Proving that $-\frac{\pi}{2}\le\int_{-1}^{1}\arctan (x)dx\le \frac{\pi}{2}$ I have the following question:

Prove that: $$-\frac{\pi}{2}\le\int_{-1}^{1}\arctan (x)dx\le \frac{\pi}{2}.$$


I know that the function is odd and therefore, the given integral is 0, and the inequality therefore holds. However, showing and proving this theorem in which the integral of an odd function from $-a$ to $a$ is zero, isn't what this question intended to. It is likely in this question that we should show that via Darboux's sums, or Riemann sums. Unfortunately, I have no idea how to deal with such trigonometric sums. Hence, I will be glad to get some help with this question, and I am sorry that I haven't written an attempt for this question. Thanks!

If $x\in[-1,1]$, $-\frac\pi4\leqslant\arctan(x)\leqslant\frac\pi4$, and therefore$$2\times\left(-\frac\pi4\right)\leqslant\int_{-1}^1\arctan x\,\mathrm dx\leqslant2\times\frac\pi4.$$
