Showing $\int_0 ^{x_1}(1-z)^{1/2}z^{1/\nu -1}dz = \frac{\Gamma(1/\nu)\Gamma(3/2) }{\Gamma(1/\nu +3/2)}$

Question

I encountered the following integral: $$ \int_0 ^{x_1}(1-z)^{1/2}z^{1/\nu -1}dz. $$ My textbook claims that the solution can be found on a integral table, and is $$ \frac{\Gamma(1/\nu)\Gamma(3/2) }{\Gamma(1/\nu +3/2)}. $$ However, I would love to be able to get to this answer somehow. I know that the Gamma function is defined as $$ \Gamma(z)=\int_0^\infty e^{-x}x^{z-1}dx. $$ While there are some similarities, I can't really think of how to connect these two. I have also been unable to find the integral on any table of integrals, so any help would be greatly appreciated

Edit:

Based on a suggestion, I tried using the incomplete Beta function: $$ B_x(z,y)=\int_0 ^x t^{z-1}(1-t)^{y-1}dt=\frac{\Gamma(z)\Gamma(y}{\Gamma(z+y)}. $$ And with $z=1/\nu$ and $y=3/2$, I get $$ B_x(z,y)=\int_0 ^1 t^{1/\nu-1}(1-t)^{1/2}dt=\frac{\Gamma(1/\nu)\Gamma(3/2)}{\Gamma(1/\nu+3/2)}, $$ which is what I am looking for.

Answer 1

If $x=1$, you obtain the result using the beta function.

Now, for $0 < x <1$ and $\nu >0$ the result is given by $$\int_0 ^{x}(1-z)^{1/2}z^{1/\nu -1}\,dz=B_x\left(\frac{1}{\nu },\frac{3}{2}\right)$$ where appears the incomplete beta function