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

Jun 1, 2018

#### Answer:

$\approx 0.41636$

#### Explanation:

we have
$r \left(x\right) = x \cdot \sin \left(3 x\right) - {\sec}^{2} \left(x\right)$
so
$\frac{\mathrm{dr}}{\mathrm{dx}} = \sin \left(3 x\right) + x \cdot \cos \left(3 x\right) \cdot 3 - 2 {\sec}^{2} \left(x\right) \cdot \tan \left(x\right)$
I have only found a numerical value for this integral.