I have problems to prove, that for $X$, $Y$ independent standard normal distributed random variables, $\frac{X^2}{X^2 +Y^2}$ is Arcsine distributed?
I try to compute the cdf:
$$P\left[\frac{X^2}{X^2 + Y^2} \leq a\right]=P\left[|X| \leq \sqrt{\frac{a}{1-a}}|Y|\right]=\frac{2}{\pi}\int_0^{\infty}\int_0^{\sqrt{\frac{a}{1-a}}y}e^{-\frac{x^2}{2}}e^{-\frac{y^2}{2}}\,dx\,dy$$
Is this correct?
Since $X^2$ and $Y^2$ are independently distributed $\chi^2_1$ random variables, you are seeking the distribution of $$Z=\frac{U}{U+V}$$, where $$U=X^2\quad\text{ and }\quad V=Y^2$$
To derive the distribution of $Z$, you can use the usual change of variables method.
Joint density of $(U,V)$ is
\begin{align} f_{U,V}(u,v)&=\frac{e^{-u/2}}{\sqrt{2\pi u}}\mathbf1_{u>0}\frac{e^{-v/2}}{\sqrt{2\pi v}}\mathbf1_{v>0} \\\\&=\frac{e^{-(u+v)/2}}{2\pi\sqrt{uv}}\mathbf1_{u>0,v>0} \end{align}
Consider the transformation $$(U,V)\to(Z,W)$$ such that $$Z=\frac{U}{U+V}\quad\text{ and }\quad W=U+V$$
This implies $$u=zw\quad,\quad v=w(1-z)$$
Clearly, $$u>0,v>0\implies 0<z<1,w>0$$
Absolute value of jacobian of transformation is $$|J|=w$$
So the joint density of $(Z,W)$ is
\begin{align} f_{Z,W}(z,w)&=f_{U,V}(zw,w(1-z))|J| \\\\&=\frac{e^{-w/2}}{2\pi\sqrt{z(1-z)}}\mathbf1_{0<z<1,w>0} \\\\&=\frac{1}{\pi\sqrt{z(1-z)}}\mathbf1_{0<z<1}\,\frac{e^{-w/2}}{2}\mathbf1_{w>0} \end{align}
Hence we find that $Z$ and $W$ are independently distributed.
In particular, $$Z\sim \text{Beta}\left(\frac{1}{2},\frac{1}{2}\right)$$
, which is the so-called Arcsine distribution.
The above procedure is all too routine and nothing special about the derivation.
The final result is not surprising considering the well-known relationship between the Gamma (a chi-squared variable is also a gamma variable) and Beta distribution.