$(2Y-1)\sqrt X\sim\mathcal N(0,1)$ when $X\sim\chi^2_{n-1}$ and $Y\sim\text{Beta}\left(\frac{n}{2}-1,\frac{n}{2}-1\right)$ independently 
$X$ and $Y$ are independently distributed random variables where $X\sim\chi^2_{(n-1)}$ and $Y\sim\text{Beta}\left(\frac{n}{2}-1,\frac{n}{2}-1\right)$. What is the distribution of $Z=(2Y-1)\sqrt X$ ?

Joint density of $(X,Y)$ is given by
$$f_{X,Y}(x,y)=f_X(x)f_Y(y)=\frac{e^{-\frac{x}{2}}x^{\frac{n-1}{2}-1}}{2^{\frac{n-1}{2}}\Gamma\left(\frac{n-1}{2}\right)}\cdot\frac{y^{\frac{n}{2}-2}(1-y)^{\frac{n}{2}-2}}{B\left(\frac{n}{2}-1,\frac{n}{2}-1\right)}\mathbf1_{\{x>0\,,\,0<y<1\}}$$
Using the change of variables $(X,Y)\mapsto(Z,W)$ such that $Z=(2Y-1)\sqrt X$ and $W=\sqrt X$, 
I get the joint density of $(Z,W)$ as
$$f_{Z,W}(z,w)=\frac{e^{-\frac{w^2}{2}}w^{n-3}\left(\frac{1}{4}-\frac{z^2}{4w^2}\right)^{\frac{n}{2}-2}}{2^{\frac{n-1}{2}}\Gamma\left(\frac{n-1}{2}\right)B\left(\frac{n}{2}-1,\frac{n}{2}-1\right)}\mathbf1_{\{w>0\,,\,|z|<w\}}$$
Marginal pdf of $Z$ is then 
$f_Z(z)=\displaystyle\int_{|z|}^\infty f_{Z,W}(z,w)\,\mathrm{d}w$, which does not lead me anywhere.
Again, while finding the distribution function of $Z$, an incomplete beta/gamma function shows up:
$F_Z(z)=\Pr(Z\le z)$
$\quad\qquad=\Pr((2Y-1)\sqrt X\le z)=\displaystyle\iint_{(2y-1)\sqrt{x}\le z}f_{X,Y}(x,y)\,\mathrm{d}x\,\mathrm{d}y$
What is an appropriate change of variables here? Is there another way to find the distribution of $Z$?
I tried using different relations between Chi-Squared, Beta, 'F' and 't' distributions but nothing seems to work. Perhaps I am missing something obvious.

As mentioned by @Francis, this transformation is a generalization of the Box-Müller transform.

As user @Chaconne has already done, I was able to provide an algebraic proof with this particular transformation. I have not skipped any details.

(We already have $n>2$ for the density of $Y$ to be valid).
Let us consider the transformation $(X,Y)\mapsto (U,V)$ such that $U=(2Y-1)\sqrt{X}$ and $V=X$.
This implies $x=v$ and $y=\frac{1}{2}\left(\frac{u}{\sqrt{v}}+1\right)$. 
Now, $x>0\implies v>0$ and $0<y<1\implies-\sqrt{v}<u<\sqrt{v}$, 
so that the  bivariate support of $(U,V)$ is simply $S=\{(u,v):0<u^2<v<\infty,\,u\in\mathbb{R}\}$.
Absolute value of the Jacobian of transformation is $|J|=\frac{1}{2\sqrt{v}}$.
Joint density of $(U,V)$ is thus
$$f_{U,V}(u,v)=\frac{e^{-\frac{v}{2}}v^{\frac{n-1}{2}-1}\left(\frac{u}{\sqrt{v}}+1\right)^{\frac{n}{2}-2}\left(\frac{1}{2}-\frac{u}{2\sqrt{v}}\right)^{\frac{n}{2}-2}\Gamma(n-2)}{(2\sqrt{v})\,2^{\frac{n-1}{2}+\frac{n}{2}-2}\,\Gamma\left(\frac{n-1}{2}\right)\left(\Gamma\left(\frac{n}{2}-1\right)\right)^2}\mathbf1_{S}$$
$$=\frac{e^{-\frac{v}{2}}v^{\frac{n-4}{2}}(\sqrt{v}+u)^{\frac{n}{2}-2}(\sqrt{v}-u)^{\frac{n}{2}-2}\,\Gamma(n-2)}{2^{\frac{2n-3}{2}+\frac{n}{2}-2}\,(\sqrt{v})^{n-4}\,\Gamma\left(\frac{n-1}{2}\right)\left(\Gamma\left(\frac{n-2}{2}\right)\right)^2}\mathbf1_{S}$$
Now, using Legendre's duplication formula,
$\Gamma(n-2)=\frac{2^{n-3}}{\sqrt \pi}\Gamma\left(\frac{n-2}{2}\right)\Gamma\left(\frac{n-2}{2}+\frac{1}{2}\right)=\frac{2^{n-3}}{\sqrt \pi}\Gamma\left(\frac{n-2}{2}\right)\Gamma\left(\frac{n-1}{2}\right)$ where $n>2$.
So for $n>2$, $$f_{U,V}(u,v)=\frac{2^{n-3}\,e^{-\frac{v}{2}}(v-u^2)^{\frac{n}{2}-2}}{\sqrt \pi\,2^{\frac{3n-7}{2}}\,\Gamma\left(\frac{n}{2}-1\right)}\mathbf1_{S}$$
Marginal pdf of $U$ is then given by
$$f_U(u)=\frac{1}{2^{\frac{n-1}{2}}\sqrt \pi\,\Gamma\left(\frac{n}{2}-1\right)}\int_{u^2}^\infty e^{-\frac{v}{2}}(v-u^2)^{\frac{n}{2}-2}\,\mathrm{d}v$$
$$=\frac{e^{-\frac{u^2}{2}}}{2^{\frac{n-1}{2}}\sqrt \pi\,\Gamma\left(\frac{n}{2}-1\right)}\int_0^\infty e^{-\frac{t}{2}}\,t^{(\frac{n}{2}-1-1)}\,\mathrm{d}t$$
$$=\frac{1}{2^{\frac{n-1}{2}}\sqrt \pi\,\left(\frac{1}{2}\right)^{\frac{n}{2}-1}}e^{-\frac{u^2}{2}}$$
$$=\frac{1}{\sqrt{2\pi}}e^{-u^2/2}\,,u\in\mathbb{R}$$
