1:54 That face when you solve an integral after a few hours of hard work. This is, in fact, a great video, never really thought you could break it down this "easily". I've only solved this integral with some weird four-way-inception substitution a long time ago. Might give it another shot in a while actually.
Because you get 1/(u^2-1) after the substitution, isn't that also the same as -1/(1-u^2) meaning the integral of csc x can also be written as -tanh^-1(cos x) + c? Just thought I'd point out an interesting relationship between the trig and the hyperbolic trig functions.
Use the power rule for logs on that last expression and you get ln ( sqrt( abs((cos x -1)/(cos x +1)))) which is the same as ln ( sqrt((1 - cos x)/(cos x +1))) which is = ln(tan(x/2)). A little easier on the eyes.
Oh, I'm just a pony from Equestria, visiting Human World and learning some of your stuff ;) Here's how I look in my entirety: goo . gl / LBS1WR BTW you can also see my Human friend, Sci Twi, somewhere around here in the comments ;)
you new this is #Biochemical rule this halp so mush in Integral of cos an sin an ton if you new Ruler of bioche you can do all integral #f(x)=F(cos(x).sin(x))
Hi, I have got a question: I dont quite understand how to deal with absolute values in calculus. I know they are there and they must be there but in your videos you often just magically get rid of them or you treat them like regular parentheses. For example now: ln (|u+1|) can be bigger than ln (|u-1|) so I think you should simplify the expression as ln(|u-1|/|u+1|) not ln(|(u-1)/(u+1)|). Or am I wrong ? I know integrals practically only from your videos, I dont have real background, so please correct me if I am wrong :) .. thanks
Lukáš Baláž Since there is an absolute value on the top and the bottom of that expression, you know the result will be positive because +/+=+ so that's why he moved the absolute value bars outside. To be clear, you totally can leave the absolute bars around the top and bottom of the fraction, it's just that people tend to re write that in a different way because it simply looks better.
Peter Chatain The better explanation is that the absolute value function is multiplicative, so if you have |a|*|b| or |a|/|b|, then you are canonically allowed to write |ab| and |a/b|.
1/sinx = 1/(2sin(x/2) cos(x/2) = 1/2((sec(1/x))^2/(tan^x/2). So we have the integral of d((tanx/2)/(tanx/2) = ln( (tan(x/2)) + C. We forget the "U business" and it makes simpler
Since this method involves memorization in some way (either by knowing that the antiderivative involves a log, so you’ll want to get a form of du/u, or just by rote trial and error), you can always rewrite csc(x) = 1/sin(x) = sin(x)/sin^2(x) = sin(x)/(1 - cos^2(x)) and setting u = cos(x), du = -sin(x)dx. From there, you can use partial fractions and simple algebraic manipulation with trig identities to arrive at the same answer. :)
I used a different method. take u=cscx and use identities to rearrange equation in terms of u to -1/sqrt(u^2-1) and from there its easy to get -arcosh(cscx)
no, the integral of du/1+u^2 is arctan(u), but nevertheless he had du/(u^2 - 1), which would use the inverse hyperbolic tangent instead since the sign is reversed.
using substitution, cant you say that for u=1/x, integral(u)=u^2/2 therefore integral(1/x)=1/(2x^2)? isn't this a problem with substitution? because not all powers of x integrate the same, and depend on how you treat them.
You must not forget about the differential dx. So if you use a u-sub where u=1/x, then du =(-1/x^2)dx = (-u^2)dx, so dx = -du/(u^2). Plugging back into (1/x)dx we get u(-du/(u^2)) = (-1/u)du. In the end, nothing changed, so the substitution method is not applicable for the integral of 1/x
You could do something similar to that and go straight to negative inverse hyperbolic cotangent from 1/(1-u^2). It's not as easily understood as the ln function you get from partial fractions though.
Me neither but it's just zeroing out 1 of the factors in the denominator at a time. In this example, (u-1) goes away when you plug in 1 leaving 1/(1+1) as the coefficient for that linear factor. Hope this helped of you didn't get it.
Nice vid, but please try not to write the solved answer on the board BEFORE you actually solve it, it seems to take away from the intuition of figuring the problem out.
Yeah yeah, that's all well and good, but can he hold 2 pens in one hand?
hes just redpen
Damn those integral symbols are sexy
why do you think the integral symbol look like a letter S for?
ASMR CSGO The word "sum"
it was a rhetorical question and kind of was a joke.
LMAO 😂
1:54 That face when you solve an integral after a few hours of hard work.
This is, in fact, a great video, never really thought you could break it down this "easily". I've only solved this integral with some weird four-way-inception substitution a long time ago. Might give it another shot in a while actually.
