parametric equations - tangents and length of the arc

Hi

x = integral sign (S) a= 1 b=t (cos u)/u du

y = S a=1; b=t (sin u)/u du

question is find the length of arc of the curve from the origin to the nearest point where there is a vertical tangent line

So I am thinking I can find dx/du and where it equals 0 is where the vertical tangent is

however when I try and figure out dx/du it gets ugly

dx/du = (1/u)(-sin u) + (cos u)(-1/sqrt u)

= - sin u/u - cos u/sqrt u

= - sin u - cos^2 u

And I don't have a clue what u is going to equal so that dx/du = 0 so I have a vertical tangent. And then the question becomes do I do that # to 1 or to 0 for the length of the arc?

Thanks for any guidance you can provide.

Calculus beginner