# What is the arclength of r=costheta-3sin2theta  on theta in [-pi/4,pi]?

Jun 3, 2018

$\approx 16.7916$

#### Explanation:

From
$r \left(\theta\right) = \cos \left(\theta\right) - 3 \sin \left(3 \theta\right)$
we get by differentiating

$r ' \left(\theta\right) = - \sin \left(\theta\right) - 6 \cos \left(2 \theta\right)$
so we get

${\int}_{- \frac{\pi}{4}}^{\pi} \sqrt{{\left(\cos \left(\theta\right) - 3 \cdot \sin \left(2 \theta\right)\right)}^{2} + {\left(- \sin \left(\theta\right) - 6 \cos \left(2 \theta\right)\right)}^{2}} d \theta$

By a numerical method we obtain

$16.7916$