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!
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
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
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!
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.
@@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
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|
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).
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
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
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).
Quite literally full circle. You can also use the unit circle and geometric interpretation of tangent and cotangent to see why the sum of arctan and arccotan is pi/2 as well. It all goes back to the unit circle! :)
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
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
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
Can you solve Integral from +inf to - inf of {Cos x/x^2+1 dx} using standard calculus? Please see attached video. ruclips.net/video/4ti74hewhf8/видео.html
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)
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.
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
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.
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!
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
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!
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!
@@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
@@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.
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.
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.
Beautiful solution and nice cover! 😁 I like the following manipulation (Manipulate expressions, not the people!) Note that (x+i)-(x-i)=2i 1/[(x-i)(x+i)]=[(x+i)-(x-i)]/[(x+i)(x-i)]*(1/2i) =(1/2i){[(x+i)/(x+i)(x-i)]-[(x-i)/(x+i)(x-i)]} =(1/2i){[1/(x-i)]-[1/(x+i)]}...
עקרון ההוכחה שarctanx+arccotx=pi/2 הוא הזהות cot(pi\2-x)=tan(x) (אותה ניתן להוכיח מהתבוננות על משולש ישר זווית, זהויות טריגונומטריות ועוד...). עקרון ההוכחה שarccot(1/x)=arctan(x) הוא שtanx=1/cotx.
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.
If we dont take the branch of Ln that is not defined in the positive numbers or the one that is not defined in the negative numbers the result function is not even continuos so is not an antiderivative.
...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
Hey bprp I love ur vids im 16 and I really enjoy calculus :) I have a question for u: could u do integral from -♾ to ♾ of sin(x^2)? Maybe with some Feynman’s I don’t know… Good luck for ur channel!
sine is an even function, so that integral is equal to 2 times the integral of sin(x^2) from 0 to inf. To evaluate an improper integral, we have to evaluate the limit of it as it approaches infinity. Sine diverges at infinity, so this integral also diverges.
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
Hmm, this doesn't always work, though, does it? Only analytical functions should have a complex integral, but all piecewise continuous real functions have a real integral.
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.
This is my favorite integral of all time. It amazes me that the scary inverse tangent function has such a simplistic looking derivative. I used to prove it the hard way by finding the derivative of the inverse tangent. Tan(y) = [e^(2iy)-1]/[i(e^(2iy)+1)]. Let x = tan(y). x[i(e^(2iy)+1)]=[e^(2iy)-1]. e^(2iy)[ix-1]= -1-ix. e^(2iy) = [-1-ix]/[ix-1]. y = 1/(2i)ln[[-1-ix]/[ix-1]] = tan-1(x). You can then use the quotient rule to calculate the derivative and you get 1/(1+x²)
@@diegoc.8518 yes ive seen simpler ways to do it. They way i did was just the way i first discovered. And the derivative of that crazy equation being so nice and compact amazes me.
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.
I'm not sure if it's really your type of content but you could do ln(x)^log(10,x) = log(10,x)^ln(x) and then all of this equals ln(log(10,x)) = log(10,ln(x)) I did it as a fourteen yo so it probably won't take you long ;-)
this is cool but, if you graph -tan^1(1/x) , it isn’t the same as cot^1(x). also, -tan^1(1/x) is a very different graph to tan^1(x). It’s lines are not connected and are translated in opposite directions. what does this mean?
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.
But the définitions of the integral affirms that the calculs are done in the R world ,it doesnt have a meaning in C in fact in the test of algebra i used limits in C world ,it payed 1 point/20 for that
Why is the coefficient that he writes for the two fractions at 54:00 1/2i? Shouldn't it be 1/2? I mean 1/2 * [1/(x+i) + 1/(x-i)] does equal 1/(1+x^2) so I don't get it. 1/2i would give a different result. I've never heard of the cover method before so I don't know what it's about, when writing partial fractions I've always figured out the coefficients by solving a system or something like that.
I already made a reply on this and I got 1/2X. He seems to be doing some sleight of hand here. see my replies above using simple fractions rule. And besides, when you separate into two fractions with variable X, it should give the same squared X components.
I was following along nicely until you threw out inverse tan(x) + inverse cot(x) = pi/2 Where did that come from please? I can't see any good explanations of that, some say that you have to use tan(pi/2) which .. how?? haha
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 😅😅
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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.
Thanos endgame ahh phrase
Should've added you know you know too!
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
Nice
Amazing explanation
Thanks.
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
Beautiful, thanks a lot for the proof
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!
They just mad they can't harness the power of complex world 😎
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.
@@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
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 was mind blowing. Imagine if this was a bonus question on a test before you get to the inverse trig functions
Good lucking knowing that identity without having studied inverse trig functions first
@@fix5072
It’s not knowing the identity if it asks you to “integrate 1/(x^2 + 1)” as the test question
@@fanamatakecick97 yeah that'd be fair
It would not have worked in my class because they taught us trig sub before partial fractions.
