We have $\begin{equation} \int^{t=2 \pi}_{t=0} \frac{5}{2} |\text{sin} 2t | \, dt = \frac{5}{2} \int^{t=2 \pi}_{t=0} |\text{sin} 2t | \, dt \end{equation}$ Now use the substitution $y = 2t$, $dy = 2dt$. The boundaries shift as follows: if $t = 2\pi$ then $y = 4\pi$ and if $t = 0$ then $y = 0$
We plug this into the integral to get $\begin{equation} \frac{5}{2} \int^{t=2 \pi}_{t=0} |\text{sin} 2t | \, dt = \frac{5}{2} \int^{y=4 \pi}_{y=0} |\text{sin } y | \, \frac{dy}{2} = \frac{5}{4} \int^{y=4 \pi}_{y=0} |\text{sin } y | \, dy \end{equation}$
It remains to remember that sin$x \geq 0$ for $x \in [0,\pi]$, and sin$x \leq 0$ when $ x \in [\pi, 2\pi]$. Finally, we can also use that sin$x$ is periodic, which means that sin$(x + 2\pi) = $ sin$x$.
So we get $\begin{equation} \frac{5}{4} \int^{y=4 \pi}_{y=0} |\text{sin } y | \, dy = 2 \left(\frac{5}{4} \int^{y=2 \pi}_{y=0} |\text{sin } y | \, dy \right) = 2 \left(\frac{5}{4} \int^{y= \pi}_{y=0} \text{sin } y \, dy - \frac{5}{4} \int^{y= 2 \pi}_{y=\pi} \text{sin } y \, dy \right) \end{equation}$
You can now compute the integral to obtain the desired result.