Question

What is the arclength of `f(t) = (sin^2t/sin(2t),tant-sec^2t)` on `t in [pi/12,pi/4]`?

Answer 1

`approx 0.5007429256`

Explanation:

We have

`x(t)=sin^2(t)/sin(2t)`
then
`x'(t)=1/(2cos(t))^2`

`y(t)=tan(t).sec(t)^2`
then
`y'(t)=-(cos(t)+2sin(t))/cos^3(t)`
so we get the integral

`int_(pi/12)^(pi/4)sqrt((1/(2cos^2(t))^2+(-(cos(t)+2sin(t))/cos(t)^3)dt`
which gives `approx 0.5007429256`