<s>[INST] Explaining $\int_{-1}^1\frac{1}{1+x^2}\,dx = \frac{\pi}{2}$. How would you explain to a student that
$$ \int_{-1}^1\frac{1}{1+x^2}\,dx = \arctan(1) - \arctan(-1) = \frac{\pi}{2} $$
and not
$$ \int_{-1}^1\frac{1}{1+x^2}\,dx = \arctan(1) - \arctan(-1) \neq \frac{\pi}{4} - \frac{3\pi}{4} = -\frac{\pi}{2}$$
besides the obvious fact that $\arctan x$ cannot map to two distinct values?
 [/INST] Well, you can also use the even function 'card',
i.e, since $f(x) = f(-x)$,
$$ \int_{-1}^{1}\frac{1}{1+x^2}=2\int_{0}^{1}\frac{1}{1+x^2}=2(\arctan 1 +\arctan 0) = 2 \, \left(\frac{\pi}{4} \right) =\frac{\pi}{2} $$
</s>