# 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 $$\frac{x\sin(x)}{1+\cos(x)^2}\geq x\sin(x)/2$$ ?

• Are asking whether those two inequalities in your last sentence are true for all $x \in [0,\pi/2]$? Apr 13 '21 at 14:12
• Any particular reason for not wanting to simply calculate the integral? Integration by parts and a nod to symmetry shows the value is $\pi^2/4$. Apr 13 '21 at 14:50

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 .

• In the inequality we want to prove $π/2≤\int_{0}^{\pi} \frac{xsin(x)}{1+cos(x)^2} ≤π$ Can we substitute any value of x in the interval? Like 0 or π ? @Vivaan Daga Apr 13 '21 at 15:05
• I don’t understand substitute $x$ where ? @user726608 Apr 13 '21 at 15:07
• In the inequality we want to prove, the one i mentioned in the comment.. Apr 13 '21 at 15:14
• Yes because of the reason mentioned in the answer it holds for $[0,/pi]$ not only $[0,/pi/2]$ this fact is important since the integral of $xsin(x)$ from $0$ to pi is equal to pi and since $f>0$ the inequality you want to prove immediately follows .@user726608 Apr 13 '21 at 15:16
• @user726608 yes exactly Apr 13 '21 at 16:07

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}

• Hi @A-Level Student nice approach! Can we show the inequality, by finding the derivative of $f$ and the stational points in the interval (if exist) ? It is problematic since in the interval the function is increasing so 0 and $\pi$ will be its stational points! And we also know that $f\geq 0, f(0)=f(\pi)=0$.. Apr 13 '21 at 21:37

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.

• Since your question has $\le$ signs --- which stands for less than OR EQUAL TO --- there is no problem substituting $x=0$. Apr 13 '21 at 14:21
• But we also have $\geq$ so $0 \geq \pi/2$..i mean substituting in the inequality we want to show Apr 13 '21 at 14:23
• I do not see $0 \ge \pi/2$ occurring anywhere in your post. Where do you see it occurring? Apr 13 '21 at 14:24
• In the inequality we want to prove $\pi/2 \leq \int_{0}^{\pi} \frac{x\sin(x)}{1+\cos(x)^2} \leq \pi$ Can we substitute any value of x in the interval? Like 0 or $\pi$ ? Apr 13 '21 at 14:34
• @LeeMosher the inequalities also hold for $0<x<\pi$ this helps because the integral of $xsinx$ from $0$ to $\pi$ is $pi$ and so the problem mentioned in the question can be answered as mentioned in my answer Apr 13 '21 at 15:22

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.

• How can you say that the integral is not elementary? I am just curious. Apr 13 '21 at 14:59
• Why do you put "$>$" sign? How do you justify this?
– user
Apr 13 '21 at 15:02
• @Myst1cal: Many ways. First at all it does not figure in any "table" of elementary integrals. Onother way is asking about this integral in a computer calculator so you can see what "complicated" is the answer.Regards. Apr 13 '21 at 15:03
• @Piquito, I agree that the indefinite integral of $x\sin x/(1+\cos^2x)$ is not elementary, but the definite integral is not hard to evaluate exactly with first-year calculus techniques. (See my comment below the OP.) Apr 13 '21 at 15:07
• @user: below the integral $x$ is a positive number which can give "problems" from $0$ to $1$ but itis compensatory from $1$ to $\pi=3.16159.....$ Actually the graph is good to prefer. Apr 13 '21 at 15:07