Explain integration by parts

[fast gyro]

I'm trying to understand integration by parts. I know it's used when you want to integrate the product of two functions, but in the formula \int u dv = u v - \int v du, I can't wrap my head around what dv and du really mean. all I know is they're differentials or something but I don't get why they're relevant or what they mean

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[plain magnet]

let's derive it

let $u$ and $v$ be functions in x, i.e., $y=u(x)$ and $y=v(x)$, thus $\dv{x}[u\cdot v]=u\cdot v'+u'\cdot v$, so $u\cdot v=\int u\cdot v'+u'\cdot vdx=\int u\cdot v'dx+\int u'\cdot vdx$, thus $$\boxed{\int u v'dx=\int udv=uv-\int v\cdot u'dx=uv-\int vdu}$$

[daring torrent]

ooaa

[hazy pivot]

When we integrate for example 3x^2*2y we can represent 3x^2 as u and 2y as dv/dx. Hence du/dx = 6x and v = y^2. Thats how it usually works and then the formula is applied.

[fast gyro]

Sorry to keep you guys waiting

But when the formula says when you integrate udv

What is dv to v?

Since dv/dx = v'

And dv = v'dx

What does it mean when you multiply v' and dx?

[hazy pivot]

dv = v.dx if v is in terms of x

[hazy pivot]

can you show like a question so that I can explain through it

[fast gyro]

Like an integration by parts question?

I have this one:

It wants the integral of (x^2)(lnx)

[hazy pivot]

so we can represent $$u = lnx$$ and $$\frac{dv}{dx} = x^2$$

[daring torrent]

Huang

[fast gyro]

Ok that makes sense

[hazy pivot]

so now we need to find du/dx and v

to use the formula $$uv-\int v\frac{du}{dx} dx$$

[daring torrent]

Huang

[hazy pivot]

$$\frac{du}{dx} = \frac{1}{x}$$ and $$v=\frac{x^3}{3}$$

[fast gyro]

Ok I understand so far

[daring torrent]

Huang

[hazy pivot]

so that we know have u,v and du/dx we can use this formula

[fast gyro]

Ohhh

So du is just du/dx times the dx?

[hazy pivot]

yes dx is cancelled out

[hazy pivot]

I suggest using this as you can understand better

[fast gyro]

Yeah I think that's why I was confused, since I was only seeing the du

I'll try another question on my own. One sec

[hazy pivot]

also when deciding what to choose as u follow this order

L - logarithmic
I - inverse trig
A - algebric
T - trig
E - exponential

some ppl even use ILATE where inverse trig is given more priority so check the qs to be sure

[fast gyro]

I think I did something wrong when I tried this

I'll send a photo of my work

I ended up having to integrate (x^2)(e^2x) which I don't think I'm supposed to do

[hazy pivot]

this is gonna get really complex which is why I said to use LIATE when choosing u. It will be easier to choose u as x so then du/dx will be 1

and dv/dx as e^2x

[fast gyro]

Oh. I think I misremembered something I heard. I thought you should choose based on which is easier to integrate

Which is why I made u e^2x and dv/dx x

[hazy pivot]

you could solve it that way too but I think it uses substituition and its like a loop

[hazy pivot]

just choose u according to LIATE

[fast gyro]

Wait so is liate the priority for u or dv?

[hazy pivot]

its for u

[fast gyro]

And the algebraic is polynomials and stuff, right?

So my u should be the x

[hazy pivot]

yes

[hazy pivot]

yep

[fast gyro]

Ok I did it and got the right answer

If it were (x^2)(e^2x) instead, would we end up using integration by parts twice?

Since the derivative of x^2 is 2x

And then the derivative of that is a constant

[hazy pivot]

yes

[hazy pivot]

you mean in your working right

[plain magnet]

yes

[fast gyro]

Yeah to get the final answer

[plain magnet]

if you need to you can follow the rule called LIATE

[hazy pivot]

yea but its gonna be a hassle and sometimes u hv to keep on using it

[plain magnet]

L-Logarithms
I-Inverse trig
A-Algebraic
T-Trigonometric
E-Exponential

but you dont even need it and i generally dislike it

just use intuition on what will simplify first, or create a cleaner integral

[plain magnet]

so here you would have $\begin{matrix}\quad &D&I\+&x&e^{2x}\-&1&2e^{2x}\end{matrix}$

[daring torrent]

ooaa

[plain magnet]

then combine the diagonal terms (multiply), and put the one on the bottom row under an integral sign

and the diagonal stuff only applies to the top-to-bottom one

so here we would have $2xe^{2x}-\int 2e^{2x}dx$

[daring torrent]

ooaa

[plain magnet]

this is what is called tabular method

more efficient than choosing a u or v

[plain magnet]

you could extend this to $$\begin{matrix}\quad &D&I\+&x&e^{2x}\-&1&2e^{2x}\+&0&4e^{2x}\end{matrix}$$ and directly get your answer

[daring torrent]

ooaa

[plain magnet]

integral of 0 is just 0

[fast gyro]

Oh yeah I think I saw that in my textbook

So I should learn that to make it easier to do?

[plain magnet]

yes

[hazy pivot]

its another method you can use

[plain magnet]

once you understand integration by parts fundamentals

it is essentially uv method but better

i will give a small example

[fast gyro]

Ok 👍

[plain magnet]

proving $\mathrm{Li}_2(z)+\mathrm{Li}_2(1-z)=\frac{\pi^2}{6}-\ln(z)\ln(1-z)$, where $$\mathrm{Li}_2(z)=\int_0^z\frac{\ln(1-t)}{t}dt$$

[daring torrent]

ooaa

[plain magnet]

all we need to do is use integration by parts

i will demonstrate

$\begin{matrix}\quad &D&I\+&\ln(1-t)&\frac{1}{t}\-&-\frac{1}{1-t}&\ln(t)\end{matrix}$

[daring torrent]

ooaa

[plain magnet]

see how i chose these based on intuition

i knew i needed to get an ln(t)ln(1-t) term for the definition given

LIATE does so-so here

[fast gyro]

Wait can you remind me what Li means? I remember knowing it before but now I forgot

Actually nevermind, I looked it up

So you need the ln(t)ln(1-t) for the proof, and setting them to u and dv correctly gave you the needed term?

[plain magnet]

yep

[plain magnet]

(well i defined it already but that doesnt matter so anyways)

[plain magnet]

^

[fast gyro]

Yeah I just saw the 2 subscript and thought that maybe made it a special case or something

But anyway

Just out of curiosity, though you wouldn't ever do this

would it be impossible to use integration by parts to solve the integral of (e^2x)(e^2x) since the derivative never goes to 0?

Nevermind, the answer is yes if you never simplify to e^4x at any point

Well thanks for the help anyway

I think I understand now

[weary quarry]

Actually, no, I don't think it is impossible.

That is to say. I think it is possible.

[fast gyro]

Wow really?

[weary quarry]

I think so.

Have you tried it?

[fast gyro]

Well when I did it I ended up having to use the integral of (2e^2x) times (1/2)e^2x

Both inside the integral

Which would just simplify back to (e^2x)(e^2x)

Maybe I'm wrong though

[weary quarry]

That's correct, but remember that's not the only term you get.

[plain magnet]

nope

you can use some sort of "cyclic" IBP or just rewrite $\frac{1}{e^{2x}}=e^{-2x}$

[daring torrent]

ooaa

[plain magnet]

ah wait

yeah there is some sort of cyclic anyways

[weary quarry]

...rewrite what as what?

[north notch]

@fast gyro

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[fast gyro]

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