Finding $ \int_{0}^{24} \left| \sin \left(\frac{\pi x}{12}\right) \right| dx$. 
Find $$ \int_{0}^{24} \left| \sin \left(\frac{\pi x}{12}\right) \right| dx.$$

My own calculations give me
$$\left[ \left| -12\cos(2\pi) \right|  \right]_{0}^{24}.$$
And here I'm stuck. Because if I evaluate this I get $24$, but it should be $48$. That's because of the $\lvert\cdot\rvert$ around the $\cos$. How do I calculate that?
 A: Hint: Let $y=\dfrac{\pi x}{12}$. Then
$$ \int_{0}^{24} \left| \sin \left(\frac{\pi x}{12}\right) \right| dx=\frac{12}{\pi}\int_0^{2\pi}|\sin(y)|dy=\frac{12}{\pi}\left[\int_0^{\pi}\sin(y)dy+\int_\pi^{2\pi}-\sin(y)dy\right]$$
as $\sin(y)\ge 0$ for $y\in[0,\pi]$ and $\sin(y)\le0$ for $y\in[\pi,2\pi]$.
A: Hint: Plot the curve, or imagine plotting it.  You know what the sine function looks like. In the second half, the sine is negative.
But if you take the absolute value, the curve of the first half repeats.
Exploit  the symmetry! Integrate up to $12$, and double the result.
Actually, the area under your given curve is $4$ times the area of the part up to $6$. I would probably integrate to $6$, and multiply the result by $4$.
Remark: Please note that an antiderivative of $\sin(\pi x/12)$ is $-\frac{12}{\pi}\cos(\pi x/12)$. In the OP, the $\pi$ in the denominator is missing.
A: I suggest splitting the integral into two parts: one where the integrand is positive and one where it is negative.  This allows you to remove the absolute value.  Note also that the integral of the absolute value of a function is not the same as the absolute value of the function's integral, which is probably where you went wrong here.
