I am trying to teach myself how to compute the arc length of a function through a textbook, and I am stuck on how they did a particular integral substitution in a worked example.
I'll provide the specific line of working I am stuck to provide relevant information quickly. I'll follow with my working, and then provide the context of the question below if necessary.
Where I am stuck
I am unsure how to get from:
$$\int_1^3\sqrt{1+4t^2}\,dt$$
to, as the book puts it exactly:
$$\int{\frac{1}{2}\cosh^2u \,du}$$
They say they are putting $2t$ = $\sinh u$.
I am unsure of how to get to this line using that substitution. (I am also unsure why the bounds of integration are gone, I originally assumed out of convenience but different bounds due to the substitution are being used later in the example.)
My attempt
$\int_1^3\sqrt{1+4t^2}\,dt$
Using $2t$ = $\sinh u\Rightarrow2\,dt = \cosh u \Rightarrow dt = \frac{\cosh u}{2}\,du$.
$\int_1^3\frac{1}{2}\sqrt{1+\sinh^2 u} \cosh u \,du$
Using $\sqrt{1+\sinh^2 u} \equiv \cosh^2 u$:
$\int_1^3\frac{1}{2}\cosh^2u \cosh u \,du$
$\frac{1}{2}\int_1^3\cosh^3u \,du$
Clearly there is an issue here, as I have arrived at $\cosh^3u$ instead of $\cosh^2u$ (and potentially different bounds of integration.) I'm not sure if there's an error in my working and/or if there's some different integration technique that I need to apply here.
More context
The example relates to finding an arc length of the function $y = x^2$ from $(1, 1)$ to $(3, 9)$. This length $s$is being worked out using: $$\frac{ds}{dt} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$$
The function is parameterised: $x = t$, y = $t^2$. I am then confident in using this to find that:
$$s = \int_1^3\sqrt{1+4t^2} \,dt$$
If there is any more information I should provide please do not hesitate to answer. (Also, I've not posted here much so if there is any error in my MathJax formatting please correct it as well as telling me what I should have done instead. Thank you.)
Note the correction: $$\sqrt{1+\sinh^2(u)} \equiv \cosh^{\color{red}{1}} (u)$$as $\cosh^2(u)-\sinh^2(u)=1$.