The name's a bit obscure^{1}, but it basically means this:
If you give me a function f(x), I can give you a function g(x)+c (where c is any real number), which will make a right angle wherever it intersects your function f(x).

Perhaps that was even more obscure... Let's take an example. Think of the functions f(x)=x and g(x)=-x+c

The function g(x) will always make a right angle with the function f(x), no matter what c.

But I can do this for ~~any~~ most functions you give me.

`x² -> -0.5Ln|x|+c`

`x³ -> 1/(3x)+c`

`3x²+5x-3 -> (-1/6)Ln|-6x-5|+c`

`Sin(x) -> Ln|Cos(x/2)-Sin(x/2)|-Ln|Cos(x/2)+Sin(x/2)|+c`

So how do I do it? Well, like this:

Say you take 1 point on a function. That point has a certain slope. If you want a line perpendicular to the tangent line in that point, do -1 divided by the slope (since the slope of a line times the slope of the perpendicular line = -1 so -1 divided by a slope gives you the perpendicular one). This will give you the slope of the line making a right angle with the tangent line (where the function of that line will be (slope)x+b).

The slope is given by the derivative. For f(x)=x² the slope is given by f'(x)=2x. This function f'(x)=2x, is actually the slope in every point on the original graph. Now, if we do -1 divided by that slope in every point, we get g'(x)=-1/2x which is the slope in every point of some other function. (Actually, of the function we want)

If we integrate this function again, we get a function which is perpendicular to the original function everywhere, in this case, g(x)=-0.5Ln|x|+c

So, to recap:
The primitive of -1 divided by the derivative of the original function, +c, gives you a function which makes a right angle on every intersection with the original function.

Or

**g(x)+c= ∫ -1/f'(x) dx**

I've been asked what makes my set of functions so special since it's possible to define a function to almost be whatever you want it to be except near the points of intersection with the other function and still have it perpendicular. But the fact is, that it's possible to move g(x) 1 place up, which would give you different intersections. But then the intersections would still make right angles. In other words, the tangent of every point of f(x) makes a right angle with the tangent of every corresponding point of g(x)

Or, when x is the same, the tangents of the functions in x make right angles, no matter what x.

If you have a graphical calculator or some graph program, you can try some of the functions above, or calculate your own one. Make sure to set the screen to square (On the TI-83: Zoom->ZSquare) to get a visual impression of the right angles.

Now, I understand getting the integral of some functions may be troublesome. For example, I wouldn't have a clue on how to integrate -1/cos(x). But there *is* an answer! The wolfram integrator!

http://integrals.wolfram.com
This gives you an actual *function* which is the integrated function of the one you submit, instead of the usual numeric approximations (if an answer actually exists). There are 2 things to note here though:

- It notes Ln(x) as Log[x]. I've learned that Log(x) means
^{10}Log(x), and so does my TI-83 interpret it too. That confused me at first. Ln(x) is ^{e}Log(x).

Note: Swap told me that this is a common notation of Ln(x), since the natural log is the only thing important to mathematics at a later stage. "There is but One True Log". Thanks swap.
- I've learned the integral of 1/x is Ln|x| (the bars mean Absolute(x)), but the integrator forgets the bars. I'm not sure what causes this, but you may want to keep this in mind, as you'll be running into it a lot with the -1/f'(x)

For example the perpendicular function of Sin(x) (look above), if you leave out the absolute bars, you only get a result on intervals [-0.5π+k2π , 0.5π+k2π] where k is any integer. If I put in the absolute bars however, I get a result for any real number, which is still correct. (Where correct in this case means: Perpendicular to Sin(x))

Taking note of that, it's a pretty good way to integrate difficult functions. (I'm not affiliated with them in any way)

1. I made it up