Haha! Same! Have to try again tomorrow. Took me an hour and couldn't remember partial fractions decomp 😖
no
Because you get 1/(u^2-1) after the substitution, isn't that also the same as -1/(1-u^2) meaning the integral of csc x can also be written as -tanh^-1(cos x) + c?
Just thought I'd point out an interesting relationship between the trig and the hyperbolic trig functions.
o shit he ascended
No hablo Inglés pero he entendido a la perfección, muchas gracias!
That fella looks baked AF
He's the last person on earth that i expect is a math teacher
Where's Steve? I believe his name was Steve.
But, great video anyway. Thank you
Marian P. Gajda behind the camera lol
Elda Afc I was literally behind the camera tho! Loll
Use the power rule for logs on that last expression and you get ln ( sqrt( abs((cos x -1)/(cos x +1)))) which is the same as ln ( sqrt((1 - cos x)/(cos x +1))) which is = ln(tan(x/2)).
A little easier on the eyes.
Great video, been looking for something like this. I have a bigger expression sinxln(tanx))dx and I ended up with the integral of 1/sinx
What happened to you? Weren't you Asian in previous videos? :q
Bon Bon!!!!!!! It's you again!!!! Apple sauce!!!!!
Yeah, it's me ;) Seems that you recognize me every time I visit your channel, but I can't remember why. Did I do something goofy here in the past? :)
Don't' worry about it! Cheers!
Btw, what are you?
Oh, I'm just a pony from Equestria, visiting Human World and learning some of your stuff ;)
Here's how I look in my entirety: goo . gl / LBS1WR
BTW you can also see my Human friend, Sci Twi, somewhere around here in the comments ;)
Hi, thanks a lot for this video. I have one question though, when I solve the sum of the fractions, on the upper side I got -1
Why don't they use a pin mic? Free hands is better
because then he wont have the need to hold two pens in one hand and the whole concept of this channel will be lost you evil
I am certainly more natural with the mic in my hand. Check out this old video then u will see the diff ruclips.net/video/aVDOvr6NNM4/видео.html
You look different in this video. You had a different hair cut here?
What are you even teaching this crap for, because your answer isn't in standard form. Teach the TRICK.
You seem younger lately, and Aryan
Stoned but smart #WOKE
Good job, but now I have to look up the "cover up method" for partial fractions ... But dat OK ...
you new this is #Biochemical rule this halp so mush in Integral of cos an sin an ton if you new Ruler of bioche you can do all integral #f(x)=F(cos(x).sin(x))
Dont know if its worth learning this way theres more to remember than the orginal method and its just a bit messy...
Nice job, Chase! :-)
Nice job Kylie!
Nice job, ASMR CSGO! :-)
nice job, everyone! :)
good saludos desde
UNTELS -LIMA -PERU
Can't figure out how the presenter managed to do the partial fraction?
No entendí ni shit lo que dijo pero ví tu pizarra y me mojé xd
I've worked it and it came -ln|cotx+cscx|+c and I can show how it comes
Hi, I have got a question: I dont quite understand how to deal with absolute values in calculus. I know they are there and they must be there but in your videos you often just magically get rid of them or you treat them like regular parentheses. For example now: ln (|u+1|) can be bigger than ln (|u-1|) so I think you should simplify the expression as ln(|u-1|/|u+1|) not ln(|(u-1)/(u+1)|). Or am I wrong ? I know integrals practically only from your videos, I dont have real background, so please correct me if I am wrong :) .. thanks
Lukáš Baláž Since there is an absolute value on the top and the bottom of that expression, you know the result will be positive because +/+=+ so that's why he moved the absolute value bars outside. To be clear, you totally can leave the absolute bars around the top and bottom of the fraction, it's just that people tend to re write that in a different way because it simply looks better.
Peter Chatain The better explanation is that the absolute value function is multiplicative, so if you have |a|*|b| or |a|/|b|, then you are canonically allowed to write |ab| and |a/b|.
Slicker than snot.
1/sinx = 1/(2sin(x/2) cos(x/2) = 1/2((sec(1/x))^2/(tan^x/2).
