This is from the 100 integrals part 2. See the full video here kzbin.info/www/bejne/oILdYpqHZ5mCfsU
@lorenzosaudito2 жыл бұрын
I was wondering why you looked so tired, now I understand 😂
@wonghonkongjames4495 Жыл бұрын
YES MU PRIME THINKS OUTSIDE THE BOX ALSO DO THE MERITAVIEN AND MIND YOUR DECISION TOO THEY ARE OUTSTANDINGLY DIFFERENT AND DR PEYAM TOO
@RaKeShCHauHAN28021 Жыл бұрын
I Watch your video from India
@lordofhunger51752 жыл бұрын
We need a tutorial about where to use each pen
@donkeybros87342 жыл бұрын
He has a KZbin short
@geneyoungdho2 жыл бұрын
Characteristic part
@patrickinocencio2305 Жыл бұрын
agree
@mathssucker Жыл бұрын
hahaha I have a tutorial!
@Mehmetbilendegilypo Жыл бұрын
youtube.com/@mehmetturk6583
@weinihao36322 жыл бұрын
This kind of video, where you show your thought process and consider which route to go and even hit a dead end is very very nice as it teaches how to tackle the problem instead of simply presenting a deus ex machina solution.
@maalikserebryakov Жыл бұрын
@julw9138 who asked?
@CurryMuncher2 Жыл бұрын
@@maalikserebryakovhuh?
@LuigiElettrico2 жыл бұрын
Looking at the clock and hearing it being synchronized with my own wall clock makes me feel like I am in the class :D Great integral!
@vatsalmalav4402 жыл бұрын
I like how you say "this guy" making numbers look like living things that make your life easier and many times Hard. This is a great integral you solved I loved it.
@yutaj52962 жыл бұрын
Expressing sin³(𝑥) in terms of sin(3𝑥) and sin(𝑥) using the triple angle formula in the first place seems helps.
@drpeyam2 жыл бұрын
Reminds me of the Borwein integrals a bit
@jamiewalker3292 жыл бұрын
The integral - after using the fact that the integrand is even, using the triple angle formula can be written as 1/8 Im{ integral (e^i3x - 3e^ix)/x^3 dx } where the integral runs from -infinty to infinity. We can analytically continue into the complex plane, separate the two integrals, run a contour along an infinite semi-circle in upper half plane, and a small-semi circle in upper half plate, circulating the singlarities at z = 0 of the function. Using Jordan's lemma to determine that the integral around the large semi-circle is 0, and using Cauchy residue (no poles within contour) means that the integral is equivalent to integrating in the complex plane the above integral around an infinitely small semi-circle, centred at z = 0. The result is 1/8 Im(0.5*2*i*pi*residue at z = 0). The residue, of the above integrand at z = 0 is -3 (which can be quickly checked by expanding the exponential numerator to quadratic term. Plugging this in gives the answer...no Feynman...
@pashaw8380 Жыл бұрын
Indeed.
@chayanaggarwal3431 Жыл бұрын
Yes I did thought of the same way whenever the limits are till infinity with some power of x in denominator I always first try to use the residue theorem
@peamutbubber Жыл бұрын
Except u can do this without any of that, u overcomplicate the simple
@CliffSedge-nu5fv6 ай бұрын
Well, yes, _obviously_ any Calc 1 student already knows how to do that, so why not challenge yourself to trying a different method?
@sngash Жыл бұрын
This is excellent and the video led me to Math 505's generalized version of sin^n(x)/x^n which looks like a great beast for you to work your magic on and possibly make understandable at around calc 2 level :). I struggled following the differentiation portion
@h10r60v8 ай бұрын
14:30 man i know that happiness and you have to experience it atleast once in a lifetime!
@zunaidparker2 жыл бұрын
If you Laplace transform this integral you'll see why the value for the 3rd power is different from the first two powers. Essentially you're doing a convolution, which amounts to taking a moving average over a sliding window of a rectangular function. For the first 2 powers, the window isn't wide enough to affect the value of the average over the moving window, but for the 3rd power, eventually we are averaging zero contributions from outside the rectangle which brings the moving average down. 3blue1brown did an AWESOME video into this: m.kzbin.info/www/bejne/bmaUhmhrbM9pfqc
@sikf2 ай бұрын
Everytime I watch one of your videos, it feels like an emotional rollercoaster.
