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!

Notice that 
$$\int_0^{\pi} x \cos xdx =\int_0^{\frac{\pi}{2}}x\cos xdx+\int_{\frac{\pi}{2}}^{\pi} \cos x dx$$
Now, the first integral is positive and the second one is negative. But for every $x\in [0, \pi/2)$ we have
$$x<\pi -x$$
Because of the monotonicity of the integral
$$\int_0^{\frac{\pi}{2}} x\cos x dx<\int_0^{\frac{\pi}{2}} (\pi-x)\cos xdx$$
Substituting $u=\pi-x$ on the right yields
$$\int_0^{\frac{\pi}{2}} x\cos xdx< -\int_{\frac{\pi}{2}}^{\pi} u\cos udu$$
$$\implies \color {red}{\int_0^{\frac{\pi}{2}} x\cos xdx +\int_{\frac{\pi}{2}}^{\pi} u\cos udu<0}$$
