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The Weierstrass substitution is great for transforming complex trig integrals into simpler rational functions. Wikipedia suggests that it wasn't invented by Weierstrass, since Euler was already familiar with it.

So,

  1. Who invented this substitution?
  2. More importantly, what intuition can be used to get at this result? It's clear that it "works", but it's not clear what the thought process was that led to it's discovery.
  3. Is there a "family" of such substitutions that lead to simplification of trig integrals, or this the only known one?
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    $\begingroup$ One can think of it as a rational parametrization of the unit circle, using $(2t/(1+t^2), (1-t^2)/(1+t^2)$. In that guise, it is quite old, going back to Diophantus. $\endgroup$ – André Nicolas Aug 7 '13 at 2:32
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This comes from the rational parameterization of the unit circle: namely, the circle $x^2 + y^2 = 1$ can be parameterized by (put in bijection with) the $t$-line via the formula $x = (1-t^2)/(1+t^2),$ $y = 2t/(1+t^2).$

This substitution goes back in some form to Euclid, at least, who used it to generate Pythagorean triples. (If we take $t$ to be a rational number $m/n$ and clear demoninators, we find that $(m^2 - n^2)^2 + (2 m n)^2 = (m^2 + n^2)^2$, and all Pythagorean triples arise in this way.)

The intuition for this formula is as follows: the curve is cut out by a degree two (i.e. quadratic) equation in $x$ and $y$, so a line will meet it in two points. So, if we fix one of the points and let the other point vary, we can describe all the points on the circle as being obtained by intersecting the circle with a varying family of lines passing through a fixed point.

More precisely and concretely, take the base point to be $(-1,0)$. Then if we consider the line of slope $t$ through this point, it meets the circle in the point $(0,1)$ (obviously) and one other point, namely $\bigl((1-t^2)/(1+t^2), 2t/(1+t^2) \bigr).$ This gives the formula.


The trignometric point of view, which historically came later, just comes from writing $x = \cos \theta,$ $y = \sin \theta$. If you draw the triangle whose vertices are $(-1,0),$ $(1,0),$ and $(x,y)$, then this a right triangle with angle $\theta/2$ at the vertex $(-1,0)$, so its slope $t$ is equal to $\tan \theta/2$.


Note that this trick of fixing one point and then parameterizing the other points by drawing a line joining the fixed point to another point on the curve only works when the curve has degree $2$. If the degree is larger, than a line through the fixed point will meet the curve in more than one other point. Thus there is no obvious way to parameterize higher degree curves by a single variable, and in general this can't be done (except in some degenerate singular cases, like parameterizing $y^2 = x^3$ via $(t^2,t^3)$).

This means that there is no rational substitution available to convert an integral like (just to give one famous example) $$\displaystyle \int \frac{dx}{\sqrt{(1-x^2)(1 - k^2 x^2)}}$$ (which is related to the degree $4$ curve $y^2 = (1- x^2)(1-k^2 x^2)$) into an integral of rational functions.

This integral is known as an elliptic integral, and studying them and trying to understand the algebraic curves that underly them, led to the development of much of modern algebraic geometry and algebraic topology (the key names, that transformed their study from a branch of analysis to geomery/topology, are Abel, Jacobi, and Riemann).

I have a few different answers here related to this theme.

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  • $\begingroup$ +1. This is a good answer. I would've upvoted it had I not exhausted all of my votes for today. :/ $\endgroup$ – user66733 Aug 7 '13 at 3:01
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    $\begingroup$ @some1.new4u: Dear some1.new4u, Thanks! Cheers, $\endgroup$ – Matt E Aug 7 '13 at 3:34
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    $\begingroup$ It might be worth noting that the reason the "lines through a fixed point" idea works for $y^2=x^3$ is that we can choose the fixed point to be "really" two points that happen to coincide (at the cusp of the curve), so a line through this point ("really" two points) meets the curve only once more. $\endgroup$ – Andreas Blass Aug 7 '13 at 4:20
  • $\begingroup$ @AndreasBlass: Dear Andreas, Indeed. I had thought about being more specific about this; now I don't need to! Thanks, and best wishes, $\endgroup$ – Matt E Aug 7 '13 at 4:26
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The effect of the $t=\tan(x/2)$ substitution is to produce a rational parametrization of the unit circle. As $t$ travels over the reals, the ordered pair $$\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right)$$ almost travels over the unit circle. The only exception is the point $(-1.0)$, which can be thought of as corresponding to "$t=\infty$."

In not quite algebraic form, this parametrization was known by Diophantus, and came before the tangent function appeared in the mathematical literature.

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  • $\begingroup$ And how did Diophantus come up with it? $\endgroup$ – nbubis Aug 7 '13 at 2:50
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    $\begingroup$ I don't know, if one looks at his substitutions he uses it, but, because there is no explicit algebraic notation, he does not state it. Presumably it comes from experience in constructing right triangles with rational sides. $\endgroup$ – André Nicolas Aug 7 '13 at 2:57
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It's merely a parametrization of the unit circle with the line that passes through (-1,0) on the circle. This is a fairly standard trick of parametrizing curves in Algebraic geometry. In fact, you're relating the slope of the line to a point on the circle by fixing one point on the line and changing the slope to get the other.

You can get the parametrization in this way:

1: Write the equation of the line that passes through the point $(-1,0)$ on the circle with the slope t. (Other points are just fine too, you could use $(1,0)$ with ease, but using some points might make your calculations more difficult)

2: When you do this, notice that you're relating the slope of the line, namely $t$, to the point $(x,y)$ on the circle.

3: Now you need to find $x$ and $y$ as functions of $t$. Plug in your equation of the line in the equation of the unit circle and solve for $t$. You must get two values, because the line would intersect the circle at two points. One point is $(-1,0)$ for sure, the other one will be the point $(x,y)$ that is related to your parameter $t$ and this is your desired parametrization.

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    $\begingroup$ Dear some1.new4u, Actually, in this particular case, the base-point is $(-1,0)$, not $(1,0)$. Regards, $\endgroup$ – Matt E Aug 7 '13 at 2:51
  • $\begingroup$ @MattE: Thanks. Actually I didn't check it, but I assumed it was (1,0) because this is how I usually parametrize the unit circle. I'll correct it. $\endgroup$ – user66733 Aug 7 '13 at 2:52
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    $\begingroup$ Dear some1.new4u, I usually do the same, so I was confused for a bit when I was writing my answer too! Cheers, $\endgroup$ – Matt E Aug 7 '13 at 2:54

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