When I was younger, I used to always hear from my teachers that "Math is magic". And by the end of this video, I can say that math IS magic... or rather, math is magical. It's amazing how everything comes together at the end. It all makes perfectly logical sense that I am just amazed.
Professor Organic Chemistry Tutor, thank you for a lengthy three product term Integration by Parts problem in Calculus Two. This type of Integral can be long, messy and highly problematic from start to finish. I cannot remember seeing this kind of problem in a Calculus book. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
Pls if anyone has watched the organic chemistry tutor video on integration by parts and has understood it well, can he or she pls help me to identify u and dv in a given problem
Watch the former integration by part video before u watch this one sooo ull grab beta😉.and when u do , ull be able to understand the identification of u and dv beta😊😉😉
Integration - Formula Sheet: bit.ly/3XCT6oz
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Thank you for your hard work! this is a heavy computation problem! This video is so valuable.
When I was younger, I used to always hear from my teachers that "Math is magic". And by the end of this video, I can say that math IS magic... or rather, math is magical. It's amazing how everything comes together at the end. It all makes perfectly logical sense that I am just amazed.
I love how you have been posting hard calculus problems lately, similar to those youd find on exams.
Wow sir. You're an absolute genius. I love your techniques. More insight
Merry Christmas, Organic Chemistry Tutor. I hope you have Christmas-related math problems for us!
jeeeeesuuuuuuus! how many marks is the problem worth.......immediately i solve this ...i'll just ask for my degree
Love that last problem. This is what mathematicians had to do for hundreds of years...before computers
This will help me alot since there is a calculus 1 final for me next week
Thanks :)
The best channel I've ever watched! thank you so much for explaining so well and so complete
The explanation was fantastic and I love how you also showed us how the formula came!
Professor Organic Chemistry Tutor, thank you for a lengthy three product term Integration by Parts problem in Calculus Two. This type of Integral can be long, messy and highly problematic from start to finish. I cannot remember seeing this kind of problem in a Calculus book. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
Scary how one integral can take a half hour to compute by hand
Thank you for help to do my homework problem!
Solution:
∫x*sin(x)*e^(2x)*dx =
I solve the integral in 3 steps:
At first solution this partial integral by double partial integration:
∫sin(x)*e^(2x)*dx =
------------------------------------
Solution by partial integration:
Partial integration can be derived from the product rule of differential calculus. The product rule of differential calculus states:
(u*v)' = u'*v+u*v' |-u'*v ⟹
(u*v)'-u'*v = u*v' |∫() ⟹
∫u*v'*dx = u*v-∫u'*v*dx
------------------------------------
∫sin(x)*e^(2x)*dx = sin(x)*1/2*e^(2x)-∫cos(x)*1/2*e^(2x)*dx
= 1/2*sin(x)*e^(2x)-1/2*[∫cos(x)*e^(2x)*dx]
= 1/2*sin(x)*e^(2x)-1/2*[cos(x)*1/2*e^(2x)+∫sin(x)*1/2*e^(2x )*dx]
= 1/2*sin(x)*e^(2x)-1/4*cos(x)*e^(2x)-1/4*∫sin(x)*e^(2x)*dx
|+1/4*∫sin(x)*e^(2x)*dx ⟹
5/4*∫sin(x)*e^(2x)*dx = 1/2*sin(x)*e^(2x)-1/4*cos(x)*e^(2x) |*4 /5 ⟹
∫sin(x)*e^(2x)*dx = 2/5*sin(x)*e^(2x)-1/5*cos(x)*e^(2x)
