I can't really go through the whole method because the DI method is a notekeeping technique for integration by parts that's difficult to render in typing.
#DI method (integration by parts)
frail ridge
indigo copper
I can't really go through the *whole* method because the DI method is a *notekee...
oh
hmmm
i watched the video you gave me
frail ridge
Okay, and were you able to follow it?
Really, the DI method isn't really required here, because it's mostly for repeated integration by parts.
indigo copper
Okay, and were you able to follow it?
most of it
indigo copper
Really, the DI method isn't *really* required here, because it's mostly for *rep...
is integration by parts easier
frail ridge
is integration by parts easier
It's literally the same thing.
DI is just a way of writing it.
frail ridge
I mean.
frail ridge
well ill try watch another video
I can explain the actual integration by parts method.
frail ridge
So like I said, we want to differentiate arcsin(x), and integrate 1 dx.
d/dx arcsin(x) = 1/sqrt(1 - x^2).
int 1 dx = x.
So int arcsin(x) dx = x arcsin(x) - int x/sqrt(1 - x^2) dx.
indigo copper
So `int arcsin(x) dx = x arcsin(x) - int x/sqrt(1 - x^2) dx`.
and what was the formula for that?
frail ridge
int u dv = uv - int v du.
That's why we integrate one function and differentiate the other in the DI.
indigo copper
That's why we integrate one function and differentiate the other in the DI.
but in DI you have to do more integration and differentiation
sometimes more than just twice
indigo copper
but sometimes?
frail ridge
DI is literally just a way of writing integration by parts.
If you have to do more than one step of DI method, it's because you have to integrate by parts more than once.
Which is something you often have to do, which is why the DI method was invented.
indigo copper
DI is *literally* just a way of *writing* integration by parts.
so when do you know when to stop integrating or differentiating?
frail ridge
so when do you know when to stop integrating or differentiating?
Was that not covered in the video?
indigo copper
Was that not covered in the video?
oh yea it was
I had to do 1\sqrt/1-x^2.x
what is that
frail ridge
...really? Because I get x/sqrt(1 - x^2).
frail ridge
I get int arcsin(x) dx = x arcsin(x) - int x/sqrt(1 - x^2) dx, like I said.
Use u substitution.
indigo copper
Was that not covered in the video?
the guy said that if the product of the integral and derivative could be integrated then that means you can stop.
frail ridge
...yes.
frail ridge
Okay, well, that's... not... what?
That's not... a thing.
Dude, x/sqrt(1 - x^2) is integrable.
I told you how to do it.
Or at least gave you a hint.
worthy geyser
well inverse functions can be integrated by a formula, int f inverse x = f inv(x) * x -F•(f inv (x))
worthy geyser
the guy said that if the product of the integral and derivative could be integra...
u watched black pen red pen?
indigo copper
u watched black pen red pen?
yes
indigo copper
Dude, `x/sqrt(1 - x^2)` is integrable.
so now that we know that the product can be integrated what next?
frail ridge
so now that we know that the product can be integrated what next?
...we... integrate it?
indigo copper
...we... integrate it?
oh
indigo copper
...we... integrate it?
ok
indigo copper
...we... integrate it?
and would that be it when we are done integrating that?
frail ridge
...look.
You have to take it step by step from the beginning.
Or else you might just straight up forget something you did earlier.
That said, integrating x/(1 - x^2) is the last step we have to deal with here.
frail ridge
Is that right for what?
indigo copper
Dude, `x/sqrt(1 - x^2)` is integrable.
integral of this
frail ridge
Yes.
indigo copper
so what after that?
indigo copper
Is that right for what?
also how do I tell if something cant be integrated?
frail ridge
so what after that?
...what do you mean, "what after that"?
indigo copper
...what do you mean, "what after that"?
what after we found the integral of x/sqrt(1-x^2)
frail ridge
also how do I tell if something cant be integrated?
There's not like a rule or anything. You just have to look at it and see if it's something you can work on with any of the techniques you know.
frail ridge
what after we found the integral of x/sqrt(1-x^2)
I still don't understand the question.
indigo copper
I still don't understand the question.
the product of the two things we found the derivative and integral of's product is x/sqrt(1-x^2) then you said we integrate that which is -(1-x^2)^1/2 is that the answer or what do we have to do next to find the final integral
frail ridge
the product of the two things we found the derivative and integral of's product ...
You have to read back.
indigo copper
You have to read back.
wdym?
frail ridge
wdym?
I mean the reason you don't know what to do next is because you've forgotten what we did before. So read back through the channel to figure it out.
indigo copper
I mean the reason you don't know what to do *next* is because you've forgotten w...
x arcsin(x)- -(1-x^2)^(1/2)+c?
frail ridge
Almost.
indigo copper
yay
frail ridge
And that's why showing your work is important.
indigo copper
and what was the x at the front for?
frail ridge
So that it's all on the page so you can just look at it.
indigo copper
And that's why showing your work is important.
ye
frail ridge
and what was the x at the front for?
What do you mean, what's it for?
indigo copper
What do you mean, what's it *for*?
how did it get there?
frail ridge
By... being the integral of 1 dx...
frail ridge
@indigo copper Why are you deleting your first post?
indigo copper
<@953140475693514804> Why are you deleting your first post?
what will I use this channel for?
frail ridge
what will I use this channel for?
That's... not an answer to the question I asked.
indigo copper
That's... not an answer to the question I asked.
because I dont know what this channel will do
frail ridge
Nothing if you close it.
indigo copper
Nothing if you close it.
what can it do
frail ridge
What can what do?
indigo copper
what can the channel do
frail ridge
When?
indigo copper
well why should I keep it open?
oh it might be useful to look at the working
how do I reopen it?
frail ridge
well why should I keep it open?
You shouldn't. I'm telling you to close it, which deleting the top level post does not do.
indigo copper
You shouldn't. I'm telling you to close it, *which deleting the top level post d...
how do I close it?
frail ridge
Type "+close".
indigo copper
+close
hazy prairieBOT
✅ This help channel has been closed
indigo copper
thanks
frail ridge
And now it's open again.
indigo copper
+close