Absolute Maximum of the integral Function  $A(t)=\displaystyle\int_t^{t+2\pi}\!\dfrac{|\sin x|}{x^2+x+1}\mathrm{d}x$ attains absolute maximum for some $t\in\mathbb{R}$. I can not prove it. How can I do it? I have proved that attains a local maximum at $t=(-2\pi-1)/2$ but why is this absolute maximum of function $A$?
 A: The integrand
$$f(x) = \frac{\left|\sin(x)\right|}{x^2+x+1}$$
is composed of a positive, bounded, periodic numerator divided by a positive denominator that grows without bound in each direction.  Thus, I think it's reasonably clear that
$$A(t) = \int_{t}^{t+2\pi} f(x) \, dx$$
tends to zero as $t\rightarrow\pm\infty$.  Furthermore, $A$ is a positive, continuous function so it must attain a maximum by the extreme value theorem.  In fact, the situation looks like so:

It's probably worth pointing out that 
\begin{align}
A'(t) &= f(t+2\pi)-f(t) \\
 &= \frac{\left|\sin(t)\right|}{t^2+t+1} - \frac{\left|\sin(t+2\pi)\right|}{(t+2\pi)^2+(t+2\pi)+1} \\
 &= -\frac{2 \left(2 \pi  t+2 \pi ^2+\pi \right) \left|
   \sin (t)\right| }{\left(t^2+t+1\right)
   \left(t^2+4 \pi  t+t+4 \pi ^2+2 \pi +1\right)}
\end{align}
and that the critical point of interest is exactly $t=-(2\pi+1)/2 \approx -3.64$.

A: Hint: what is the derivative of $A(t)$?
A: HINT: It is positive, bounded and continuous. On compact sets it attains maxima and for large $|t|$ is arbitrary small.