= 1/5*[2*sin(x)*e^(2x)-cos(x)*e^(2x)]
Now solution of the whole integral again with partial integration:
∫x*sin(x)*e^(2x)*dx =
= x*1/5*[2*sin(x)*e^(2x)-cos(x)*e^(2x)]
-1/5*∫[2*sin(x)*e^(2x)-cos(x)*e^(2x)]*dx =
= x*1/5*[2*sin(x)*e^(2x)-cos(x)*e^(2x)]
-2/5*∫sin(x)*e^(2x)*dx+1/5*∫cos(x)*e^(2x)*dx =
= x*1/5*[2*sin(x)*e^(2x)-cos(x)*e^(2x)]
-2/5*1/5*[2*sin(x)*e^(2x)-cos(x)*e^(2x)]
+1/5*∫cos(x)*e^(2x)*dx =
= x*1/5*[2*sin(x)*e^(2x)-cos(x)*e^(2x)]
-2/25*[2*sin(x)*e^(2x)-cos(x)*e^(2x)]
+1/5*∫cos(x)*e^(2x)*dx =
------------------------------
Additional calculation:
∫cos(x)*e^(2x)*dx =
∫cos(x)*e^(2x)*dx = cos(x)*1/2*e^(2x)+∫sin(x)*1/2*e^(2x)*dx
= 1/2*cos(x)*e^(2x)+1/2*[∫sin(x)*e^(2x)*dx]
= 1/2*cos(x)*e^(2x)+1/2*[sin(x)*1/2*e^(2x)-∫cos(x)*1/2*e^(2x)*dx]
= 1/2*cos(x)*e^(2x)+1/4*sin(x)*e^(2x)-1/4*∫cos(x)*e^(2x)*dx
|+1/4*∫cos(x)*e^(2x)*dx ⟹
5/4*∫cos(x)*e^(2x)*dx = 1/2*cos(x)*e^(2x)+1/4*sin(x)*e^(2x) |*4/5 ⟹
∫cos(x)*e^(2x)*dx = 2/5*cos(x)*e^(2x)+1/5*sin(x)*e^(2x)
= 1/5*[2*cos(x)*e^(2x)+sin(x)*e^(2x)]
-----------------------------
∫x*sin(x)*e^(2x)*dx
= x*1/5*[2*sin(x)*e^(2x)-cos(x)*e^(2x)]
-2/25*[2*sin(x)*e^(2x)-cos(x)*e^(2x)]
+1/5*∫cos(x)*e^(2x)*dx =
= x*1/5*[2*sin(x)*e^(2x)-cos(x)*e^(2x)]
-2/25*[2*sin(x)*e^(2x)-cos(x)*e^(2x)]
+1/25*[2*cos(x)*e^(2x)+sin(x)*e^(2x)] =
= x*1/5*[2*sin(x)*e^(2x)-cos(x)*e^(2x)]
-4/25*sin(x)*e^(2x)+2/25*cos(x)*e^(2x)
+2/25*cos(x)*e^(2x)+1/25*sin(x)*e^(2x) =
= 1/5*x*[2*sin(x)*e^(2x)-cos(x)*e^(2x)]-3/25*sin(x)*e^(2x)+4/25*cos(x)*e^(2x)
Can you do a video about modelling with differentiation ?
Can u please do a video about hyperbolic functions?
Love the way you say produc
The intro man 😂😂
thank you for existing
Now you post after people finished calculus in first semester 😂 what a smart guy...
He has an older video than explains the same topic. Could’ve checked that out.
You both have valid points actually.
I thought to decide u, v, dw is to follow LoPET rule, Log>poly>exp>trigo like for 2 function, we need to follow that.
Bless you abundantly TOCT!
Bastante interesante, nunca se me hubiera ocurrido. Gracias por compartirlo.
Love u sir💙💙💙
Excellent work.
What is this? Its so lengthy. And i have to do 2 these kinds of integration and others for a differential equation problem.
How will you know which function to choose as your u and dv, or it that you can choose any function to be your u and dv
It gets easier as you work through the homework problems. You will begin to see a pattern.
Merry Christmas
Pls if anyone has watched the organic chemistry tutor video on integration by parts and has understood it well, can he or she pls help me to identify u and dv in a given problem
Watch the former integration by part video before u watch this one sooo ull grab beta😉.and when u do , ull be able to understand the identification of u and dv beta😊😉😉
@@somtonwadiaro8076 ok thanks 😊
Complex analysis one shots this integral
Amazing sir
Make videos on bjt and mosfets
What will be the integration of underroot cosx
I guess cosx power 3/2 Divided 3/2
@@AbishekRajkumar2kDid you forget the chain rule exists
Lovely, this!
Thanks sir
Why cant you just set v=e^2xsinx and pretend its one function?
when you realize it took him 24min to solve this:
💀
Int x*exp(2+i)x dx 取虛部。
6:29
In this link you will find an interesting integral:ua-cam.com/video/xCrKsJfkwsY/v-deo.html
I love you
(no homo)?
Great
Wow
But actually, letting “(x)(sin x)” as “u” is much easier
True, but still impressive.
Cool
🫡🫡🫡🫡🫡🫡🫡🫡🫡
o_0 waytoodank
8O
Bruh...
Wow