You’re such a great teacher. What I love most of what you do is: you repeat yourself, you don’t skip steps, you explain with a degree of patience and professionalism.
I also first thought to use the complex form of trigo functions. Either expressing the cos as the half sum e^ix and e^-ix or simply taking the real part of the result of integrating e^x.e^ix is slightly faster than IPP IMHO. Thanks for continuing on this channel. I haven't done much maths since graduation but I keep stumbling on your channel and I love jogging my memory. I would love more prime number related brain twisters.
This integral can be calculated by parts directly (without substitution) There are two ways to choose the parts The obvious one is du=dx , u=x v=cos(ln(x)), dv=-sin(ln(x))/xdx
Anyone who has used integration by parts has used this method. DI is just a terminology that is used to impress people, but he had invented nothing. D-I stands for Differentiate-Integrate. Example let's consider a fonction f(x) you want to find the integral. Sometimes when f(x) isn't easy to integrate, in this case you can write f(x) as the product of two functions v(x) and u(x) such as f(x)=u(x)*v(x) therefore by Differentiation of f(x) we get Df(x)=D(uv)=u'v+UV' Thus u'v=D(uv)-uv' Assume you has to integrate x*ln(x) In this case for example you set u=ln(x)-----Differentiate it------ u'=1/x v'=x. -------Integrate it--------'---v=(x^2)/2 This what he calls D I to impress. Finally you get I=x^2)*ln(x)/2-(x^2)/4 +C
Your second way is easier, but if you'd continued the DI method there you could get it following the quotient rule and so x^2 and 1/(x^2) cancel and so on... well, complex analysis is also nice.
i just got this in my calc 2 exam today and didn't know how to do it. bruh you need to do u sub and make x the subject and do by parts and bring the integral to the left side, i would not have gotten it even if you gave me the whole day to do
@@mariahamilton6096 the name completely escaped me, but that is what I meant. I think it is only possible in the complex plane using the Lambert W function, but I'm not sure.
Yeah I don't know too much about it, except that it probably won't be elementary. Well actually you can integrate x^(-x) and x^x from 0 to 1 and get an infinite sum as a result without using the complex plane too much (you use the definition of the gamma function at one point, but that's it).
There is an other method, you integrate x^i that is (x^i)/(i+1) and you write like a+i*b (it is easy). This is equal to integrate e^(i*ln(x)), that is equal to the integation of cos(ln(x))+i*sin(ln(x)) and you can split in the integal of cos(ln(x)) and i times the integal of sin(ln(x)), all this stuff is eqal to a+i*b. Well done! Can I have an
While i probably would have come up with the first method, the second one is much more elegant and quicker so i like that better. (Usually a u sub is the first thing i try if i see a function inside a function. And given the u was ln x it was not hard of dealing with the x. And that was a more thsn doable integral so the first one is pretty much an intuitive way, but the second as i said much smoother) Although i like to use the classic formula for IBP simply because it really is rinse and repeat until you have either a solution or got the original back and is just muscle memory after a time, while the DI set up if you know, you won't get a repeat might be tempting to go crazy. I was staring horror at the D column "you don't want to differentiate that sucker!" Was my first thought. (The d column as you stated would have gotten worse and worse with every iteration so it really needs that "sense of danger". The classic formula as annoying as it is, doesn't have that. And this clearly was one of the integral where classic formula and dv=1dx is just smooth sailing without thinking. You just need to repeat until you see "wait, that looks familiar")
why did u introduce e in that u subsitutiion? . or when do you know that i must introduce e ? bcz what i know is that when you have a fuction raise to a fuction u introduce ln now im confused with e
I mean, this one could be done just looking at what could be derived to get that function. To get cos(ln(x)) there must have been a sin(ln(x)) to be derived in the integral. But deriving that, we get an extra 1/x. Multiplying that by x and deriving it will get rid of the 1/x, but add an extra sin(ln(x)). We could then add something that derived would get rid of that sin, which would be cos(ln(x))x. The other term of the derivative of that will just be cos(ln(x)), so you end up with 2cos(ln(x)). The true integral is, then: 1/2[sin(ln(x))x + cos(ln(x))x] Still my favorite method for these simple integrals, just so much quicker.
