Calculate the integral for the arc length of $y = \frac{e^x + e^{-x}}2$ from $-1$ to $1$

Question

I know to set up this integral you find the derivative: $$y' = \frac{e^x - e^{-x}}2dx$$ Then you set up the arc length function: $$L = \int^1_{-1}\sqrt{1 + (\frac{e^x - e^{-x}}2)^2 }dx $$

However I don't know how to solve this integral. Please help.

(Note: No knowledge about $\sinh$ and $\cosh$ and the related stuff.)

Answer 1

Hint:

$$1+(\frac{e^x-e^{-x}}2)^2=1+\frac{e^{2x}+e^{-2x}-2}4=\frac{e^{2x}+e^{-2x}+2}4=[g(x)]^2$$

Then compute

$$\int^1_{-1}g(x)dx$$