Haha... nice black-red marker. I look at it the same way as I look at calculators... makes life much easier, but reduces the need for skill (mental arithmetic in the case of the calculator, and your previously demonstrated marker-switching skills, in the case of this dual colour marker.) Great video!
Hi, just a suggestion it's off topic but our Calc 2 course covers a section on using integration tables a d you're suppose to use substitution methods I think tonsolve. Were you ever going to do a hw set on those? It's the ones with complex integrals that can't be solved by hand. Ty, Dan
@@happyjohn1656 Nah I ended up taking a different course of life paths, ended up taking Calc 3, DIff eq, physics w/ calc and linear algebra and washed out hard that semester and said fuck it I'll go back to the work force. One day I'll relearn math for funzies and hopefully pass the classes I failed at.
You can make u=cos x for problem 2 if you turn tan x into (sin x)/(cos x) because then you end up with the integral of (-1/u)(ln u) du and if you do another substitution where w=ln u you just get the integral of -w dw and after using the power rule and substituting u and x back in you will still end up with -1/2 (ln(cos x))^2 +C just takes a little longer.
Hey man, your videos are really good man. I couldn't understand u substitution for a long time, but you made me understand it within the first 10 minutes. Thank you!:)
In the question no. 2, we can write tanx as sinx/cosx and then put cosx = u and after substitution we have a very nice example of integration by parts! Haha
I have just been taught all 3 methods for integrals and im watching ALL your integral videos to learn. They are being extremely good, i now comprehend better how to do them! As always thank you and keep uploading videos, i love them! [And you too :)]
Hello, so for problem number 2, I got a different method but same answer. Set cosx equal to u, then change tanx to sinx over cosx, since they are the same. Derivative of cosx is -sinx, then it will be du/-sinx. This will cancel out the sinx (from sinx/cosx which is the same as tanx). Then it will become ln(u)/u, do substitution again, v equal ln(u), derivative is 1/u. Cancel out you at the bottom and so on.
You can do the second one with integration by parts. It's a looper and regrouper in one. Given: integral tan(x) ln(cos(x)) dx Let tangent be integrated, and the log composition be differentiated. d/dx ln(cos(x)) = 1/cos(x) * -sin(x) = -tan(x) Construct IBP table: S ____ D _________ I + ____ln(cos(x)) __ tan(x) - ____-tan(x) _____ -ln(cos(x)) Attach S-column signs, construct along diagonal. Then construct an integral along the bottom row. -ln(cos(x))^2 - integral tan(x) ln(cos(x)) dx Spot the original integral, and call it I. Set the whole expression equal to I. I = -ln(cos(x))^2 - I Solve for I: 2*I = -ln(cos(x))^2 I = -1/2*ln(cos(x))^2 Solution: -1/2*ln(cos(x))^2 + C
br oyou´re by far the best maths teacher ever. i never got that substitution thing but now it makes perfectly sense. do you know a site with some training examples?
blackmarkerredmarker um can't you just put du right away and just replace the terms which match in the du? Example: I = int((3+6x^2)(3x+2x^3)dx) u = 3x+2x^3 du = 1*3x^0 + 3*2x^2 = 3 + 6x^2 dx *Which means... * I = int(udu) = (u^2)/2 + C = (3x+2x^3)^2/2 + C
Very great video! My first day of calculus 3 was monday, and this was a great refresher for me on previous sections! Thanks! And if you can post some calculus 3 sections 11 and up, that would be greatly appreciated :) Nice marker btw haha
The integral is taken in terms of u substituting in for x, so when taking the integral/antiderivative of 1 it would be u rather than x. Hope this helped!
A bit late for exams but the integral was in terms of "u" (notice the "du" at the end of the equation) so the antiderivative of 1 was u. And for at 8:50 he multiplied both sides by 2sqrt(x) so that du=(1/2sqrt(x))dx simplifies to 2sqrt(x)du=dx (again notice the difference in "du" and "dx")
10:50 Use the parentheses, because without them, it looks like as if you were integrating only the first term and then subtracting the 1/u from the result.
