For reference, take a look at this demonstration of $\cosh(x)$ and the definitions of $\cosh(x)$ and $\sinh(x)$ below:
$$
\begin{array}{l}
\sinh x:=\frac{e^{x}-e^{-x}}{2} \\
\cosh x:=\frac{e^{x}+e^{-x}}{2}
\end{array}
$$
From the demonstration, we see that the arclength $MN$ originates at $(0,1)$ and terminates at $(x,r)$ as per the definitions above. Notice that $r=\frac{e^{x}+e^{-x}}{2}=\cosh(x)$ and is the radius of the circle. To answer your question, it must be shown that the arclength of $\cosh(x)$ is the same as the bottom leg of the right triangle shown in the demonstration. Notice that this triangle has a vertical (smaller) leg of $1$ and a hypotenuse of $r$. From simple trigonometry, the long leg, or the distance along the $x$-axis, is $\sqrt{r^2-1}$.
Mathematically, this can be summarized as (using the arclength formula):
$$
\int_0^x \sqrt{1+(\frac{d}{du}\cosh(u))^2}\;du=\sqrt{r^2-1}
$$
Taking the arclength expression first, note that $\frac{d}{dx}\cosh(x)=\sinh(x)$. This makes the argument of the square root $1+\sinh^2(x)$. To continue integrating this, this argument can be simplified to $\cosh^2(x)=1+\sinh^2(x)$ from the identity $\cosh^2(x)-\sinh^2(x)=1$. Our integral then becomes:
$$
\int_0^x \sqrt{\cosh^2(u)}\;du=\int_0^x \cosh(u)\;du
$$
From inspection, also note that $\int \cosh(x)=\sinh(x)+C$. Thus, our result is:
$$
\sinh(x)-\sinh(0)=\sinh(x)
$$
Now, we must show that $\sinh(x)=\sqrt{r^2-1}$:
$$
\begin{align*}
r^2-1&=\frac{e^{2x}+2e^xe^{-x}+e^{-2x}}{4}-\frac{4}{4}\\
&=\frac{e^{2x}+e^{-2x}-2}{4}
\end{align*}
$$
$$
\begin{align*}
\sinh^2(x)&=\frac{e^{2x}-2e^xe^{-x}+e^{-2x}}{4}\\
&=\frac{e^{2x}+e^{-2x}-2}{4}
\end{align*}
$$
$$\therefore r^2-1=\sinh^2(x)\implies \boxed{\sqrt{r^2-1}=\sinh(x)}$$
This shows that the length of the segment along the $x$-axis intersecting the circle ($\sqrt{r^2-1}$) is the same as the arclength of $\cosh(x)$.
Best Answer
From what I understand from the video, it's only a heuristic technique of computing the derivative of the integral $I(t) := \int_{x_1}^{x_2} f(x, t) dx.$ Thus, it is meant to be taken with a grain (or rather a full tablespoon) of salt. It is not rigorous. In his video, $dt$ signifies a finite quantity, however small, thus a constant, it is not to be taken as the $1-$form $dx$ (or measure, what have you, choose your preferred way of looking at integrals), thus can be safely moved outside the integral.