What is the arclength of r=-cos(theta/2-(3pi)/8)/theta  on theta in [(3pi)/4,(7pi)/4]?

Arc length $s = 0.757882$ units

Explanation:

r=−cos(θ/2−3π/8)/θ on θ∈[3π/4,7π/4]

there is a need to determine first $\frac{\mathrm{dr}}{d \theta}$

$\frac{\mathrm{dr}}{d \theta} = \frac{\theta \cdot \sin \left(\frac{\theta}{2} - 3 \frac{\pi}{8}\right) \left(\frac{1}{2}\right) - \left(- \cos \left(\frac{\theta}{2} - 3 \frac{\pi}{8}\right)\right) \cdot 1}{\theta} ^ 2$

$\frac{\mathrm{dr}}{d \theta} = \frac{\frac{\theta}{2} \cdot \sin \left(\frac{\theta}{2} - 3 \frac{\pi}{8}\right) + \cos \left(\frac{\theta}{2} - 3 \frac{\pi}{8}\right)}{\theta} ^ 2$

$\frac{\mathrm{dr}}{d \theta} = \left(\frac{1}{2 \theta} \cdot \sin \left(\frac{\theta}{2} - 3 \frac{\pi}{8}\right) + \frac{1}{\theta} ^ 2 \cdot \cos \left(\frac{\theta}{2} - 3 \frac{\pi}{8}\right)\right)$

The formula for arc length s in Polar coordinates is

$s = {\int}_{a}^{b} \sqrt{{r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2}} d \theta$

s=int_(3 pi/4)^(7 pi/4) sqrt((−cos(θ/2−3π/8)/θ)^2+
(1/(2 theta)*sin(theta/2-3 pi/8)+1/theta^2*cos(theta/2-3 pi/8))^2) d theta

#s=0.757882 units