Question

How do you find the arc length of the curve `y= ln(sin(x)+2)` over the interval [1,5]?

Answer 1

You'll need numerical methods.

Explanation:

Note that for `y=ln(sin(x)+2)`, we have `dy/dx = cosx/(sin(x)+2)`

You need to find

`int_1^5 sqrt(1+(dy/dx)^2) dx = int_1^5 sqrt(1+(cosx/(sin(x)+2))^2) dx`

I can't find a closed form antiderivative (Wolfram Alpha gives one that uses elliptic integrals of the first and third kind. http://www.wolframalpha.com/input/?i=int++sqrt%281%2B%28cosx%2F%28sinx%2B2%29%29%5E2%29+dx )

Use your favorite approximation technique to get an answer close to 4.23701 (also from WolframAlpha).