@abdulmalek1118 Жыл бұрын
Hello ! I hope you see my comment I saw this nice question so that I recommend it The question is : solve the system of equations a = exp (a) . cos (b) b = exp (a) . sin (b) It can be nicely solved by using Lambert W function after letting z = a + ib Hope you the best ... your loyal fan from Syria
@kelvin312724 ай бұрын
Email him!
@ritvikg2 жыл бұрын
10:16 I didn't get this step. Firstly how did he replaced the sin(3tx) with just sin(tx) and in the next step after substituting tx as u, he should get a 't' after integration which he missed as well. It should have been -3πt/8 + 9πt/8 considering his previous step of omitting 3 from the sin is right. Can anyone help me with this.
@ernestschoenmakers81812 жыл бұрын
I didn't get that either, he switched from sin(3tx) to sin(tx), must be a mistake over there and maybe despite of that the result is the same by coincidence.
@digbycrankshaft7572 Жыл бұрын
The first issue is definitely a mistake. With regard to the u=tx substitution this gives dx=du/t. When the substitution is performed it gives integral of sinu/(tx) du and as u=tx this gives integral of sinu/u du with the limits of integration being u=t×0=0 and u=t×infinity=infinity. This then is just the straightforward known integral with u instead of x with the same limits of integration giving pi/2.
@amirbasson532 Жыл бұрын
There was no mistake, The integral from the type: integral from zero to infinity of sin(Ax)/x dx always equal to π/2, because: (Integral from zero to infinity of sin(tx)/x dx) let tx = u x = u/t dx = du/t and then: the integral of sin(u)/(u/t) × du/t equal to: t×sin(u)/u × du/t t and t cancel out (the integral from zero to infinity of sin(u)/u du)= π/2 and because of this: the integral from zero to infinity of sin(3tx)/x = the integral from zero to infinity of sin(tx)/x = π/2 Hope I helped you :)
@digbycrankshaft7572 Жыл бұрын
@@amirbasson532 it was a mistake not making any reference to this fact as it was an assumption which has evidently caused confusion to several people.
@shoto206 Жыл бұрын
@@amirbasson532 ohhhh that explains it, thanks!
@AntimatterBeam8954 Жыл бұрын
I just bought calculus clothing off your store. My weird maths science wardrobe is increasing. I am happier than I was 30 mins ago now.
@omograbi2 жыл бұрын
10:14 shouldn't it be sin(3tx)/x? Or it's anyway the same answer?
@ianfowler93402 жыл бұрын
I would say it's a different answer.
@MarkPaul13162 жыл бұрын
@omograbi gives the same answer, but he could have continued with sin(3tx)/x, making the substitution a = 3tx, arriving at the integral of from 0 to infinity of (sina)/a which gives pi/2.
@yoyoezzijr2 жыл бұрын
its the same answer, integral of sin(tx) / x from 0 to ∞ is π/2, so putting 3t instead of t will be the same
@wolliwolfsen2912 жыл бұрын
Yes, it‘s the same result, but it is confusing
@krisbrandenberger5442 жыл бұрын
Both integrals will have the same exact value. Performing the substitution u=t*x for the first one will imply that 1/x=t/u and dx=(1/t)du, which makes the t's cancel out. Likewise, for the second one, if you let w=3*t*x, that will imply that 1/x=3*t/w and dx=(1/(3*t))dw, which makes the 3*t's cancel out.
@-fai74852 жыл бұрын
Hey sir, Feynman's technique is mad cool but... Where should I set the parameter? Is there any "rule" to follow? I mean, you are supposed to put the parameter on a place in which after deriving, the integral is easier to solve, but it would be marvelous if you have a structured guide that tells you where to put it depending on the situation. It would be great a video like... "When and where to use Feynman's technique" Thanks sir.
@mumilala9940 Жыл бұрын
The alternative way is Fourier transform, split it into (sinx/x)(sin²x/x²) then convolution time!
