Understanding a particular transformation of an integral given in a proof Using the theorem of mean values find the sign of the integral...
  $$\int_{0}^{2 \pi}{\sin x \over x}dx= \int_{0}^{\pi}{\sin x \over x}dx+\int_{\pi}^{2 \pi}{\sin x \over x}dx$$
  Then: $[x-\pi=t ; dx=dt]$
  $$\int_{0}^{\pi}{\sin x \over x}dx+\int_{0}^{\pi}{\sin(t+ \pi) \over t+ \pi}dt \, \,\color{red}{\fbox{=}} \int_{0}^{\pi}{\sin x \over x}dx-\int_{0}^{\pi}{\sin(x)\over x+ \pi }dx$$


How does one see the (boxed, red) equality?
Hint:

if you have $f(x)=F'(x)$, you can change the ''symbol'' for the variable, i.e write $t=x$, and you have: $f(t)=F'(t)$.

Than:

$$
\int_a^b f(x)=F(x)|_a^b=F(b)-F(a)
$$

$$
\int_a^b f(t)=F(t)|_a^b=F(b)-F(a)
$$

In your case you have:
$$
\int_0^\pi\dfrac{\sin (t+\pi)}{t+\pi}dt=-\int_0^\pi\dfrac{\sin t}{t+\pi}dt=-\int_0^\pi\dfrac{\sin x}{x+\pi}dt
$$
Hint:

if you have $f(x)=F'(x)$, you can change the ''symbol'' for the variable, i.e write $t=x$, and you have: $f(t)=F'(t)$.

Than:

$$
\int_a^b f(x)=F(x)|_a^b=F(b)-F(a)
$$

$$
\int_a^b f(t)=F(t)|_a^b=F(b)-F(a)
$$

In your case you have:
$$
\int_0^\pi\dfrac{\sin (t+\pi)}{t+\pi}dt=-\int_0^\pi\dfrac{\sin t}{t+\pi}dt=-\int_0^\pi\dfrac{\sin x}{x+\pi}dt
$$