$$\int_{-5}^5 \frac{1}{1+2^{\arctan(x)}} dx$$
In the solution of this example authors do such thing:
$$\int_{-5}^5 \frac{1}{1+2^{\arctan(x)}} dx = [x = -t; dx = -dt; t = -x] = \int_5^{-5} \frac{-1}{1+2^{-{\arctan(t)}}} dt = \int_{-5}^5 \frac{1}{1+2^{-{\arctan(x)}}} dx$$
Then we just sum this two integrals, everything is great, the answer is 5. But I can't understand how we came to the final integral. If I just replace back t, I will get the first integral. Why does this last step work?
$$ \int_5^{-5} \frac{-1}{1+2^{-{\arctan(t)}}} dt\\=\int_{-5}^5 \frac{1}{1+2^{-{\arctan(t)}}} dt\\= \int_{-5}^5 \frac{1}{1+2^{-{\arctan(x)}}} dx$$
That last step (renaming $t$ to $x$) works because $t$ and $x$ here are just dummy variables.
After all, $$\int_0^\pi \sin(x)\,\mathrm dx=2=\int_0^\pi\sin(\theta)\,\mathrm d\theta.$$
(On the other hand, $\large\int \sin(x)\,\mathrm dx=-\cos(x)\ne-\cos(\theta)=\int\sin(\theta)\,\mathrm d\theta.)$