Learn more calculus by visiting Brilliant 👉 brilliant.org/blackpenredpen/ (20% off with this link!)

@diaa5355

2 жыл бұрын

Goodbye head of math 🇨🇳🇨🇳🇨🇳🇨🇳

@ImranKhan-tu7ix

2 жыл бұрын

In (x+1 )^2 intigration why we use u = x+1 ans is (x+1)^3/3 = x^3/3 + x^2 + x +1/3 Why not use u = x or if we dont use u why answer different Ans is x^3/3 + x^2 + x Difference is 1/3 one have 1/3 other Don't have 1/3

Also, if you did not use the principal branch of the complex log, the only difference would be a shift of the integration constant, so using other branches also works!

"I know I know, I know you know, and you know I know." The greatest phrase spoken by this man.

This was mind blowing. Imagine if this was a bonus question on a test before you get to the inverse trig functions

@fix5072

2 жыл бұрын

Good lucking knowing that identity without having studied inverse trig functions first

@fanamatakecick97

2 жыл бұрын

@@fix5072 It’s not knowing the identity if it asks you to “integrate 1/(x^2 + 1)” as the test question

@fix5072

2 жыл бұрын

@@fanamatakecick97 yeah that'd be fair

@edgarb.6187

Жыл бұрын

It would not have worked in my class because they taught us trig sub before partial fractions.

all my friends thought i was doing it wrong by using complex numbers to solve integrals albeit getting the correct results now I'll show this video to them if it ever happens again! thanks bprp!

@himanshu6002

2 жыл бұрын

They just mad they can't harness the power of complex world 😎

@sniperwolf50

2 жыл бұрын

The reason you can get away with this method is that the complex integral of 1/(z² + 1) is path independent, which is not generally true for any complex function f(z). Path independent integrals can be computed as if they were integrals in the reals.

@krishna2803

2 жыл бұрын

@@sniperwolf50 woah thats cool! i think i saw something like this in a Michael Penn video in which he talked about complex integrals and the fact that the result changes when the path is changed but it didn't ring any bell here. thanks! (1) here, Steve mentioned that 'x' is real, so i kinda thought that this is just a 'real' integral with some goofy tricks, in the complex integral case, we would be integrating over a complex variable (idk if thats how to say it) does that make any difference or it's still the same? (2) also, more generally, if we can reduce an expression involving complex numbers to an expression having only reals, will this still hold? like (x+i)(x-i) can be reduced to (1+x²) which is real for all real x if (2) is true, then it would mean that this should work for all real integrals as we are just creating a complex expression from a real one, so it can always be reduced a real expression

Love it. It clearly demonstrates that we classify/identify function based on how they relate to each other and not necessarily their explicit details. Like how sin and cos are just the same function with a phase shift.

Here's a simple explanation for why tan-1(x) + tan-1(1/x) = pi/2: Imagine a right triangle, "a" units long and "b" units tall. Its angles will be 90°, tan-1(a/b), and tan-1(b/a). As with any triangle, these must add to 180°. 90° + tan-1(a/b) + tan-1(b/a) = 180° tan-1(a/b) + tan-1(b/a) = 90° (= pi/2) Now sub in x = a/b to derive the identity: tan-1(x) + tan-1(1/x) = pi/2

@TheCrunchyGum

2 жыл бұрын

Nice

@moviesadda24x79

2 жыл бұрын

Amazing explanation

@rafaelstv

2 жыл бұрын

Thanks.

@YoutubeModeratorsSuckMyBalls

2 жыл бұрын

No, it is longer than in video, there is simpler explanation. Tan(theta) = x, cot(theta) =1/tan(theta)=1/x, but cot(theta)=tan(pi/2-theta)=1/x. So Theta= arctan(x)=pi/2-arctan(1/x). So arctan(1/x) + arctan(x) =pi/2

@pedroribeiro1536

2 жыл бұрын

Beautiful, thanks a lot for the proof

Hahah, today I was doing a lecture about Partial Fractions and one of the students actually tried to do this. I said, "Usually I would tell you don't memorize, but in this case it's faster just to know that integral if 1/(x^2+1) = arctan(c).

@pi_xi

Жыл бұрын

I am pretty sure that the integral is arctan(x) + c and not arctan(c).

