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Let $U$ and $V$ be independent random variables that are uniformly distributed on $[0,1]$. Define $$X := \sqrt{−2\log(U)}\, \cos(2\pi V)$$

Prove that $X$ follows normal distribution.

I'm attempting to prove that without defining $Y := \sqrt{−2\log(U)}\, \sin(2\pi V)$ (which would be out-of-the-blue).

I tried something along the lines of product distribution but the computation is quite messy. After computing the densities of $\sqrt{−2\log(U)}$ and $\cos(2\pi V)$ which are respectively $$x\mapsto \chi_{(0,\infty)}(x)\cdot x \exp(-\frac{x^2}2)$$ and $$x\mapsto \chi_{(0,1)}(x)\frac{1}{2\pi \sqrt{1-x^2}}$$

I get the following integral representation of the density of $X$:

$$f(z)=\int_z^\infty x\frac{\exp(-\frac{x^2}2)}{2\pi \sqrt{x^2-z^2}} dx$$ which is quite ugly...

I'd like some suggestions about computing this integral, or any faster method that doesn't involve Box-Muller method.

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    http://www.wolframalpha.com/input/?i=integrate%5Bx+Exp%5B-x%5E2%2F2%5D%2FSqrt%5Bx%5E2-1%5D,%7Bx,1,inf%7D%5D – 2017-01-26
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    this integral screams for a hyperbolic sub – 2017-01-26
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    $$2\pi f(z)=\int_1^{\infty}y\frac{e^{-z^2y^2/2}}{\sqrt{y^2-1}}=e^{-z^2/2}\int_0^{\infty}d(\sqrt{y^2-1})e^{-z^2(y^2-1)/2}=\sqrt{2\pi}e^{-z^2/2}$$ – 2017-01-27

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