Masterfully presented without any pretense - allowing his quiet, dignified charisma to speak for itself. Few are lucky enough to get a teacher like this and how desperately they are needed. Especially in presenting math on this level and showing such intuitive reasoning and passion for the subject. When I had a go I ignored the fact that it's a definite integral and ended in a mess.
Solution before watching: I = int_0^(π/2) sin^k x / (sin^k x + cos^k x) dx Substitute x -> π/2-x, swapping cos and sin: I = int_0^(π/2) cos^k x / (sin^k x + cos^k x) dx Adding both together gives 2I = int_0^(π/2) (sin^k x + cos^k x) / (sin^k x + cos^k x) dx = int_0^(π/2) 1 dx = π/2 Therefore, I = π/4
That is pretty neat, for someone who dont know that method can leverage the conditions Since the problem states for all integers k , we can try and replace k with -k, this should not change the solution since we get the answer in terms of -k if it shud depend on that, then I = (sinx)^-k/((sinx)^-k + (cosx)^-k) dx After simplication it reduces to I = (cosx)^k/((sinx)^k + (cosx)^k) dx Add both, 2*I = 1 dx I = pi/4 which is the same thing as the substitution method!
I just divided the top and bottom by (cosx)^k and then used tan^2x + 1 = sec^2x to arrive at the integral of sin^2x from 0 to pi/2 and got the same answer. I am wondering if this is fine because at pi/2 cosx equals 0 so I probably should not be dividing by it, but on the other hand I get an integral in terms of sin which is clearly defined within the limits of the integral. Thank you for this video though.
I like the way you explored special cases to come up with a hypothesis. Actually, the identity is true not just for all integers, but at least for all real numbers greater than or equal to one. The exponent doesn't matter. You can come up with even more pi/4-integrals by dividing top and bottom of the integrand by sin^k(x) or cos^k(x), respectively.
Nice ! I really didn't expect that the result is independent of integer k ! But, really, have k to be an integer ? Could k be any real number ? Could k be any complex number or anything ?
This is very nice. Though, I think when k = 0 and x -> 0, (sin x)^k should be an interdeterminate form. So, do we still get 1 as the limit. I checked Wolfram Alpha and yes we did but I'm not sure why. PS: I'm not in touch with my Mathematical side so can't figure it out that easily myself.
ATTEMPT: Tried, for sanity check, the k = 0, 1, and 2 cases. Every time I got back out pi/4. When I hit k = 3 I initially tried to do something recursive, expressing them in terms of previous integrals. But that wouldn't work because there's a nasty sum in the denominator. And then it hit me like a truck. King's property. Of course, how could I have missed it? It's so painfully obvious! With that, I now had 2I = the integral of 1 dx from 0 to pi / 2. Should be clear that I = pi / 4 for all integer cases of k (maybe even all real cases too if the integral doesn't diverge)
Hello sir Can you solve this questions integrate (sin x + cos x)/(9 + 16sin 2x) dx from 0 to pi/4 This type of questions asked my exam just few days l gave jee main exam these questions was asked
I looked at the problem for a bit and tought why not factor sin^k(x) so we get the integral of 1/(1+ctg^2(x)) if u=ctg(x) then du=-(sin^2x+cos^2(x))/sin^2(x)dx du=-(1+u^2)dx becomes -1/(1+u^k)*1/(1+u^2) I think for small values of k we can solve the integral easily and try mathematical induction later
Make a video showing all integers solutions for n which (3n+16) / (4-n) = k² | k € Q I made this question by myself and the solutions for n are n = -5, -3, 0, 3 I even put the equation on a C compiler and these are the solutions fr
Masterfully presented without any pretense - allowing his quiet, dignified charisma to speak for itself. Few are lucky enough to get a teacher like this and how desperately they are needed. Especially in presenting math on this level and showing such intuitive reasoning and passion for the subject. When I had a go I ignored the fact that it's a definite integral and ended in a mess.
I used kings rule and immediately got it in my head the moment I saw the thumbnail.
Same
I Love your teaching style！
Solution before watching:
I = int_0^(π/2) sin^k x / (sin^k x + cos^k x) dx
Substitute x -> π/2-x, swapping cos and sin:
I = int_0^(π/2) cos^k x / (sin^k x + cos^k x) dx
Adding both together gives
2I = int_0^(π/2) (sin^k x + cos^k x) / (sin^k x + cos^k x) dx
= int_0^(π/2) 1 dx
= π/2
Therefore, I = π/4
@@kappasphere This is precisely what is done in the video.
never stop learning ! much love from Iran
I dont know any of this it just came to my recomendations and acidently pressed😭
the magic is done ! Again. Congrats & Standing Applause !
You should do more videos about combinatorics , especially teach people about derangements!
