Originally Posted by

**SlipEternal** Another way to do this problem:

$$x = \cosh y$$ so $$\dfrac{dx}{dy} = \sinh y$$. This simplifies the formula to:

$$S = 2\pi \int_0^{\ln 2}\cosh y \sqrt{1+\sinh^2 y}dy$$

Since $$\cosh^2 y = 1 + \sinh^2 y$$, you can rewrite:

$$S = 2\pi \int_0^{\ln 2} \cosh^2 y dy$$

Using the double angle formula for hyperbolic cosine: $$\cosh^2 y = \dfrac{1}{2} + \dfrac{1}{2}\cosh(2y)$$

$$S = \left. \pi y + \dfrac{\pi}{2}\sinh(2y) \right]_0^{\ln 2} = \left. \pi y + \dfrac{\pi}{4}\left(e^{2y} - e^{-2y}\right) \right]_0^{\ln 2}$$