Explanation of the steps on calculating the length of the catenary

Can someone please explain to me what identities have been used in the calculation of the below example and why the limits of integration change from $$0$$ to $$a$$? I have tried $$\sinh^2{ t} = [-1 + \cosh (2 t)]/2$$ and end up with nothing like the form below.

Example:

The catenary, is the shape of a wire hanging under its own weight and is given by

$$x(t) = t{\bf{e}}_1 + \cosh{t} {\bf{e}}_2,$$ with $$I = [−a,a]$$.

The total length of this catenary is

\begin{align} L &=\int_ {a}^{−a} {\sqrt{1 + \sinh^2{ t}}} \quad dt \\ &=2\int_0^a \cosh{t} \quad dt \\ &=2\sinh{a}. \end{align}

Just as we have the identity $$\sin^2t+\cos^2t=1$$, so do we have the identity $$\cosh^2t-\sinh^2t=1$$ Thus $$\sqrt{1+\sinh^2t}=\cosh t$$ and $$L=\int_{-a}^a\sqrt{1+\sinh^2t}\,dt=\int_{-a}^a\cosh t\,dt$$ $$\cosh$$ is an even function, and for even functions the integral from $$-a$$ to $$a$$ is twice the integral from 0 to $$a$$. $$\int_{-a}^a\cosh t\,dt=2\int_0^a\cosh t\,dt$$ The antiderivative of $$\cosh$$ is $$\sinh$$, like how the antiderivative of $$\cos$$ is $$\sin$$: $$2\int_0^a\cosh t\,dt=2[\sinh t]_0^a=2\sinh a$$
• For any function $$f:[-a,a] \to \mathbb{R}$$ which is both even and integrable, $$\int_{-a}^a f(x) dx = 2 \int_{0}^af(x) dx$$. This explains the change in the limits of integration.
• For any real number $$t$$, $$1+ \sinh ^2(t) = \cosh ^2 (t)$$.
• For any non-negative real number $$x$$, $$|x| = \sqrt{x^2} = x$$, used in conjunction with the fact that $$\cosh ^2 (t) > 0$$, for all $$t \in \mathbb{R}$$.
$$\textit{Side note:}$$ The last point is important to be aware of. We cannot always make the simplification $$\sqrt{x^2} = x$$, as it fails when $$x < 0$$.