you just watched: integral of sqrt(1+x^2) by hyperbolic substitution now by trig sub: ua-cam.com/video/O6i5zeoIlsM/v-deo.html and by Euler's sub: ua-cam.com/video/7lPb89DqhVY/v-deo.html
Really enjoy your channel. Question: the above integral can be done using regular trig substitution. When would it be advantageous to use hyperbolic substitution instead of regular trig substitution? I would imagine it would be if you knew the shape you are working with is hyperbolic? also, are you on Patreon? Would like to contribute as a supporter.
Never do these root problems using hyperbolic or trigonometric substitution. The slick way is to do it by parts because it is so much easier. Take u =(1+x^2)^(1/2) and dv =dx. The original integral will return itself after adding and subtracting 1 in the numerator of the second term and splitting it up Then it can be transposed to the left side and both sides divided by 2 to get the final answer. If you can not follow this , I can rereply and go through the steps
We can do it by trig substitution also because sec²x=1+tan²x. Answer will be ln|√(x²+1)+x|+c Well awesome video for the introduction of hyperbolic substitution in calculus
In this method, I can't understand how you can say "Let x = sinh(t)". How can you say that? What if I say x = b^2 - 4*ac, would that be a correct substitution mathematically? Using u-substitution makes sense since you are just "Naming" the expression with a variable, but using a mathematically defined function does not make sense to me. It seems like the whole integral has changed.
Since sinh(t) is a one-to-one function, you can do that. You are essentially defining t as arcsinh(x), but implicitly instead of explicitly. You could define x as b^2 - 4*a*c, but it won't help you very much, since you now have an unconstrained system with 2 degrees of freedom when defining your new variables.
I used the substitution x=isint. Then result was (isin-¹(-xi) /2)+x(1+x²)½/2 Then I used the formula cosr+isinr=e to the power ir. I set r=sin-¹(-xi) to get the value of isin-1(-xi) and got the same answer.
Dude I just got into uni and I have no slightest idea about this "cosh t" "sinh t" thing, care to explain? I know trigs ok for a high school graduate but I've never seen those
In a simple way, while trig functions are on the circunference, the ones with "h" that means hyperbolic are a similar thing, but in the hyperbole. Look at that image and you'll understand: upload.wikimedia.org/wikipedia/commons/thumb/b/bc/Hyperbolic_functions-2.svg/1200px-Hyperbolic_functions-2.svg.png
Let me show you a method that requires little calculation Let √（x^2+1)=y, then xdx=ydy, dx/y = dy/x, which is also equal to d(x+y)/(y+x) The given integral is ∫ydx = xy - ∫x(dy/dx)dx = xy - ∫x(x/y)dx =xy - ∫(y^2-1)/ydx = xy - ∫ydx + ∫dx/y therefore , 2∫ydx = xy +∫dx/y As mentioned before, dx/y = dy/x = d(x+y)/(y+x) So ∫dx/y = ∫d(x+y)/(y+x) = log(x+y) Finally, ∫ydx = 1/2(xy +log(x+y))= 1/2 ( x√(x^2+1) + log (x + √(x^2+1))
You're not wrong. Here he's taking the hyperbolic substitutions, which involve sinh and cosh functions. You'll study it in your higher classes don't worry
Thanks BPRP, as one of my math professors had said: "Derivation is a technique, integration is an art."
Maybe he meant differentiation?
probably
@@chessandmathguy They're both used interchangeably
I think he stole it from the Norwegian mathematicain called Viggo Brun, he said «differentiation is a craft, integration is an art»
you just watched: integral of sqrt(1+x^2) by hyperbolic substitution
now by trig sub: ua-cam.com/video/O6i5zeoIlsM/v-deo.html
and by Euler's sub: ua-cam.com/video/7lPb89DqhVY/v-deo.html
into the "t world" gets me every time
Holy crap, I forgot how to do this. I gotta refresh before my semester starts.
Really enjoy your channel. Question: the above integral can be done using regular trig substitution. When would it be advantageous to use hyperbolic substitution instead of regular trig substitution? I would imagine it would be if you knew the shape you are working with is hyperbolic? also, are you on Patreon? Would like to contribute as a supporter.
For me, it is easier to deal with hiperbolic stuff if I use their definition interms of exponentials
sinch
Ty man that just saved me never heard of sin hyperbolicus before in school suddenly university kicks in :D
I never learned how to do hyperbolic functions when I took calc 2.
Easier to do by parts with unity as the second function.
Thank you, you help me a lot for my internship :)
Never do these root problems using hyperbolic or trigonometric substitution. The slick way is to do it by parts because it is so much easier. Take u =(1+x^2)^(1/2) and dv =dx. The original integral will return itself after adding and subtracting 1 in the numerator of the second term and splitting it up Then it can be transposed to the left side and both sides divided by 2 to get the final answer. If you can not follow this , I can rereply and go through the steps
Hi, @johngreen3543.
Would you mind going through the steps for me? intrigued by your methodology, just slightly struggling to follow.
Thanks!
But sqrt((cosh(t))²) is abs(cosh(t)) right? And cosh(t) * abs(cosh(t)) can be negative, so isn't it different from cosh(t) ² ?
Djdjcjcjcj Jfnfjfidnf that must be in the vid
Cosh t can never be negative, if you look on the graph the range is greater than 1, so you take the positive value of the square root
why was I expecting him to draw a catenary just like how you draw a right triangle with trig sub?
thank you sir
Sir,what about this one:; cosh²z=(½(e^2z+e^-2z))². Where is it applied coz my teacher used it in the same question??
We can do it by trig substitution also because sec²x=1+tan²x.
