Q:

Find the length of a curve $\gamma(t)=(t, ch(t))$

I need to find the length of a curve that is parametrized like this:
$$\gamma(t)=(t, ch(t))=\left(t,\frac{e^t+e^{-t}}{2}\right)$$
So I use formula and I get:
$$s(t)=\int_0^t\sqrt{1+\frac{(e^t-e^{-t})^2}{4}}dt=\frac{1}{2}\int_0^t\sqrt{2+e^{2t}+e^{-2t}}dt$$
But I don't know how to evaluate this integral.

A:

$\sqrt{1+\left(\frac{e^t-e^{-t}}{2}\right)^2} = \sqrt{1+\sinh^2(t)} = \sqrt{\cosh^2(t)} = \cosh(t)$ 
$\int\limits_0^t\cosh tdt = \sinh t|_0^t = \sinh t$

