Question

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

Answer 1

`approx 0.41636`

Explanation:

we have
`r(x)=x*sin(3x)-sec^2(x)`
so
`(dr)/dx=sin(3x)+x*cos(3x)*3-2sec^2(x)*tan(x)`
I have only found a numerical value for this integral.