Period and probability density for a harmonic oscillator

[Another]

$E = \frac{1}{2}(kx^2+m \dot{x}^2)$
$\frac{2E - kx^2}{m}=\dot{x}^2$
$\frac{dx}{dt} = \sqrt{\frac{2E - kx^2}{m}}$ or $dt = \sqrt{\frac{m}{2E - kx^2}}dx$ ⇒$= \frac{1}{\sqrt{\frac{2E - kx^2}{m}}}dx$

My Question please help me.
1. I know $T = 2\pi\sqrt{\frac{m}{k}} .$ but i don't understand why $T = 2 \int_{-l}^{l}\frac{1}{\sqrt{\frac{2E - kx^2}{m}}}dx$2.In this case. Why do we choose the probability density function, $p(x) = \frac{2dt}{T}$ ?

 

[Another]

I don't understand how to define the probability density function.

 

[mjc123]

1. The denominator is the speed at position x (see the equation for E). The time dt spent in the interval dx is dx/v_x.
2. The probability of being between x and x+dx equals the fraction of the time that is spent in this interval, i.e. 2dt/T (2 because it traverses this interval twice in a full period), where dt is related to dx as above.