Question

Y=x(sin(x)), 0 ≤ x ≤ 2π?

Answer 1

Careful; (dy/dx)^2 = x^2 cos^2(x) + 2x sin x cos x + sin^2(x).

So, the arc length equals 
∫(x = 0 to 2π) √[1 + (x^2 cos^2(x) + 2x sin x cos x + sin^2(x))] dx 
= ∫(x = 0 to 2π) √[1 + x^2 cos^2(x) + x sin(2x) + sin^2(x)] dx, via double angle identity. 

Let Δx = (2π - 0)/10 = π/5. 
Using Simpson's Rule with n = 10, this integral approximately equals 
((π/5)/3) * [f(0) + 4 f(π/5) + 2 f(2π/5) + 4 f(3π/5) + 2 f(4π/5) + 4 f(π) + 2 f(6π/5) + 4 f(7π/5) + 2 f(8π/5) + 4 f(9π/5) + f(2π)], 

where f(x) = √[1 + x^2 cos^2(x) + x sin(2x) + sin^2(x)]. 
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I hope this helps!