We present a nice trick involving the integral of trigonometric functions. Playlist: • Interesting Integrals www.michael-penn.net www.randolphcollege.edu/mathem...
That lemma is very useful for some integrals, and its a case of a general property: integral from a to b f(x)dx= integral of a to b f(a+b-x)dx. The proof is quite simple: consider the integral from a to b f(a+b-x). let u=a+b-x. The limits become b to a , but then du = -dx, so using the negative sign , the limits become a to b again: and so we end up with int(a to b) f(u)du!! I find this sort of "symmetry" of a+b-x to give a and b again quite nice!!
Yeahh Indian refers this property of integrals as King's property. As We are told to just use this property inside your head and see whether it generates a useful result or not.if not then and then proceeding to other methods of evaluation.
The derivation was one of the first parts in a STEP II question (the next three parts being three integrals which exploited this trick in different ways).
Awesome !!! Congrats !! :-) Thank you very much for this lemma... I teach to make a polynomial division for making the expansion of P(x)/Q(x) when degree(P(x))>=degree(Q(x)) (Improper Fraction). This, splits into a C(x) + R(x)/Q(x), where C(x) is de quotient and R(x) is the reminder. The interest of that is the new fraction R(x)/Q(x) is a proper fraction (degree(R(x))
sin(pi-u) = sin(u) doesn't need a big explanation, you can simply combine the two simple identities sin(x+pi) = -sin(x) and sin(-x)=-sin(x) and therefore, sin(pi-u)=-sin(-u)=sin(u)
Integral from 0 to Pi of x*f(sin(x)) = Integral from 0 to Pi of (x - Pi/2)*f(sin(x)) plus Pi/2 * Integral from 0 to Pi of f(sin(x)). Since (x - Pi/2) is antisymmetric around x=Pi/2 while sin(x) and, therefore, f(sin(x)) is symmetric around x=Pi/2 we conclude that Integral from concludes
Hi Michael. Very nice proof. My question is, is there maybe a similar trick but when the integral goes from 0 to pi/2? This trick that spu proved is commonly used on my university, but also it is frequent to see integral from 0 to pi/2 xf(sinx).
As was stated earlier, there is a property that the integral from a to b of f(x)dxis equal to the integral from a to b of f(a+b-x)dx. The lemma in the video is a special case of this property.
I'm currently brooding over primitive functions to logs of trigonometric functions. Maybe this trick can be used to shed some light over that problem? I'm thinking partial integration here.
It also works if the representative curve of f(x), let’s call it (C), admits an axis of symmetry x=a. Then we will have: integral of xf(x) from a-x0 to a+x0 is equal to 2a(integral of f(x) from a-x0 to a+x0). Its application on the sine function is a corollary.
Since we're dealing with a definite integral, the integral will have a certain value which can be figured out (we don't need to know the exact value tho). We can change the u to an x here simply because it doesnt matter whether the variable x runs from 0 to pi or the variable u does. If the functions is the same and the bounds are identical, the variable's name doesnt matter since the value of the integral wont change. Hope this helps
This is epic 😆 but it would more beautiful if the integral were 2+2x instead 1+2x because we would get I = pi² - 4 which looks more satisfying then (pi+1)(pi-2) I suppose 🤔
He didn't say "and that's a good place to stop" :(
That lemma is very useful for some integrals, and its a case of a general property: integral from a to b f(x)dx= integral of a to b f(a+b-x)dx. The proof is quite simple: consider the integral from a to b f(a+b-x). let u=a+b-x. The limits become b to a , but then du = -dx, so using the negative sign , the limits become a to b again: and so we end up with int(a to b) f(u)du!! I find this sort of "symmetry" of a+b-x to give a and b again quite nice!!
Thanks for that, I can't believe no one ever taught me that (and also that I never noticed it on my own)!
Yeahh Indian refers this property of integrals as King's property. As We are told to just use this property inside your head and see whether it generates a useful result or not.if not then and then proceeding to other methods of evaluation.
Thanks
who knows how many times i watched the moment when he checked the box...
The derivation was one of the first parts in a STEP II question (the next three parts being three integrals which exploited this trick in different ways).
The STEP exams certainly have some interesting questions ... it'd be cool to see Michael do a few on the channel
Please, make more videos, Michael!
Awesome !!! Congrats !! :-) Thank you very much for this lemma... I teach to make a polynomial division for making the expansion of P(x)/Q(x) when degree(P(x))>=degree(Q(x)) (Improper Fraction). This, splits into a C(x) + R(x)/Q(x), where C(x) is de quotient and R(x) is the reminder. The interest of that is the new fraction R(x)/Q(x) is a proper fraction (degree(R(x))
sin(pi-u) = sin(u) doesn't need a big explanation, you can simply combine the two simple identities sin(x+pi) = -sin(x) and sin(-x)=-sin(x) and therefore, sin(pi-u)=-sin(-u)=sin(u)
I keep thinking that the mic is a bug on my phone, amazing vid btw
Integral from 0 to Pi of x*f(sin(x)) = Integral from 0 to Pi of (x - Pi/2)*f(sin(x)) plus
Pi/2 * Integral from 0 to Pi of f(sin(x)). Since (x - Pi/2) is antisymmetric around x=Pi/2 while sin(x) and, therefore, f(sin(x)) is symmetric around x=Pi/2 we conclude that
Integral from concludes
9:57 that look of disappointment
Hi Michael. Very nice proof. My question is, is there maybe a similar trick but when the integral goes from 0 to pi/2? This trick that spu proved is commonly used on my university, but also it is frequent to see integral from 0 to pi/2 xf(sinx).
As was stated earlier, there is a property that the integral from a to b of f(x)dxis equal to the integral from a to b of f(a+b-x)dx. The lemma in the video is a special case of this property.
I don't think there is such a formula as the form of π/2. Because the proof of this relays heavily on the symmetry of trigonometric identity.
I'm currently brooding over primitive functions to logs of trigonometric functions. Maybe this trick can be used to shed some light over that problem? I'm thinking partial integration here.
It also works if the representative curve of f(x), let’s call it (C), admits an axis of symmetry x=a.
Then we will have: integral of xf(x) from a-x0 to a+x0 is equal to 2a(integral of f(x) from a-x0 to a+x0).
Its application on the sine function is a corollary.
This is just as neat as the Laplace Transition.
10:00 why are we still here, just to suffer
XD
I looked for a comment like this as soon as I saw that haha
👍👍
arctan(1) = pi/4 not pi/2
Why is it that we don't need to do double return in terms of variables? "new X" -> u -> original X? (In the Lema I mean)
Since we're dealing with a definite integral, the integral will have a certain value which can be figured out (we don't need to know the exact value tho).
We can change the u to an x here simply because it doesnt matter whether the variable x runs from 0 to pi or the variable u does.
If the functions is the same and the bounds are identical, the variable's name doesnt matter since the value of the integral wont change.
Hope this helps
This means that you can rename the main variable of any definite integral.
@@reeeeeplease1178 Yes, this helped, thank you.
Simple king property
This is epic 😆 but it would more beautiful if the integral were 2+2x instead 1+2x because we would get I = pi² - 4 which looks more satisfying then (pi+1)(pi-2) I suppose 🤔
dédicace a M.fiszka
You remind me of Sheldon Cooper from The Big Bang Theory, I don't know why... Nice vid!!