0
$\begingroup$

Is it possible to reduce the following Z to Legendre form Elliptic integral of the First kind?

$$ \int \dfrac {\sec u\; du } {\sqrt{ (1- {(\nu \tan u)}^2 }} ..(1*)$$

With substitution of $ \tan u = t, $ I get it into ArcSin type:

$$ \int \dfrac {dt } {\sqrt{ (1- {(\nu \; t)}^2)} } ...(2*) $$

EDIT 2:

After the above error pointed out by bbgodfrey is corrected, (1*) assumes form:

$$ \int \dfrac {dt } {\sqrt{(1+ t^2) (1- {(\nu t)}^2)} } ...(2*) $$

which still needs to be adjusted to get to classical form.

Mathematica evaluation

DSolve[ Z'[u] == Sec[u]/Sqrt[ 1 - (nu Tan[u])^2 ], Z, u]

results in $ EllipticF\; [u, 1 + \nu^2] $. Obviously I am missing something but cannot

find what.

$\endgroup$
4
$\begingroup$

The two integrals are not equivalent, as can be seen from

D[Tan[u], u]
(* Sec[u]^2 *)

The transformation from t to u takes the second integral from

Integrate[1/Sqrt[1 - (nu t)^2], t]

to

Integrate[Sec[u]^2/Sqrt[1 - (nu Tan[u])^2], u]
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.