How do you find the arc length of $\dfrac{e^x + e^{-x}}{2}$ from [0, 2] without using hyperbolic functions?
Using the formula $$\int_{a}^{b} \sqrt{1+(y')^2}dx$$ I run into a problem where I can't get from
$$y' = \left(\frac{1}{2}\right)\left({e^x - e^{-x}}\right)$$
to
$$1 + (y')^2 = \biggr(\frac{1}{2}\left({e^x + e^{-x}}\right)\biggr)^2$$
The book and other solutions I've seen make this step, but I don't follow. Shouldn't I end up with,
$$1 + (y')^2 = \biggr(\frac{1}{2}\left({e^x - e^{-x}}\right)\biggr)^2 + 1$$
What am I missing?
$$\quad{\int_{a}^{b} \sqrt{1+(y')^2}dx=\\ \int_{a}^{b} \sqrt{(\frac{1}{2}\left({e^x - e^{-x}}\right))^2 + 1}dx= \\\int_{a}^{b} \sqrt{\frac{1}{4}(\left({e^x - e^{-x}}\right)^2 +4)}dx= \\\int_{a}^{b} \sqrt{\frac{1}{4}(({e^{2x} +e^{-2x}-2e^x e^{-x}} +4)}dx= \\\int_{a}^{b} \sqrt{\frac{1}{4}(({e^{2x} +e^{-2x}-2} +4)}dx= \\\int_{a}^{b} \sqrt{\frac{1}{4}((e^{2x} +e^{-2x}+2)}dx= \\\int_{a}^{b} \sqrt{\frac{1}{4}((e^{2x} +e^{-2x}+2e^x e^{-x})}dx= \\\int_{a}^{b} \sqrt{\frac{1}{4}(e^{x} +e^{-x})^2}dx= \\=\int_{a}^{b} \sqrt{(\frac{e^{x} +e^{-x}}{2})^2}dx \\=\int_{a}^{b} (\frac{e^{x} +e^{-x}}{2})dx}$$