Answer: \(\int x\sqrt{1-x^2}\,\mathrm dx=-\frac13(1-x^2)^{3/2}+C.\)

\(\displaystyle (n+2)I_n=(n-1)I_{n-2}.\)

\(\displaystyle I_0=\frac\pi4,\qquad I_2=\frac\pi{16},\qquad I_4=\frac\pi{32}.\)

First, for the elementary integral, put \(u=1-x^2\). Then \(\mathrm du=-2x\,\mathrm dx\), so

\(\int x\sqrt{1-x^2}\,\mathrm dx=-\frac12\int u^{1/2}\,\mathrm du=-\frac13u^{3/2}+C.\)

Hence

\(\int x\sqrt{1-x^2}\,\mathrm dx=-\frac13(1-x^2)^{3/2}+C.\)

Let

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

For \(n\geq2\), write

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

Use integration by parts with

\(u=x^{n-1},\qquad \mathrm dv=x\sqrt{1-x^2}\,\mathrm dx.\)

Then

\(\mathrm du=(n-1)x^{n-2}\,\mathrm dx,\qquad v=-\frac13(1-x^2)^{3/2}.\)

The boundary term is zero at \(x=0\) and \(x=1\), so

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

Since \((1-x^2)^{3/2}=(1-x^2)\sqrt{1-x^2}\),

\(I_n=\frac{n-1}{3}\int_0^1x^{n-2}(1-x^2)\sqrt{1-x^2}\,\mathrm dx.\)

Thus

\(I_n=\frac{n-1}{3}\left(I_{n-2}-I_n\right).\)

Rearranging gives

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

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

Now

\(I_0=\int_0^1\sqrt{1-x^2}\,\mathrm dx=\frac\pi4,\)

since this is the area of a quarter-circle of radius \(1\). Therefore

\(4I_2=I_0,\qquad I_2=\frac\pi{16},\)

and

\(6I_4=3I_2,\qquad I_4=\frac12I_2=\frac\pi{32}.\)