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

Sep 26, 2015

You'll need numerical methods.

#### Explanation:

Note that for $y = \ln \left(\sin \left(x\right) + 2\right)$, we have $\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \frac{x}{\sin \left(x\right) + 2}$

You need to find

${\int}_{1}^{5} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx} = {\int}_{1}^{5} \sqrt{1 + {\left(\cos \frac{x}{\sin \left(x\right) + 2}\right)}^{2}} \mathrm{dx}$

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).