Editing Elliptic integral (section)

==Incomplete elliptic integral of the first kind==
The '''incomplete elliptic integral of the first kind''' {{mvar|F}} is defined as

<math display="block"> F(\varphi,k) = F\left(\varphi \mid k^2\right) = F(\sin \varphi ; k) = \int_0^\varphi \frac {d\theta}{\sqrt{1 - k^2 \sin^2 \theta}}.</math>

This is Legendre's trigonometric form of the elliptic integral; substituting {{math|1=''t'' = sin ''θ''}} and {{math|1=''x'' = sin ''φ''}}, one obtains Jacobi's algebraic form:

<math display="block"> F(x ; k) = \int_{0}^{x} \frac{dt}{\sqrt{\left(1 - t^2\right)\left(1 - k^2 t^2\right)}}.</math>

Equivalently, in terms of the amplitude and modular angle one has:
<math display="block"> F(\varphi \setminus \alpha) = F(\varphi, \sin \alpha) = \int_0^\varphi \frac{d\theta}{\sqrt{1-\left(\sin \theta \sin \alpha\right)^2}}.</math>

With {{math|1=''x'' = sn(''u'', ''k'')}} one has:
<math display="block">F(x;k) = u;</math>
demonstrating that this [[Jacobian elliptic function]] is a simple inverse of the incomplete elliptic integral of the first kind.

The incomplete elliptic integral of the first kind has following addition theorem{{Citation needed|date=February 2024}}:
<math display="block">F\bigl[\arctan(x),k\bigr] + F\bigl[\arctan(y),k\bigr] = F\left[\arctan\left(\frac{x\sqrt{k'^2y^2+1}}{\sqrt{y^2+1}}\right) + \arctan\left(\frac{y\sqrt{k'^2x^2+1}}{\sqrt{x^2+1}}\right),k\right] </math>

The elliptic modulus can be transformed that way:
<math display="block">F\bigl[\arcsin(x),k\bigr] = \frac{2}{1+\sqrt{1-k^2}}F\left[\arcsin\left(\frac{\left(1+\sqrt{1-k^2}\right)x}{1+\sqrt{1-k^2x^2}}\right),\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}}\right] </math>