Determine the sign of the integral $\int_0^\pi x\cos x\,dx$ without calculating it 
Without explicitly evaluating, determine the sign of the integral $$\int_0^{\pi} x\cos(x)dx$$

I know $x\cos(x) > 0$ when $0 < x < {\pi}/2$ and $x\cos(x) < 0$ when $\pi/2 < x < \pi$, and, in fact, the end result is negative, but I'm unsure of where to go to show this. Do I now need to split the integral up into two regions and manipulate the function? 
Thanks!

$\begin{array}\\
\int_0^{\pi} x\cos(x)dx
&=\int_0^{\pi/2} x\cos(x)dx+\int_{\pi/2}^{\pi} x\cos(x)dx\\
&=\int_0^{\pi/2} x\cos(x)dx+\int_{0}^{\pi/2} (x+\pi/2)\cos(x+\pi/2)dx\\
&=\int_0^{\pi/2} (\pi/2-x)\cos(\pi/2-x)dx+\int_{0}^{\pi/2} (x+\pi/2)\cos(x+\pi/2)dx\\
&=\int_0^{\pi/2} (\pi/2-x)\cos(\pi/2-x)dx-\int_{0}^{\pi/2} (x+\pi/2)\cos(x-\pi/2)dx\\
&=\int_0^{\pi/2} ((\pi/2-x)\cos(\pi/2-x)-(x+\pi/2)\cos(x-\pi/2))dx\\
&=\int_0^{\pi/2} ((\pi/2-x)-(x+\pi/2))\cos(x-\pi/2)dx\\
&=\int_0^{\pi/2} -2x\cos(x-\pi/2)dx\\
&< 0
\end{array}
$
