Definite integral of $|\sin x|$ over $[0, 10\pi]$ I'm trying to solve this and when I add both of the different signs together I ended by a result of 2, but it's not one of the choices, any help would be appreciated
$$\int_0^{10\pi}\ |\sin x| \, \mathrm{d}x$$
 A: $$
 \lvert \sin x \rvert
$$

You are making $10$ copies
$$
 \int_{0}^{10\pi} \lvert \sin x \rvert dx = 10 \int_{0}^{\pi} \sin x dx = 10 \cdot 2 = 20
$$
A: Note that $\sin $ is negative on $[\pi, 2\pi] \cup [3\pi, 4\pi] \cup\cdots \cup  [9\pi, 10\pi]$ and positive otherwise.  So $$\int_0^{10\pi} |\sin x| \, \mathrm{d}x = 5\int_0^{\pi} \sin x - 5 \int_{\pi}^{2\pi}\sin x \, \mathrm{d}x $$ by periodicity. But this is $5\cdot 2 - 5\cdot (-2) = 20$. 

Alternatively, if you're fine with a more intuitive geometrical reasoning then note $|\sin x|$ consists of $10$ identical humps, each of area $\int_0^{\pi} \sin x \, \mathrm{d}x = 2$. So the total area is $20$.
A: Notice from a quick look on a graph and recalling the period of sine that we have
$$\int_0^{10\pi}|\sin x|\ dx=10\int_0^\pi\sin x\ dx=10(\cos0-\cos\pi)=20$$
A: hint
$$\int_0^{2\pi}|\sin (x)|dx=$$
$$\int_0^\pi \sin (x)dx-\int_\pi^{2\pi}\sin (x)dx=$$
$$1+1+1+1=4$$
$$\int_0^{10\pi}|\sin (x)|dx=5\times 4=20$$
A: The integral 
$$\int_0^{10\pi} \sin |x|dx= \int_0^{10\pi} \sin x dx=-\cos(10\pi)-(-\cos(0))=0.$$
On the other hand, 
$$\int_0^{10\pi} |\sin x|dx = 10\int_0^\pi \sin (x) dx= 10(1+1)= 20$$
