Pretty straight forward, isn't it? Considering the other problems you have posted on here, you should be able to do this. The length of the graph of y= f(x), from x= a to x= b, is given by [tex]\int_{x=a}^b \sqrt{1+ f'(x)^2}dx[/tex] With y= f(x)= sin(x), f'(x)= cos(x) so that becomes [tex]\int_{x=0}^{2\pi} \sqrt{1+ cos^2(x)}dx[/tex] However, that looks to me like a version of an elliptical integral which cannot be done in terms of elementary functions. Hey, no fair posting while I'm typing!
to get a numerical value try numerical integration, simpson's rule? etc... this is no worse than finding the area under the curve from 0 to 1. i.e. both are approximations. (nobody knows what cos(1) is.)