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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}

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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 $\sin$: $$2\int_0^a\cosh t\,dt=2[\sinh t]_0^a=2\sinh a$$

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  • $\begingroup$ Excellent. Thank you very much :) $\endgroup$ – user606466 Jan 13 '19 at 16:24
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The following identities/facts have been used:

  • 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 $x < 0$.

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