Work out the arc length of the curve $y = \frac12\left(e^x-e^{-x}\right)$ between $\left[-\ln|2| , \ln|2|\right]$

Question

Work out the arc length of the curve $y = \dfrac12\left(e^x-e^{-x}\right)$ between $\left[-\ln|2| , \ln|2|\right]$.

I got this question in a math exam and I felt very defeated. Please can someone shed some light?

Thanks, Nick

Answer 1

WolframAlpha gives the solution as

$$\int_{-\log(2)}^{\log(2)} \sqrt{1 + \cosh^2(x)}\ dx = -2 i \sqrt{2} E\biggr(i \log(2) \biggr|1/2\biggr)≈2.0437$$

On the other hand, following Paul Sinclair's suggestion ($e^x+e^{-x}$), then the solution is

$$\int_{-\log(2)}^{\log(2)} \sqrt{\cosh^2(x)}\ dx = \frac{3}{2}$$

I have verified these solutions numerically.