Calculus intergration word problem

[cricket21]

In electromagnetic theory, the magnetic potential u at a point on the axis of a circular coil is given by

u = Ar ∫ 1/(r^2 + x^2)^(3/2) dx
going from a to infinity and A, r, a are constants

u=r*tany
du= r*sec^2(y) dy

r^2 + x^2 = r^2 + r^2 * tan^2 y
= r^2 (1 + tan^2 y) = r^2 * sec^2 y
so then i integrated the r*sec^2(y)/ (r)^(3/2)*sec^3(y) and got 1/sqrt(r)*cos(y)
I need help finishing this problem, thank you

 

[DrPhil]

I tried to re-format your integral - did I do it right?

In electromagnetic theory, the magnetic potential u at a point a on the axis of a circular coil is given by

\(\displaystyle \displaystyle u = A\ r \int_a^{\infty}\dfrac{dx}{(r^2 + x^2)^{3/2}}\)

going from a to infinity and A, r, a are constants

u=r*tany
du= r*sec^2(y) dy

r^2 + x^2 = r^2 + r^2 * tan^2 y
= r^2 (1 + tan^2 y) = r^2 * sec^2 y
so then i integrated the r*sec^2(y)/ (r)^(3/2)*sec^3(y) and got 1/sqrt(r)*cos(y)
I need help finishing this problem, thank you

I didn't follow your change of variables, but your problem may be that you neglected to change the limits of integration. I understand that \(\displaystyle \tan{\gamma} = x/r\). If that is correct, then the lower limit is \(\displaystyle \tan^{-1}(a/r)\) and the upper limit is \(\displaystyle \pi /2\).

However, a trig substitution may not be required. Did you try \(\displaystyle v = \sqrt{x^2 + r^2}\)?

Please work a bit more, and show us in detail how you have substituted variables. Thanks!