Never saw this strange D I + - integration by parts schema before. For me it much easier to use direct formula - INT(f*d(g)) = f*g - INT(d(f)*g). So e^u*cos(u)*du is e^u*d(sin(u)). Then you just apply formula twice =)
Hmmm.... Well. Differentiating the inside and see: -sin(ln(x))/x. The simple way to avoid this X is to multiply by it. x cos(ln(x)) becomes cos(ln(x))-sin(ln(x)) For x sin(ln(x)) the derivative is sin(ln(x)) + cos(ln(x)), summing the two gets twice the original arg, divide by two to adjust.
The first method is definitely better, but I disagree with your second point. By the time students would encounter such an integral, they would have developed that "sense of danger" skill. It is a skill. In integration, the convention for integration by parts is to pick a u whose derivative is "simple" and a dv where you know the anti-derivative. If taking the derivative of u(DI method is integration by parts in tabular form) would lead to a complicated function, then it'd be time to rethink things.
Why can't it be like this- ¶cos(lnx) = -sin(lnx)/the derivative of what's inside the function ,ie - lnx = -sin(lnx)/1/x = - sin(lnx) . x I used the chain rule for integration
Let's do the following: The integrand is cos(ln x); since cos(y)=cos(y+2nPi), where n is any integer and for any y, then cos(ln x)=cos(ln x +2nPi)=cos(ln(exp(2nPi) x)). Let's integrate both sides with respect to x; then Int cos(ln x) dx=Int cos(ln(exp(2nPi) x)) dx. Now in the right-hand-side we make the change of variable u=exp(2nPi) x, therefore, one gets Int cos(ln x) dx=exp(-2nPi) Int cos(ln u) du What's happening?
why such complex explanation for integration by parts? Simply write it as ∫e^x d(sinx) = e^x sin x - ∫ sin x d(e^x) dx. Also why such difficult to understand english diction?
By complex analysis ua-cam.com/video/WgOzdcvPb-k/v-deo.html
@Rajneesh Mishra its a menorisation mechanic, which does not always work.
@@japotillor here, we have U=ln(X), and so we know what ought to be v, and what your dw ought to be.
@@byronrobbins8834 u = ln x, du = (1/x)dx, dx = x du = integral of cos u (xdu) = integral of e^u cos u du, then u = e^u and dv = cos u du :)
You’re such a great teacher. What I love most of what you do is: you repeat yourself, you don’t skip steps, you explain with a degree of patience and professionalism.
that transition from one way to another was so smooth 😂
I also first thought to use the complex form of trigo functions. Either expressing the cos as the half sum e^ix and e^-ix or simply taking the real part of the result of integrating e^x.e^ix is slightly faster than IPP IMHO. Thanks for continuing on this channel. I haven't done much maths since graduation but I keep stumbling on your channel and I love jogging my memory. I would love more prime number related brain twisters.
That transition was so smooth;)
Thanks! : )
Finally something I got right from the first time. It has been 25 years ago I studied this. Your videos are excellent for refreshing my knowledge.
You should have done it with complex analysis, it's very easy.
Logan Hargrove agree! I did that last year already ua-cam.com/video/WgOzdcvPb-k/v-deo.html
Yessss!! specifically wanted to see integral of x^i very nice! thank you BPRP you are awesomeeee!!!
Yupp very easy.
x/2 (cos(ln(x)) + sin(ln(x))) + c
This integral can be calculated by parts directly (without substitution)
There are two ways to choose the parts
The obvious one is
du=dx , u=x
v=cos(ln(x)), dv=-sin(ln(x))/xdx
We can also choose parts in this way
du = cos(ln(x))/x dx, u = sin(ln(x))
v = x, dv = dx
I have not seen the DI method in class but I like it a lot. This is neat!
But I was lucky that I have been taught.
And believe me it worked like a weapon for me.🙂🙂🙂🙂
Anyone who has used integration by parts has used this method. DI is just a terminology that is used to impress people, but he had invented nothing.
D-I stands for Differentiate-Integrate.
Example let's consider a fonction f(x) you want to find the integral. Sometimes when f(x) isn't easy to integrate, in this case you can write f(x) as the product of two functions v(x) and u(x) such as
f(x)=u(x)*v(x) therefore by Differentiation of f(x) we get
Df(x)=D(uv)=u'v+UV'
Thus u'v=D(uv)-uv'
Assume you has to integrate x*ln(x)
In this case for example you set
u=ln(x)-----Differentiate it------ u'=1/x
v'=x. -------Integrate it--------'---v=(x^2)/2
This what he calls D I to impress.
Finally you get
I=x^2)*ln(x)/2-(x^2)/4 +C
Great you helped me alot with integral of sec^1412 dx in my calculus exam hope you're well ♥
This question came on WBJEE 2022 literally the best explanation
this is 10 times more useful than listening to my math teacher...
