Integral $ \int_a^b ((b-x)(x-a))^{n-1}x^{\frac12-n}dx $ How can I prove the integral equality below
$$ \int_a^b ((b-x)(x-a))^{n-1}x^{\frac{1}{2}-n}dx = \frac{2^{2n-1}( (n-1)!)^2}{(2n-1)!}\left(\sqrt b-\sqrt a\right)^{2n-1} $$
I have tried to find a recurrence integrating by parts:
$$I_n:= \int_a^b ((b-x)(x-a))^{n-1}x^{\frac{1}{2}-n}dx\\=\int_a^b \frac{2}{3-2n} ((b-x)(x-a))^{n-1}(x^{\frac{3}{2}-n})'dx\\=\int_a^b \frac{2(n-1)}{2n-3} ((b-x)(x-a))^{n-2}(a+b-2x)x^{\frac{3}{2}-n} dx\\=\frac{2(a+b)(n-1)}{2n-3}I_{n-1}-\frac{4(n-1)}{2n-3} \int_a^b ((b-x)(x-a))^{n-2}x^{\frac{5}{2}-n}dx$$
But, I don't know how to continue.
Maybe it is related to Beta function :
$$B(n,1/2)=\frac{2^{2n-1}}{(2n-1)!}((n-1)!)^2$$

Indeed, the integral can be recast into the beta function $B(n,1/2)$ as shown below. Substitute $x=\sqrt{ab}\>t^2$ to reexpress the integral as
\begin{align}
I= & \int_a^b ((b-x)(x-a))^{n-1}x^{\frac12-n}dx \\
=& 2 \sqrt[4]{ab}\int_{\sqrt[4]\frac ab }^{\sqrt[4]\frac ba } \left(b+a-\sqrt{ab}(t^2+\frac1{t^2})\right)^{n-1}dt \\
 \overset{t\to\frac1t}=& \sqrt[4]{ab}\int_{\sqrt[4]\frac ab }^{\sqrt[4]\frac ba } \left(b+a-\sqrt{ab}(t^2+\frac1{t^2})\right)^{n-1}(1+\frac1{t^2})dt \\
=& \sqrt[4]{ab}\int_{\sqrt[4]\frac ab }^{\sqrt[4]\frac ba } \left((\sqrt b-\sqrt a)^2-\sqrt{ab}(t-\frac1{t})^2\right)^{n-1}d(t-\frac1{t})\\
=& \>(\sqrt b-\sqrt a)^{2n-1}\int_{-1}^1 \left(1-y^2\right)^{n-1}dy\\
\end{align}
where the substitute $y= \frac{\sqrt[4]{ab}\>(t-\frac1t)}{\sqrt b-\sqrt a}$ is used in the last step. Note that $B(1,1/2)=2$
\begin{align}
\int_{-1}^1 \left(1-y^2\right)^{n-1}dy
& = \int_{0}^1 \left(1-t\right)^{n-1}\>t^{-\frac12}\>dt\\
&=B\left(n, \frac12\right)
= \frac{2(n-1)}{2n-1} B\left(n-1, \frac12\right)=\cdots\\
&= \frac{2^{2n-1}\> ((n-1)!)^2 }{(2n-1)!}
\end{align}
Thus
\begin{align}
I=\frac{2^{2n-1}\> ((n-1)!)^2 }{(2n-1)!}\left(\sqrt b-\sqrt a\right)^{2n-1}
\end{align}
