+blackpenredpen I'm doing A Level Maths. We do U-Substitution but this trig for algebra substitution looks more advanced and the reasoning isn't known to me. Why can't you do U=x^2-9?
Fantastic proof, but there could a little inaccuracy. First of all, the domain of the function f(x) in the integral is D = {x=3} In D, f(x) is positive when x is positive, negative when x is negative. In the "teta world", sqrt(tan^2(teta)) = |tan(teta)|, so your solution is correct if and only if tan (teta) = tan(arcsec(x/3)) is >=0, and this happens if and only if x>=3. So, the correct solution of the integral is your solution multiplicated by sgn (x).
You can avoid trigonometric functions here First calculate by parts D I + sqrt(x^2-9) 1/x^3 - x/sqrt(x^2-9) -1/2*1/x^2 -1/2sqrt(x^2-9)/x^2 +1/2Int(x/sqrt(x^2-9)*1/x^2,x) Let u = sqrt(x^2-9) u^2 = x^2-9 2udu = 2xdx udu = xdx x^2=u^2+9 -1/2sqrt(x^2-9)/x^2 +1/2Int(u/u*1/(u^2+9),u) -1/2sqrt(x^2-9)/x^2 +1/2*1/3*Int(1/3/(1+(u/3)^2),x) -1/2sqrt(x^2-9)/x^2 +1/6arctan(u/3) -1/2sqrt(x^2-9)/x^2 +1/6arctan(sqrt(x^2-9)/3)+C
+AlchemistOfNirnroot When you let u = (X^2-9) and take the derivative, you get (du/2)=xdx which doesn't show up in your numerator so it doesn't help simplify things because you can't integrate x with respect to u. Hope that's helpful, I am no expert!
Επειδή ο εκθέτης του χ ( που είναι εκτός της ρίζας) είναι περιττός δεν χρειάζεται η τριγωνομετρική αντικατάσταση. Απλώς αντικαθιστούμε την ρίζα ίσον u !!! Όσο αφορα το χ στον παρονομαστή, απλώς πολλαπλασιάζουμε πάνω και κάτω επί χ!!!
You can and there’s nothing wrong with that, but sin(2(sec^-1/(x/3))) looks much more intimidating than the answer he got, if you can simplify your answer you should
Well if you let x=3sin(u) where u=arcsin(x/3) and dx=3cos(u)du then after doing all the steps for solving the integral you'll get I=sqrt(3-x^2) + 3lnIx/(3+sqrt(3-x^2)I + C.
Lai Linder , the reason for this is when we perform this substitution, we use x=3sec(theta), because of this relationship, theta is given as the arcsec(x/3). The reason arccos and arc sin are used is that they define a relationship in theta, and the original integral is with respect to x and not theta
since dx is the derivative of x, and x equals to sec(u), then dx equals to the derivative of sec(u), which is d/du (sec(u)) = sec(u)tan(u), which means that dx = sec(u)tan(u)du. Sorry for using u, don't have a key for theta
The term power reduction formula refers to the formula which you derived in the video ruclips.net/video/y-9iRPENXoU/видео.html and using it in this context is confusing
No es un inglés muy complicado, además, si pones atención a lo que hace en cuanto a la parte matemática, podría hablar en cualquier idioma y no habría problema alguno.
unnecessary comment, clearly English is not his first language, however he is still trying and helping us learn. He is doing something for us (which he doesn't have to do) and you are being offensive.
I love how much reduction and simplification you have to get to sin^2
Thank you very much. I had a very similar problem but just different numbers. Take care!
There is an exercise involving right triangle wich helps derive u subsitution for these integrals
After that long integral imagine forgetting + C.
Yes
The teacher: minus 5 points for forgetting +C
this is what im looking for. thank you soooo much
you're welcome
Thank you so much. I was super confused on the part where we change from theta to x again and inserting it back into the function.
Been doing trig sub as an adult who's teaching himself for fun. I got this right first Try. Time to learn application of integrals!
you are the best!!!!!!
You are awesome!!!!!! well done all around.
YOU SAVED MY LIFE THANK YOU
i got stuck after sin^2(x) thank you :D
you are the major reason why I know trigonometry substitution :D
He has no idea what he is doing, the ball knows alll!
Yup. The ball is my spirit!
+blackpenredpen I'm doing A Level Maths. We do U-Substitution but this trig for algebra substitution looks more advanced and the reasoning isn't known to me. Why can't you do U=x^2-9?
+AlchemistOfNirnroot. if you use u=x^2-9 then du would be 2x and you wouldnt be able to plug in du because it has a variable in it
JohnsonDirtBikes Oh, thanks.
You can still plug in du for dx, you just have to solve the equation for x and plug in into the equation
The only reason I could do this is because of you, dr peyam, flammable maths and fematika (although he does higher level maths)
Thank you sooooo much!
saved my day!
Thank you so much! Your videos are great to check where I've gone wrong when I get stuck.
Yay!!
@@blackpenredpen try this one out...integral of 20x(x^3+3)^9
@@alexkaranja9981 Just expand the integrand and then do the integration, it's gonna be a very long answer.
*Nobody* : *Integration* *is* *difficult* .
*Blackpenredpen* : *Hold* *my* *pens* .
I knew how to do this yay!
