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 $\frac{x\sin(x)}{1+\cos(x)^2}\geq x\sin(x)/2$ ?
 A: 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 .
A: You have some excellent answers about proving your inequality now. However, some seem to think that we cannot evaluate this integral directly. I will show that in fact this integral can be evaluated directly.

Firstly, we need to know the following very useful result, which can be seen by making the substitution $u=a+b-x$:
$$\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx$$
In our case, if $I$ is our integral, then
$$\begin{align}I&=\int_{0}^{\pi} \frac{x\sin(x)}{1+\cos^2(x)}dx=\int_0^{\pi}\frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}dx\\
&=\int_0^{\pi}\frac{(\pi-x)\sin(x)}{1+\cos^2(x)}dx=\pi\int_0^{\pi}\frac{\sin(x)}{1+\cos^2(x)}-\int_0^{\pi}\frac{x\sin(x)}{1+\cos^2(x)}dx\\
&=\pi\int_0^{\pi}\frac{\sin(x)}{1+\cos^2(x)}dx-I\\
\implies 2I&=\pi\int_0^{\pi}\frac{\sin(x)}{1+\cos^2(x)}dx\\
\implies I&=\frac{\pi}{2}\int_0^{\pi}\frac{\sin(x)}{1+\cos^2(x)}dx
\end{align}$$
Now make the substitution $y=\cos x$, so that $\sin x~dx=-dy$. Dealing with the bounds now: when $x=\pi$, $y=-1$ and when $x=0$, $y=1$. Hence,
$$\begin{align}I&=\frac{\pi}{2}\int_0^{\pi}\frac{1}{1+\cos^2(x)}\sin x~dx\\
&=\frac{\pi}{2}\int_1^{-1}\frac{-1}{1+y^2}dy=\frac{\pi}{2}\int_{-1}^1\frac{1}{1+y^2}dy\\
&=\frac{\pi}{2}\left[\arctan y\right]_{-1}^1=\frac{\pi}{2}(\arctan 1-\arctan(-1))\\
&=\frac{\pi}{2}(2\arctan 1)=\pi\arctan 1\\
&=\frac{\pi^2}{4}
\end{align}$$
A: Those inequalities in your last sentence are both obviously true when $x=0$.
And if $0 < x \le \pi/2$ then all factors occurring in those inequalities are positive (the factor $x$, the factor $\sin(x)$, and the factor $1 + \cos(x)^2$), and so after some rearrangement those inequalities are equivalent to $1 + \cos(x)^2 \ge 1$ and $1 + \cos(x)^2 \le 2$ which together are equivalent to $0 \le \cos(x)^2 \le 1$ which is true.
A: COMMENT.-This integral is not elementary so it is not question of direct calculation. However we have
$$\int_{0}^{\pi} \frac{x\sin(x)}{1+\cos(x)^2}\gt \int_{0}^{\pi} \frac{\sin(x)}{1+\cos(x)^2}=\frac{\pi}{2} $$ and something similar coul be apply maybe to the majorant $\pi$. This is just a comment.