+D4v30r Yes, even then, and _especially_ then. The `du` is not there just to "enclose the integral" - it's actually a factor. The integral is secretly a sum of the differentials multiplied by the values of the integrated function. You can think of it as a fancy summation symbol ∑ f(x)·dx taken in the limit (the `∫` symbol is an elongated "S" from Latin "summa" = total, whole, entirety). If you saw a summation like this: ∑ something + somethingElse would you consider only the first term as being summed over, or both? How about this one?: ∑ (something + somethingElse) Exactly the same argument applies to integrals. +mrBorkD Yes, it doesn't make any sense, and that's why I'm nitpicking it. It only makes any sense if the entire thing is a factor of `du`, and since this factor itself is a sum, it has to be enclosed in parentheses.
I understand that. However, I have never seen an integral without a du,dx etc and it is common practice to place it at the end. Therefore your parentheses are implied.
Tejero Life, square root of x is an expression in terms of x, we cannot integrate more than 1 variable in a single variable integral, the variable depends d(?)
This 2 pointed marker is very 1 dimensional... What if we up it to 2 dimensions... a cross (think x-y axis)... with 4 marker tips, red, black, blue and green each at right angles to each other?... or 3 dimensions (think x-y-z axis), 6 points, red, black, blue, green, purple and yellow all at right angles?? ... Could we extend this into 4 dimensions? NOTE: If we consider a 3-d marker projected into a 2-d world... The 2-d world essentially "sees" the 3-d marker only 4 colors at a time.. when we rotate the 3-d, the 2-d world sees it as simply 2 of the 4 tips changing color... SO... on our 3-d marker set (6 points), if we could set the tips to change colors, we could essentially model a 4-d marker set where each 4-d orientation is "seen" in the 3-d world as the marker tips not only revolving but changing colors!... Could we model higher dimensions?... Would LOVE to see someone work out the details on this! ....Am I over thinking this black-red marker thing??
3rd part can also be solved by substituting x as (tan@)^2........I admit the fact that it requires further substitution of tan@ as t but the ans is same!!!
Hello thanks for your nice video, I have one question : at 13'48'' : 1 + square root of x is positive so there is no need to use absolute value...no problem if x > 0 but what happens if x < 0?
I love your videos. My passion has always been for discrete maths but at the age of nearly 60 you've awoken a love of calculus. In this one you state rt(x) is always positive. Surely not.
Elliott Manley rt(x) is always positive because the definition of a finding a root is basically just finding the number that has been squared. As you know, a number squared is positive e.g. -2^2 is +4 Therefore the reason for his statement is because you cannot square a number to get a negative number and nor can you do it in reverse (Unless you take into account complex numbers) As an example rt(4) = both +2 and -2 for the reason explained making rt(-x) not true for all real integers for x
+Elliott Manley Rt(4) is not +-2. Rather, the solution to the equation x^2=4 is +-2. Rt(4) written on its own is strictly the positive result. I tell students: whenever you see a square root, it is positive. Whenever you have to take a square root to solve an equation, add plus or minus.
@@stephenbeck7222 , thanks. I was wondering about the same thing! Could you please tell me why we have this convention? Is it to preserve a one-to-one mapping?
For problem 3 I did the u sub u = ×^1/2, so I ended up with the correct answer without the +2. Since this was an indefinite integral that appears to not have mattered, but if this was a definite integral would that deeply affect the answer?
the way I would've done the second one is I would've done u = cos (x) because tan(x) is just sin(x)/cos(x) so you have du/u. Then you have integral of 1/u ln u which I'd recognise already, but you can also take it to the v world with v = ln u Now of course this is essentially a more complicated way to do the same thing because v = ln u = ln (cos x) which is what you did, but I think it's a more understandable method to get there
A non-universal method for integrating with the chain rule is to say that int f(g(x))g’(x)dx= int f(g(x))d(g(x)). You don’t need to do any differentiation, and you can integrate x/(1+x^4) using this method.
it's a standard entry in the integration table. But the way it's arrived at is through Trig substitution, let u=tan(t). du=sec^2(t)dt. After the substitution, the resulting denominator of (1+tan^2(t)) can be rewritten as sec^2(t), by a trig identity. So you end up getting the integral of [sec^2(t)/sec^2(t)]dt, or integral of 1dt. This results in t (plus constant of integration). To go back to "u", use the original substitution equation and take the inverse tangent on both sides to get t=tan^-1(u). done.
