# $\mathbf{P}\left(2<\left|X\right|+\left|Y\right|<3\right)$ where the random variables are independent $N(0,1)$?

If $$X$$ and $$Y$$ are independent $$N(0,1)$$ random variables, then how can I calculate the following probability: $$\mathbf{P}\left(2<\left|X\right|+\left|Y\right|<3\right)?$$ I calculated the density function of $$\left|X\right|$$ and $$\left|Y\right|$$. I tried to write the density functions of there sums, but I couldn't continue where I had to integrate $$e^{-t^{2}}$$. I had to acknowledge this path I followed leads nowhere to solve this problem. Is there any easier way? Anyway, I got $$f_{\left|X\right|+\left|Y\right|}\left(y\right)=\frac{4}{\pi}e^{-\frac{y^{2}}{2}}e^{\frac{1}{4}y^{2}}\sqrt{2\pi}\frac{1}{\sqrt[4]{2}}\left[\Phi\left(\frac{1}{\sqrt{2}}y\right)-\frac{1}{2}\right]$$ if it means something, but I'm not sure it is a good solution. Any easier way?

Hint: Consider the integral $$I=\int_a^b f'(x)f(x)\,dx$$. Then by integration by parts with $$dv=f'(x)\,dx$$ and $$u=f(x),$$ we have $$I=f(b)^2-f(a)^2-I$$ so $$I=\frac{f(b)^2-f(a)^2}{2}$$.
Your problem is very similar; try performing integration by parts. I didn't check your pdf of $$|X|+|Y|$$ too closely, but it's definitely at least the correct form.