For question 3 e^x * sinx, in the final part where you do - integral of final components multiplied, why do you get - instead of - integral -...? Of course you get e^x * (-sinx), so youd have ... - integral e^x * (-sinx), and when you take the minus out of the integral it becomes + integral e^x * sinx ?
@@Manuel71 Yes, I had realized and learnt it in the months that came by. I still appreciate your patience to help random students on the internet! Thank you!
@@Anmol_Sinha It really is the same thing as integration by parts, just organized in a much more practical way than the traditional method. The traditional method works by assigning one component function as u, and the other (with the original differential, e.g. dx) as dv. It then constructs the formula integral u dv = u*v - integral v*du. You then pick a u to differentiate, and pick a dv to integrate, and repeat. This method structures what you usually do with this procedure, such that you don't even need to think about u and v. Set up alternating signs, to account for the minus in the formula, differentiate down the D-column, and integrate down the I-column. Stop when you get to one of the three stops. Then, connect each sign with the D-column, and the next row down in the I-column, and integrate across the final row. Stop 1: the ender. Stop when you differentiate the D-column down to zero. Common for polynomial factors. Stop 2: the looper. Stop when you recognize that you'd otherwise be in an infinite loop, because you recognize the original integral. This is common for exponentials mixed with trig, or trig mixed with other trig. It doesn't work for sine mixed with cosine, unless they have different frequencies. Stop 3: the regrouper. Stop, when you recognize you can regroup along a row, into something you can integrate by another method, or start a new IBP table with different functions in each column. This is common for logs and inverse trig.
It's called integration by parts, and the tabular method. He puts one component function in the D-column, and the other in the I-column. Then differentiates down the D-column, and integrates down the I-column, along with alternating signs in front of the D-column. Connect along diagonals, until you get to the final row, which you connect along the row with a new integral.
In example 1, he's integrating x^2*e^(3*x) dx. The algebraic function x^2 is in the D-column, because it differentiates to zero. The exponential is in the I-column. This one is an ender, because x^2 eventually differentiates to zero. In example 2, the logarithm is in the D-column (because we can't integrate it without IBP), and the algebraic function x^3 is in the I-column. Logarithms become algebraic after differentiating, which means we can regroup, and form a new integral we can do directly. In example 3, we have a looper, because we are in an infinite loop when both differentiating and integrating exponentials and sine & cosine trig. To get out of the infinite loop, we spot the original integral, and solve for it as an algebraic unknown.
Generally only one will make sense depending on the scenario. If you’re deriving an algebraic function then it’s guaranteed to eventually be zero on the table, so just do that. When you’re deriving transcendental functions it’ll just end up being whichever thing shows up first. The only way it’ll ever repeat like it does in stop 3 is with trig functions, logs and algebraic functions aren’t gonna be perfect like that(not that I’ve seen anyway)
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when you have 10 seconds left on a test
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This really is the calm before the storm. Next time’s gonna be a rollercoaster!
It is just amazing....
so satisfying
For question 3 e^x * sinx, in the final part where you do - integral of final components multiplied, why do you get - instead of - integral -...? Of course you get e^x * (-sinx), so youd have ... - integral e^x * (-sinx), and when you take the minus out of the integral it becomes + integral e^x * sinx ?
Man you a freak fr god bless
Do they teach these methods in college? Can we get an explanation on how this method works the way it does?
It's just a shortcut for integration by parts, which is why it's often skipped in college.
@@Manuel71 Yes, I had realized and learnt it in the months that came by. I still appreciate your patience to help random students on the internet! Thank you!
@@Anmol_Sinha It really is the same thing as integration by parts, just organized in a much more practical way than the traditional method. The traditional method works by assigning one component function as u, and the other (with the original differential, e.g. dx) as dv. It then constructs the formula integral u dv = u*v - integral v*du. You then pick a u to differentiate, and pick a dv to integrate, and repeat.
This method structures what you usually do with this procedure, such that you don't even need to think about u and v. Set up alternating signs, to account for the minus in the formula, differentiate down the D-column, and integrate down the I-column. Stop when you get to one of the three stops. Then, connect each sign with the D-column, and the next row down in the I-column, and integrate across the final row.
Stop 1: the ender. Stop when you differentiate the D-column down to zero. Common for polynomial factors.
Stop 2: the looper. Stop when you recognize that you'd otherwise be in an infinite loop, because you recognize the original integral. This is common for exponentials mixed with trig, or trig mixed with other trig. It doesn't work for sine mixed with cosine, unless they have different frequencies.
Stop 3: the regrouper. Stop, when you recognize you can regroup along a row, into something you can integrate by another method, or start a new IBP table with different functions in each column. This is common for logs and inverse trig.
@@carultch thank you
I wanna see you do construction. Pwease?
Off by 1 sec lol
I thought i put the video on 2x by accident lol
The only time I made a video play slower 💀
Please can you explain this method in details
It's called integration by parts, and the tabular method. He puts one component function in the D-column, and the other in the I-column. Then differentiates down the D-column, and integrates down the I-column, along with alternating signs in front of the D-column. Connect along diagonals, until you get to the final row, which you connect along the row with a new integral.
In example 1, he's integrating x^2*e^(3*x) dx. The algebraic function x^2 is in the D-column, because it differentiates to zero. The exponential is in the I-column. This one is an ender, because x^2 eventually differentiates to zero.
In example 2, the logarithm is in the D-column (because we can't integrate it without IBP), and the algebraic function x^3 is in the I-column. Logarithms become algebraic after differentiating, which means we can regroup, and form a new integral we can do directly.
In example 3, we have a looper, because we are in an infinite loop when both differentiating and integrating exponentials and sine & cosine trig. To get out of the infinite loop, we spot the original integral, and solve for it as an algebraic unknown.
Do you get to choose which one you do
Generally only one will make sense depending on the scenario. If you’re deriving an algebraic function then it’s guaranteed to eventually be zero on the table, so just do that. When you’re deriving transcendental functions it’ll just end up being whichever thing shows up first. The only way it’ll ever repeat like it does in stop 3 is with trig functions, logs and algebraic functions aren’t gonna be perfect like that(not that I’ve seen anyway)
@@evefroggo4755 thx
I solve "by parts" by ILATE
It’s not always true for every cases
LIATE*
@@justreplied5090In India , it's ILATE
wthell r u doing no sense