$\int\sqrt{1+t^2}\, dt$ using the substitution $t = \frac{1}{2}(e^u-e^{-u})$ This is what I have done so far:
$$\frac{dt}{du}=\frac{d}{du}\left(\frac{1}{2}(e^{u}-e^{-u})\right)=\frac{1}{2}(e^{u}+e^{-u}) \implies dt=\frac{1}{2}(e^{u}+e^{-u})\,du$$
$$L=\frac{1}{2}\int_{0}^{2}\sqrt{1+t^2}\, dt=\frac{1}{2}\int_{u(0)}^{u(2)}\frac{1}{2}(e^{u}+e^{-u})\sqrt{1+\left(\frac{1}{2}(e^{u}-e^{-u})\right)^2}\, du$$
The integral I get is quite complicated and ugly so I don't know were to go from here.
Solving for $dt$ seems like a step one should take, but what should I do with the $t^2$ under the square root? Should I try to solve for $u$ perhaps?
The integral should be solved using the substitution above. With other words: no integration by parts unless it is an intermediate step. Also no use of hyperbolic trig please.

Starting with
$$L=\frac{1}{2}\int_{a}^{b}\sqrt{1+t^2}\, dt=\frac{1}{4} \, \int_{u(a)}^{u(b)} (e^{u}+e^{-u}) \, \sqrt{1+\left(\frac{1}{2}(e^{u}-e^{-u})\right)^2} \, du$$
and then using
$$ 1 + \frac{(e^{u} - e^{-u})^2}{4} = \frac{e^{2 u} + 2 + e^{- 2 u}}{4} = \left(\frac{e^{u} + e^{-u}}{2}\right)^2 $$
the integral $L$ becomes
\begin{align}
L &= \frac{1}{4} \, \int (e^{u} + e^{-u}) \, \sqrt{\left(\frac{e^{u} + e^{-u}}{2}\right)^2} \, du \\
&= \frac{1}{8} \, \int (e^{u} + e^{-u})^2 \, du \\
&= \frac{1}{8} \, \left[ 2 \, u + \frac{e^{2 u} - e^{-2 u}}{2} \right]
\end{align}
Now, the hard part. Find what $u$ is to invert the variables.
\begin{align}
2 t &= e^{u} - e^{-u} \\
e^{2 u} - 2 t \, e^{u} - 1 &= 0 \\
e^{u} &= t \pm \sqrt{t^2 + 1} \\
u &= \ln(t \pm \sqrt{t^2 + 1}).
\end{align}
The use of this can be seen in:
\begin{align}
e^{2 u} &= e^{2 \, \ln(t + \sqrt{t^2 + 1})} = (t + \sqrt{t^2 + 1})^2 \\
e^{-2 u} &= (t + \sqrt{t^2 + 1})^{-2} = (t - \sqrt{t^2 + 1})^2 \\
\frac{1}{2} \, \left( e^{2 u} - e^{-2 u} \right) &= t \, \sqrt{t^2 +1}
\end{align}
and gives $L$ as
$$ L = \frac{1}{4} \, \left( \ln(t + \sqrt{t^2 +1}) + t \, \sqrt{t^2 +1} \right). $$
Using the limits the value desired should be obtained.
