The -1 we obtained from integration by parts is evaluated from pi/4 to pi/2, but evaluating the "function" -1 at everywhere is -1, so they cancel down to 0. That's probably why dropping it gave us the correct result
There is a simpler way to solve using only 2 u-subs. first step is let x = 1/u, dx= -1/u^2 , then after inserting everything, you can simplify and factor stuff. Then you can see that you can do another u-sub on the factored (arctan(1/u) +u) in the denominator by letting v = (arctan(1/u) + u), therefore dv = u^2/(u^2+1) du, but because of how stuff got factored, it cancels out and from there you will get the the result you want after you integrate and evaluate
Hi great solution I do have a question though, I understood everything you did in the video except in the first substitution I didn't understand why you changed the intervals on the integral, I know how you got the numbers them self but not why you changed them, please explain. Btw in that first substitution instead of using your logic of the derivative of arctan being this expression you see (x^2+1) and using rules of trig sub x=1tan(t) I know it's the same just wanted to point that out. Again amazing video definitely an interesting integral😄
When doing in an integral what you're actually finding is the area under a curve. So when you have a bounded integral which is the one with numbers and the one that we have got, essentially the entire thing just represents a number, and it's our job to work out what that number is. When doing a substitution you are fundamentally changing what the curve that you are finding the area under, looks like. Hence we get something that looks different inside the integral. Changing the numbers (bounds) of the integral is essential here because the area from A to B under one curve, could be completely different to another curve from the same places A to B. So what changing the numbers (bounds) does is it ensures that our new curve will give us the same value for the area. Think, the area of a square will be different to the area of a rectangle, however if you change where you look in the rectangle (how far and how high up you look), you would still be able to find the area of the square. Hope this helped!
@@Jagoalexander I understand now, so for every substitution in a bounded integral the new bounds are inputting the previous bounds of the integral into the function x or any variable. So if I have an integral of dx with the bounds 0 and 5, and x=sint Then the new bounds are: sin0, sin5 Am I right?
Bro, why did you say anything about cotangent function being weird while you use cosecans and probably scans, which are literally just as useless as cotangent and you go back to the definition of them at some point either way? I really don't understand why in the US you use those functions
Thank you, and oo well spotted, I shall investigate…. What’s peculiar is the solution is correct and I’ve been double checking on my calculator as well as wolfram alpha, I’ll figure out where that -1 has gone!
What did people think about the music in this video? Yay or Nay?
I went for Clair De Lune, then Shostakovich’s Piano Concerto No.2 Andante
Yay
yay!!
Yay
Yay
:)
yay but maybe use some pokemon music too!
I'd be more impressed if somebody could find a physical system that can be described by this equation
The -1 we obtained from integration by parts is evaluated from pi/4 to pi/2, but evaluating the "function" -1 at everywhere is -1, so they cancel down to 0. That's probably why dropping it gave us the correct result
Yes exactly. A constant evaluated between two limits is zero.
There is a simpler way to solve using only 2 u-subs. first step is let x = 1/u, dx= -1/u^2 , then after inserting everything, you can simplify and factor stuff. Then you can see that you can do another u-sub on the factored (arctan(1/u) +u) in the denominator by letting v = (arctan(1/u) + u), therefore dv = u^2/(u^2+1) du, but because of how stuff got factored, it cancels out and from there you will get the the result you want after you integrate and evaluate
You dropped the -1, didn’t you? Otherwise, great job.
This has been pointed out by someone else too, I shall investigate… what is funny is my answer is correct without the -1? 🤔
And thank you :)
Very underrated channel!
love the videos
Thanks mate 👍
Thank you so much!!
Great video, it looked impossible at first 😂
For the third time Finally!
It kept getting copyright strikes due to a 200 year old piece of music in the background. It’s sorted now… I think 😅
Hi great solution I do have a question though, I understood everything you did in the video except in the first substitution I didn't understand why you changed the intervals on the integral, I know how you got the numbers them self but not why you changed them, please explain. Btw in that first substitution instead of using your logic of the derivative of arctan being this expression you see (x^2+1) and using rules of trig sub x=1tan(t) I know it's the same just wanted to point that out.
Again amazing video definitely an interesting integral😄
When doing in an integral what you're actually finding is the area under a curve. So when you have a bounded integral which is the one with numbers and the one that we have got, essentially the entire thing just represents a number, and it's our job to work out what that number is.
When doing a substitution you are fundamentally changing what the curve that you are finding the area under, looks like. Hence we get something that looks different inside the integral. Changing the numbers (bounds) of the integral is essential here because the area from A to B under one curve, could be completely different to another curve from the same places A to B.
So what changing the numbers (bounds) does is it ensures that our new curve will give us the same value for the area.
Think, the area of a square will be different to the area of a rectangle, however if you change where you look in the rectangle (how far and how high up you look), you would still be able to find the area of the square.
Hope this helped!
@@Jagoalexander I understand now, so for every substitution in a bounded integral the new bounds are inputting the previous bounds of the integral into the function x or any variable.
So if I have an integral of dx with the bounds 0 and 5, and x=sint
Then the new bounds are: sin0, sin5
Am I right?
@gabyshavit8539 yes precisely, however when doing a substitution always use a new letter, the most common are u and t
@gabyshavit8539 so you would say u = sinx for example, also your substitution should be something you see in the integral most of the time
@@Jagoalexander yeah I know I typed wrong its x=sint for example, thank you for the answer! 🙂
Amazing!!!
i thought of factoring out an x and x^3 and using partial fractions😂😂
I'm sure there are multiple ways of doing it! However you would run into the issue of the Arctans, you can't do much with those
Great vid big bro
Thank you :) ❤
You should correct the lower limit 0 in the integral in the title (this would not converge ).
Thank you, sorted now!
Bro, why did you say anything about cotangent function being weird while you use cosecans and probably scans, which are literally just as useless as cotangent and you go back to the definition of them at some point either way? I really don't understand why in the US you use those functions
In what world do I sound like I am from the US? I have an accent as British as it gets. Also what's your point exactly?
What about the -1 🤔 great job btw
Thank you, and oo well spotted, I shall investigate…. What’s peculiar is the solution is correct and I’ve been double checking on my calculator as well as wolfram alpha, I’ll figure out where that -1 has gone!
Why did you write 'math' in the title? 👎👎👎
Literally most of my subscribers are from the US
Also ain’t that deep brother