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).
I am pretty sure that the integral is arctan(x) + c and not arctan(c).
@@pi_xiyessss
@@pi_xi bruhh
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
Bro me too, bro me too.
this is definitely a more intuitive approach as compared to substituting x = tan theta
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
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).
My last college math class was an intro to complex analysis class. I really struggled in that class, but this demo really inspires me.
1:37 : Not only is the absolute value not required (when integrating a reciprocal) in the complex numbers, it is *forbidden.*
I actually forgot about tangent being an odd function. This also helps with re^itheta practice. It all comes back into full circle.
Quite literally full circle. You can also use the unit circle and geometric interpretation of tangent and cotangent to see why the sum of arctan and arccotan is pi/2 as well. It all goes back to the unit circle! :)
Really cool! That green marker part in the end really made me say “wow” out loud
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
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
It's correct a think, because the function 1+x^2 is finite and bounded everywhere.
@@anshumanagrawal346 also complex functions works extremely different from normal functions, since they require 4D to work
@@anshumanagrawal346
The arctan(1/0) is undefined yes but with limits it's arctan(infinity) which is π/2 + πn (n is an integer)
Excellent comprehensive problem that involves many aspects of Calculus. Thank you for showing and explaining this problem!
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
Can you solve Integral from +inf to - inf of {Cos x/x^2+1 dx} using standard calculus? Please see attached video.
ruclips.net/video/4ti74hewhf8/видео.html
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)
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.
J'ai découvert cette chaîne il y a peu de temps et je la trouve vraiment incroyable
This is such a cool way to do this integral. Really helps you understand complex numbers and their applications
Great job BPRP!. It's nice to know partial fraction decomposition works in the complex world.
but arctan(x) and -arctan(1/x) are not a constant apart... only when x>0 and then another constant apart when 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
A question similar to this that i did recently was the indefinite integral of (ln x)/(1+x^2).
You should definitely try it!
Can't be done by parts... Very interesting
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.
Complex analysis is so fun
This channel is like a loophole for math problems.
It's a very nice videos.
It proves the beauty of complex numbers to it's ultimate level......
Thanks.
arctan(1/x)+arctan(x) is equal to pi/2 if x>0 and equal to -pi/2 if x
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!
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
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!
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!
All of ab is taught in bc anyways so just skip to bc
Always happy with your great content ! 1M reached soon :)
Thank you so much 😀
@@blackpenredpen Omg you replied !!! Hello !!!
It does reveal a fascinating relationship between ln and arctan.
You just blew my mind
Nice, I remember stumbling across that same question some years ago, but I could not continue from the ln .
Really great explanation. Thank you.
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.
7:00 The Green Marker!! :O :O :O
You could also aply that cosx=(e^ix+e^-ix)/2 and sinx=(e^ix-e^-ix)/2i
isn't that formulat of coshx and sinhx? .. i mean hyperbolic functions
@@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
@@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.
@@affapple3214 ohh okay mate thanks for the reply. I learnt how to remember that formula in a easy way today ...
@@affapple3214 yeah i was meaning that only.. isn't that a formula of sinhx without i's
We simply learnt substitute x to tan θ in \int \frac{1}{x^2+1} dx before learning euler formula.
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.
Can we use this trick for all rational functions? Ex. for 1/(x^4-1) we use the 4-roots of 1.
yes it works fine for that function too
but I wouldn't really recommend it because there is better options
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.
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.
Beautiful solution and nice cover! 😁
I like the following manipulation (Manipulate expressions, not the people!)
Note that (x+i)-(x-i)=2i
1/[(x-i)(x+i)]=[(x+i)-(x-i)]/[(x+i)(x-i)]*(1/2i)
=(1/2i){[(x+i)/(x+i)(x-i)]-[(x-i)/(x+i)(x-i)]}
=(1/2i){[1/(x-i)]-[1/(x+i)]}...
Mathematics are indeed exact sciences !!!! Brilliant!
My major is history, i dont know anything he say and this is the most beautiful gibberish for me I have ever heard...
I'm a stats guy and this all went right over my head. I still had fun watching, though.
עקרון ההוכחה שarctanx+arccotx=pi/2 הוא הזהות cot(pi\2-x)=tan(x) (אותה ניתן להוכיח מהתבוננות על משולש ישר זווית, זהויות טריגונומטריות ועוד...). עקרון ההוכחה שarccot(1/x)=arctan(x) הוא שtanx=1/cotx.
Today after a long period of time ur video made me really happy from inside from these type of question
Your videos are such a gem! Excellent teacher!
when blackpenredpen introduces bluepengreenpen… it’s gonna be awesome
i always wondered if you could do this! this is amazing!
