What is the arclength of f(t) = (sin2t-tcsct,t^2-1) on t in [pi/12,(5pi)/12]?

Arc length $s = 2.095350308 \text{ }$units

Explanation:

For arc length for which x is in terms of t and y also in terms of t, the formula is

$s = \int \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

From the given $f \left(t\right) = \left(\sin 2 t - t \csc t , {t}^{2} - 1\right)$ from $t = \frac{\pi}{12}$ to
$t = \frac{5 \pi}{12}$

this means $x = \sin 2 t - t \csc t$ and $y = {t}^{2} - 1$

find $\frac{\mathrm{dx}}{\mathrm{dt}}$ and $\frac{\mathrm{dy}}{\mathrm{dt}} \text{ }$first

solve $\frac{\mathrm{dx}}{\mathrm{dt}}$
$x = \sin 2 t - t \csc t$

$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 \cos 2 t - \left(- t \cdot \csc t \cdot \cot t + \csc t\right) = 2 \cos 2 t + t \csc t \cdot \cot t - \csc t$

solve $\frac{\mathrm{dy}}{\mathrm{dt}}$
$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 t$

Solve for the length of arc $s$

$s = {\int}_{\frac{\pi}{12}}^{\frac{5 \pi}{12}} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

$s = {\int}_{\frac{\pi}{12}}^{\frac{5 \pi}{12}} \sqrt{{\left(2 \cos 2 t + t \csc t \cdot \cot t - \csc t\right)}^{2} + {\left(2 t\right)}^{2}} \mathrm{dt}$

Using Simpson's Rule

$s = {\int}_{\frac{\pi}{12}}^{\frac{5 \pi}{12}} \sqrt{{\left(2 \cos 2 t + t \csc t \cdot \cot t - \csc t\right)}^{2} + {\left(2 t\right)}^{2}} \mathrm{dt}$

$s = 2.095350308 \text{ }$units

God bless....I hope the explanation is useful.