Circumference by integral?

So,I was playing with integrals of sin(x),and I noticed that the integral of sin(x)+1 form -pi/2 to 3pi/2 is 2pi.So I assumed that this result has something to do with the circumference of the unit circle.Then I noticed that you can remove the sinus form the function,because the integrals cancel out (-cos(-pi/2)-(-cos(3pi/2))=0),so i was left with the integral of 1.And becAuse the radius of the unit circle is 1,I put in other number,and got the correct circumferences.Next,I tried to compute the length of the archimedean spiral doing a single revolution,putting x instead of a constant in the integrated function,and got pi^2.I can't find any info on what this length really is,so I'm here now and asking this-is it the correct length,and is this formula (integral of r from -pi/2 to 3pi/2) of any value,for example,in calculating the circumferences of elipses?

>>1
First things first, definitely do it from 0 to 2pi, absolutely no point doing it between the limits you did and it just complicates the matter.

So what you're trying to work out is arc length.
I'm not going to go into derivations here, but reference http://en.wikipedia.org/wiki/Arc_length

Basically for a function defined in polars co-ordinates, which makes your two examples simpler, the arc length between two angles w is the integral of sqrt(r^2 + (dr/dw)) dw

luckily in your two cases this works out easily.

For 1. r=1 for all w and so you get the integeral of 1 from 0 to 2pi

For your second case r=w so (dr/dw)=1.


therefore you want to calculate the integral of sqrt(w^2+1) dw, which you should probably be easily able to do by a substitution of w=cosh(y) or something similar [maybe tan(y) would be better, you decide]