<s>[INST] parametric spiral arc length I am attempting to calculate the arc length of a spiral. Given an Archamedean spiral of the parametric form:
$x(t) = \sqrt{t}\cos\left(\omega\sqrt{t} \right)$ and $y(t) = \sqrt{t}\sin\left(\omega\sqrt{t} \right)$, the arc length $L$ is calculated as
\begin{align}\label{parametricArcLength}
 L & = \int_a^b \sqrt{1 + \left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \right)^2}\cdot \frac{dx}{dt}dt \\
 & = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ \ dt 
\end{align}
with 
\begin{align}
 \frac{dx}{dt} & = \frac{1}{2\sqrt{t{}}} \cdot \cos (\omega \sqrt{t}) - \sqrt{t} \cdot \frac{\omega}{2\sqrt{t}} \sin (\omega \sqrt{t}) \\
 & = \frac{\cos (\omega \sqrt{t})}{2\sqrt{t}} - \frac{\omega \sin (\omega \sqrt{t})}{2} \nonumber
\end{align}
\begin{align} \label{dy_dt}
\frac{dy}{dt} & = \frac{1}{2\sqrt{t}} \cdot \sin (\omega \sqrt{t}) + \sqrt{t} \cdot \frac{\omega}{2\sqrt{t}} \cos (\omega \sqrt{t}) \\
 & = \frac{\sin (\omega \sqrt{t})}{2 \sqrt{t}} + \frac{\omega \cos (\omega \sqrt{t})}{2} \nonumber
\end{align}
yielding
\begin{split}
L = &\int_{t_a}^{t_b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ \ dt \\
   = & \int_{t_a}^{t_b} \left[\frac{1}{4}\left(\frac{\cos^2 \left(\omega \sqrt{t}\right)}{t} - \frac{2\omega\cos\left(\omega\sqrt{t} \right) \sin \left(\omega \sqrt{t}\right)}{\sqrt{t}} + \omega^2 \sin^2\left(\omega \sqrt{t}\right)\right) \right.\\
 & \left. + \frac{1}{4}\left( \frac{\sin^2\left( \omega\sqrt{t} \right)}{t} + \frac{2\omega \sin \left(\omega \sqrt{t}\right) \cos\left(\omega \sqrt{t} \right)}{\sqrt{t}} + \omega^2 \cos\left(\omega \sqrt{t} \right)  \right) \right]^{1/2}dt\\
 = & \int_{t_a}^{t_b} \left[\frac{1}{4}\left( \frac{\sin^2\left(\omega\sqrt{t}\right) + \cos^2\left(\omega\sqrt{t}\right)}{t}   + \omega^2\sin^2\left(\omega\sqrt{t}\right) + \omega^2\cos^2\left(\omega\sqrt{t}\right)\right)\right]^{1/2}dt\\
 = & \int_{t_a}^{t_b} \left[ \frac{1}{4} \left( \frac{1}{t} + \omega^2 \right)\right]^{1/2}dt\\
 = &  \frac{1}{2}\int_{t_a}^{t_b}\left[\frac{1}{t} + \omega^2 \right]^{1/2}dt
\end{split}
Is the above integral the correct solution? If so, what method should be used to solve it?
 [/INST] Parametric curves are just screaming out to be solved in the complex plane. Consider that
$$z=x+iy=\sqrt{t}~e^{i\omega\sqrt{t}}$$
The arc length is given by
$$s=\int |\dot z|dt$$
Thus,
$$
\dot z=\left(\frac{1}{2\sqrt{t}}+\frac{i\omega}{2} \right)e^{i\omega\sqrt{t}}\\
|\dot z|=\frac{1}{2}\sqrt{\frac{1}{t}+\omega^2}
$$
Then
$$
\begin{align}
s=\int |\dot z|dt
&=\frac{1}{2}\int \sqrt{\frac{1}{t}+\omega^2}~dt\quad \text{(same as in the OP)}\\
&=\frac{1}{\omega}\int \sqrt{1+u^2}~du,\quad u=\omega\sqrt{t}\\
&=\frac{1}{2\omega}\left[u\sqrt{1+u^2}+\sinh^{-1}u \right]\\
&=\frac{1}{2\omega}\left[\omega\sqrt{t}\sqrt{1+\omega^2t}+\sinh^{-1}(\omega\sqrt{t}) \right]\\
\end{align}
$$
Notice, in particular, how this method circumvents all of the trigonometry in order to arrive at the same expression for the arc length integral as in the OP.
I have verified this result numerically for randomly selected values of $\omega$.
</s>