How to find the unit tangent vector of a curve in R^3

Question

I have this curve defined by : \begin{align} x & =\int_0^t \frac{(1+\cosh^3 u)\cos u \,du}{\cosh^2 u}; \\[10pt] y & =\int_0^t \frac{(1+\cosh^3 u)\sin u \,du}{\cosh^2 u}; \\[10pt] z & =\int_0^t \frac{(1+\cosh^3 u)\sinh u \, du}{cosh^2 u} \end{align} I should find $\vec{T}$,the unit tangent vector to(C) in M,so I have to use t as a parameter,is there any easy way to integrate x,y,z?

Answer 1

$$\begin{align} x & =\int_0^t \frac{(1+\cosh^3 u)\cos u \,du}{\cosh^2 u}; \\[10pt] y & =\int_0^t \frac{(1+\cosh^3 u)\sin u \,du}{\cosh^2 u}; \\[10pt] z & =\int_0^t \frac{(1+\cosh^3 u)\sinh u \, du}{cosh^2 u} \end{align}$$ implies that $$ \frac {dx}{dt}=\frac{(1+\cosh^3 t)\cos t }{\cosh^2 t }$$$$ \frac {dy}{dt}=\frac{(1+\cosh^3 t)\sin t }{\cosh^2 t }$$$$ \frac {dy}{dt}=\frac{(1+\cosh^3 t)\sinh t }{\cosh^2 t }$$ We scale the results to get a tangent vector $$ V = (cost,sint,sinht)$$ Upon normalizing, we get a unit tangent vector $$T=\frac {1}{cosht} (cost,sint,sinht)$$