That is pretty neat, for someone who dont know that method can leverage the conditions
Since the problem states for all integers k , we can try and replace k with -k, this should not change the solution since we get the answer in terms of -k if it shud depend on that, then
I = (sinx)^-k/((sinx)^-k + (cosx)^-k) dx
After simplication it reduces to
I = (cosx)^k/((sinx)^k + (cosx)^k) dx
Add both,
2*I = 1 dx
I = pi/4
which is the same thing as the substitution method!
Here you assume that I(k)=I(-k) so that their sum gives 2I. And why is it so?
I just divided the top and bottom by (cosx)^k and then used tan^2x + 1 = sec^2x to arrive at the integral of sin^2x from 0 to pi/2 and got the same answer. I am wondering if this is fine because at pi/2 cosx equals 0 so I probably should not be dividing by it, but on the other hand I get an integral in terms of sin which is clearly defined within the limits of the integral. Thank you for this video though.
I like the way you explored special cases to come up with a hypothesis.
Actually, the identity is true not just for all integers, but at least for all real numbers greater than or equal to one. The exponent doesn't matter.
You can come up with even more pi/4-integrals by dividing top and bottom of the integrand by sin^k(x) or cos^k(x), respectively.
Seen a question very similar to this during my Cambridge STEP preparation. The substitution of pi/2 -x is so useful for trig integrals
Thank you, teacher!
I will never stop learning.
Is it possible to proof this, not by the mentioned trick (very nice by the way) but by investigating the graphics of the function?
You have so many tricks up your sleeve!
Nice ! I really didn't expect that the result is independent of integer k !
But, really, have k to be an integer ? Could k be any real number ? Could k be any complex number or anything ?
By the way, the trick is called the "king's property" or "king's rule"
When k is an integer.
@@RyanLewis-Johnson-wq6xs Never used, in fact. k can be any real number.
Great! Enjoied your viedo very much! And a wonderful unexpected result!
Beautiful!! Greetings from Italy!!🎉🎉🎉
Very relevant ❤ second to watch my fellow Mathematician ❤
King's rule 💘
This is very nice. Though, I think when k = 0 and x -> 0, (sin x)^k should be an interdeterminate form. So, do we still get 1 as the limit. I checked Wolfram Alpha and yes we did but I'm not sure why.
PS: I'm not in touch with my Mathematical side so can't figure it out that easily myself.
Very simple. For 0
Great solution . Thanks !
I told you , ...Newton, you...are very smart Teachers. Thanks!
What is if k=1 it will the same answer pi/4 can you please give a hint
@primenewtons just so u know there is a huge market in india for this kind of concepts ...so u got a nice escuse to learn learning hindi now
Lol 😆. Hindi?
ATTEMPT:
Tried, for sanity check, the k = 0, 1, and 2 cases. Every time I got back out pi/4.
When I hit k = 3 I initially tried to do something recursive, expressing them in terms of previous integrals. But that wouldn't work because there's a nasty sum in the denominator.
And then it hit me like a truck. King's property. Of course, how could I have missed it? It's so painfully obvious!
With that, I now had 2I = the integral of 1 dx from 0 to pi / 2. Should be clear that I = pi / 4 for all integer cases of k (maybe even all real cases too if the integral doesn't diverge)
Excellent video!
Hats off to you! But still keep YOUR hat on 🙂
Beautiful❤
Spettacolo❤
I do not fully agree with you because sin and cos can be zero so 0^0 is indeterminate….the first part of this problem is already inconsistent
Thanks Sir 👍
I know the rule and have intuitive understanding.
Will try to prove it by myself.
G
very nice property
Hello sir
Can you solve this questions integrate (sin x + cos x)/(9 + 16sin 2x) dx from 0 to pi/4
This type of questions asked my exam just few days l gave jee main exam these questions was asked
I sent you proposition for video
I gave you product and sum
and I claim that for all natural n they are equal
Thanks. I will look at it. I need to get a rhythm for now.
Very interesting!
Great great exercise, very interesting love you !!!
🙏👏
pi/4
I looked at the problem for a bit and tought why not factor sin^k(x) so we get the integral of
1/(1+ctg^2(x)) if u=ctg(x) then du=-(sin^2x+cos^2(x))/sin^2(x)dx
du=-(1+u^2)dx becomes
-1/(1+u^k)*1/(1+u^2)
I think for small values of k we can solve the integral easily and try mathematical induction later
That's really nice!
bro you are the best
Make a video showing all integers solutions for n which (3n+16) / (4-n) = k² | k € Q
I made this question by myself and the solutions for n are n = -5, -3, 0, 3
I even put the equation on a C compiler and these are the solutions fr
Integrate[((Sin[x])^k)/((Sin[x])^k+(Cos[x])^k),{x,0,π/2}]=π/4
@@RyanLewis-Johnson-wq6xs And where is the calculation?
Ok, cool!
Nice
Yes but when do i use them in real life rather than in abstract math?
@@shomshay Abstract Maths is a part of a real life. Such formulas appear in fluid dynamics, for instance.
best
nice
No, i wont! lmaooo jk jk great video!!
Already knew it before seeing the video