@md2perpe Жыл бұрын
I used that technique for the integral of sin²x/(x²(1+x²)), seen in kzbin.info/www/bejne/iWaVdZSmpK1kba8
@CDChester2 жыл бұрын
Another integral for the collection!
@vogelvogeltje Жыл бұрын
You have 60hz hum coming from your microphone JSYK. Try to turn your microphone up more without clipping over 0dbFS, or look and see if any wires are crossing over a power wire from your interface.
@DanielCohen-d4v Жыл бұрын
Because sin^3(x)/x^3 is an even func you can write the integral to be 1/2 of the same integral over the real line. Then after you take the d/dt and use the sin^2(x)=1-cos^2(x) you get an odd function over a simatric interval, so it's 0. So you don't need to take the second d/dt you did
@God-ld6ll2 жыл бұрын
use infinity & beyond. works wonders
@fantastic1046 Жыл бұрын
We can get the same result by a sampled function and sampled squared in frequency domain then taking area at freq = 0
@АлександрКузнецов-р6ю Жыл бұрын
bro u are insane pls make more viedos like this, i realy like it
@prollysine Жыл бұрын
Hi bprp, thank you, the complicated calculation can be followed. Yes, Mr. Feynman could not only joke... I will study a lot...
@boomgmr64032 жыл бұрын
At 11:54 you dont write sin3tx again, is that a mistake?
@boomgmr64032 жыл бұрын
dont you then get integral sin3tx/x dx? Does that change something?
@Gamedolf2 жыл бұрын
At 10:50 he says you can have any constant multiple and it will always be pi/2
@djsmeguk2 жыл бұрын
Yes, but also no. The result is always pi/2 so it's not significant.
@boomgmr64032 жыл бұрын
@@Gamedolf I see
@MarkPaul13162 жыл бұрын
@@boomgmr6403 he should have written it as sin3tx and shown that making the substitution u = 3tx would take the integral from 0 to infinity of (sinu)/u which gives pi/2.
@wcottee2 жыл бұрын
Maybe I missed it but at 10:16 how did the second term go from sin(3tx) to just sin(tx)?
@ActALCOCERBONILLAARTUROAZAEL2 жыл бұрын
Same doubt
@noopcode2 жыл бұрын
it was a mistake but the integral is still pi/2
@cristofer68062 жыл бұрын
yeah It should be 3tx but as he already mentioned, the result is π/2 regardless of the input.
@selectname97902 жыл бұрын
I don't think we need to do the u-substitution separately for the sin(3tx) to see that it is also π/2. We can reason directly from the sin(tx)/x integral. Since it's result is π/2 for any t value that means even 3t is a value that works. So t=1,2,'3',4,5,'6',7... work which includes the multiples of 3 ie. t=3,6,9...
@selectname97902 жыл бұрын
@@noopcode but yeah I think he just missed writing the 3
@premdeepkhatri14415 ай бұрын
Thank You for this video.
@maalikserebryakov Жыл бұрын
you can use de moivre angular expansion to write trig^n as a sum of linear trig
@scottleung95872 жыл бұрын
Damn, that's hardcore - nice going!
@mikefochtman71649 ай бұрын
I lost something at 10:16. You got 9/4 int(0,inf) sin(tx)/x. But the previous step it was 3/4 sin(3tx)3x/x^2 ?? Shouldn't the argument to the sin function still be 3tx? How did that 3 disappear?
@kelvin312724 ай бұрын
So here's what I think: the missing 3 was a copying mistake, which luckily turns out not to have any consequence on the answer, since the definite integral of the form sin(ax)/x evaluated between 0 and inf, is always equal to π/2, for all a > 0 (I think!) That means no matter if it is sin(3t) or sin(t) on the top, both integrals are still of the form sin(ax)/x (assuming t> 0 so that a>0, which was indeed stated earlier), and so both integrals are still π/2. If you want to prove that this is true, solve by subbing u=ax, and thus dx=du/a, into the integral of sin(ax)/x evaluated between 0 and inf, simplify, and you'll see it becomes the integral of sin(u)/u from 0 to inf, which is already known to be π/2. Thus the integral in that line where he makes the mistake, with or without the 3, is still equivalent to π/2, and there is no impact on the final answer. I hope this helped!
