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!

$$\int_0^{\pi} x\cos(x)dx=\int_0^{\frac \pi 2} x\cos(x)dx+\int_{\frac \pi 2} ^{\pi} x\cos(x)dx$$
let $u=\pi-x$ in the second integral
$$\int_0^{\pi} x\cos(x)dx=\int_0^{\frac \pi 2} x\cos(x)dx - \int_0^{\frac \pi 2}  (\pi-u)\cos(u)du$$
$$=\int_0^{\frac \pi 2}  (2u-\pi)\cos(u)du$$
which is negative because $\cos(u)\ge 0$ and $(2u-\pi )\le 0 $ whenever $0 \le u \le \frac \pi 2$ 