The way he switches markers is kinda mesmerizing
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.
do you need an ambulance
No.
Such a fun video, thanks again! ❤️
I loved the video, hope it will help me in my exam tomorrow.
If we dont take the branch of Ln that is not defined in the positive numbers or the one that is not defined in the negative numbers the result function is not even continuos so is not an antiderivative.
...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
Beautifully done ! this is pure perfection 👏🏻
Fascinating stuff! Thank you!
Hey bprp I love ur vids im 16 and I really enjoy calculus :) I have a question for u: could u do integral from -♾ to ♾ of sin(x^2)? Maybe with some Feynman’s I don’t know…
Good luck for ur channel!
sine is an even function, so that integral is equal to 2 times the integral of sin(x^2) from 0 to inf. To evaluate an improper integral, we have to evaluate the limit of it as it approaches infinity. Sine diverges at infinity, so this integral also diverges.
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.....
Thank you Euler.
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
Maybe because they don't teach this in calculus. As per the video title.
Arctanx+C
Hmm, this doesn't always work, though, does it? Only analytical functions should have a complex integral, but all piecewise continuous real functions have a real integral.
As you saw, in the end all of the complex terms canceled out.
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.
Can you do another video where you solve this using residues? Complex integration
Wow nice! That is taking the scenic route but it works ha ha
This was very amusing and sublime!
Thank you so much❤
This is my favorite integral of all time. It amazes me that the scary inverse tangent function has such a simplistic looking derivative. I used to prove it the hard way by finding the derivative of the inverse tangent. Tan(y) = [e^(2iy)-1]/[i(e^(2iy)+1)]. Let x = tan(y). x[i(e^(2iy)+1)]=[e^(2iy)-1]. e^(2iy)[ix-1]= -1-ix. e^(2iy) = [-1-ix]/[ix-1]. y = 1/(2i)ln[[-1-ix]/[ix-1]] = tan-1(x). You can then use the quotient rule to calculate the derivative and you get 1/(1+x²)
i calculate derivative by (f^-1)'(x)=1/f'(f^-1(x)) in this case (f^-1)'(x)= 1/tan'(arctan(x)) = 1/1+tan(x)^2(arctan(x) = 1/1+x^2 problem solved easily
@@diegoc.8518 yes ive seen simpler ways to do it. They way i did was just the way i first discovered. And the derivative of that crazy equation being so nice and compact amazes me.
@@JSSTyger oh ok nice
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.
Lets generalise Productlog(k,(a+bi)e^(c+di))
Absolutely amazing as usual
I'm not sure if it's really your type of content but you could do ln(x)^log(10,x) = log(10,x)^ln(x) and then all of this equals ln(log(10,x)) = log(10,ln(x)) I did it as a fourteen yo so it probably won't take you long ;-)
I used this to prove Euler's formula without power series.
Untying a fine thread demonstrated mathematically.
Solve for x and y
x^2 +y^2 = a
x^3+y^3 = b
this is cool but, if you graph -tan^1(1/x) , it isn’t the same as cot^1(x). also, -tan^1(1/x) is a very different graph to tan^1(x). It’s lines are not connected and are translated in opposite directions. what does this mean?
Interesting how the imaginary world intersects the real world in unexpected ways, but still following rules.
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.
@@dalisabe62 If you think of it, even the negative and irrational numbers have got bad names. We're just so used to it we never notice.
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
Blackpenredpen, I have an indefinite integral for you : integral of ln(csc(x)+tan(x)).
Hey, how about dx/[(x+1+sqrt(2x))(x+1-sqrt(2x)) ?
It was beautiful
How are you so good at math
7:27 that's only when x is positive. The second formula that is.
You ignored the fact that both ln and arctan are multi-valued functions in complex analysis.
waiting for your video related to IMO problem
quick question : how do you integrate ln(lnx)?
Eu gosto disso! Boa explicação detalhada!
But the définitions of the integral affirms that the calculs are done in the R world ,it doesnt have a meaning in C in fact in the test of algebra i used limits in C world ,it payed 1 point/20 for that
So good!
Why is the coefficient that he writes for the two fractions at 54:00 1/2i? Shouldn't it be 1/2? I mean 1/2 * [1/(x+i) + 1/(x-i)] does equal 1/(1+x^2) so I don't get it. 1/2i would give a different result.
I've never heard of the cover method before so I don't know what it's about, when writing partial fractions I've always figured out the coefficients by solving a system or something like that.
I already made a reply on this and I got 1/2X. He seems to be doing some sleight of hand here. see my replies above using simple fractions rule. And besides, when you separate into two fractions with variable X, it should give the same squared X components.
I was following along nicely until you threw out inverse tan(x) + inverse cot(x) = pi/2
Where did that come from please? I can't see any good explanations of that, some say that you have to use tan(pi/2) which .. how?? haha
that's a excelent exercise
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 😅😅