Showing without calculating the integral that: $\pi/2 \leq \int_{0}^{\pi} \frac{x\sin(x)}{1+\cos(x)^2} \leq \pi$ I want to prove without calculating the integral that:
$\pi/2 \leq \int_{0}^{\pi} \frac{x\sin(x)}{1+\cos(x)^2} \leq \pi$.
I have some problem, since if my function is:
$f:[0,\pi]\to R$ defined by
$f(x)=\frac{x\sin(x)}{1+\cos(x)^2}$ then $f(0)=f(\pi)=0$ and since f is non-negative in this intervel then the given integral is $\geq 0$ so is it legal to use the integral monotone theorem?
I mean can i say that for every $x\in [0,\pi]$, $\frac{x\sin(x)}{1+\cos(x)^2} \leq x\sin(x)$ and $\frac{x\sin(x)}{1+\cos(x)^2}\geq x\sin(x)/2$ ?

For  $x\in [0,\pi]$, $\frac{x\sin(x)}{1+\cos(x)^2} \leq x\sin(x)$ and $\frac{x\sin(x)}{1+\cos(x)^2}\geq x\sin(x)/2$  , this is true because ,the first inequality is true because the denominator is $>1$ for all $x$ in the domain and the numerator is positive and the second because the denominator is always less than equal to $2$ for all $x$ in the domain and the numerator is always positive  .
Since $f(x)>0$ you can conclude that the inequality also holds true for their integrals from $0$ to $\pi$
And since the integral of $x\sin(x)$ from $0$ to $\pi$ is $\pi$ the first inequality in the question is true .