@josdurkstraful Жыл бұрын
I understood absolutely nothing of all this but still watched the whole video...
@ericknutson83102 ай бұрын
10:22 i believe you miss the 3tx in the argument of the right most sin integrand. however a similar substitution of w = 3tx for the right integrand at this step gives the exact same conclusions.
@fordtimelord8673 Жыл бұрын
I know it’s not the same integral, but I recommend checking out the use of complex contour integration to come up with a general formula for the integral of sin (x^n)/x^n on the same interval. kzbin.info/www/bejne/pafNaGSnpZx2as0
@worldnotworld2 ай бұрын
Fantastic. Need to rehearse my trig identities!
@Master_mind__2352 жыл бұрын
To take out √-1 put e^iπ in the place of -1 Please do it !
@pacifyplayer9 ай бұрын
After you simplified I''(t), where you split the integral into two integrals, how did you get rid of the 3tx inside of the sin function? In the next step, there is just a tx and no 3tx, how can you do this? Can someone explain, please?
@kelvin312724 ай бұрын
So here's what I think: the missing 3 was a copying mistake, which luckily turns out not to have any consequence on the answer, since the definite integral of the form sin(ax)/x evaluated between 0 and inf, is always equal to π/2, for all a > 0 (I think!) That means no matter if it is sin(3t) or sin(t) on the top, both integrals are still of the form sin(ax)/x (assuming t> 0 so that a>0, which was indeed stated earlier), and so both integrals are still π/2. If you want to prove that this is true, solve by subbing u=ax, and thus dx=du/a, into the integral of sin(ax)/x evaluated between 0 and inf, simplify, and you'll see it becomes the integral of sin(u)/u from 0 to inf, which is already known to be π/2. Thus the integral in that line where he makes the mistake, with or without the 3, is still equivalent to π/2, and there is no impact on the final answer. I hope this helped!
@gooball2005 Жыл бұрын
I like to think that he's just in somebody else's office at 12:30 in the morning doing integrals
@imsengky Жыл бұрын
It is so good. I am really happy to see that solution. Thank
@AbhishekSachans2 жыл бұрын
At 10:20, in the sex nd integral, it should be sin(3t).
@PunmasterSTP Жыл бұрын
That was some Feynmania!
@armanavagyan18762 жыл бұрын
I adore your videos i watched the half of 100 integrals)
@Manuel_Gestal Жыл бұрын
10:13 wouldn't it be sin(3tx) instead of sin(tx) ??
@Ron-pe4bp2 ай бұрын
Good catch but kzbin.info/www/bejne/fp7Yip2umLd8jtE So I think it doesn't matter.
@syamantagogoi4 ай бұрын
10:14 In the second integral it should have been Sin(3tx) in numerator and I think it would be difficult to get the final answer in this context.
@melstadevosyan Жыл бұрын
How did sin(3tx) became sin(tx)? Are they equal?
@fadihamed4826 Жыл бұрын
I've the same question also ... that it makes my brain explode
@amtep Жыл бұрын
I think it was a mistake that didn't matter to the answer, because the integral of sin(ax)/x dx is the same for any nonzero a
@ee-prakalyadav Жыл бұрын
I solve these question in my paper . But your techniques are quite impressive
@rogerdudra1782 жыл бұрын
Greetings from the BIG SKY. Nothing like a bit of calc to end the day.
@boombam5589 Жыл бұрын
7:25 Evil laughter 🤣
@abdulmalek1118 Жыл бұрын
I have found a pretty nice question that I will suggest Solve the system a = exp (a) . cos (b) b= exp (a) . sin (b) Can be solved easily using Lambert "W" function by computing ( a+ib ) And thanks
@frederickwong4390 Жыл бұрын
I think using the triple angle formula sin(3x)=3sin(x)-4(sin(x))^3 is easier. Unlike what other have said, there is no need to use contour integration.
