Motion of a pendulum equation in the George Simmons book on differential equations

Question

I just can't understand the transition between this two formulas, why $dt$ becomes $T/4$. Can anybody help me with that?enter image description here

Answer 1

Because we have:

$1$ period=(time from $\alpha$ to $0)+$ (time from $0$ to $-\alpha)+$ (time from $-\alpha$ to $0$)+ (time from $0$ to $\alpha)$

and, by symmetry:

$T=(1$ period)=$4($time from $\alpha$ to $0)$ so: $$ \int _\alpha ^0 dt=T/4 $$