simple real analysis question on integration it is trivial that $\int_0^{2\pi} \cos(x)\,dx = 0$. 
Intuitively, it is clear that for a strictly decreasing
positive function $f(x)$, 
$$
\int_0^{2\pi} f(x) \cos(x)\,dx \ge 0
$$
but I have no glue how to prove that one. Any hints?
Cheers, Eric
edit: I checked it for power-type functions $(a+t)^{-\alpha}$ numerically.
 A: This is not true. For $\epsilon>0$ small, consider the function whose graph consists of the straight line segments:
$\ \ \bullet\ $from $[0, 1+\epsilon]$ to $[{3\pi\over2}-\epsilon,1]$ 
$\ \ \bullet\ $from $[{3\pi\over2}-\epsilon,1]$ to $[{3\pi\over2},\epsilon]$ 
and 
$\ \ \bullet\ $from $[{3\pi\over2},\epsilon]$ to $[2\pi,{\epsilon\over2}]$.
Here, $\int_0^{3\pi/2} f(x)\cos x\,dx\approx -1$ and $\int_{3\pi/2}^{2\pi}f(x)\cos x\,dx\approx 0$.
 

A: For real numbers $a>b>c>d>0$ consider the function
$$
f:[0,2\pi]\ni x
\mapsto\left\{\begin{array}{ll}
\frac{2}{3\pi}\left[a\left(\frac{3\pi}{2}-x\right)+bx\right] & \text{, }x\in\left[0,\frac{3\pi}{2}\right] \\
\frac{2}{\pi}\left[c\left(2\pi-x\right)+d\left(x-\frac{3\pi}{2}\right)\right] & \text{, }x\in\left(\frac{3\pi}{2},2\pi\right]
\end{array}\right\}
\in\mathbb{R}\text{.}
$$
By construction, $f$ is positive and stricly decreasing.
Furthermore, it holds
$$
\int_0^{\frac{3\pi}{2}} f(x) \cos(x)\,\mathrm{d} x
=-\frac{(3\pi+2) b-2 a}{3\pi}
$$
and
$$
\int_{\frac{3\pi}{2}}^{2\pi} f(x) \cos(x)\,\mathrm{d} x
=\frac{2 d+(\pi-2) c}{\pi}\text{.}
$$
Choosing $b>\frac{2 a}{3\pi+2}$ makes the value of the first integral negative.
Clearly, $c$ and $d$ can be chosen such that the value of the second integral becomes arbitrarily small.
Thus, there exists a function violating the original conjecture.
A: As the answers by precarious and David Mitra show, your intuited result is false.  On the other hand, for a function $f(x)$ that is strictly decreasing and positive on $[0,2\pi)$, it is certainly true that 
$$\int_0^{2\pi} f(x)\sin(x)\mathrm dx > 0$$
and this can be shown by the method suggested to you by Fabian, viz.,
$$\begin{align*}
\int_0^{2\pi} f(x)\sin(x)\mathrm dx 
&= \int_0^{\pi} f(x)\sin(x)\mathrm dx +  \int_{\pi}^{2\pi} f(x)\sin(x)\mathrm dx\\
&=  \int_0^{\pi} f(x)\sin(x)\mathrm dx -  \int_0^{\pi} f(x+\pi)\sin(x)\mathrm dx\\
&= \int_0^{\pi} [f(x) - f(x+\pi)]\sin(x)\mathrm dx\\
&> 0.
\end{align*}$$
Perhaps you were thinking sine but absent-mindedly wrote cosine instead?
A: How about integration by parts...
$$ \int_0^{2\pi} f(x) \cos x\,{\rm d}x = \left[ f(x)\sin x \right]_{x=0}^{x=2\pi}-\int_0^{2\pi} \sin x\,f'(x)\,{\rm d}x$$
$$ \left[ f(x)\sin x \right]_{x=0}^{x=2\pi} = 0$$
$$ f'(x) \leq 0 $$
So now you have to prove that
$$ -\int_0^{2\pi} \sin x\,f'(x)\,{\rm d}x \ge 0$$
or expanded as
$$ \int_0^{2\pi} \sin x\,f'(x)\,{\rm d}x = \int_0^{\pi} \sin x\,f'(x)\,{\rm d}x +\int_\pi^{2\pi} \sin x\,f'(x)\,{\rm d}x \leq 0$$
Since $\sin x$ is positive between $x=0\ldots\pi$ and negative otherwise the above can be written as
$$ \int_0^{\pi} \sin x\,f'(x)\,{\rm d}x -\int_0^{\pi} \sin x\,f'(x+\pi)\,{\rm d}x \leq 0$$
or
$$ \int_0^{\pi} \sin x\,\left(f'(x)-f'(x+\pi)\right)\,{\rm d}x \leq 0 $$
Which is true only if $f'(x)-f'(x+\pi)\leq0$ for $x=0\ldots\pi$. If $f(x)$ is decreasing with negative slope then $|f'(\pi)|\leq|f'(0)|$ and thus the above is true.
