Integral $ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ Question
$ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ = ?
we tried to bound it from both sides using $x$ and $(x-1)$, which yield nice estimation ($\frac {24}\pi$) - ($\frac {36}\pi$) but not a precise one. 
we also tried to split it to 6 integrals using $\lfloor x\rfloor  $ as a different constant each time. $(0\cdot (\cos(...)-\cos(...))+1(\cos(\frac \pi3)-\cos(\frac\pi6))+2\cdot...$ etc
it produced the right result ($\frac {30}\pi$), but the number is not important as achieving the general function.

x is real number, m is integer, and $\mathbb{Z}$ is the set of integers (positive, negative, and zero).
$$\lfloor x \rfloor=\max\, \{m\in\mathbb{Z}\mid m\le x\}$$
So
$$\begin{alignat*}{1}
\int_{0}^{6}\lfloor x\rfloor\sin(\frac{6x}{\pi})\ \mathrm{d}x= & \int_{0}^{1}0.\sin(\frac{6x}{\pi})\ \mathrm{d}x+\int_{1}^{2}1.\sin(\frac{6x}{\pi})\ \mathrm{d}x+\int_{2}^{3}2.\sin(\frac{6x}{\pi})\ \mathrm{d}x\\
 & +\int_{3}^{4}3.\sin(\frac{6x}{\pi})\ \mathrm{d}x+\int_{4}^{5}4.\sin(\frac{6x}{\pi})\ \mathrm{d}x+\int_{5}^{6}5.\sin(\frac{6x}{\pi})\ \mathrm{d}x\\
= & \frac{1}{6}\pi(\cos(\frac{6}{\pi})-\cos(\frac{12}{\pi}))+\frac{1}{3}\pi(\cos(\frac{12}{\pi})-\cos(\frac{18}{\pi}))\\
 & +\frac{1}{2}\pi(\cos(\frac{18}{\pi})-\cos(\frac{24}{\pi}))+\frac{2}{3}\pi(\cos(\frac{24}{\pi})-\cos(\frac{30}{\pi}))\\
 & +\frac{5}{6}\pi(\cos(\frac{30}{\pi})-\cos(\frac{36}{\pi}))\\
= & \frac{1}{6}\pi\left(\cos(\frac{6}{\pi})+\cos(\frac{12}{\pi})+\cos(\frac{18}{\pi})+\cos(\frac{24}{\pi})+\cos(\frac{30}{\pi})-5\cos(\frac{36}{\pi})\right)
\end{alignat*}$$
