I think it's amazing that one can come to the correct answer despite performing the steps to get there incorrectly. No hate, just curiosity as to why that can happen.
@tervalas
7 ай бұрын
oddly, there is one glaring mistake in their final answer (not withstanding that they just stopped without reducing the right side of the answer). There is a mysterious x that wasn't in the original problem. Another response points out what they missed to have that.
@Th3OneWhoWaits
7 ай бұрын
@@tervalas credit to @neun_owo for catching the mistake (deleted dv by accident)
@jamescollier3
7 ай бұрын
It's called: working with a friend
@sidgar1
7 ай бұрын
It happens when you're working with too much in memory and are not careful with what you're writing down. You've already made the calculation in your head so once you arrive at the answer, you write down the answer you solved mentally, even though you didn't write everything down physically. The opposite can also happen, when you subconsciously write down the correct solution steps, but your conscious mind leaves something out of the equation and you come to an incorrect answer even though every step before that was written correctly. It shows holes in your mental awareness and it's good to see and catch those things so that you can be more mindful of them in the future. Sometimes your own mind can out-think itself so it's important to keep it from getting too far ahead of the solution.
@Mike__B
7 ай бұрын
3 lefts make a right? :)
@jannegrey593
7 ай бұрын
Yeah. That's a difficult one - most annoying mistakes often come from omission of some number or forgetting a sign. Sometimes it is very hard to track them. And people have it harder when they are checking their own work, because they often subconsciously "agree" with what they did so they don't notice the mistake. Mind you - I am not a professional teacher, though I used to be math tutor for highschool.
@stephenbeck7222
7 ай бұрын
Brilliant video. This is a good student that probably watched a lot of your videos already (excellent command of u sub setup and also the DI method), but just with some understandable mistakes here and there.
@bprpcalculusbasics
7 ай бұрын
Thank you! In fact I made a very similar mistake myself what I was teaching the integral of e^x*sin(x)cos(x) = 1/2*e^x*sin(2x) many years ago. So I was passionated to answer the question for the person!
@phiefer3
7 ай бұрын
Your suggestion at the end doesn't actually make things any simpler. Sure, u = ln2 +lnx, and xdu = dx, but when you convert that x into terms of u you just end up with dx=(e^u)du/2. So you still end up with the 1/2 to deal with. In fact, you end up with the exact same integral, just with u's instead of v's: half the integral of sin(u)*e^u du The only difference is that you basically did both substitutions at the same time. But you're still going to end up with a 1/4 out front before sub back from u's to x's.
@Mike__B
7 ай бұрын
It looks like he(or she) did the integration by parts (minus the 1/2), but then when he wrote the original integration WITH the 1/2 and incorrectly said that equaled the stuff on the right side. Then he CORRECTLY had the right side divided by 2 as that's what the integral WITH the 1/2 should be.... but then incorrectly removed the 1/2 from the integral. Honestly those two lines, the integrals on the left are swapped around, the 1/2 integral sin(v)... should be the 2nd line, and the, integral sin(v)... (without the 1/2) should be the 1st line. Then he just did algebra correctly for the rest of it, but it was correct algebra with incorrect terms. I see these little issues way too often for my liking in my class.
@tervalas
7 ай бұрын
Two errors. Line 2 they forgot the 1/2 from line 1, so when they accounted for it they simply took half the right side. But they should have caught themselves on line 3, when they added the integral on both sides and didn't account for the fact that it was part of the numerator therefore they should have been adding 1/2 the integral. Maybe this would have keyed them off that something wasn't quite right, even though you could have handled 3/2 of the integral and still just been off by a factor.
@christianvanderstap6257
7 ай бұрын
Missing closing brace at the end :)
@Neun_owo
7 ай бұрын
Wait, OP in the video actually missed the dv, that's why there's a extra 2x! Since dv=1/u, if OP included the dv, he gets the correct answer. he got very close !
@tervalas
7 ай бұрын
Yeah, incorrectly correctly got to a mostly correct answer except for that 2x.
@Th3OneWhoWaits
7 ай бұрын
Yes in step 5 in OP's work, the dv just randomly despawned.
@abhishekbhattach5431
7 ай бұрын
Great video. But one thing: there is a mistake in the 'mistake' spelling in the thumbnail.
@luisaleman9512
7 ай бұрын
That's a recurring joke in this channel... can you spot the MISTEAK? xD
@abhishekbhattach5431
7 ай бұрын
@@luisaleman9512 😂
@DarkNight768
7 ай бұрын
great video can you make a video on how to find A,B and C in partial fractions thank you
@stephenbeck7222
7 ай бұрын
Just search (probably on his regular blackpenredpen channel more of them) for partial fractions. He’s done a lot of them.
@bprpcalculusbasics
7 ай бұрын
Here it is: how to solve partial fractions kzitem.info/news/bejne/uKh9vn2gkYSErKg
@DarkNight768
7 ай бұрын
@@bprpcalculusbasics thanks
@GGFrosty69
7 ай бұрын
hello
@nemanjalazarevic9249
7 ай бұрын
hi
@broytingaravsol
7 ай бұрын
just done with D-operators by me
@VivBrodock
7 ай бұрын
In the first u sub just do u=ln(2x) That eliminates the floating 2s in the math which seems like his main problem.
@phiefer3
7 ай бұрын
That doesn't eliminate the floating 2s at all. Because you still end up with du = dx/x, which gives us dx=xdu, and then when you convert that leftover x you get dx=(e^u)du/2. You get the same integral as in the video, just with u's instead of v's, but everything else will be the same.
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