Output of square root of x not always positive. He can be zero and positive, so he always non-negative. But certainly square root of x+1 is always positive
I tried to separate the problem: 1dx + 1dx/sqrt(x) and got x+2sqrt(x) is it the same?? I don’t think I violated any laws... Can someone enlighten me... Edit: I’m talking about the last problem
Great video @blackredpen It does remember when I was studying systems engineering in a course called Mathematics II. It follow the success ando greetings from Venezuela
Spoilers: he doens't keep using this. So it must not have worked out too well.
Too bad. He could have had all 3 colors in one hand easily if it did.
or he lost it
I saved it because it was a gift. I
@@blackpenredpen I kinda assumed you'd get more if it worked for you.
@Brooks Archer bro that is so toxic
I'm a first-year mechanical engineering major and let me just say you have SAVED my life.
the way i felt this 😭
How you doing?
@@damian4091 well thanks to this channel. 2.0 to a 3.8 gpa
@@unholykraut1107 jeeeeez good work
What
Haha... nice black-red marker. I look at it the same way as I look at calculators... makes life much easier, but reduces the need for skill (mental arithmetic in the case of the calculator, and your previously demonstrated marker-switching skills, in the case of this dual colour marker.)
Great video!
Thank you thank you!
Hi, just a suggestion it's off topic but our Calc 2 course covers a section on using integration tables a d you're suppose to use substitution methods I think tonsolve. Were you ever going to do a hw set on those? It's the ones with complex integrals that can't be solved by hand.
Ty,
Dan
@@dmorgan0628 Ever figure those out? 😄
1:55 AM
1/21/2022
@@happyjohn1656 Nah I ended up taking a different course of life paths, ended up taking Calc 3, DIff eq, physics w/ calc and linear algebra and washed out hard that semester and said fuck it I'll go back to the work force. One day I'll relearn math for funzies and hopefully pass the classes I failed at.
@@dmorgan0628 can't wait for another life update after the next 4 years
You can make u=cos x for problem 2 if you turn tan x into (sin x)/(cos x) because then you end up with the integral of (-1/u)(ln u) du and if you do another substitution where w=ln u you just get the integral of -w dw and after using the power rule and substituting u and x back in you will still end up with -1/2 (ln(cos x))^2 +C just takes a little longer.
yeah this was my first instinct as well
best channel of math
Dude you are a Genius, actually my favourite RUclipsr And my inspiration
Yo did anyone else wake up at 9:31?
Hey man, your videos are really good man. I couldn't understand u substitution for a long time, but you made me understand it within the first 10 minutes. Thank you!:)
Absolutely amazing video!!
I'm learning so much:)
And that marker is so perfect for you!
He only uses the marker in this video...
Thank you for this amazing challenge!
In the question no. 2, we can write tanx as sinx/cosx and then put cosx = u and after substitution we have a very nice example of integration by parts! Haha
Second example scared the sizzle out of me.
thank you for the videos. They are very helpful
I have just been taught all 3 methods for integrals and im watching ALL your integral videos to learn. They are being extremely good, i now comprehend better how to do them! As always thank you and keep uploading videos, i love them! [And you too :)]
this guy is the GOAT hands down
Hello, so for problem number 2, I got a different method but same answer. Set cosx equal to u, then change tanx to sinx over cosx, since they are the same. Derivative of cosx is -sinx, then it will be du/-sinx. This will cancel out the sinx (from sinx/cosx which is the same as tanx). Then it will become ln(u)/u, do substitution again, v equal ln(u), derivative is 1/u. Cancel out you at the bottom and so on.
Please don't forget to like the video. I watch all of these videos and they are so good, sometimes I forget to like them
I like the marker. Good marketing out there.
Much better than the chalkboard and the two-headed marker sure looks easier to use.
I like it ,but what about using integration by parts on the second question
You can do the second one with integration by parts. It's a looper and regrouper in one.
