Пікірлер

• 
  @VaghavVarnaСағат бұрын

  this is in A DAS gupta

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  @owl3math20 минут бұрын

  Hi. Is that a book? I know a few books from Vikas Gupta

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  @adandap12 сағат бұрын

  Ah... my previous comment was made too early. 😂

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  @owl3math2 сағат бұрын

  🤣 yes I figured this was what you were referring to :)

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  @owl3math16 сағат бұрын

  Here's the video with the other method: kzread.info/dash/bejne/i3h8l6V8Yb3Mkdo.html

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  @nezby394519 сағат бұрын

  Wow you’re awesome! Perfect explanation I always thought mit bee integrals are waaay too difficult for me, but this problem is pretty nice. Will you be making a video about integrating using partial fractions? I can never integrate more difficult problems when partial fractions are involved. Still thank you for this awesome video!

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  @owl3math18 сағат бұрын

  Hey Nezby. Thanks! I’ve done some videos on partial fractions before. Have you watched my other channel “Owls School of Math”? I think there’s quite a few over there but let me know if you need some help finding it

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  @owl3math18 сағат бұрын

  youtube.com/@owlsschoolofmath9732

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  @iehudimКүн бұрын

  makes me sad

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  @owl3math20 сағат бұрын

  Hello. This problem makes you sad?

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  @iehudim20 сағат бұрын

  @@owl3math Sad because i dont know How to solve It, till now

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  @owl3math16 сағат бұрын

  @@iehudim ok good news! :)

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  @usualhacker2584Күн бұрын

  in india we use this standard equation if integral a to b f(x) + integral f(a) to f(b) f inverse (x) is given then the answer is bf(b)-af(a)

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  @owl3mathКүн бұрын

  Oh that’s interesting. Don’t think I’ve seen the formula like that before. 👍 thanks

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  @user-lg6fq1yt4g2 күн бұрын

  Could, we solve this problem by using integration?

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  @owl3mathКүн бұрын

  For the series convergence yes you can use the integral test here. But that does not provide an exact solution to the series

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  @user-lg6fq1yt4gКүн бұрын

  @@owl3math 🙏🏽🙌🏽

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  @felipefred12792 күн бұрын

  I really don’t remember those hyperbolic identities

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  @owl3mathКүн бұрын

  Right! Many people have this issue. Use my trig cheat sheet! owlsmath.neocities.org/Trig%20Identities%20and%20Formulas/trig%20identities Free plug for me 😄

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  @owl3math2 күн бұрын

  Alternate method video: kzread.info/dash/bejne/f4ap0qx8aaauZNI.html

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  @ryanng19054 күн бұрын

  Beta function is so funny because every video that uses it is using a different representation each time

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  @owl3math4 күн бұрын

  That’s true! 4 or more formulas for the same thing makes it hard to remember

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  @theupson4 күн бұрын

  u= 2019(t^-2)/4 is a higher-odds thing to try- you can easily see it will give you a gamma integral, whereas your downfield vision has to be much sharper to see OPs approach will yield the gaussian integral- as opposed to something similar but intractable.

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  @owl3math4 күн бұрын

  ooooh interesting!

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  @hemavathirajesh80124 күн бұрын

  The integral is very good, but i tried to do by parts, by which i got 0. Nonetheless, thanks for the good integral !!

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  @owl3math4 күн бұрын

  Nice thanks!

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  @felipefred12795 күн бұрын

  You make it so simple

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  @owl3math4 күн бұрын

  Thanks Felipe!

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  @adandap5 күн бұрын

  Cool. I wouldn't have thought to do that. My solution was to write it as Re[ Int( x e^(- x (1 - i) ) dx)] and then IBP is easier.

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  @owl3math5 күн бұрын

  Ah complex method is nice here. Then it’s just like the IBP of 2 things 👍

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  @VeteranVandal3 күн бұрын

  The method I thought was this plus the Feynman trick.

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  @dkravitz785 күн бұрын

  What you could have done instead is Feynman's trick, you end up doing almost exactly the same calculation but it is less rules/machinery. Set F(s) = integral ( exp(-sx) cos(x) , x=0..infinity) and assume s>0. F'(s) = - integral( x exp(-sx) cos(x) ) and so we want F'(1) F(s) is an easier integral, it is 1/(s^2+1) times exp(-sx) times (sin(x) - s*cos(x) ) From x=0 to x=infinity (assuming s>0) gives -s/(s^2+1) Derivative of this with respect to s is F'(s) = (s^2-1) / (s^2+1)^2 And so we quickly see F'(1)=0. Similar to your path but less "rules" needed.

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  @owl3math5 күн бұрын

  Nice! Yep Feynman works nice here👍👍

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  @MikeT101015 күн бұрын

  Nice job!...and...it's ZERO! ; )

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  @owl3math5 күн бұрын

  Good timing Mike! You posted like just as I was switching pages and it popped right up 🤣

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  @antoinehedin66085 күн бұрын

  Maybe a little bit more efficient than the classical way I initially chose, implying 4 consecutive IBP ^^.

