Robert J. did great work.
For John let me explain more detail processes.
x = t cost dx/dt = cost  tsint
y = t sint dy/dt = sint + tcost
(dβ/dt)^{2} = (dx/dt)^{2} + (dy/dt)^{2} ; in here β is a arc length and dβ is a small segment of this
dβ/dt = √((dx/dt)2 + (dy/dt)2 )
= √( ( cost^2 2tcost sint + t^{2} sint^2 ) + (sint^2 +2tsint cost +t^{2}cost^2) )
= √( (cost^2 + sint^2) + t^{2}(sint^2 +cost^2) )
= √( 1 + t^{2} )
β = ∫(from 1 to 1) √( 1 + t^{2} ) dt
to solve above equation, let t = tanx , ( this is an easier way to solve this type's of problem. If you solve many problems you will know it by experiences )
t = tanx dt = secx^2 dx (and when t = 1, x = inverse tan 1, therefore x = pi/4)
so, when t varies from 1 to 1 it is better to calculate 2 times of integration of x from 0 to pi/4.
therefore, β = 2∫(from 0 to pi/4) √(1+tant^2) secx^2 dx = 2∫(from 0 to pi/4) √( secx^3 ) dx
after this step, you have to do the partial integration. (this process is also hard).
The result of Robert J. is correct.
Good luck.
11/30/2012

Sung taee L.
Comments
First off, thanks for the help! I'm taking calc 2 online and I think there is a big gap in my curriculum... where did sqrt(1+t^2) come from? (not in my text), why did you choose t= tanx? How did you know to change the limits of integration to (0, pi/4)?
(cost  tsint)^2 + (sint + tcost)^2 = 1 + t^2, since /+2tsintcost terms cancelled, and cos^2t + sin^2t = 1
Pick t = tanx because 1+tan^2x = sec^2x. In this way, you can get rid of radical.
Using symmetry, integral [1, 1] sqrt(1+t^2) dt = 2integral [0, 1] sqrt(1+t^2) dt. When t = 0, tanx = 0, x = 0, the lower limit; and when t = 1, tanx = 1, x = pi/4, the upper limit.
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