@scienceresearchwithishan6965 Жыл бұрын
Bro u should also make a video doing this from contour integration using residue theorem and Jordan's lemma 😁😁
@danmart18792 жыл бұрын
I was lost for most of the video !! I have a long way to go.
@yabaminozomi Жыл бұрын
10:14 why the sin (3tx) suddenly becomes sin(tx)
@Ron-pe4bp2 ай бұрын
Good catch but kzbin.info/www/bejne/fp7Yip2umLd8jtE So I think it doesn't matter.
@holyshit922 Жыл бұрын
Integrate by parts twice with D I sin^3(x) 1/x^3 then substitute u=3x finally i got 3/4Int(sin(x)/x,x=0..infinity) then i calculated Laplace transform L(sin(t)/t) and plugged in s = 0
@jeanmaxcoransoni2183 Жыл бұрын
At 10:14 : error sin(3tx) not sin(tx)
@tahsintarif6864 Жыл бұрын
make a video on solving 100 Putnam Calc 2 Problems
@maalikserebryakov Жыл бұрын
5:20 this is the main weakness in your approach ive noticed When you arrive at two viable techniques to modify the integration you force yourself to choose. Just do both!
@ayoubelouafy61742 жыл бұрын
There's a mistake in the 2nd derivative of I(t) in the 2nd line you got sin(3tx) in the 2nd term. All respect to u it's a hard integral .
@jatingupta6198 Жыл бұрын
I didn't understand what u did but i would have used product rule of integration using ILATE😅
@khoozu78022 жыл бұрын
A fastest way is applying the formula sin^3(x)=1/4(3sinx-sin3x) And using integrate by parts
@dipankarbanerjee1130 Жыл бұрын
Actually what is Feynman's trick ? I am a high school student and this isn't in my syllabus but I am eager to know
@KingGisInDaHouse Жыл бұрын
Wouldn’t complex analysis work here?
@uncelesteperro8258 Жыл бұрын
10:12 how did he get rid of the 3 on sine's angle?
@Ron-pe4bp2 ай бұрын
Good catch but kzbin.info/www/bejne/fp7Yip2umLd8jtE So I think it doesn't matter.
@darkknight32920 Жыл бұрын
Sorry for the naive question, but when solving for c, what if you let t equal any multiple of pi? Wouldn't that change what c is? Why is it possible to choose the "easiest" value?
@aadisankar.s4449 Жыл бұрын
Sir, please explain why there exists two types of vector products...
@ES-qe1nh Жыл бұрын
There's three, actually
@lilsourmango Жыл бұрын
At 6:20 why do we have cos(tx)-cos(2tx)?Where does the minus come from? And why do we have a plus afterwards?
@jackychan4640 Жыл бұрын
我想祝福你新年快樂happy Lunar New Year
@yokoyapen2 жыл бұрын
9:32 the 2 appears like magic
@mcalkis57712 жыл бұрын
After the 100-x series I am surprised you want to even SEE another integral ever again.
@wydadiyoun2 жыл бұрын
10:55 proof pleaaaaaaaaaaaaaaaaaaaaaaaaaase! why it always give pi/2 with any constant???
@wydadiyoun2 жыл бұрын
ok nevermind, I figured it out with my effort
@leonig1002 ай бұрын
at 10.19 sin(3tx) becomes sin(tx) after the separation of the 2 integrals. Is this correct?
@INSANITY335 Жыл бұрын
we can even use sin3theta here right?
@Sbombaflex Жыл бұрын
sin(3tx) right high corner , where did he go?
@romanbykov59222 жыл бұрын
9:45 why did you differentiate the numerator here but didn't you differentiate the denominator of x^2?
@PaoloCasillo2 жыл бұрын
Because x^2 is a constant in t world.
@arsh.008 Жыл бұрын
On the right half of the board, the second step where you took negative common, shouldn't it be -(9/4) and so on?
@amtep Жыл бұрын
No he split the expression at the +, so the left half is for -sin(tx)•x and the right half is for sin(3tx)•3x
@manishkumardeep2230 Жыл бұрын
PLEASE MAKE A VIDEO ON FORBENIUS METHOD OF SPECIAL FUNCTIONS
@Mini_Wolf.6 ай бұрын
Can you do the taylor series of sin and work from there?
