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Consider the given function \[y = \cosh x\] which is further defined as the \[\cosh \left( x \right) = \dfrac{{{e^x} + {e^{ - x}}}}{2}\].

Now, find the derivative of the given function with respect to \[x\].

Thus, we get that,

\[\dfrac{{dy}}{{dx}} = \sinh x\]

Thus, substitute the value in the formula of the arc length that is \[S = \int_a^b {\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} dx} \]

From this, we get that,

\[ \Rightarrow S = \int_a^b {\sqrt {1 + {{\left( {\sinh x} \right)}^2}} dx} \]

As we know the trigonometric identity that \[1 + {\sinh ^2}x = {\cosh ^2}x\]. So, we will apply it here and simplify the value.

\[

\Rightarrow S = \int_a^b {\sqrt {{{\cosh }^2}x} dx} \\

\Rightarrow S = \int_a^b {\cosh xdx} \\

\]

Now, apply the limits given in the question, that is \[\left[ { - \ln 2,\ln 2} \right]\].

\[ \Rightarrow S = \int_{ - \ln 2}^{\ln 2} {\cosh xdx} \]

Now, as we see here that the limits are symmetric so we can take the limits as \[\left[ {0,\ln 2} \right]\] and multiply with 2 and integration of \[\cosh x = \sinh x\]

\[

\Rightarrow S = 2\int_0^{\ln 2} {\cosh xdx} \\

\Rightarrow S = 2\left[ {\sinh x} \right]_0^{\ln 2} \\

\]

Now, we are given that \[y = \cosh x\] is defined as the \[\cosh \left( x \right) = \dfrac{{{e^x} + {e^{ - x}}}}{2}\]. Similarly, we know that \[y = \sinh x\] is defined as the\[\sinh \left( x \right) = \dfrac{{{e^x} - {e^{ - x}}}}{2}\].

Substitute the value of \[\sinh \left( x \right) = \dfrac{{{e^x} - {e^{ - x}}}}{2}\] in the integral part.

Thus, we get,

\[

\Rightarrow S = 2\left[ {\dfrac{{{e^x} - {e^{ - x}}}}{2}} \right]_0^{\ln 2} \\

\Rightarrow S = 2\left[ {\dfrac{{{e^{\ln 2}} - {e^{ - \ln 2}}}}{2} - \dfrac{{{e^0} - {e^{ - 0}}}}{2}} \right] \\

\Rightarrow S = 2\left[ {\left( {\dfrac{{2 - \dfrac{1}{2}}}{2}} \right) - \left( {\dfrac{{1 - 1}}{2}} \right)} \right] \\

\Rightarrow S = 2\left[ {\dfrac{{\dfrac{3}{2}}}{2} - 0} \right] \\

\Rightarrow S = 2\left[ {\dfrac{3}{4}} \right] \\

\Rightarrow S = \dfrac{3}{2} \\

\]

Thus, the arc length of the given function \[y = \cosh \left( x \right)\] is \[\dfrac{3}{2}\].

Also, values of \[{e^0} = 1\] and \[{e^{ - 0}} = 1\]. The derivative of \[\cosh x = \sinh x\] and integration of \[\sinh x = \cosh x\]. As the given interval is symmetric, we can divide it into half and the interval get changes into \[\left[ {0,\ln 2} \right]\]. Also use the identity \[1 + {\sinh ^2}x = {\cosh ^2}x\] which makes the calculation easier.

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