx=\left[-\dfrac{2}{3}(1-x)^{3/2}\right]_0^1=\dfrac{2}{3}\).

Using the recurrence,

\(I_1=\dfrac{2}{5}I_0=\dfrac{4}{15}\),

\(I_2=\dfrac{4}{7}I_1=\dfrac{16}{105}\),

and

\(I_3=\dfrac{6}{9}I_2=\dfrac{2}{3}\cdot\dfrac{16}{105}=\dfrac{32}{315}\).