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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 \[\left[ { - \ln 2,\ln 2} \right]\].

Last updated date: 17th Apr 2024
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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.