@bisheshshakya3838 Жыл бұрын
10:13 If I'm not mistaken, it's supposed to be sin(3tx) but you wrote sin(tx)....please clarify?
@EntropicNightmare2 жыл бұрын
When you do the u substitution when computing I''(t), you get constant 3pi/4, which is fine for all t>0, but how do you justify that at t=0 where the integral appears to give zero when you go to fix your constants of integration?
@physicsmath8293 Жыл бұрын
14:42 sin(3tX) ---》 sin(tX) ?!!! I have a problem here 😕
@thelittlesillystar Жыл бұрын
what if d/dx((sinh(x*ln(x)))^(2x*cos(x)))
@alibekturashev62512 жыл бұрын
i have never seen you that tired🥺
@nickharrison3748 Жыл бұрын
what is the practical use of this equation?
@fadiel-riachi66752 жыл бұрын
I am confused by the evaluation of I'(0) and I(0) at 12:14 and 13:26 respectively. Shouldn't (sin(x)/x)^n be equal to 1 at x=0 for positive integers n? Is there something I am missing? In order to have a nice value for I'(t), we need cos(tx) to be 0. In other words, we need to evaluate at tx= pi(k-1)/2, which changes the expression a lot and affects the next integration that finds I(t).
@amtep Жыл бұрын
If t = 0 then the sin(tx) term is 0 and the whole expression goes to 0. Remember that I is a function over t not over x, so x = 0 doesn't need to be considered anywhere.
@fadiel-riachi6675 Жыл бұрын
@@amtep Right, of course! Thank you!
@emmagutielmejor2 жыл бұрын
Borwein integrals ?
@tarentinobg Жыл бұрын
I'd use brackets [ ] instead of multiple parentheses. Love your work. Thank you.
@ThreePointOneFou Жыл бұрын
So would I. Unfortunately, the old rule about surrounding parentheses with brackets (and brackets with braces) is becoming increasingly out of vogue. Current standards of mathematical notation lean toward using all parentheses, even though it can obscure nested expressions.
@غازيالغزوان Жыл бұрын
Can you solve the integral of : ln(sinx+cosx)/(cosx-sinx) dx
@Johnny-tw5pr2 жыл бұрын
Would this work? I(t)=(integral)sin^tx/x^t
@crescenzosimeolisimeoli8756 Жыл бұрын
Is this integral generalizable for n?
@krishnankuttyp4478 Жыл бұрын
Sin3xt/x put 3xt=u x=u/3t 3tdx=du dx=du/3t integral sin3xt/xdx=(sinu×du/3t)/u/3t =integral sinu/udu =pie/2
@alankuo5579 Жыл бұрын
What about Nth power
@GauravKumar-vl7rt2 жыл бұрын
Love from India
@nirvikthapa94362 жыл бұрын
Me trying to figure out is it 12 Am or Pm in the clock
@saurabhkatiyar2704 Жыл бұрын
Very very easy question
@mrwest9840 Жыл бұрын
How can we write d/dx( sin^3x) = 3sin^2x cosx ?????? 😏
@SakshamSiwa4 ай бұрын
Bro what is your educational qualifications?
@stevenowensweanxen24062 жыл бұрын
Sum of n=1 to infinity of 1/x^x =..?
@krishnanadityan201710 ай бұрын
The second term in I'''(t) expression on RHS is not correct
@shubhadipmondal7868 Жыл бұрын
I noticed the clock and found that the problem took 22 mins (12:07 to 12:29)approx to explain and solve...by looking at the wall clock 😅...while video editing cut it short to 14mins+😂.. results of Watching Sherlock Holmes 😂
@Wout680 Жыл бұрын
Hey blackpenredpen, I can't figure it out, but why is x^(1/log_b(x)) equal to b (the base of the logarithm)?
@spaghetti1383 Жыл бұрын
Assume that identity is true. Then take log base b on both sides. Each side simplifies to 1. Logarithms are increasing and 1=1 so the identity must be true.
@Wout680 Жыл бұрын
@@spaghetti1383 That was a very clear explanation, thank you :)