Find arc length of $ y=( e^{x/2} + e^{-x/2} ) $ over this interval $[-2,2]$

Question

Question : Find arc length of $ y=( e^{x/2} + e^{-x/2} ) $ over this interval $[-2,2]$.

My Try : I found the derivative of $y$ and substituted it in the formula but after that I am stuck.

$$\int_{-2}^{2} \sqrt{\frac{(1+(e^{-x/2}(e^x-1))^2}{2}}dx$$

Answer 1

Note that $$y'=\frac{1}{2}(e^{x/2}-e^{-x/2})$$ so $$y'^2=\frac{1}{4}(e^x+e^{-x}-2)$$ It gives $$2 e-\frac{2}{e}$$ Note that $$1+\frac{1}{4}(e^x+x^{-x}-2)=\frac{4+e^x+e^{-x}-2}{4}=\frac{(e^{x/2}-e^{-x/2})^2}{4}$$

Answer 2

An easier way to do this is recognize that $y=2\cosh(x/2)$. Then $y'=\sinh(x/2)$ and

$$ (y')^2=\sinh^2(x/2)\\ 1-(y')^2=\cosh^2(x/2)\\ s=\int_{-2}^2 \sqrt{\cosh^2(x/2)} dx=2\sinh(x/2)\big|_{-2}^2=2(\sinh(1)-\sinh(-1))\\ s=4\sinh(1)\approx4.70080 $$