@cillipill

5 ай бұрын

​@@pi_xiyessss

@kornelviktor6985

3 ай бұрын

​@@pi_xi bruhh

this is definitely a more intuitive approach as compared to substituting x = tan theta

Fun fact, that also explains why the Maclaurin series for the arctangent only has a radius of convergence of 1. If you come back to 1/2i (ln(x-i)-ln(x+i)), you get 1/2i ln((x-i)/(x+i)), which is undefined when x = i (because you get ln(0)) and x=-i (because you get smth divided by 0). So the series centered around 0 will only converge when |x| < 1... because it only converges when |z|

This is quite helpful in computer algebra systems, because integration of rational functions can be reduced to the case of linear, rather than quadratic denominators

I’ve done this very integral the same way before! One might consider the particular solution satisfying y(0)=0 and you get tan⁻¹(x) and (1/2i)ln((x-i)/(x+i))-π/2

@Batshitcrazy6942

3 ай бұрын

Bro me too, bro me too.

This channel is like a loophole for math problems.

Excellent comprehensive problem that involves many aspects of Calculus. Thank you for showing and explaining this problem!

1:37 : Not only is the absolute value not required (when integrating a reciprocal) in the complex numbers, it is *forbidden.*

One great thing about this approach is that you can get a complex definition of arctan and arccot (like how the complex definitions of sin and cos are well-known).

Really cool! That green marker part in the end really made me say “wow” out loud

My last college math class was an intro to complex analysis class. I really struggled in that class, but this demo really inspires me.

Really great explanation. Thank you.

Great job BPRP!. It's nice to know partial fraction decomposition works in the complex world.

i always wondered if you could do this! this is amazing!

J'ai découvert cette chaîne il y a peu de temps et je la trouve vraiment incroyable

It's a very nice videos. It proves the beauty of complex numbers to it's ultimate level...... Thanks.

It’s funny, next semester I’m going into calc 1 but I already know how to find the derivative of certain functions and know some basics in integration, love this channel tbh

A few months ago I asked a few teachers at my school and no one was able to give me a proper answer if this is correct or not. I thank you so much for this video qwq

@anshumanagrawal346

2 жыл бұрын

It's not correct, strictly speaking. There's a lot of theory surrounding this, and you can't just do what he did in the video without knowing all that, it's good for having fun, but to make it rigorous you need to know rules about how complex numbers and Calculus works. If you don't believe me, you can notice a few things right off the bat, first the logarithm function is multi valued, second the argument of a+bi is not always equal to tan^-1(b/a), also the identity tan^-1(1/x) = cot^-1(x) only holds for certain values of x, i.e., not always. For example you can see that the original function is perfectly well defined at x=0, but in this way not because we have a 1/x

@midas-holysmoke7642

2 жыл бұрын

It's correct a think, because the function 1+x^2 is finite and bounded everywhere.

@kepler4192

2 жыл бұрын

@@anshumanagrawal346 also complex functions works extremely different from normal functions, since they require 4D to work

@motazfawzi2504

Жыл бұрын

@@anshumanagrawal346 The arctan(1/0) is undefined yes but with limits it's arctan(infinity) which is π/2 + πn (n is an integer)

What's funny is that this was the first thing I did when I learned how to do partial fractions, but instead of trying to prove that they are similar using polar coordinates (which I had not thought about at all,) I decided to use the taylor series for sin(x), cos(x), and e^x in order to try to set arctan(x) equal to that ln function. I started on Friday and got a decent chunk of the way there but then school ended and I haven't finished. This sort of spoiled it but at least I know it's possible with my other approach now lol (or at least it should be)

Beautifully done ! this is pure perfection 👏🏻

Such a fun video, thanks again! ❤️

Great video !! I was trying to figure out how to go from the imaginary world to the real one and this answered the exact question I had!! Thank you very much for making it ! It’s beautiful.

Your videos are such a gem! Excellent teacher!

Fascinating stuff! Thank you!

This was very amusing and sublime! Thank you so much❤

Nice, I remember stumbling across that same question some years ago, but I could not continue from the ln .

It does reveal a fascinating relationship between ln and arctan.

Always happy with your great content ! 1M reached soon :)

@blackpenredpen

2 жыл бұрын

Thank you so much 😀

@kiaruna

2 жыл бұрын

@@blackpenredpen Omg you replied !!! Hello !!!

You just blew my mind

Absolutely amazing as usual

Today after a long period of time ur video made me really happy from inside from these type of question

Eu gosto disso! Boa explicação detalhada!

Thank you Euler.

A question similar to this that i did recently was the indefinite integral of (ln x)/(1+x^2). You should definitely try it!

