Confused by a solution given by professor $X$ and $Y$ are two continuous i.i.d random variables. They are both symmetric about zero. The problem is to show that
$P(|X+Y|<2|X|) > 0.5$
The model solution is the following:
$\iint\limits_{|x+y|<2|x|} f(x)f(y) \ dxdy = $
$\int_0^{\infty} (\int_{-3x}^x f(y) \ dy) \ f(x) dx \ + \ \int_{-\infty}^0 (\int_x^{-3x} f(y) \ dy) \ f(x) dx  = $
$2\int_0^{\infty} (\int_{-3x}^{-x} f(y) \ dy) \ f(x) dx \ + \ 2\int_{0}^{\infty} (\int_{-x}^{x} f(y) \ dy) \ f(x) dx$    (by  symmetry)
$= 2\int_0^{\infty} (\int_{-3x}^{-x} f(y) \ dy) \ f(x) dx \ + 0.5 > 0.5$
What I don't understand is how he got from line 2 to line 3 'by symmetry'.

$f(-x)=f(x)$ and $f(-y)=f(y)$. Substitute $-x$ and $-y$ for $x$ and $y$ in the second integral, and it becomes the same as the first, which is where the $2$ comes from. Then just split the integral into two parts. 
