Q) Given that y(x) = log(1 + cos x) on the interval (0, π)
Taking signed arclength s=0 at x=0 show that:
x(s) = 4arctan(tanh(s/4))
Attempt)
I tried using (ds/dx)^2= 1 + (dy/dx)^2 rule to get dx/ds = (cosx+1)/root(2cosx+2)
Then integrating dx/ds gives me x(s) = 2sin(x/2)+ c
Im really stuck on the next step on what to do next to get x(s) = 4arctan(tanh(s/4)) and how do I calculate y(s) from this?
The problem is a bit obscure. Recalling what you did, for the sake of public understanding:
$$\ell = \frac{\text{d}s}{\text{d}x} = \sqrt{1 + \left(\frac{\text{d}y}{\text{d}x}\right)^2}$$
$$y(x) = \ln(1 + \cos(x)) \longrightarrow 1 + \left(\frac{\text{d}y}{\text{d}x}\right)^2 = \frac{2}{1 + \cos(x)}$$
$$s(x) = \int \sqrt{\frac{2}{1 + \cos(x)}}\ \text{d}x = 2 \ \text{arctanh}\left(\sin \left(\frac{x}{2}\right)\right) + C$$
On the other hand we have too:
$$\frac{\text{d}x}{\text{d}s} = \frac{1}{\frac{\text{d}s}{\text{d}x}} = \frac{1}{\sqrt{\frac{2}{1 + \cos(x)}}} = \cos\left(\frac{x}{2}\right)$$
Yet from here I cannot understand the statement of the problem and most of all how it gets to that result. Do you have a copy of the homework, for even finding the inverse function I do not think there is a way to come to that result...