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:

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


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.


1 Answer 1


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]


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

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