# What is the arclength of the polar curve f(theta) = 2thetasin(5theta)-thetacot2theta  over theta in [pi/12,pi/3] ?

Jun 3, 2018

$\approx 2.73636$

#### Explanation:

From
$r \left(\theta\right) = 2 \theta \cdot \sin \left(5 \theta\right) - \theta \cot \left(2 \theta\right)$

we get by differentiating with respect to $\theta$:

$r ' \left(\theta\right) = 10 \theta \cos \left(5 \theta\right) - \cot \left(2 \theta\right) + 2 \theta {\csc}^{2} \left(2 \theta\right) + 2 \sin \left(5 \theta\right)$
and our integral is given by

${\int}_{\frac{\pi}{12}}^{\frac{\pi}{3}} \sqrt{{\left(2 \theta \cdot \sin \left(5 \theta\right) - \theta \cot \left(2 \theta\right)\right)}^{2} + {\left(10 \theta \cos \left(5 \theta\right) - \cot \left(2 \theta\right) + 2 \theta {\csc}^{2} \left(2 \theta\right) + 2 \sin \left(5 \theta\right)\right)}^{2}} d \theta$