Searching for an elegant, high school level, technique for solving an integral by hand I would appreciate some assistance finding alternative approaches to solving the integral posed in the link below.
https://www.desmos.com/calculator/rstvj2v3cs
It describes and graphs a progression for decomposing the integral
$\int_0^{2\pi}\frac{x\sin x}{2+\left|\cos x\right|}dx$
I am hoping for some insight into a simplified approach.
Much appreciated, 
**
Thanks to all for the feedback.
I have completed the extended solution in desmos and welcome any additional feedback:
https://www.desmos.com/calculator/g2lj84ofdu
 A: It can be done in a simple way. Split the integral into four parts and substitute
$$x=y,\qquad x={\pi\over2}+y,\qquad x=\pi+y,\qquad x={3\pi\over 2}+y$$ in the four parts. This leads to the integrals
$$\eqalign{&\int_0^{\pi/2}{y\sin y\over2+\cos y}\>dy,\qquad \int_0^{\pi/2}{(\pi/2+y)\cos y\over2+\sin y}\>dy,\cr &\int_0^{\pi/2}{(\pi+y)(-\sin y)\over2+\cos y}\>dy,\qquad\int_0^{\pi/2}{(3\pi/2+y)(-\cos) y\over2+\sin y}\>dy .\cr} $$
The integrals with the same denominator can be collected, whereby the disagreeable parts disappear, and we are left with
$$-\pi\int_0^{\pi/2}{\cos y\over 2+\sin y}\>dy-\pi\int_0^{\pi/2}{\sin y\over 2+\cos y}\>dy=2\pi\log{2\over3}=-2.5476\ .$$
A: \begin{align}
&\int_0^{2\pi}\frac{x\sin x}{2+\left|\cos x\right|}dx
\\[1em]
&=\int_0^{\pi}\frac{x\sin x}{2+\left|\cos x\right|}dx 
  + \int_{\pi}^{2\pi}\frac{x\sin x}{2+\left|\cos x\right|}dx
\end{align}
In the second integral, put $\pi+u=x$. This yields
\begin{align}
&\int_0^{\pi}\frac{x\sin x}{2+\left|\cos x\right|}dx + \int_0^{\pi}\frac{(u+\pi)(-\sin u)}{2+\left|\cos u\right|}du
\\[1em]
&=\int_0^{\pi}\frac{x\sin x}{2+\left|\cos x\right|}+\frac{(x+\pi)(-\sin x)}{2+\left|\cos x\right|}dx
\\[1em]
&=-\pi \int_0^{\pi}\frac{\sin x}{2+\left|\cos x\right|}
\end{align}
You should be able to solve it from here by putting $\cos x=u$.
I'm getting the final answer as $2\pi \ln \left(\frac{2}{3}\right)=−2.54761241$.
