What is the arclength of the polar curve f(theta) = 2thetasin(5theta)-thetacot2theta f(θ)=2θsin(5θ)θcot2θ over theta in [pi/12,pi/3] θ[π12,π3]?

1 Answer
Jun 3, 2018

approx 2.736362.73636

Explanation:

From
r(theta)=2theta*sin(5theta)-thetacot(2theta)r(θ)=2θsin(5θ)θcot(2θ)

we get by differentiating with respect to thetaθ:

r'(theta)=10thetacos(5theta)-cot(2theta)+2thetacsc^2(2theta)+2sin(5theta)
and our integral is given by

int_(pi/12)^(pi/3)sqrt((2theta*sin(5theta)-thetacot(2theta))^2+(10thetacos(5theta)-cot(2theta)+2thetacsc^2(2theta)+2sin(5theta))^2)d theta