@hmm.2013

2 жыл бұрын

Can't be done by parts... Very interesting

The end is brilliant. Without knowing that tan^(-1)(x) + tan^(-1)(1/x) = pi/2, I would have thought that I made something wrong.

Interesting how the imaginary world intersects the real world in unexpected ways, but still following rules.

@dalisabe62

2 жыл бұрын

The imaginary world is a bad name for the complex plane. There is nothing imaginary about it. Only because we no better definition for the square root of -1, that we are forced to call it imaginary. We defined the square root of a number in such away that it is the quantity that if multiplied by itself, it would give the number itself. There is no such scenario for negative numbers, and it was a good thing we had none because that prompted the introduction of the complex plane.

Nice approach sir

Dude I took AP calc ab and this all went over my head but I loved every second of it even though I did not get any of this apart from the first integration

@Spongebob-lf5dn

6 ай бұрын

Maybe because they don't teach this in calculus. As per the video title.

arctan(1/x)+arctan(x) is equal to pi/2 if x>0 and equal to -pi/2 if x

well the method was new to me and I really liked it. but one thing in last step where you put tan^(-1)(1/x) = cot^(-1)(x), you should have preferably made two cases x>0 and x0 tan^(-1)(1/x) = cot^(-1)(x) and for x

We simply learnt substitute x to tan θ in \int \frac{1}{x^2+1} dx before learning euler formula.

...Good day Teacher Steve, How come I can follow your clear presentation very well, but that I would never come up with this in my entire life... Mission impossible for me... Thank you for again another educative presentation, Jan-W

Haha, I literally did this once out of blu and put limits 0 to pi/4 to find out what e raised to itheta was equal to, I was so happy.

Technically, the transition to exponential form is not correct for cases where x < 0. The angle should be derived differently for (x, i), x < 0 and (x, -i), x < 0. Argument is not just simply arctan(1/x) and arctan(-1/x) correspondingly. I believe there should be an everywhere continuous piecewise function defined first to get a correct primitive and then a proof provided that it's off by a constant with arctan(x). Conclusions in the video are only valid for positive values of x. Please correct me if I'm wrong.

So good!

Complex analysis is so fun

omfg, this was amazing. in my graduate electromagnetism class we were applying conformal mapping theory to solve electrostatic boundary value problems. there was one particular transformation that involved the logarithms with complex numbers in their arguments and i had no fucking idea how the final answer reduced to some tan^-1. watching you actually go through the algebra has cleared so much up for me. keep at it my guy, college sophomores aren't the only beneficiaries here!

@diegoc.8518

2 жыл бұрын

u usually know when it's about arctan or arccos or arccsin because depending on your math class u know these are their derivatives. d(arrcos)/dx= -1/sqrt(1-x^2) , d(arcsin)/dx=1/sqrt(1-x^2) and d(arctan)/dx= 1/1+x^2

Thanks!

I'm a stats guy and this all went right over my head. I still had fun watching, though.

I loved the video, hope it will help me in my exam tomorrow.

Thank you

Mathematics are indeed exact sciences !!!! Brilliant!

Who suggested you the name black pen red pen. How did it came to your mind. Is their any story behind this name. Please make a video on it.....

My major is history, i dont know anything he say and this is the most beautiful gibberish for me I have ever heard...

Can we use this trick for all rational functions? Ex. for 1/(x^4-1) we use the 4-roots of 1.

@user-sk5zz5cq9y

2 жыл бұрын

yes it works fine for that function too

@user-sk5zz5cq9y

2 жыл бұрын

but I wouldn't really recommend it because there is better options

@skylardeslypere9909

2 жыл бұрын

It would generally, but for this example of 1/(x⁴-1) it's particularly easy to see, since the 4th roots of 1 are just 1,-1,i,-i. I.e., you would just decompose it as 1/(x+1), 1/(x-1), 1/(x+i) and 1/(x-i) We already saw that the last two yield the same result as 1/(x²+1) when integrated.

@ExplosiveBrohoof

2 жыл бұрын

1/(x^4-1) can work exactly the same way, because the fourth roots of 1 are just ±1 and ±i. So you get the same things under integration as you do in this video, plus some extra ln terms. I'd be wary of generalizations though, because you need to be diligent with your choice of constant term and choice of branch sometimes before your solution resolves to a real valued function.

