how Richard Feynman would integrate 1/(1+x^2)^2

when we see the integral of 1/(1+x^2) of course  we know the answer for this is just equal to   tan^-1(x)+C and if you want to show work of  course we can just do trig sub let x equal to   tangent Theta and then we can continue from here  and this is usually done in calculus 2 but last   time also showed you guys how we can use complex  numbers to do this and that is to factor 1 plus   x squared as I plus x times negative I plus X  and then two partial fractions in the complex   world but I did that for you guys last time  already so if you haven't seen that video go   ahead and watch that after this video the link  will be in the description for your convenience   so what exactly are we doing here today well  don't worry I'm going to square the denominator   easy right because we can just  Square this no of course not   don't do that the question is can we  still do tricks up the answer for this yes   can we do this also yes but in fact we are not  going to do this anymore instead I will show you   guys a very powerful technique for integration and  that's called the finance technique of integration   AKA differentiation under the integral sign AKA  that means true for integration so how exactly   do we do it though well first we have to start  with something that we know so check this out but what if this number is now one let's say it's  17. well in that case what do we do on the right   hand side right so as you can see this formula is  not so complete yet but instead this formula is   better that is when we have a a squared right  here when we have that the right hand side is   going to be 1 over a times the inverse tangent of  X over a and you can do tricks up for this as well   so I'm going to delete that to you guys I'm just  going to use this formula to show you this a is   called the parameter because it kind of generates  like the old complete version of this formula   for you guys right so how exactly can we utilize  this though five minutes is going to tell us that   whenever you have a formula with parameter just go  ahead and differentiate that with respect to the   parameter and hope for the good things to happen  and it will happen trust me he didn't say that but   I made up so let's go ahead and check it out I'm  just going to differentiate this with respect to a   like so well on the left hand side we are  differentiating an integral so what do we do well   this is the time that we have to use the so-called  differentiation on the dental cosine that tells us   that we can bring this inside but change the D to  the partial derivative symbol because we have the   A and also the X inside here so this definite side  becomes the integral and then partial derivative   and right here to differentiate that  let's just go ahead and write this   as a square plus x square raised to a  negative one and we still have the DX   on the right hand side I'm just going to keep yes  take the derivative and we have this right here   so it's the one over a but this right here will  be our first function because a is the variable   here and then times the inverse tangent of X  over a and of course for the plastic right here   Okay so let's go ahead and work out the derivative  for this use the power rule first so bring   the power to the front and then minus one so we  have negative a squared plus x squared raised to   the negative two but don't forget the chain rule  multiply by derivative inside with respect to a so   that is just going to be too high the derivative  x squared is zero because we are in the a world   so that's the derivative from the inside but we  still have the integral so go ahead put down the   integral right here and see some of you guys  see what we are doing on the right hand side   if we can figure out the derivative we pretty  much will have the formula well you have to use   the product rule we are going to keep the first  function which is 1 over a times the derivative   inverse tangent it will give us 1 over 1 plus  whatever this is a square so X over a square   that's not yellow we will have to use the  chain rule so X is just a constant multiple   in the a world so let's put that down and then  differentiating one over eight write that as a   to the negative one will get Negative a to the  negative two so it's just negative one over a   squared good and then we will continue keep the  second function so plus inverse tangent of X over   a times Theta derivative of that which is negative  one over a squared derivative C is just zero   can we differentiate both sides with  respect to C don't do that you could but why   yeah anyway though that's pretty much it now we  just have to clean things up so on the left hand   side let me just put this down as the integral  negative 2A over and then here already as a   squared plus x squared and then square and then  d a this right here to clean this up okay we have   investable parentheses on the bottom here I'm just  going to take the a square multiplying multiplying   on the top we have negative 1 times x so that's  Negative X over this a is just chilling at the   front right here this time start is a squared this  times that is x squared so that's the first part   and for the second part we'll just say minus 1  over a squared times inverse tangent of X over a   yeah we're pretty much done but I  would like to divide Everybody by   2A here so that it's cleaner Jose so let's  go ahead oh sorry it should be a negative   right here negative 2A so let's divide the  value by negative 2A so 1 over negative 2A   so finally integral 1 over a squared plus x  squared and then Square DX no partial fraction no tricks up yeah differentiation under the  interco side give us that negative negative   becomes positive we have just X on the top over 2  a squared times a squared plus x squared and then   negative negative becomes positive 1 over 2 a to  the third power and then inverse tangent of X over   a wow how cool is this right and because we are  talking about formula and this is the intervene   integral so at the end here I will put down  plus C so ladies and gentlemen this is a very   nice formula for you guys with this approach and  of course we should finish this question right   so this and that is just when the case a  is equal to 1. can you put negative one yes   ah interesting but why don't give us negative  negatives and passes so one's okay so I'm just   going to say that a equal to 1. so that's that  and then on the right hand side we'll just have   X over 2 and then 1 plus x squared and then right  here just one over two and we have inverse tangent   of X over a which is just over one so we have  this and finally put a plus c just like that so   what do you guys think go up hey I hope you  liked the video so far and if you enjoying   problem solving and want to know more then  you should check our sponsor today brilliant   brilliant is an excellent online learning platform  with a big focus on 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Channel: blackpenredpen

Views: 492,959

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Id: KA6foxmx3kU

Length: 8min 53sec (533 seconds)

Published: Thu Sep 29 2022