Answer will be ln|√(x²+1)+x|+c
Well awesome video for the introduction of hyperbolic substitution in calculus
So is this going to be your new classroom? :P
I will be gone soon, unfortunately.
Gone from that classroom, or gone from UA-cam ??
From that classroom in Berkeley.
I will have to go back to LA next week and be with my usual whiteboards
Ah cool cool.Well man wish you a good summer.
Zylo Sol thanks
Nice video man, you'd make an awesome teacher :)
Thanks!
Btw, I have been a teacher for many years already : )
Finally got the video I was searching for!!!!❤️❤️❤️❤️🇧🇩
well done but this is the simple case - what if the 1 was replaced by 4 ?
Thank you for another great video /lecture on Integration using Hyperbolic Functions.
Great Video! Solvin integrals is allways fun!
Thank you so much 🌹
In this method, I can't understand how you can say "Let x = sinh(t)". How can you say that? What if I say x = b^2 - 4*ac, would that be a correct substitution mathematically?
Using u-substitution makes sense since you are just "Naming" the expression with a variable, but using a mathematically defined function does not make sense to me. It seems like the whole integral has changed.
Since sinh(t) is a one-to-one function, you can do that. You are essentially defining t as arcsinh(x), but implicitly instead of explicitly. You could define x as b^2 - 4*a*c, but it won't help you very much, since you now have an unconstrained system with 2 degrees of freedom when defining your new variables.
Damn that's fancy! I just got a BS in math, but never heard of doing it like this.
I hope to get mine in a few years.
I love your videos, they help in math and even to have fun, hello from Mexico
This is great! Thanks for watching them too!
Can we do x=tan theta, and then use the identity and get rid of square root and move on from there
Very nice.
I've known this for a while but thanks for finally talking about it
Sorry, Goddard yay!! I know. It's overdue..
The solutions picture is not clear.
You always impress me
Mustafa Jassim thank you
wow this is great!!!! thank you much
My pleasure
Thnks 😍😍😍
Great workshop!
Just work
coooool!!!! thanks!
I used the substitution x=isint. Then result was (isin-¹(-xi) /2)+x(1+x²)½/2 Then I used the formula cosr+isinr=e to the power ir. I set r=sin-¹(-xi) to get the value of isin-1(-xi) and got the same answer.
isint=sinht
@@taidghusflynnius that's not true
why is sqrt(cosh^2(t)) is not |cosht| but just cosht ?
Because cosh t cannot less than 0 for t ∈ ℝ
Shouldn't it have been the absolute value of cosh(t) when it was the square root of cosh²(t)
To anyone possibly reading this: the absolute value of cosh(t) is just cosh(t), since its reach is from and including +1 to +infinity.
You should change your name to WhitechalkWhitechalk
Thx
Can you also solve it my using trig sub: x=tant?
I tried, cant seem to get it tho. Ik its possible.
S'il vous plaît : primitive de (x^4)/√(x^10-2) et primitive de √[(x^2)+c] où c = constante
Legend
I love math
Isn't sqr(cosh(t) ^2)=abs(cosh(t)l?
Cosh(t) is always greater or equal to 1 :)
this is so fucking lit bro! This vid got me fuckin domed nigga! FUCK YEAH!
Dude I just got into uni and I have no slightest idea about this "cosh t" "sinh t" thing, care to explain? I know trigs ok for a high school graduate but I've never seen those
In a simple way, while trig functions are on the circunference, the ones with "h" that means hyperbolic are a similar thing, but in the hyperbole. Look at that image and you'll understand:
upload.wikimedia.org/wikipedia/commons/thumb/b/bc/Hyperbolic_functions-2.svg/1200px-Hyperbolic_functions-2.svg.png
Thanks for nice presentation. black board , white chalk .DrRahul
I thought everyone did this kind of integral like this like i genuinely dont know whst other method id use
So,dx=cosh(t)dt ...but you need to reduce it,i think.. 1/cosh(t)*integral(cosh^2(t)dt).
1+(sinx)^2 not= cosx
Sinx)^2-1=cos^2xx
Let x=tan(t)
Am i a joke to u?
coated
Vay ak 👍
lmao then why we doing trigo substitution??? this one is speed
this integral is a single line problem it can be done without any derivation or substitution bruh
Let me show you a method that requires little calculation
Let √（x^2+1)=y, then xdx=ydy, dx/y = dy/x, which is also equal to d(x+y)/(y+x)
The given integral is ∫ydx = xy - ∫x(dy/dx)dx = xy - ∫x(x/y)dx =xy - ∫(y^2-1)/ydx
= xy - ∫ydx + ∫dx/y
therefore , 2∫ydx = xy +∫dx/y
As mentioned before, dx/y = dy/x = d(x+y)/(y+x)
So ∫dx/y = ∫d(x+y)/(y+x) = log(x+y)
Finally, ∫ydx = 1/2(xy +log(x+y))= 1/2 ( x√(x^2+1) + log (x + √(x^2+1))
am I the only one who has no sound at your videos?
I think the sound is working ok
It’s wrong it’s cos^2 + sin^2 = 1
Your head
You're not wrong. Here he's taking the hyperbolic substitutions, which involve sinh and cosh functions. You'll study it in your higher classes don't worry
Wrong method
But hyperbolic sin(2t).
coshit!
r u ok?
Sorry, blackpenredpen! Too many ad interruptions! Disgraceful commercialism for an educational feature. Capitalism gone to the dogs! Goodbye.
Stfu you probably think everyone can do shit for free to let you live your comfortable life.
Finally got the video I was searching for!!!!❤️❤️❤️❤️🇧🇩