I love your videos man! Very educational helped me a lot
I feel like the u sub is more fundamental, so I would choose that.
I did it the second way and also by taking the real part of the integral of x^i.
Just bought your merch. Love the design.
arequina thank you!!!!
D-I tabular method for integral by parts is what I got from the video, thanks.
Your so brilliant at explaining
Most of the times i prefer the u substitution.
I did exactly same like 1st method 😋😋. I just love integration
That was exactly what was just asked in the exam!
Well it was sin instead of cos but still great timing!
You could use X=e^x and resolve the problem with imaginary numbers (real part of the primitiv of e^X(i+1)). Three ways !
U r so smart
Thank you 💗
I like the 2nd method, but can appreciate the first for the sake of knowing another way, just to have in my toolbox
Oh man this is great information about integrals, in wich I was very poor at school.
Nobody can ever forget the complex analysis way.
I have solved this problem earlier....Thank you for video
"the reason i knew to stop is 'cause i had a sense of danger" hahahaha :'v
Excellent explanation.
Thank you! Loved your explanation
We can use complex for calculation
Your second way is easier, but if you'd continued the DI method there you could get it following the quotient rule and so x^2 and 1/(x^2) cancel and so on... well, complex analysis is also nice.
God Is Great, May God Bless them all and you all and your families
i just got this in my calc 2 exam today and didn't know how to do it. bruh you need to do u sub and make x the subject and do by parts and bring the integral to the left side, i would not have gotten it even if you gave me the whole day to do
Why you integrate and differentiate the symbol 2 time (e^u) and (cosu)??
Is there a video dedicated to your D I + - chart?
yes, here: ua-cam.com/video/matDV3XL2J8/v-deo.html
Appreciated
Can we see an integration of a function that uses up arrow notation?
Yeah that'd be interesting (if it's possible, assuming you mean like tetration).
@@mariahamilton6096 the name completely escaped me, but that is what I meant. I think it is only possible in the complex plane using the Lambert W function, but I'm not sure.
Yeah I don't know too much about it, except that it probably won't be elementary. Well actually you can integrate x^(-x) and x^x from 0 to 1 and get an infinite sum as a result without using the complex plane too much (you use the definition of the gamma function at one point, but that's it).
There is an other method, you integrate x^i that is (x^i)/(i+1) and you write like a+i*b (it is easy). This is equal to integrate e^(i*ln(x)), that is equal to the integation of cos(ln(x))+i*sin(ln(x)) and you can split in the integal of cos(ln(x)) and i times the integal of sin(ln(x)), all this stuff is eqal to a+i*b. Well done!
Can I have an
While i probably would have come up with the first method, the second one is much more elegant and quicker so i like that better. (Usually a u sub is the first thing i try if i see a function inside a function. And given the u was ln x it was not hard of dealing with the x. And that was a more thsn doable integral so the first one is pretty much an intuitive way, but the second as i said much smoother) Although i like to use the classic formula for IBP simply because it really is rinse and repeat until you have either a solution or got the original back and is just muscle memory after a time, while the DI set up if you know, you won't get a repeat might be tempting to go crazy. I was staring horror at the D column "you don't want to differentiate that sucker!" Was my first thought. (The d column as you stated would have gotten worse and worse with every iteration so it really needs that "sense of danger". The classic formula as annoying as it is, doesn't have that. And this clearly was one of the integral where classic formula and dv=1dx is just smooth sailing without thinking. You just need to repeat until you see "wait, that looks familiar")
How to solve integral of A*cos(C + 4*Ln(cosh(sqrt(A)*x))) ? or A*cos(C - 4*Ln(cosh(sqrt(A)*x))) ?
Appetiser and Entrées are both starters by the way. Except in the USA where they mess everything up.
Dear bprp, a másodikat nem ismertem, köszönöm !
u = cos(lnx) dv=dx
du = -sin(lnx)/x. v = x
Final answer: x/2 [ cos(lnx) + sin(lnx)] +c
Sir how about the integral of e^x overr x^2+1?
why did u introduce e in that u subsitutiion? . or when do you know that i must introduce e ? bcz what i know is that when you have a fuction raise to a fuction u introduce ln now im confused with e
I mean, this one could be done just looking at what could be derived to get that function.
To get cos(ln(x)) there must have been a sin(ln(x)) to be derived in the integral. But deriving that, we get an extra 1/x. Multiplying that by x and deriving it will get rid of the 1/x, but add an extra sin(ln(x)).
