Probability distribution of a coordinate of the random point on a hypersphere with given radius If $(x_1,x_2,...,x_N)$ is a uniformly randomly chosen point on a hypersphere of a dimension $N$ with the radius $R$ (center in origin). What is the probability distribution of any coordinate?
Done so far (2-D):
From coordinates in polar form, and angle chosen uniformly I've calculated the distribution function, which is equal to $$f(x) = \frac{1}{\pi \sqrt{r^2-x^2}}$$ (here the radius is denoted with $r$, and coordinate with $x$). Cumulative distribution function is $$F(x)=1-\frac{cos^{-1}\left(\frac{x}{r}\right)}{\pi}\text{ ,}$$ if it should be of any use.
I've tried to calcualte probability distribution of coordinate in 3-D by denoting $f(x)$ given above as $f(x|r)$, and knowing that (calculated similarly as the distribution above) $$f(r) = \frac{1}{\pi \sqrt{R^2-(R-r)^2}}\text{ ,}$$I've tried to calculate the distribution $f(x)$ in 3-D by integrating $f(x,r)=f(x|r) f(r)$ from $0$ to $R$, but Wolfram Alpha timed out (link). Mathematica returned me the inputed integral for
Integrate[(2/Pi^2)*(1/Sqrt[(r^2 - x^2)*(R^2 - (R - r)^2)]), {r, 0, R}]
 A: 
Each coordinate $x_k$ is distributed like $R$ times a random variable with PDF $f_N$, where
  $$
f_N(s)=\frac{\Gamma(\tfrac{N}2)}{\sqrt{\pi}\,\Gamma(\tfrac{N-1}2)}\cdot(1-s^2)^{\frac{N-3}2}\cdot[-1\leqslant s\leqslant1].
$$

The simplest approach to this result might be to remember that a standard normal vector $z=(z_k)$ is such that $z/\|z\|$ is uniformly distributed on the unit sphere. Hence, each $x_k$ is distributed like $R\cdot\xi$ where $\xi=z_1/\|z\|$. 
Now, $\|z\|^2=z_1^2+r^2$, where $r\gt0$ is independent of $z_1$ with density proportional to $r^{N-1}\mathrm e^{-r^2/2}$. Forgetting some irrelevant normalizing constants, this yields that, for every bounded measurable function $u$,
$$
\mathrm E(u(\xi))\propto\iint u\left(\frac{w}{\sqrt{w^2+r^2}}\right)\mathrm e^{-w^2/2}r^{N-1}\mathrm e^{-r^2/2}\mathrm dr\mathrm dw.
$$
The change of variables $(w,r)\to(s,r)$ with $s=w/\sqrt{r^2+w^2}$ yields $w=rs/\sqrt{1-s^2}$ and $\mathrm dw\mathrm dr=(1-s^2)^{-3/2}\mathrm ds\mathrm dr$ hence
$$
\mathrm E(u(\xi))\propto\iint u\left(s\right)\mathrm e^{-r^2s^2/(2(1-s^2))}r^{N-1}\mathrm e^{-r^2/2}(1-s^2)^{-3/2}\mathrm dr\mathrm ds,
$$
which shows that the density of $\xi$ is $f_N$ with
$$
f_N(s)\propto\int\mathrm e^{-r^2s^2/(2(1-s^2))}r^{N-1}\mathrm e^{-r^2/2}(1-s^2)^{-3/2}\mathrm dr=(1-s^2)^{-3/2}\int\mathrm e^{-r^2/(2(1-s^2))}r^{N-1}\mathrm dr.
$$
Using the change of variable $r=(1-s^2)^{1/2}x$ in the last integral shows that it is proportional to $(1-s^2)^{N/2}$ hence $f_N(s)\propto(1-s^2)^{(N-3)/2}$.
