Split (1)

train · 116k rows

role stringclasses 2 values	content stringlengths 0 2.1k	session_id int64 10 21.7k	sequence_id int64 0 2.38k	annotations listlengths 0 8
volunteer	uh inner role when you have it defined as your parameter. Your parametric integration.	15,912	729	[]
volunteer	And that's basically what the um	15,912	730	[]
volunteer	not the next section, but the section after that integration by parts. That's when we start getting into	15,912	731	[]
volunteer	um	15,912	732	[]
volunteer	you know, how do we introduce	15,912	733	[]
volunteer	you know, how do we deal with problems like that you can't just	15,912	734	[]
volunteer	directly do	15,912	735	[]
volunteer	one parameter.	15,912	736	[]
volunteer	They would if you have to do, you have, would have to use two parameters instead.	15,912	737	[]
volunteer	or more specifically, what if your um	15,912	738	[]
volunteer	if you're integral kind of takes the form like X	15,912	739	[]
volunteer	Y DY	15,912	740	[]
volunteer	where you can create some separate um	15,912	741	[]
volunteer	parameters to say U is equal to X.	15,912	742	[]
volunteer	um, and YDY is equal to DV.	15,912	743	[]
volunteer	Right? Cause you already know what this	15,912	744	[]
volunteer	kind of like did this why crime and stuff is a derivative of.	15,912	745	[]
volunteer	So you take the innergral of it	15,912	746	[]
volunteer	I'm sorry, that's not right. That's	15,912	747	[]
volunteer	we'll just wait till the section to get there.	15,912	748	[]
volunteer	That's where it is supposed to be the integral of uv.	15,912	749	[]
volunteer	That's what makes sense	15,912	750	[]
volunteer	Yeah.	15,912	751	[]
volunteer	where Y is some function of X.	15,912	752	[]
volunteer	So that when you do this	15,912	753	[]
volunteer	you get you equal to X.	15,912	754	[]
volunteer	or, or whatever X term it is could be X and could be cosine x and then you have your own separate DUDX versus your second term, which we define as DV equals DY.	15,912	755	[]
volunteer	Then you integrate that, you get V equal to Y of X.	15,912	756	[]
volunteer	and then that has its own um kind of reverse chain rule.	15,912	757	[]
volunteer	Um, it's proven, it'll, it'll show it that it proves that the, that the inner role	15,912	758	[]
volunteer	of sort of um	15,912	759	[]
volunteer	what is it? UDV.	15,912	760	[]
volunteer	Yeah.	15,912	761	[]
volunteer	is equal to	15,912	762	[]
volunteer	UV +.	15,912	763	[]
volunteer	the integral	15,912	764	[]
volunteer	of VDU	15,912	765	[]
volunteer	and this becomes like	15,912	766	[]
volunteer	very important going on.	15,912	767	[]
volunteer	this little property.	15,912	768	[]
volunteer	because you're more likely to get problems, I think that, that you need	15,912	769	[]
volunteer	um this sort of integration by parts	15,912	770	[]
volunteer	then you will just a simple substitution.	15,912	771	[]
volunteer	Anyway, um, hope I didn't lose you there.	15,912	772	[]
volunteer	Hopefully you're still with me	15,912	773	[]
student	im here. chat not working. i can hear you	15,912	774	[]
volunteer	OK, good	15,912	775	[]
volunteer	Um	15,912	776	[]
volunteer	right, so we can move, um, are you OK with 7 section 7.3.	15,912	777	[]
volunteer	with the um	15,912	778	[]
volunteer	trigonometric for uh functions.	15,912	779	[]
volunteer	identities	15,912	780	[]
volunteer	You seem pretty strong in trig identities. This one was just more of a um	15,912	781	[]
volunteer	just kind of remembering certain aspects, like certain identities.	15,912	782	[]
volunteer	the phase shift and	15,912	783	[]
student	yes! can we review #19, what was wrong with it?	15,912	784	[]
volunteer	and how to do your, how to do your um inverse trig triangle.	15,912	785	[]
volunteer	Oh yes, of course	15,912	786	[]
volunteer	So	15,912	787	[]
volunteer	let me clear this problem.	15,912	788	[]
volunteer	So, 19, 1 over sign	15,912	789	[]
volunteer	X cosine cut.	15,912	790	[]
volunteer	X	15,912	791	[]
volunteer	So, I see what you did. You divide, the first thing you did was to	15,912	792	[]
volunteer	use the identity, um	15,912	793	[]
volunteer	like the original tri I did a sin X square, sine square plus cosine squared equals 1 and you separated terms. That makes sense.	15,912	794	[]
volunteer	Then you use substitution to solve the um	15,912	795	[]
volunteer	kind of definite integral.	15,912	796	[]
volunteer	which makes sense. It should just be C and D	15,912	797	[]
student	ohh no! i just drag whole thing	15,912	798	[]
volunteer	However	15,912	799	[]
volunteer	well, you just dragged it where? You lost it?	15,912	800	[]
volunteer	I still see it	15,912	801	[]
volunteer	You, you probably hit undo	15,912	802	[]
volunteer	Let me see if I can drag everything up.	15,912	803	[]
volunteer	Yeah, I moved everything up.	15,912	804	[]
volunteer	OK	15,912	805	[]
student	no, I mean like i didn't see tanxsecx	15,912	806	[]
volunteer	Oh, I see what you're saying	15,912	807	[]
volunteer	kind of the square root of	15,912	808	[]
volunteer	the derivative of, um,	15,912	809	[]
volunteer	Siganex. I'm pretty sure Seconex is just	15,912	810	[]
volunteer	yeah, it's just Econex TNX.	15,912	811	[]
volunteer	Is there a sign difference	15,912	812	[]
volunteer	I mean, you proved the derivative of seinex tanix just by using your integration uh substitution, so that's fine.	15,912	813	[]
volunteer	There's no.	15,912	814	[]
volunteer	problem with that	15,912	815	[]
volunteer	I just don't know where they got	15,912	816	[]
volunteer	tan squared.	15,912	817	[]
volunteer	1/2 10 squared. So they probably use the property.	15,912	818	[]
volunteer	unless there's some tea.	15,912	819	[]
volunteer	that they probably look, actually it looks like a	15,912	820	[]
student	what's the answer?	15,912	821	[]
volunteer	so you said sine makes you go to TE and sine X D X you go to negative BT.	15,912	822	[]
volunteer	So they said	15,912	823	[]
volunteer	oh sorry	15,912	824	[]
volunteer	I'm not showing it	15,912	825	[]
volunteer	Uh, let me show this to.	15,912	826	[]
volunteer	Can you see it	15,912	827	[]
volunteer	So I agree with you. I, I agree with them for the log tan. I don't know where they got the 1/10 squared.	15,912	828	[]

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