this would have been legit ASMR if you did it on parchment paper with a parker pen or a fountain pen, with a mic right up close for those scratchy pen noises. mmmm...Sec(X)
1:15 am, Monday morning here in the UK. I used to be quite popular maybe 13, 14 years ago, now I'm a bit of a hermit, studying maths, with diminishing social skills. Being up at night watching math videos feels much more satisfying than traveling home after a night out, on a night-bus next to drunk party-ers, arguing couples, or other guys needing to show how 'tough' they are. After watching and learning math I usually fall asleep with a little buzzing sensation going through my body; I can live with that.
Good idea, but I would have substituted immediately after you obtained the integrand in the form cos(x)/(1 - sin^2(x)). Let u = sin(x) du= cos(x) and your new integrand would be 1/(1-u^2) integrated with respect to u. I think it simplifies the computations.
And then the integral of 1/(1 - u^2) is equal to arctanh x + c and you or done!! (or artanh x or argtanh x as you may also run across) The inverse hyperbolic tangent function is very useful for integration. Learn it!!
If you multiple secx for (secx + tgx) above and below the integral and consider ( secx + tgx) as “u” in substitution method, you gonna get the same result as presented in the video much more cleaner and faster
There is a clever-er way. You multiply the integral by sec(x)+Tan(x)/sec(x)+Tan(x). You get integral of sec^2(x) times Sec(x)Tan(x)/sec(x)+Tan(x). U=Sec(x)+Tan(x) and du=sec(x)Tan(x) times sec^2(x). That gives us integral of 1/u du = ln|u| +c = ln|sec(x)+Tan(x)| +C. Done
Actually there is an easier solution to this problem, simply stated that the integral of sec x is = to arctanh (sin x) + c. ( or artanh (sin x) + C as it is sometimes written) No need to do partial fractions, the integral is done once you get it into the form of the integral of //(1/(1-u^2))du where u = sin x and du = cos x dx. No need to diddle around with partial fractions when you can just use the inverse hyperbolic tangent function!! Just remember that //( 1/ (1 + x^2) ) dx = arctan x + C -- the inverse trigonometric tangent -- and that //(1/(1 - x^2) )= arctanh x + C which is the inverse hyperbolic tangent. Simple and straight forward! Note that I use a double slash for the integration sign. “//“. (In some text the inverse hyperbolic tangent function is called the inverse argument hyperbolic function and abbreviated as “argtanh”.) IF YOU WERE TO GET THIS INTEGRAL ON A TEST DO IT THIS WAY AND NOT AS SHOWN ABOVE, YOU WILL BE SAVING LOTS OF TIME DOING IT THIS WAY.
Way back in my University days, I memorized this one (not the derivation!) and int ln x dx because they both are a PIA ! ( although int ln x dx is less of a pain). I used to have these 2 and 18 other common forms of Integration memorized. But after the TI 89 came out way back when, I was of the opinion: why bother...takes up too much of my brain's memory which could be devoted to other things.That was around the time I started using Mathematica and Maple as well. Remarkably when I saw the titles of these two videos int sec(x) dx & int ln(x) dx I still knew the answers! Maybe my Alzheimer's is quite as bad as I thought!
The integral //sec x dx = /(cos x)/((cos x)^2) or /cos x)/(1 - (sin x)^2}. Now let sin x = u and then cos x dx = du and we have the integral //(1/(1 - u^2)) which just happens to be artanh(x) + C. Done and done! Get to learn the inverse hyperbolic functions!! Saves u time n effort!! And you do not have to do any partial fraction this way. LEARN THE HYPERBOLIC TRIGONOMETRIC IDENTITIES AND THEY WILL BE VERY GOOD TO YOU IN SOLVING INTEGRALS.
if u mutiplies it for (secx+tgx). u gonna have (sec^2(x) + secxtgx)/(secx + tgx) then say that U=secx + tgx, and derivating this you gonna have du=(sec^2(x) + secx.tgx) dx ( exctly what u have on upper part of fraction, now its just an interger 1/u du.) much less work haha.
Man, this was fun to watch. However, there's a much easier way to integrate this in just a few minutes, just multiply the integral of sec(x) by (sec(x) + tan(x) / sec(x) + tan(x)) and just integrate by parts from there! Gives the same solution, and gets you through the integral in under 3 minutes!
@@hadiyaebrahim1680 OK it's like scratching the bottom of your left foot, where the left heel has been placed on the back of your neck !! try it and tell us if it is easy or not.