Lets generalise Productlog(k,(a+bi)e^(c+di))

Can you do another video where you solve this using residues? Complex integration

7:00 The Green Marker!! :O :O :O

It was beautiful

Stunned 🥶🤩

that's a excelent exercise

AMAZING!

Delightful!

This is super fun

loved it

Awesome!

Did you make a shortcut in getting 1/2i and -1/2i? By law of fractions, this should be 1/2x. Or am I mistaken? it is as simple as 1/A + 1/B = (A+B)/AB.

very good video ☺👍❤

Very nice man

Awesome video! I wonder if taking the derivative of tan-1(x) is correct as well… It’s easier for me at least

Brilliant!!

waiting for your video related to IMO problem

Wow nice! That is taking the scenic route but it works ha ha

when blackpenredpen introduces bluepengreenpen… it’s gonna be awesome

Amazing 👏

Love it

Vecteur nice method thanks sir

Nice Thank you teacher

@blackpenredpen

2 жыл бұрын

You are welcome

imho: The partial fraction substitution step (2nd line) is incorrect as it makes the numerator equal zero. Interestingly the final result is correct, but i suspect when done correctly using integration by parts those 1/x terms in the correct numerator cancel.

The way he switches markers is kinda mesmerizing

quick question : how do you integrate ln(lnx)?

Hi I have a question about "Convergence of a Power Series" Can you help me solve that?

You could also aply that cosx=(e^ix+e^-ix)/2 and sinx=(e^ix-e^-ix)/2i

@saitama1830

2 жыл бұрын

isn't that formulat of coshx and sinhx? .. i mean hyperbolic functions

@affapple3214

2 жыл бұрын

@@saitama1830 yes but you have to just remove all the i's, example: sinh(x) = (e^x-e^-x)/2 That's how I remember sinh formulas, I derive the complex form of sin X and cos X and remove the i's

@user-vj6ty5lb1l

2 жыл бұрын

@@affapple3214 Cool!! We can also view the connections between the hyperbolic trigs funcs and the complex trigs funcs: e^(ix) = cos(x) + isin(x) (1) e^(-ix) = cos(x) - isin(x) (2) (1)-(2) /2i ⇒ sin(x) = [e^(ix) - e(-ix)]/2i (3) (1)+(2) /2 ⇒ cos(x) = [e^(ix) + e(-ix)]/2 (4) (3) ⇒ sin(ix) = [e^(-x) - e(x)]/2i = i[e^(x) - e(-x)]/2 = isinh(x) (4) ⇒ cos(ix) = [e^(-x)+ e(x)]/2 = [e^(x) + e(-x)]/2 = cosh(x) → sin(ix) = isinh(x) → cos(ix) = cosh(x) In order to remember that we can think sin and cos as odd an even functions. sin(-x) = -sin(x) cos(-x) = cos(x) This isn't just a coincidence. Think about the taylor expantion series.

@saitama1830

2 жыл бұрын

@@affapple3214 ohh okay mate thanks for the reply. I learnt how to remember that formula in a easy way today ...

@saitama1830

2 жыл бұрын

@@affapple3214 yeah i was meaning that only.. isn't that a formula of sinhx without i's

Brilliant!

great sir❤❤

Hey bprp, I'm a high school freshman trying to skip Precalc and Calc AB - everything's in place for me to skip Precalc, but AB is a bit more challenging. Thankfully, your videos make solving these problems so much easier and incredibly intuitive! Thank you so much for your videos, they help me so much. Keep it up!

@Reallycoolguy1369

2 жыл бұрын

The Khan Academy lessons on Calculus are free and excellent. They are also have questions for mastery. Combine with BPRP and a couple other math for fun channels and you will be set!

@mr.twicks3009

2 жыл бұрын

All of ab is taught in bc anyways so just skip to bc

If we write the lower bound of this integral as 0 and the upper bound as infinity we get the equation e^(i×pi)+1=0

what did you do at 4:45

Very interesting

How do I integrate √(x). with limits from -5 to -1.

Beautiful

Even tough i is not a real numbers so when calculating that area, there's no place for i, when going from 0 to +oo, we'll find π/2 because after integrating we have arctanx and when having (1/2i)ln|(x-i)/(x+i)| we'll have π/2 too. Ok let's watch the video.

@gasun1274

2 жыл бұрын

do you need an ambulance

@lazaremoanang3116

2 жыл бұрын

No.

I remember when I tried to solve it the way you did but I failed to continue I just compared the logharithmic result with the tangent thinking that I can come with the identity of "i" or something new 😅😅