I wanted to come up with a formula for the total distance travelled by a projectile with some initial velocity $\langle v_x,v_y\rangle$ in $\mathbb R^2$.
Its parametrization should be the following:
$$v(t)=\left\langle v_xt,\frac{1}{2}gt^2+v_y\right\rangle$$
Its component derivatives are as follows:
$$\frac{dx}{dt}=v_x$$ $$\frac{dy}{dt}=gt+v_y$$
The arclength $\ell$ of a parametrized curve is given by
$$\ell = \int\limits_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\;dt$$
To find the total distance travelled between times $t=0$ and $t=T$, we have the following:
$$\ell=\int\limits_0^T\sqrt{v_x^2+(gt+v_y)^2}\;dt$$
Now for a $u$-substitution:
$$u=gt+v_y$$ $$du=g\;dt$$ $$(0,T)\mapsto (v_y,gT+v_y)$$
$$\ell=\frac{1}{g}\int\limits_{v_y}^{gT+v_y}\sqrt{u^2+v_x^2}\;du$$
Now for a hyperbolic substitution:
$$u=v_x\sinh(\theta)$$ $$du=v_x\cosh(\theta)\;d\theta$$ $$(v_y,gT+v_y)\mapsto(a_0,a_T)$$
We declare the following constants:
$$a_0=\text{asinh}\left(\frac{v_y}{v_x}\right)$$ $$a_T=\text{asinh}\left(\frac{gT+v_y}{v_x}\right)$$
Our integral becomes the following:
$$\ell=\frac{1}{g}\int\limits_{a_0}^{a_T}\sqrt{v_x^2\sinh^2(\theta)+v_x^2}\cdot v_x\cosh(\theta)\;d\theta$$
Recall $\sinh^2(\theta)+1=\cosh^2(\theta)$.
$$\ell=\frac{v_x^2}{g}\int_{a_0}^{a_T}\cosh^2(\theta)\;d\theta$$
Recall $\cosh^2\theta=\frac{1+\cosh(2\theta)}{\theta}$.
$$\ell=\frac{v_x^2}{2g}\int_{a_0}^{a_T} 1+\cosh(2\theta)\;d\theta$$
Integrating:
$$\ell=\frac{v_x^2}{2g}\left(\theta+\frac{\sinh(2\theta)}{2}\right)_{a_0}^{a_T}$$
Recall $\sinh(2\theta)=2\sinh(\theta)\cosh(\theta)$.
$$\ell=\frac{v_x^2}{2g}\left(a_T-a_0+\left.\sinh(\theta)\cosh(\theta)\right|_{a_0}^{a_T}\right)$$
Recall $\cosh(\theta)=\sqrt{1+\sinh^2(\theta)}$.
$$\ell=\frac{v_x^2}{2g}\left(a_T-a_0+\left.\sinh(\theta)\sqrt{1+\sinh^2(\theta)}\right|_{a_0}^{a_T}\right)$$
Back-substituting $a_0$ and $a_T$:
$$\tiny{\ell=\frac{v_x^2}{2g}\left(\text{asinh}\left(\frac{gT+v_y}{v_x}\right)-\text{asinh}\left(\frac{v_y}{v_x}\right)+\sinh\left(\text{asinh}\left(\frac{gT+v_y}{v_x}\right)\right)\sqrt{1+\sinh^2\left(\text{asinh}\left(\frac{gT+v_y}{v_x}\right)\right)}-\sinh\left(\text{asinh}\left(\frac{v_y}{v_x}\right)\right)\sqrt{1+\sinh^2\left(\text{asinh}\left(\frac{v_y}{v_x}\right)\right)}\right)}$$
Simplifying:
$$\scriptsize{\ell=\frac{v_x^2}{2g}\left(\text{asinh}\left(\frac{gT+v_y}{v_x}\right)-\text{asinh}\left(\frac{v_y}{v_x}\right)+\frac{gT+v_y}{v_x}\sqrt{1+\frac{(gT+v_y)^2}{v_x^2}}-\frac{v_y}{v_x}\sqrt{1+\frac{v_y^2}{v_x^2}}\right)}$$
I felt like using a hyperbolic trig sub as opposed to a regular trig sub would make integration easier as it allowed me to avoid integrating secant cubed by parts. But I'm not sure if I got all the details right with regard to changing the limits of integration and wanted to know if this formula would work. Thank you.