Given: integral tan(x) ln(cos(x)) dx
Let tangent be integrated, and the log composition be differentiated.
d/dx ln(cos(x)) = 1/cos(x) * -sin(x) = -tan(x)
Construct IBP table:
S ____ D _________ I
+ ____ln(cos(x)) __ tan(x)
- ____-tan(x) _____ -ln(cos(x))
Attach S-column signs, construct along diagonal. Then construct an integral along the bottom row.
-ln(cos(x))^2 - integral tan(x) ln(cos(x)) dx
Spot the original integral, and call it I. Set the whole expression equal to I.
I = -ln(cos(x))^2 - I
Solve for I:
2*I = -ln(cos(x))^2
I = -1/2*ln(cos(x))^2
Solution:
-1/2*ln(cos(x))^2 + C
br oyou´re by far the best maths teacher ever. i never got that substitution thing but now it makes perfectly sense. do you know a site with some training examples?
Your Integration videos are addicting haha
Thank you man i appreciate your efforts
blackmarkerredmarker um can't you just put du right away and just replace the terms which match in the du?
Example: I = int((3+6x^2)(3x+2x^3)dx)
u = 3x+2x^3
du = 1*3x^0 + 3*2x^2
= 3 + 6x^2 dx
*Which means... *
I = int(udu) = (u^2)/2 + C
= (3x+2x^3)^2/2 + C
Thanks for explanation 😸🎉
In 2 ques it will be 1/4 if u integrate udu
So funny and still clever! :D
Please make a video on Euler's substitution and feynman's
I used double substitution to solve 2 and 3 but I like how you do it with just 1 substitution
Very great video! My first day of calculus 3 was monday, and this was a great refresher for me on previous sections! Thanks! And if you can post some calculus 3 sections 11 and up, that would be greatly appreciated :) Nice marker btw haha
Hi there, sorry I am not teaching calc 3 anytime soon. (thank you for the comment regarding to the marker ^^ )
this is an amazing video
When you solve an integral, are you allowed to merge all the constant into one single ”C”?
You're just adding a bunch of different constants, it'll still be a constant afterwards so it's fine
for oexams, make it clear what you are doing
Obviously.
@@chessandmathguy Get off your high horse buddy
C is just a random constant. So one C is enough. If you multiply anything with a constant you get the constant.
11:05 - Isnt integral of 1, x ? And why at 8:50 ,there is no more 1 over 2sqrtX ? But great vid , saving me before exams :D
The integral is taken in terms of u substituting in for x, so when taking the integral/antiderivative of 1 it would be u rather than x. Hope this helped!
A bit late for exams but the integral was in terms of "u" (notice the "du" at the end of the equation) so the antiderivative of 1 was u. And for at 8:50 he multiplied both sides by 2sqrt(x) so that du=(1/2sqrt(x))dx simplifies to 2sqrt(x)du=dx (again notice the difference in "du" and "dx")
Put u = 1+x^2, du = 2xdx, xdx=du/2. u-world: Int(du/2u) = ln(u)/2 = ln(sqrt(1+x^2))+C I think.
I'm convinced Expo only created that marker to get a marketing shout-out from this guy.
1:54 AM (yep!)
1/21/2022
😆
Thank you
would it not be better notation to write arctan instead of tan^-1?
i want the pen switching back!
4:27 Please tell me why don't people write acrtan(x)? tan^-1(x) confuses some of my friends cause they sometimes think that it's power.
I personally believe that u-substitution is slightly trickier than IBP. Still, great video BPRP!
7:47 i substituted u = sqrt x and i did the integral... i got direct answer i didnt need to merge that 2 into +c ....
10:50 Use the parentheses, because without them, it looks like as if you were integrating only the first term and then subtracting the 1/u from the result.
even tho there's a du at the end?
without context it could be read as: (∫1 ) - ( 1/u du)
I don't think ∫1 makes sense though without a dx or a du or anything like that
+D4v30r Yes, even then, and _especially_ then. The `du` is not there just to "enclose the integral" - it's actually a factor. The integral is secretly a sum of the differentials multiplied by the values of the integrated function. You can think of it as a fancy summation symbol ∑ f(x)·dx taken in the limit (the `∫` symbol is an elongated "S" from Latin "summa" = total, whole, entirety). If you saw a summation like this:
∑ something + somethingElse
would you consider only the first term as being summed over, or both? How about this one?:
∑ (something + somethingElse)
Exactly the same argument applies to integrals.
