Question

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

Answer 1

`approx 2.73636`

Explanation:

From
`r(theta)=2theta*sin(5theta)-thetacot(2theta)`

we get by differentiating with respect to `theta`:

`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`