Integrate the function $\int \sqrt{2+ e^{2t} + e^{-2t} } \; dt$

Question

I'm unsure how to solve this arc length. The original problem says to find the arc length of

$r(t) = \langle \sqrt{2}t, e^t, e^{-t} \rangle $

deriving I arrive at

$r'(t) = \langle \sqrt{2}, e^t, -e^{-t} \rangle $

Using the arc length formula I arrive at

$\int \sqrt{2+ e^{2t} + e^{-2t} } \; dt$

I'm aware of

$e^{2t} + e^{-2t} = 2 \cosh \, 2t$

via WolframAlpha but I'm sure how this might help with integrating. Any help is appreciated!

Answer 1

Hint:

$$e^{2t}+e^{-2t}+2=(e^t)^2+(e^{-t})^2+2e^t\cdot e^{-t}=(e^t+e^{-t})^2$$

and for real $a, e^t,e^{-t}>0$

Finally for real $x,|x|=+x$ if $x\ge0$

Answer 2

$$\int \sqrt{2+e^{2t}+e^{-2t}}\;\mathrm dt=\int\sqrt{(2\cosh t)^2}\;\mathrm dt=\int2\cosh t\;\mathrm dt =2\sinh t.$$

Answer 3

Following your own hint,

$$2+e^{2t}+e^{-2t}=2+2\cosh2t=4\cosh^2t.$$

The rest is immediate.