So we have the integral of d((tanx/2)/(tanx/2) = ln( (tan(x/2)) + C. We forget the "U business" and it makes simpler
Since this method involves memorization in some way (either by knowing that the antiderivative involves a log, so you’ll want to get a form of du/u, or just by rote trial and error), you can always rewrite csc(x) = 1/sin(x) = sin(x)/sin^2(x) = sin(x)/(1 - cos^2(x)) and setting u = cos(x), du = -sin(x)dx. From there, you can use partial fractions and simple algebraic manipulation with trig identities to arrive at the same answer. :)
You could move the one half to the inside of the log and that makes it tan(x/2)
chase been smoking that stanky dank ganja green
I used a different method. take u=cscx and use identities to rearrange equation in terms of u to -1/sqrt(u^2-1) and from there its easy to get -arcosh(cscx)
Lost me at your technique for PFD, I don't understand why you did that.
Awesome, thanks
Thanks a lot brooo!
where is Blackpen Redpen?
After multiplying by sinx up and down can i change dx with dcosx and say that the answer is -tanh^-1(cosx) or is this a wrong approach?
At 1:20 can you write the integral as (-1) (1/(1-u^2)) du put the constant (-1) outside the integral and substitute the integral for tanh^-1(u) ???
Sebastian Cor that is correct!
blackpenredpen thanks, I remebered that identity ftom one of your videos. Love your channel
Sebastian Cor yea! This works the same as integral of sec(x)
THANKS
bro got a new skin
rather keep it up sir
you could integrate du/1+u^2 directly since is equal arcsin(u) any way great job
Where is the square in the dominstor
no, the integral of du/1+u^2 is arctan(u), but nevertheless he had du/(u^2 - 1), which would use the inverse hyperbolic tangent instead since the sign is reversed.
2:00
Instead of pastial fraction decomposition, I like better to change 1/(u^2-1) into (1+u-u)/(1-u^2). It's also work, harder but trickier
Chase SChidfar
C SC
csc
Lol
0:17 'Reciprocate' means 'take the reciprocal'; so you didn't reciprocate csc, you just wrote it down as the reciprocal of sin.
They make mics that can fit on your shirt...
Tanner Cypret we record with a MacBook so that's why I used this USB mic.
But do they look as cool as these? :q
Weierstrass substitution is also nice
Yup! I also have a vid on that too
1:48 you should just write =-arctanh(u) and continue with that
oh that's really easy!
using substitution, cant you say that for u=1/x, integral(u)=u^2/2 therefore integral(1/x)=1/(2x^2)?
isn't this a problem with substitution? because not all powers of x integrate the same, and depend on how you treat them.
You must not forget about the differential dx. So if you use a u-sub where u=1/x, then du =(-1/x^2)dx = (-u^2)dx, so dx = -du/(u^2). Plugging back into (1/x)dx we get u(-du/(u^2)) = (-1/u)du. In the end, nothing changed, so the substitution method is not applicable for the integral of 1/x
Also if you differentiate what you're saying the answer is you don't get 1/x back
Я не говорю по-английски, но я все прекрасно понял, спасибо большое!
This is just a great explanation , keep it sir
You forgot the constant in the two lines before the last one.
Use weirstrass or multiply by (csc+cot)/same thing
Not at all the same thing. How do you know that you should multiply both sides by csc+cot? It works as a proof but not as science...
At 0:00 I'm the other chase
chase marangu hi chase!
Would it be possible to pull the negative 1 out of the integral (after the u-sub) and get the negative inverse tangent of u out of it?
nevermind... just noticed its 1/(1-u^2) instead of 1/(1+u^2)
You could do something similar to that and go straight to negative inverse hyperbolic cotangent from 1/(1-u^2). It's not as easily understood as the ln function you get from partial fractions though.
That`s really nice!! Do you know a website with a proof? I would really like to see it! Also thanks for your answer ^^
Okay, I've been through a lot of math courses and I've never seen that "cover-up" method for PFD.
Chiborino I learned it in Calc BC
Me neither but it's just zeroing out 1 of the factors in the denominator at a time. In this example, (u-1) goes away when you plug in 1 leaving 1/(1+1) as the coefficient for that linear factor. Hope this helped of you didn't get it.
@@kamarinelson I don't like the method, it's more unintuitive than the original.
MAGNIFICENT
Weierstrass substitution makes it very easy
THANKS
شكررااا روعة ❤
Clever.
chris schmidt thanks!
Nice vid, but please try not to write the solved answer on the board BEFORE you actually solve it, it seems to take away from the intuition of figuring the problem out.
I'm late but could you take out the -1 and then do partial fractions with 1-u^2, and then take the -1 into the fi al logarithm?
you can also use : sin(x)=2tan(x/2)/(1+tan(x/2)^2) after switching cos(x)=sin(pi/2-x)
Yeah that's the same as the Weierstrass substitution.
How does it result the 1/2?
But isn't doing partial fractions method as u did just a big too longer than than normal derivation
Thanks.