If $\displaystyle \alpha$ has constant torsion $\displaystyle c, \ c \neq 0$ show that $\displaystyle \alpha = \int_{s_0}^sT \ dt + K$ where $\displaystyle K \in \mathbb{R}^3$ and $\displaystyle T$ represents the tangent vector from the Frenet frame.

I was thinking of representing $\displaystyle \alpha$ through it's Taylor series but that got me nowhere. So I would have $\displaystyle \alpha(s) = \alpha(0) + sT_0 + \kappa_0\frac{s^2}{2}N_0+\kappa_0\tau_0\frac{s^3}{ 6}B_0$. But this doesn't seem to help.

Know I figured that since we are dealing with constant torsion we can just express it through the formula yielding:

$\displaystyle c= \frac{\det(\alpha', \alpha'', \alpha''')}{\|\alpha' \times \alpha''\|^2}$

now rearranging and using the dot product with $\displaystyle B$ I get $\displaystyle \|\alpha' \times \alpha''\|c = {\det(\alpha', \alpha'', \alpha''')} B$

Now I don't know If I'm on the right track or if this is even correct.

2. don't need to have constant torsion. For any curve $\displaystyle \alpha$, $\displaystyle \frac{d\alpha}{ds}=T$. Intergrate both sides we get $\displaystyle \alpha(s) = \int_{s_0}^s T dt + \alpha(s_0)$