Question

Anyone have any integral/derivative tricks/interesting stuff they'd like to share?

Answer 1

The tangent half-angle substitution is a trick for integrals that I don't see very often.

Answer 2

The technique for evaluating the gaussian integral is also quite cool.

Answer 3

Sure, sounds interesting, I've seen one method for evaluating the gaussian integral but I couldn't really seem to remember it, it involved squaring and using polar coordinates...

Care to elaborate more on those?

Answer 4

The tangent substitution helps you get rid of sine and cosine from integrals that would otherwise do annoying and sometimes cyclic things.

Answer 5

The method for evaluating the gaussian integral that you mentioned recalling is the same one I was talking about (I'm not sure if there even is another way?)

Answer 6

Using hyperbolic trig functions is another nice 'trick' depending on whether or not your class treated them as 'first-class citizens'.

Answer 7

I've never been in a class where they evaluated it. It came up in diffeq 2 but they just said, here, it's the erf(x) function.

Answer 8

In most of my earlier classes hyperbolics were basically ignored for integration and you can get some 'magic' answers in terms of logs if you know the log forms of the inverse functions.

Answer 9

I can evaluate it as a power series lol. Then again I can do all integrals this way, but it's lame lol.

Answer 10

As far as misc. interesting stuff goes, calculating sums of divergent series is always a good time.

Answer 11

Sure, I'm really interested in seeing examples of stuff, I'm sort of unsure where to go from there. When might I apply a tangent half angle? What's the "technique" to seeing them?

Answer 12

Wikipedia has a nice example for the integral of secant, which is neat given the historical background associated with finding that one.

Answer 13

(In reference to the tangent substitution I mean)

Answer 14

Know how to use substitutions, polar and trig stuff.

Answer 15

For example, find the derivative of arctan(x).

Answer 16

Sure, I love trig substitution in terms of stuff like 1/1+x^2 kind of stuff.

Any time I've done integral of secant i'd just multiply by secx+tanx divided by itself to get a nice little natural log... But I never really know who came up with that clever thing or how it came to them lol.

Answer 17

I've never liked that method because it seemed like some kind of cargo-cult magic.

Answer 18

I still recommend going into the complex plane. There are some integrals that are unable to be solved without the change. Spherical, polar, and other change of base are also useful. Cauchy Integral formula and green's theorem are also vital

Answer 19

It definitely works for both sec and csc but I really hate memorizing clever things like that.

Answer 20

I don't like the multiplications. If you understand how derivatives work, stuff becomes easier.
@FibonacciChick666 Slow down there; bro/broad is a noob.

Answer 21

Yeah I'm kind of a fan of Green's thm but really I think since it's just Stoke's theorem it's really not that interesting... It sort of just makes sense though which is a recent thing in my world which I thought I'd never be saying.

Answer 22

oops

Answer 23

Who are you calling a n00b? I love complex numbers and I'm down with them.

Answer 24

\(i^i \approx 0.2\) isn't particularly useful but I think it falls under the interesting category.

Answer 25

That's a bad approximation.

Answer 26

\[e^x=e^{i(-ix)}=\cos(ix)-isin(ix)\]

Wheee... You turn the angle around that you rotate by 90 degrees and you go from a hyperbola to a circle...

Answer 27

Noob ^^^

Answer 28

I wasn't worried about the actual value as much as the fact that it's a real number.

Answer 29

But yeah that was on a \(\pi \approx 3\) level.

Answer 30

haha, more-so things like gamma and beta functions

Answer 31

Somebody's showing off......

Answer 32

not showing off, just happen to be in that class and it was first to come to mind

Answer 33

lol

Answer 34

@Kainui Stuff like knowing how to derive spherical differential pieces is also important.

Answer 35

@primeralph that's vector cal stuff... I'm looking for stuff that's not normally taught.

Answer 36

Solving \(y=x^x\) for \(x\) is moderately interesting, although I've never seen a practical use for it. I just learned about that the other day on here actually.

Answer 37

@Kainui It is not normally taught. They just throw it in your face and ask you to memorize it.

Answer 38

Oh, well I guess I had a good teacher then.

Answer 39

hahaha that is very true @primeralph

Answer 40

Okay, derive all the length pieces in a spherical system.

Answer 41

In 1 minute.

Answer 42

or the Cauchy- Riemann eq have been useful

Answer 43

residue thm too

Answer 44

Here's a fun derivation of some of the gamma function:

a is just a number greater than 0.
\[1=\int\limits_{0}^{\infty}ae^{-ax}dx\]
That equals 1, if you don't believe me, try it out. Then since a is a constant with respect to the integral so we can pull it out...

\[a^{-1}=\int\limits_{0}^{\infty}e^{-ax}dx\]
take the derivative of both sides WITH RESPECT TO A, which is linearly independent of x.
\[-a^{-2}=\int\limits_{0}^{\infty}-x*e^{-ax}dx\]
The negative signs will cancel out. Take a few more successive derivatives (cancelling out the negatives each time...
\[2*a^{-3}=\int\limits_{0}^{\infty}x^2e^{-ax}dx\]\[6*a^{-4}=\int\limits_{0}^{\infty}x^3e^{-ax}dx\]\[24*a^{-5}=\int\limits_{0}^{\infty}x^4e^{-ax}dx\]\[\frac{ n! }{ a^{n+1} }=\int\limits_{0}^{\infty}x^n e^{-ax}dx\] set a=1\[n! =\int\limits_{0}^{\infty}x^n e^{-x}dx\]

fancy

Answer 45

@primeralph don't you just mean I need to find the jacobian of a 3x3 matrix? Not sure if we're talking about the same thing.

Answer 46

@Kainui That only works when you know what system you're transforming to. I asked you to derive it from scratch using trig and geometry. If you were taught that in class, then you got a cute teacher (no homo).

Answer 47

What do you mean "system"? You want me to go from rectangular to physicist's or mathematician's spherical coordinates? I can derive the one you want if that's what you mean?

Answer 48

Without using The Jacobian.
Go for it.

Answer 49

Without using any "aftermarket" tools.

Answer 50

I don't follow. The jacobian makes sense to me. The determinant is the volume/area/size of a bunch of linearly independent vectors. So that would make the determinant of the partials the volume... I've already shown it to myself before that this is true... It's linear algebra. I'll show you how it's done in 2D So you have the x and y components there and can add them up then find the area of that box. Subtract out pieces as triangles which isn't too complicated... Similar argument for higher determinants.

Maybe I'm missing something here, but this will just translate over perfectly fine, it's just a handy shortcut to the correct answer...

Answer 51

here's a trick
PRACTICE

Answer 52

@FibonacciChick666 what's a fun integral that has to be (or is easiest) to be done by complex substitution?

Answer 53

uhm... I can't htink of any off the top of my head. None of them are particularly easy

Answer 54

anything that is a weird trig one though like \(\int \frac{SinxCosx}{e^x-e^{-x}} dx\)

Answer 55

A basic exmaple: