Question

What is the arclength of the polar curve `f(theta) = 2thetasin(5theta)-cottheta` over `theta in [pi/12,pi/6]`?

Answer 1

`approx 1.76588`

Explanation:

We have given
`r(theta)=2theta*sin(5theta)-cot(5theta)`
then
`r'(theta)=2sin(5theta)+10cos(5theta)-1/sin^2(theta)`
and we get

`int_(pi/12)^(pi/6)sqrt((2theta*sin(5theta)-cot(theta))^2+(2sin(5theta)+10cos(5theta)-1/sin(theta)^2)^2)d theta`
I have found only a numerical value of that integral
`approx 1.76588`