I like this one. It can be done much faster by multiplying the integral by (secx + tanx) on the top and bottom. From there, the integral is in the form to be anti-derived into natural log form. (Make u = secx + tanx)
The solution you outline here is the first one that is usually taught in Calculus I. But this relies on a clever trick, Nothing wrong with that but the most straightforward method is to convert the integral of secant x to the integral of cosine x over one minus (sin x)^2 and you are done, having the answer as arctanh (x) + c.
When its 3am on a Friday night and you rather get lost in a calculus youtube rabbit hole than go out with friends
It's easy when you don't have any friends to hang out with....
me now. friday night, 2:31 am
Can you go over the basics of calculus plz that would be great
But I'm having a wonderful time right here.
My friends are the numbers!
LOL!! I LOVE THIS!!!
hey
hey
hey
hey
Flammable Maths LMAOOO you’re so-
**gets back test**
**99% exactly**
“I forgot the ‘+C’ after integrating”
**suicide intensifies**
PLEASE DO MOREEEE. This is perfect for 2 am studiers
This was sooo satisfying
Never thought that math could replace a sleep meditation, but it works 🤓.
Hearing you to talk about calculus/math is so attractive.
I don’t know why. Seriously, I have no idea.
Also, you have a nice voice.
Can you do this on a Taylor series please?
Math is beautiful. And tiring apparently...
this would have been legit ASMR if you did it on parchment paper with a parker pen or a fountain pen, with a mic right up close for those scratchy pen noises. mmmm...Sec(X)
1:15 am, Monday morning here in the UK. I used to be quite popular maybe 13, 14 years ago, now I'm a bit of a hermit, studying maths, with diminishing social skills. Being up at night watching math videos feels much more satisfying than traveling home after a night out, on a night-bus next to drunk party-ers, arguing couples, or other guys needing to show how 'tough' they are. After watching and learning math I usually fall asleep with a little buzzing sensation going through my body; I can live with that.
This is bringing me back to Mrs. Miller's Pre-Calc class, and also making me fall asleep at my desk. Awesome video!
Multiply and divide (secx + tanx)
You hav got integral dt/t where t is secx + tanx. Then you get log(secx + tanx)+c.
Thank me later😂😀
Good idea, but I would have substituted immediately after you obtained the integrand in the form cos(x)/(1 - sin^2(x)). Let u = sin(x) du= cos(x) and your new integrand would be 1/(1-u^2) integrated with respect to u. I think it simplifies the computations.
Yeah i was also thinking that.
And then the integral of 1/(1 - u^2) is equal to arctanh x + c and you or done!! (or artanh x or argtanh x as you may also run across) The inverse hyperbolic tangent function is very useful for integration. Learn it!!
Had a really relaxing time working on this integral before bed! 1 minute in: SOLVED. Goodnight Britain xxox
Going into calc 2 and this stuff is what i live for in math, a math person who wants me to sleep at intergrals
Oh my god I love this.
Love it, if you ever do this again darker background would be awesome! :)
It gave me the same excitement as a horror movie would
If you multiple secx for (secx + tgx) above and below the integral and consider ( secx + tgx) as “u” in substitution method, you gonna get the same result as presented in the video much more cleaner and faster
His beginning description omg me
There is a clever-er way. You multiply the integral by sec(x)+Tan(x)/sec(x)+Tan(x). You get integral of sec^2(x) times Sec(x)Tan(x)/sec(x)+Tan(x). U=Sec(x)+Tan(x) and du=sec(x)Tan(x) times sec^2(x). That gives us integral of 1/u du = ln|u| +c = ln|sec(x)+Tan(x)| +C. Done
brilliant
j'adore !! merci
Actually there is an easier solution to this problem, simply stated that the integral of sec x is = to arctanh (sin x) + c. ( or artanh (sin x) + C as it is sometimes written) No need to do partial fractions, the integral is done once you get it into
the form of the integral of //(1/(1-u^2))du where u = sin x and du = cos x dx. No need to diddle around with partial fractions when you can just use the inverse hyperbolic tangent function!!
Just remember that //( 1/ (1 + x^2) ) dx = arctan x + C -- the inverse trigonometric tangent -- and that //(1/(1 - x^2) )= arctanh x + C which is the inverse hyperbolic tangent. Simple and straight forward! Note that I use a double slash for the integration sign. “//“.
(In some text the inverse hyperbolic tangent function is called the inverse argument hyperbolic function and abbreviated as “argtanh”.) IF YOU WERE TO GET THIS INTEGRAL ON A TEST DO IT THIS WAY AND NOT AS SHOWN ABOVE, YOU WILL BE SAVING LOTS OF TIME DOING IT THIS WAY.
