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# The hyperbolic cosine function $\cosh \left( x \right)$ is defined as $\cosh \left( x \right) = \dfrac{{{e^x} + {e^{ - x}}}}{2}$. Find the arc length of $\cosh \left( x \right)$ on the interval $\left[ { - \ln 2,\ln 2} \right]$.  Hint: First find the derivative of the given function and after that use the arc length formula $S = \int_a^b {\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} dx}$, substitute the obtained values in it and find the integral value. The interval is given as $\left[ { - \ln 2,\ln 2} \right]$ which is the limit of the integral $\left[ {a,b} \right]$.

Consider step by step solution
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 $\dfrac{3}{2}$.
Note: We have used the logarithm or exponent rules which states that ${e^{\ln x}} = x$ and ${e^{ - \ln x}} = \dfrac{1}{x}$.
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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