We could then add something that derived would get rid of that sin, which would be cos(ln(x))x. The other term of the derivative of that will just be cos(ln(x)), so you end up with 2cos(ln(x)). The true integral is, then:
1/2[sin(ln(x))x + cos(ln(x))x]
Still my favorite method for these simple integrals, just so much quicker.
How to integrate cos(lnx)/x from interval 0 to 3??
Never saw this strange D I + - integration by parts schema before. For me it much easier to use direct formula - INT(f*d(g)) = f*g - INT(d(f)*g). So e^u*cos(u)*du is e^u*d(sin(u)). Then you just apply formula twice =)
Integral of x^i
Hmmm.... Well. Differentiating the inside and see: -sin(ln(x))/x. The simple way to avoid this X is to multiply by it. x cos(ln(x)) becomes cos(ln(x))-sin(ln(x))
For x sin(ln(x)) the derivative is sin(ln(x)) + cos(ln(x)), summing the two gets twice the original arg, divide by two to adjust.
Fantastic...
Awsome explaind
I really liked the second way.
12:08 use the chen lu!
Love the nike plug!
Hello blackpenredpen, can you make a video about integrating the definite integral of [t+abs(t-x)]dt from 0 to 2x?
What about going into the complex world to then take the real part?
What about with complex analysis
Nice: A rare video where you use isn't it right ;)
As always great video man :D
Faster way: cos(lnx) = Re(x^i) just integrate a simple power function and take real part.
Both are good
why e^lnx = x? Demonstration.
I like the first method more because as a beginner you don't need to sense that differentiating sin(lnx)/x is bad.
The first method is definitely better, but I disagree with your second point. By the time students would encounter such an integral, they would have developed that "sense of danger" skill. It is a skill. In integration, the convention for integration by parts is to pick a u whose derivative is "simple" and a dv where you know the anti-derivative. If taking the derivative of u(DI method is integration by parts in tabular form) would lead to a complicated function, then it'd be time to rethink things.
Good job
Can't you use complex numbers for an easy solution?
I like the second one! :D
Stydras me too i would do this also by the second method
The first method involves two unnecessary substitution steps (x -> u and then back again) and the second method is the cleaner and easier one.
Why can't it be like this-
¶cos(lnx) = -sin(lnx)/the derivative of what's inside the function ,ie - lnx
= -sin(lnx)/1/x = - sin(lnx) . x
I used the chain rule for integration
Plz sir solve integration of cos(e^t) dt
Can you find
Integral of tan(ln(x))
'AND BE HAPPY!'
8:21 8:21 8:21
I used the method you showed us in the integral of x^i but got the wrong sign lol I should check my calculations
tnx
excellent
🔥🔥🔥🔥🔥
Integral 1/x^3-1
I which university you are studying
How you make such a thumbnail
Can’t u just use complex definition of cosine and things will start to cancel out?
Nice sir
That's majic
we have the the blue pen today 😄
must be missing something - the answer to the question includes the question itself? guess old age finally caught up with me!
Always use complex numbers of possible
Where is problem Corona Integral
Let's do the following:
The integrand is cos(ln x); since cos(y)=cos(y+2nPi), where n is any integer and for any y, then cos(ln x)=cos(ln x +2nPi)=cos(ln(exp(2nPi) x)).
Let's integrate both sides with respect to x; then Int cos(ln x) dx=Int cos(ln(exp(2nPi) x)) dx. Now in the right-hand-side we make the change of variable u=exp(2nPi) x, therefore, one gets
Int cos(ln x) dx=exp(-2nPi) Int cos(ln u) du
What's happening?
Either the integral is zero, or not valid for every n integer......
There is an even better way to solve it. By using complex numbers and Euler's identity
The first method is better 👌🏻
I think I prefer the second method.
Would like to have Thai subtitles
Both :)
Great
That second way with the ones is sooo cheating :D
That’s your own work. No instructions say you can’t do that. Nothing cheating about it.
I got it by the second way.
why such complex explanation for integration by parts? Simply write it as ∫e^x d(sinx) = e^x sin x - ∫ sin x d(e^x) dx.
Also why such difficult to understand english diction?
Jesus Christ Is Lord and Is The Only Way To Heaven🙏🏽🙏🏽🙏🏽🙏🏽🙏🏽🙏🏽🙏🏽🙏🏽🙏🏽❤️❤️
#YAY