Way back in my University days, I memorized this one (not the derivation!) and int ln x dx because they both are a PIA ! ( although int ln x dx is less of a pain). I used to have these 2 and 18 other common forms of Integration memorized. But after the TI 89 came out way back when, I was of the opinion: why bother...takes up too much of my brain's memory which could be devoted to other things.That was around the time I started using Mathematica and Maple as well. Remarkably when I saw the titles of these two videos int sec(x) dx & int ln(x) dx I still knew the answers! Maybe my Alzheimer's is quite as bad as I thought!
The integral //sec x dx = /(cos x)/((cos x)^2) or /cos x)/(1 - (sin x)^2}. Now let sin x = u and then cos x dx = du and we have the integral //(1/(1 - u^2)) which just happens to be
artanh(x) + C. Done and done! Get to learn the inverse hyperbolic functions!! Saves u time n effort!! And you do not have to do any partial fraction this way. LEARN THE HYPERBOLIC TRIGONOMETRIC IDENTITIES AND THEY WILL BE VERY GOOD TO YOU IN SOLVING INTEGRALS.
if u mutiplies it for (secx+tgx). u gonna have (sec^2(x) + secxtgx)/(secx + tgx) then say that U=secx + tgx, and derivating this you gonna have du=(sec^2(x) + secx.tgx) dx ( exctly what u have on upper part of fraction, now its just an interger 1/u du.) much less work haha.
when you've been a **** but rather than call u out bae just calms u down with some advanced mathematical proofing 😍
This is beautiful
Love your voice, love your topics. Tell me another bedtime story :-)
"brutal"
wHaT??
Oh ok it's 4am. I guess it's normal.
Seriously i loved it. Can you do more?
Just multiply the integrand by (secx+tanx)/(secx+tanx) and then use a u substitution!
Wow. just wow. What have I even stumbled upon?
Beautiful
HAHAHAHAHAH this is so funny bro!
Was this a request?
Man, this was fun to watch. However, there's a much easier way to integrate this in just a few minutes, just multiply the integral of sec(x) by (sec(x) + tan(x) / sec(x) + tan(x)) and just integrate by parts from there!
Gives the same solution, and gets you through the integral in under 3 minutes!
Can you please do more asmr vids? Thanks bud!
Lol, I didn't understand any of this. Well except from the stuff I've already learned
just put horseshoe
1239712938712 likes omg!! xd
Should have been louder, not softer, to wake 'em up!
😆😆😆😆😆😆😆😆
This bish ain’t using substitution.
Zzzzzzz Zzzzzzzz
THIS SOLUTIION IS TOO COMPLIATED. TAKES WAY TOOOO LONG! LIKE SCRATCHING YOU LEFT EAR WITH YOUR RIGHT HAND!!
Thomas Arch i just scratched my left ear with my right hand to check and its not even mildly difficult
@@hadiyaebrahim1680
OK it's like scratching the bottom of your left foot, where the left heel has been placed on the back of your neck !! try it and tell us if it is easy or not.
Thank you! Love this idea, hope you make more like it.
*whispers
Why are we whispering?
Just put a damn horseshoe there
Ahaha
It's funnier to watch when you are joyful and woke up than this "ASMR".
i used to watch this video before i was going to bed when i was repeating my university mathematics 1 course
You made it too long .it is very short.
came to find a genuine answer, found gold instead
def the weirdest video
Weird...
Bruh how does he know
😴😴😴😴😴
i just fell asleep (again)
Please stay that way and don't comment -- all of us will breathe a sigh of relief!
Great video, only tip would be watch how you draw square brackets as some may misconstrue yours as Mod signs.
were where you all this time???? you could have helped meeee ... but better late than never
Quarter to two on a Saturday night. Damn this guy isn’t just good at maths he can predict the future too.
Maths and ASMR. Definitely in love with you. Thank you sooooo much
this is amazing. Can we have some more?
I like this one. It can be done much faster by multiplying the integral by (secx + tanx) on the top and bottom. From there, the integral is in the form to be anti-derived into natural log form. (Make u = secx + tanx)
The solution you outline here is the first one that is usually taught in Calculus I. But this relies on a clever trick, Nothing wrong with that but the most straightforward method is to convert the integral of secant x to the integral of cosine x over one minus (sin x)^2 and you are done, having the answer as arctanh (x) + c.
Please make more.
more asmr
Do more :3
C'est propre