Integrating $ \int_0^{2\pi} \frac{5}{2}|\sin(2t)| dt$ I'm stuck on integration of the following function:
$$ \int_0^{2\pi} \frac{5}{2}|\sin(2t)| dt$$
I understand the few first steps with substitution, etc., but I can't get the end result which is "10".
Could someone give me a step by step solution for it?
 A: Note that $$\lvert \sin(2t) \rvert = \begin{cases} \sin(2t) & t\in[0,\pi/2]\\ - \sin(2t) & t\in[\pi/2,\pi]\\ \sin(2t) & t\in[\pi,3\pi/2]\\ -\sin(2t) & t\in[3\pi/2, 2\pi] \end{cases}$$
Also, $$\int_0^{\pi/2} \sin(2t) dt = -\int_{\pi/2}^{\pi} \sin(2t) dt = \int_{\pi}^{3\pi/2} \sin(2t) dt = -\int_{3\pi/2}^{2\pi} \sin(2t) dt$$
Hence, the integral $$I = \int_0^{2 \pi} \frac52 \lvert \sin(2t) \rvert dt = 4 \times \int_0^{\pi/2} \frac52 \sin(2t) dt = 10 \times \int_0^{\pi/2} \sin(2t) dt = 10$$
A: 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.
