From the article pendulum(mathmetics) from wikipedia.
There is a demonstration that this equation:
$$\dfrac{dt}{d\theta } = \sqrt{\dfrac{l}{2g}}\dfrac{1}{\sqrt{\cos\left(\theta\right)-\cos\left(\theta_0\right)}}$$
Has the integration over one complete cycle,
$$T = t\left(\theta_0\rightarrow0\rightarrow-\theta_0\rightarrow0\rightarrow\theta_0\right)$$
equals to the 4 times the integration over quarter-cycle.
$$T = 4t\left(\theta_0\rightarrow0\right)$$
I totally have no clue about how they are equal (in this situation), but doubt it will equivalent like this in the other situations.
So, I need a proof that the quarter (or half) cycle and complete cycle integration are (not)always equal.
It really was my failure to notice that the "cycle" was the motion of the pendulum itself.
So, what wikipedia said is the integration from where the pendulum start to where it reaches bottom, which equals 1/4 of its trajectory. Thus, the equation could be written in that form.