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  @owl3math5 күн бұрын

  Yes I did one previously like that and it’s fine but it’s a lot of IBP! 😂

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  @koennako21955 күн бұрын

  You could also do this one without doing L transform, but I hate it. Mine aswell start memorizing L transform if I want to do these competition integrals.

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  @owl3math5 күн бұрын

  I guess if it were an indefinite integral we’re forced into the IBP but I do like to avoid it in this case where we have 3 things. 👍

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  @MartinPerez-oz1nk5 күн бұрын

  EXCELLENT !!!!!

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  @owl3math5 күн бұрын

  Thanks Martin! 🙏😀

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  @theupson5 күн бұрын

  im sorry, you handled this pretty clumsily. from 3:14 (and for petes sake the evenness symmetry doesn't take so much explanation), you should partition it along the integers. THEN you use the sub u = pi (x-n). that gets you to a better place than 9:39 in the video in 2 easy steps, given that |sin(x)| has period pi that recurring crappy integral e^x*sin(x)... never fricking humor your math prof by using integration by parts. e^x*sin(x) = im e^((i+i)*x), fast clean and painless.

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  @felipefred12796 күн бұрын

  I am trying to find a video in that you use laplece to solve a mit integral. Do you remember which video is that?

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  @owl3math6 күн бұрын

  There may be more than 1 but I did find this one in the Laplace transform playlist: kzread.info/dash/bejne/mophlZWync6vpM4.htmlsi=Ol2qEoyMlNrXjrE0

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  @felipefred12796 күн бұрын

  @@owl3math is exactly this one, thanks. I wanted to show the technique to my differential equations professor

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  @owl3math5 күн бұрын

  Oh interesting. Wonder what he says 😂

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  @adandap7 күн бұрын

  I always liked this result. It easily generalises to L{t^n f(t)} as well.

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  @owl3math6 күн бұрын

  Hi adandap. Right yes nth derivative and alternating sign. :)

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  @MikeT101017 күн бұрын

  Laplace and differentiating under the integral sign...not bad! ; )

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  @owl3math7 күн бұрын

  This is a fun one! 😃

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  @OwenChiu7 күн бұрын

  What type of calc is this, i took calc ab and taking bc next year

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  @owl3math7 күн бұрын

  Good question but not quite sure just because I don’t think they cover beta function or integral of inverses in integral calculus

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  @OwenChiu7 күн бұрын

  cOol

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  @owl3math7 күн бұрын

  Thanks!

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  @wasimohammad42807 күн бұрын

  Just take out x from square root,then put 1/x=t,reduces to -Sin inverse 1/x,put the limit,get the answer-pi/6.

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  @maxgomez49098 күн бұрын

  I like your videos

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  @owl3math8 күн бұрын

  good to hear it! Thanks Max :)

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  @alexselby8028 күн бұрын

  Just multiply by (1-sinx)/(1-sinx)

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  @owl3math8 күн бұрын

  HI Alex. Yep, that's the method I mentioned early in the video. It's waaay way WAY faster that way :)

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  @henningkneller7 күн бұрын

  how would you continue from there?

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  @owl3math7 күн бұрын

  @@henningkneller Then you have 1- sin^2 x in the denominator which can be written as cos^2 x. Divide that into (1 - sin x) and split it into 2 integrals: sec^x and sec x tan x :)

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  @parthkalel67815 күн бұрын

  ​​@@owl3math when I do this I get the indefinite integral tanx - secx (+c) which is undefined at pi/2 so am I doing something wrong

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  @henningkneller5 күн бұрын

  Clever! Thank you:)

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  @adandap9 күн бұрын

  Who can go the distance? We'll find out in the long run...🎶

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  @owl3math8 күн бұрын

  a lot of music references coming up! :) And nice use of music notes in the comment :)

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  @MikeT101019 күн бұрын

  Take the long way home...

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  @owl3math9 күн бұрын

  Yep 😀. I think that’s a song 😂

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  @owl3math8 күн бұрын

  supertramp maybe

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  @MikeT101018 күн бұрын

  @@owl3math You got it! kzread.info/dash/bejne/foSG1KqyiMbOdbw.html

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  @kaftan17769 күн бұрын

  I like the power indices because in music 3/2 is a perfect 3rd and 2/3 is a perfect fourth.

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  @owl3math9 күн бұрын

  Oooh. I didn’t know that 😀

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  @KevinAPamwar10 күн бұрын

  Nice... Another solution Sin3y =3Siny-4Sin³y

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  @owl3math9 күн бұрын

  Hi Kevin. So Use triple angle on Sin 6x = 3 sin 2x - 4 sin^3 2x? Makes sense...then its easy from there.