+mrBorkD Yes, it doesn't make any sense, and that's why I'm nitpicking it. It only makes any sense if the entire thing is a factor of `du`, and since this factor itself is a sum, it has to be enclosed in parentheses.
I understand that. However, I have never seen an integral without a du,dx etc and it is common practice to place it at the end. Therefore your parentheses are implied.
Very helpful, thank you!
*Smiles*
My brain: Are you meditating or studying?
Why is it that square root of x is not invited in the u world? can you explain further
Tejero Life, square root of x is an expression in terms of x, we cannot integrate more than 1 variable in a single variable integral, the variable depends d(?)
This 2 pointed marker is very 1 dimensional... What if we up it to 2 dimensions... a cross (think x-y axis)... with 4 marker tips, red, black, blue and green each at right angles to each other?... or 3 dimensions (think x-y-z axis), 6 points, red, black, blue, green, purple and yellow all at right angles?? ... Could we extend this into 4 dimensions? NOTE: If we consider a 3-d marker projected into a 2-d world... The 2-d world essentially "sees" the 3-d marker only 4 colors at a time.. when we rotate the 3-d, the 2-d world sees it as simply 2 of the 4 tips changing color... SO... on our 3-d marker set (6 points), if we could set the tips to change colors, we could essentially model a 4-d marker set where each 4-d orientation is "seen" in the 3-d world as the marker tips not only revolving but changing colors!... Could we model higher dimensions?... Would LOVE to see someone work out the details on this! ....Am I over thinking this black-red marker thing??
This is the best channel on the internet. Like seriously I fucking love this channel. So underappreciated!
You always lose your black pen legibility and we are finding it difficult to follow your writings if you could follow somewhat big letters instead.
thank you sir, your hard work will never go unoticed...
Thank you!
great marker that you could habe made with 2 MARKERS AND A PIECE OF MASKING TAPE!!!! This guy works at an engineering school?
3rd part can also be solved by substituting x as (tan@)^2........I admit the fact that it requires further substitution of tan@ as t but the ans is same!!!
Thank you, Sherry. Here is what you were trying to read.
Im first year pre-engineering student and i can't explain in words how grateful I am
In the third integral, substituting x=t^2 is much easier
Actually nice sponsorship besides wonderful video!
In the future this will be a skill completely done by machines. This guy really plugged in his Amazon link. Plug in PayPal instead.
Hello thanks for your nice video, I have one question : at 13'48'' : 1 + square root of x is positive so there is no need to use absolute value...no problem if x > 0 but what happens if x < 0?
for x
In the last integral couldn't you factor out 2, and name C2/2=C3?
I love your videos. My passion has always been for discrete maths but at the age of nearly 60 you've awoken a love of calculus.
In this one you state rt(x) is always positive. Surely not.
Elliott Manley rt(x) is always positive because the definition of a finding a root is basically just finding the number that has been squared. As you know, a number squared is positive e.g. -2^2 is +4
Therefore the reason for his statement is because you cannot square a number to get a negative number and nor can you do it in reverse (Unless you take into account complex numbers)
As an example rt(4) = both +2 and -2 for the reason explained making rt(-x) not true for all real integers for x
That's exactly my point.
Rt(4) = +/-2.
+Elliott Manley Rt(4) is not +-2. Rather, the solution to the equation x^2=4 is +-2. Rt(4) written on its own is strictly the positive result. I tell students: whenever you see a square root, it is positive. Whenever you have to take a square root to solve an equation, add plus or minus.
@@stephenbeck7222 , thanks. I was wondering about the same thing!
Could you please tell me why we have this convention? Is it to preserve a one-to-one mapping?
@@elliottmanley5182 sqrt(x) only finds the principal or positive root.
So BlackPenRedPen is Darth Maul now?
