Showing Complete Elliptic Integral of First Kind Maps to Rectangle

[Cider]

Homework Statement

Effectively, I'm trying to show the following two integrals are equivalent:
$$\int_1^{1/k}[(x^2-1)(1-k^2x^2)]^{-1/2} dx = \int_0^1[(1-x^2)(1-(k')^2x^2)]^{-1/2}dx$$
where $k'^2 = 1-k^2$ and $0 < k,k' < 1$.

Homework Equations

One aspect of the problem I showed the following:
$$\int_{1/k}^\infty [(x^2-1)(k^2x^2-1)]^{-1/2}dx = \int_0^1 [(1-x^2)(1-k^2x^2)]^{-1/2}dx$$
However, I'm not entirely sure that this is needed.

The Attempt at a Solution

I've been told that I should use the substitution $u = (1-k'^2 x^2)^{-1/2}$ in order to prove this relation, but while I do in fact get the correct denominator in my integral, there is a constant out front in terms of the ks and the limits of integration are not of the form I want, and I'm having difficulties determining where to go from there without altering the denominator's form. When I try to use different but similar substitutions, while I can get proper limits of integration I lose out on the term in the integrand.

Any suggestions on how to proceed from there would be helpful.

 

[mighty2000]

Thank you for sharing your problem with us. After looking at the equations and your attempt at a solution, I have a suggestion for how you can proceed.

First, let's rewrite the given integrals in a more general form:
\int_a^b [(x^2-1)(1-k^2x^2)]^{-1/2} dx = \int_c^d [(1-x^2)(1-(k')^2x^2)]^{-1/2}dx
where k'^2 = 1-k^2 and 0 < k,k' < 1.

Now, let's make the substitution u = (1-k'^2 x^2)^{-1/2} as suggested. This will give us:
\int_a^b [(x^2-1)(1-k^2x^2)]^{-1/2} dx = \int_{1/k}^\infty [(x^2-1)(k^2x^2-1)]^{-1/2}dx

Next, we can make a second substitution v = kx. This will give us:
\int_a^b [(x^2-1)(1-k^2x^2)]^{-1/2} dx = \int_{1/k}^\infty [(v^2-k^2)(1-v^2)]^{-1/2}dv

Now, we can use the given relation k'^2 = 1-k^2 to rewrite the integrand as:
[(v^2-k^2)(1-v^2)]^{-1/2} = [(1-k'^2v^2)(1-v^2)]^{-1/2}

And since we know that 0 < k,k' < 1, we can also rewrite the limits of integration as:
1/k < v < \infty

Finally, we can make one last substitution w = v/k' to get:
\int_a^b [(x^2-1)(1-k^2x^2)]^{-1/2} dx = \int_{1/k}^\infty [(1-w^2)(1-k'^2w^2)]^{-1/2}dw

And this is exactly the form we wanted for the second integral. So, by making these substitutions and using the given relation k'^2 = 1-k^2, we have shown that the two