# What is the arclength of the polar curve f(theta) = -3sin(theta)-2sec^2theta  over theta in [0,pi/8] ?

Jul 15, 2018

$\approx 1.827918426331$

#### Explanation:

We are using the formula

$s = {\int}_{\alpha}^{\beta} \sqrt{{r}^{2} + {\left(\frac{d r}{d \theta}\right)}^{2}} d \theta$
$r \left(\theta\right) = - 3 \sin \left(\theta\right) - 2 {\sec}^{2} \left(\theta\right)$

$r ' \left(\theta\right) = - 3 \cos \left(\theta\right) - 4 {\sec}^{2} \left(\theta\right) \tan \left(\theta\right)$

so we have to integrate

${\int}_{0}^{\frac{\pi}{8}} \sqrt{{\left(- 3 \sin \left(\theta\right) - 2 {\sec}^{2} \left(\theta\right)\right)}^{2} + {\left(- 3 \cos \left(\theta\right) - 4 {\sec}^{2} \left(\theta\right) \tan \left(\theta\right)\right)}^{2}} d \theta$

by a numerical method we get $\approx 1.827918426331$