Find the arc length of the curve from t = 0 to t = 1 whose derivatives in parametric form are dx/dt=2-cos(t) and dy/dt=ln(t^2)

Question

Type your answer in the space below and give 2 decimal places. If your answer is less than 1, place a leading "0" before the decimal point (ex: 0.48)

So I was trying to do it:
integral from 0 to 1 √[2-cos(t)+ln(t^2)] dt
then: integral from 0 to 1 √[(sint)^2+(2/t)^2] dt
and I am stock, please help

Answer 1

recall that for parametric equations,
ds^2 = (dx/dt)^2 + (dy/dt)^2
So, for your curves, that would be (recalling that ln(t^2) = 2lnt
ds^2 = (2-cost)^2 + (2lnt)^2
So the arc length would be
∫[0,1] √((2-cost)^2 + (2lnt)^2) dt

good luck with that. Better go numeric.