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!
 A: You may just integrate by parts:
$$
\int_0^{\pi} x\cos x\:dx=[x \sin x]_0^\pi-\int_0^{\pi} \sin x\:dx=0-\int_0^{\pi} \sin x\:dx<0
$$ since $\sin x \geq0$ over $[0,\pi]$.
A: $$\int_0^{\pi} x\cos(x)dx=\int_0^{\frac \pi 2} x\cos(x)dx+\int_{\frac \pi 2} ^{\pi} x\cos(x)dx$$
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 $0 \le u \le \frac \pi 2$ 
A: $\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}
$
A: 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
$$\int_0^{\frac{\pi}{2}} x\cos xdx< -\int_{\frac{\pi}{2}}^{\pi} u\cos udu$$
$$\implies \color {red}{\int_0^{\frac{\pi}{2}} x\cos xdx +\int_{\frac{\pi}{2}}^{\pi} u\cos udu<0}$$
A: $\cos x$ goes from $1$ (at $x=0$) to $0$ (at $x=\frac{\pi}{2}$) and $-1$ (at $x=\pi$). And x goes from $0$ to $\pi$ being positive, so the signe of $x\cos x$ depends of the signe of $\cos x$ in the interval $[0, \pi]$.
We have $$\int _{\frac{\pi}{2}}^{\pi}x\cos x\approx \frac{\pi}{2}(-\pi)=-\frac{\pi^2}{2}$$ and $$\int_0^{\frac{\pi}{2}} x\cos x\lt|\int_0^{\frac{\pi}{2}}\cos x|=1 $$
Thus the integral must be negative. 