Maybe?!
@@blackpenredpen hmmmm
blackpen + redpen = pen(black+red)
appreciate u man, u helped me pass my cal 1 final
This was so helpful! Thank you so much!
Solve integrals that comes in jee advanced..... These are not harder ones
blackpenredpen sponsored by blackpenredpen basically
0:25 bprp is basically unofficially sponsored by expo lol
For problem 3 I did the u sub u = ×^1/2, so I ended up with the correct answer without the +2.
Since this was an indefinite integral that appears to not have mattered, but if this was a definite integral would that deeply affect the answer?
No because in definite integration, the constants will get subtracted and cancelled out (it's why we ignore +c when evaluating definite integrals)
For first part I took x^4 from denominator and after simplification I put 1/x=t but I am getting answer -1/4log|x^4+1/x^4| + c is it right.
Bring back the Darth Maul Expo Marker 🤝
if u wanna really learn try the examples by ur own and then check the result
the flash of thank your sharri made me spit my coffee out hahaha
the way I would've done the second one is I would've done u = cos (x) because tan(x) is just sin(x)/cos(x) so you have du/u. Then you have integral of 1/u ln u which I'd recognise already, but you can also take it to the v world with v = ln u
Now of course this is essentially a more complicated way to do the same thing because v = ln u = ln (cos x) which is what you did, but I think it's a more understandable method to get there
Why is simply taking the derivative of both sides and substituting dx in the integral justified?
I was taught u-sub differently.
How did u learned it?
So much integration…
Good video
A non-universal method for integrating with the chain rule is to say that int f(g(x))g’(x)dx= int f(g(x))d(g(x)). You don’t need to do any differentiation, and you can integrate x/(1+x^4) using this method.
I was doing alright until I forgot to turn u into x :/
some smart guy
I don't understand how 1/1+U^2 integrates to tan^-1(U)??
@blackpenredpen
Its a rule
Look for proof online
You could do trig substitution to integrate, let u = tan（t）
it's a standard entry in the integration table. But the way it's arrived at is through Trig substitution, let u=tan(t). du=sec^2(t)dt. After the substitution, the resulting denominator of (1+tan^2(t)) can be rewritten as sec^2(t), by a trig identity. So you end up getting the integral of [sec^2(t)/sec^2(t)]dt, or integral of 1dt. This results in t (plus constant of integration). To go back to "u", use the original substitution equation and take the inverse tangent on both sides to get t=tan^-1(u). done.
4:52 why don't division by sin cancell the tan
good marker hhh
god help me i love integral
Love those integrals!
my hero black red pen.
Hey
Great video and awesome marker (haha). Anyways, I wanted to ask which book do you use for the questions?
Viraj Madaan i use Stewart for my classes. But oftentimes I just come up with my questions or search online.
Output of square root of x not always positive. He can be zero and positive, so he always non-negative. But certainly square root of x+1 is always positive
I like When you say "some thing"
It sounds like some Chinese food :]
Who's Sheri?
Pls. Deep pen use
For question three, couldn't the integrand be rewritten as 1/(1 + (x^1/4)^2), and then you could just use the arctan integral?
I tried to separate the problem: 1dx + 1dx/sqrt(x) and got x+2sqrt(x) is it the same?? I don’t think I violated any laws... Can someone enlighten me...
Edit: I’m talking about the last problem
You mean in the last problem?
1/(1+sqrt(x)) is not equal to 1 + 1/sqrt(x).
1 + 1/sqrt(x) = (1 + sqrt(x)) / sqrt(x).
Vincent D Thanks for the reply. Hahaha I’ve made such a stupid error.
Thank you, Sheri😅
How do you come up with that u-1 strategy ? How do you think like that ?
bprp and 3blue 1 brown are the only channels getting me thru calculus rn
Great video @blackredpen It does remember when I was studying systems engineering in a course called Mathematics II. It follow the success ando greetings from Venezuela
U r amazing BRoooooo
Good video
9:29 lmao
Lol it’s the love for the market for me
Market or marker?
I'm french i don't understand English but I understand you learning I don't know maybe your Witcher 🤣🤣🤣