# What is the arclength of the polar curve f(theta) = 2thetasin(5theta)-cottheta  over theta in [pi/12,pi/6] ?

Jun 2, 2018

$\approx 1.76588$

#### Explanation:

We have given
$r \left(\theta\right) = 2 \theta \cdot \sin \left(5 \theta\right) - \cot \left(5 \theta\right)$
then
$r ' \left(\theta\right) = 2 \sin \left(5 \theta\right) + 10 \cos \left(5 \theta\right) - \frac{1}{\sin} ^ 2 \left(\theta\right)$
and we get

${\int}_{\frac{\pi}{12}}^{\frac{\pi}{6}} \sqrt{{\left(2 \theta \cdot \sin \left(5 \theta\right) - \cot \left(\theta\right)\right)}^{2} + {\left(2 \sin \left(5 \theta\right) + 10 \cos \left(5 \theta\right) - \frac{1}{\sin} {\left(\theta\right)}^{2}\right)}^{2}} d \theta$
I have found only a numerical value of that integral
$\approx 1.76588$