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  @KevinAPamwar10 күн бұрын

  Nice... You can skip the 1st nine minutes... Sin^4x + Cos^4x = (Sin^2x + Cos^2x)^2 - 2*Sin^2x*Cos^2x= 1- 1/2 *(Sin^2(2*x))

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  @justabunga110 күн бұрын

  The second method to do that is to let u=ln(x), so x=e^u. Then, dx=e^udu. When x=1, u=0. The limit as x goes to ∞, then u goes to ∞. The new limit of that integral should be the integral from e to ∞ of (e/3)^udu. That integration is evaluated from 0 to ∞ for the function (e/3)^u/ln(e/3). This means (e/3)^0/ln(e/3)=1/ln(e/3)=log_(e/3)(e). Note that the upper limit of integration will approach to 0. You can also look at the exponential function 1/(3^ln(x))=(1/3)^ln(x)=x^ln(1/3) and then you can go from there to integrate that function.

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  @owl3math9 күн бұрын

  nice method! thanks :)

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  @dkravitz7810 күн бұрын

  My philosophy is there's no reason not to at least start this way. It won't hurt.

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  @owl3math10 күн бұрын

  Oooh this was kind of an interesting scenario. Kind of forgot about this one. :)

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  @Hypnotic.-.10 күн бұрын

  You explained it so well that even me as a high school student could understand it 💀 I couldn’t repeat what you did but I wasn’t really confused at any point 😂

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  @owl3math10 күн бұрын

  WOW thats great! I think i'm proud of both of us 😂

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  @takyc788310 күн бұрын

  both methods so cool

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  @owl3math10 күн бұрын

  THANKS! :)

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  @Arandomguy-bi1yl10 күн бұрын

  Excellent method I haven't spent much time with beta function, I here know a very common one y=(1-x^3/2)^3/2 x=(1-y^2/3)^2/3 Just calculated the inverse function So (1-x^2/3)^2/3=F inverse x The integral of f-1(x) from 0 to 1 now, put x=f(u)=(1-u^3/2)^3/2, dx=f'(u)du The limit changes to 1 to 0, as there is a minus Infront of it we change it into positive by changing the limit The inverse integral changed into xf'(x) or uf'(u) after the transformation Now we if we take the whole integral f(x)+xf'(x) from 0 to 1, see the term in the integral is the derivative of xf(x) So the integration is xf(x) and after putting the values We get 1f(1)-0f(0)=0 f(1)=(1-1)^3/2=0 The result is 0 Oh I skipped the first part, it's actually the same thing

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  @owl3math10 күн бұрын

  Yep it's the first method but still thanks for the nice explanation!

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  @DakshSinghRawat-kr7nw10 күн бұрын

  Nice one tho

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  @owl3math10 күн бұрын

  thanks!

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  @prakharjain544810 күн бұрын

  B(f(b) - af(a) pretty standar identity tho its pretty hard striking when not dine something similar

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  @owl3math10 күн бұрын

  Hi. What is that identity? Beta subtracting 2 functions?

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  @arzaseb10 күн бұрын

  cool

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  @owl3math10 күн бұрын

  thanks!

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  @adandap10 күн бұрын

  Oh, I didn't know the variation on the beta function with the 'n' present. That's useful. I did a u sub in each term, u = x^(2/3) etc.

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  @owl3math10 күн бұрын

  Nice. Yes, it's nice to have all the forms of beta but u-sub always gets it back to the more familiar form. :)

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  @s.hjb010 күн бұрын

  bloody hell would never have thought of this

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  @owl3math10 күн бұрын

  this situation with the Inverses doesn't come up all that often so don't feel too bad :)

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  @mangakhoon4517go10 күн бұрын

  Cool

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  @owl3math10 күн бұрын

  thanks!

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  @mangakhoon4517go10 күн бұрын

  @@owl3math but I don't understand properly they didn't teach this in highschool

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  @booboobaloney10 күн бұрын

  Very cool use of inverses!

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  @mhm642112 күн бұрын

  80 * 6 - 60 * 24 + 14 * 120 - 720 just used gamma definition, no pen or paper

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  @owl3math12 күн бұрын

  Great! That works nicely 👍👍👍

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  @owl3math12 күн бұрын

  error correction: the formula I use for sinh at initially is incorrect and should be a/(s^2 - a^2) and therefore my formula for cosh is wrong as well and should be s/(s^2 - a^2). I'll work on a correction. Thanks for your patience. :)

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  @LCDL613 күн бұрын

  Isn't the Laplace of sinh at=a/(s²-a²)? In the video it is a/(s²+a²), that changes the result of the Laplace of cosh

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  @owl3math12 күн бұрын

  Yep you're right. Big mistake in this one! I'll look into making another video I think. I did have it correct in the sinh at video before it but then brought over the wrong value into this video for no particular reason.

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  @rwilson560113 күн бұрын

  i genuinely look forward to these. ty

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  @owl3math12 күн бұрын

  Glad to hear it! Thank you :)

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  @gesucristo013 күн бұрын

  Just multiply the bracket to the exponential and you get the gamma function.

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  @owl3math13 күн бұрын

  Yep good method! 👍👍👍