# What is the arclength of f(t) = (sint*cos2t,sint) on t in [-pi,0]?

Jun 10, 2018

$\approx 3.943805190587053$

#### Explanation:

We have
$x \left(t\right) = \sin \left(t\right) \cos \left(2 \cdot t\right)$
$x ' \left(t\right) = \cos \left(t\right) \cos \left(2 t\right) - 2 \sin \left(t\right) \sin \left(2 t\right)$

$y \left(t\right) = \sin \left(t\right)$
$y ' \left(t\right) = \cos \left(t\right)$
so we have to integrate

${\int}_{- \pi}^{0} \sqrt{{\left(\cos \left(t\right) \cos \left(2 t\right) - 2 \sin \left(t\right) \sin \left(2 t\right)\right)}^{2} + {\cos}^{2} \left(t\right)} \mathrm{dt}$
i got only a numerical result

$\approx 3.943805190587053$