I want to calculate the surface area of the hyperboloid $$M = \{(x,y,z) \in \mathbb R^3 : x^2 + y^2 -z^2 = 1, -1\leq z \leq 1\}.$$ Let $$k: ]-\infty, \infty[ \times [0, 2\pi] \to M; \\k(x,t) = (\cosh x \cos t, \cosh x \sin t, \sinh x)$$ be a parametrization of $M$. Now $$Dk(x,t) = \begin{pmatrix} \sinh x \cos t & \sinh x \sin t & \cosh x \\ -\cosh x \sin t & \cosh x \cos t & 0\end{pmatrix}$$ and $ \sqrt{\det Dk(x,t)^TDk(x,t)} = \cosh(x)\sqrt{1+2\sinh^2(x)}$. Now is this correct? Because the integral I need to solve is then $$\omega^M(M) = 2 \pi \int_{\text{arsinh} (-1)}^{\text{arsinh(1)}}\cosh(x)\sqrt{1+2\sinh^2(x)} dx.$$ How do I do that?
Update:
Substituting $u = \sqrt{2} \sinh(x)$ we get$$2\pi \frac{1}{\sqrt{2}}\int_{-\sqrt2}^{\sqrt2}\sqrt{1+u^2} du$$ and again by substituting $$= \frac{2\pi}{\sqrt2}\int_{\text{arsinh}(-\sqrt2)}^{\text{arsinh}(\sqrt2)}\cosh^2(x)dx$$ and now I can evaluate this using $$\cosh(x) = \frac{1}{2}(e^x + e^{-x})$$. I assume that this does not evaluate to something beautiful.
hint no.1:...substitute $\sinh u=\sqrt{2}\sinh x$
Hint no.2:...use $\cosh^2x=\frac 12(1+\cosh2x)$ and $\sinh2x=2\sinh x\cosh x$