Fantastic proof, but there could a little inaccuracy. First of all, the domain of the function f(x) in the integral is D = {x=3} In D, f(x) is positive when x is positive, negative when x is negative. In the "teta world", sqrt(tan^2(teta)) = |tan(teta)|, so your solution is correct if and only if tan (teta) = tan(arcsec(x/3)) is >=0, and this happens if and only if x>=3. So, the correct solution of the integral is your solution multiplicated by sgn (x).
Subscribed ! thanks a lot !
THANKS A LOT.
thank´s, it help me so much
thank you very much!!
Can you do a video on a Integral Problem for sqrt(9-x^2)/x^3
Thanks sir ❤
Thank you sir so much
Thank you so much !!!
You need to change your name of youtube to blackpenredpenbluepen hahaha thanks for u teach time!! u help to me so much sorry for my bad english :P
Gracias eres increíble 💕😊
Thank you Mr math guy!!!
You can avoid trigonometric functions here
First calculate by parts
D I
+ sqrt(x^2-9) 1/x^3
- x/sqrt(x^2-9) -1/2*1/x^2
-1/2sqrt(x^2-9)/x^2 +1/2Int(x/sqrt(x^2-9)*1/x^2,x)
Let u = sqrt(x^2-9)
u^2 = x^2-9
2udu = 2xdx
udu = xdx
x^2=u^2+9
-1/2sqrt(x^2-9)/x^2 +1/2Int(u/u*1/(u^2+9),u)
-1/2sqrt(x^2-9)/x^2 +1/2*1/3*Int(1/3/(1+(u/3)^2),x)
-1/2sqrt(x^2-9)/x^2 +1/6arctan(u/3)
-1/2sqrt(x^2-9)/x^2 +1/6arctan(sqrt(x^2-9)/3)+C
Thank you so much
thanks for all the help!
+AlchemistOfNirnroot When you let u = (X^2-9) and take the derivative, you get (du/2)=xdx which doesn't show up in your numerator so it doesn't help simplify things because you can't integrate x with respect to u. Hope that's helpful, I am no expert!
You will get to integral (u/x^3)(u/x) du
Επειδή ο εκθέτης του χ ( που είναι εκτός της ρίζας) είναι περιττός δεν χρειάζεται η τριγωνομετρική αντικατάσταση. Απλώς αντικαθιστούμε την ρίζα ίσον u !!! Όσο αφορα το χ στον παρονομαστή, απλώς πολλαπλασιάζουμε πάνω και κάτω επί χ!!!
Hey ! I've been watching your videos for a while, but I still cannot find in my books the inverse function of the sec x, what is it exactly?
Thank U🥺❤️.
MX
thank you bro
Why not just use theta = sec^-1(x/3) in place of sin(2theta), in other words, just plug in theta inside the sin(2theta)?
You can and there’s nothing wrong with that, but sin(2(sec^-1/(x/3))) looks much more intimidating than the answer he got, if you can simplify your answer you should
@@harrywang2795if that expression (sin(2*sec^-1(x/3))) were to have the limit as x approaches infinity, do you know what the answer would be?
@@gingermuro6388I’m not sure, sorry
@@harrywang2795 it’s okay lol, i’m just stuck on a problem, but thanks anyways!
@@gingermuro6388 No problem!
Thank you!!
can +you please make a video of x=a.sec θ and show on the triangle please?
Can you help me with the exercise #19, sect 7.3? please
thanks for the video :3
absolutely bonkers problem
Thanks
It really shows that you NEED to know all the trigonometric identities
Can you solve the same problem?the only diffrence is sqrt 9-x^2 and instead of x^3 only x?
Well if you let x=3sin(u) where u=arcsin(x/3) and dx=3cos(u)du then after doing all the steps for solving the integral you'll get I=sqrt(3-x^2) + 3lnIx/(3+sqrt(3-x^2)I + C.
Hey BPRP! Is it true that you prefer to use integration by parts now to integrate sin^2(theta) when using trigsubs?
in the final answer why did you write sec^-1 but not sin^-1 cos^-1 ?
Lai Linder , the reason for this is when we perform this substitution, we use x=3sec(theta), because of this relationship, theta is given as the arcsec(x/3). The reason arccos and arc sin are used is that they define a relationship in theta, and the original integral is with respect to x and not theta
Are not*
+Andreus Brammer thanks, I've successfully passed my calculus 2 course and this all makes sense. Practice makes perfect : )
why are we allowed to omit the absolute value when we take the square root of tan squared theta?
i think it might be that the square root is only positive while talking about functions. Don't quote me on that though.
thx
why not √(secθ)^2-1 = | tanθ | abs.
WHAT A MAN
Hey, you put a 9 after removing aswelll
how did sin^2 teta turn into 1/2 (1-cos 2teta)?
That’s a formula you memorize
I need to do reciprocal of this but 9-x²
your answer is correct if tan(theta)>0 but false when tan(theta)
How did he get value of dx?
since dx is the derivative of x, and x equals to sec(u), then dx equals to the derivative of sec(u), which is
d/du (sec(u)) = sec(u)tan(u), which means that dx = sec(u)tan(u)du. Sorry for using u, don't have a key for theta
can you do sqrt(x^2-16)dx
I’d have done it differently I’d have either used 3 cos x or 3 sin x
Hello how are you?
great great
The answer given is incorrect when x 0, SGN(x) = -1, if x < 0, SNG(0) = 0.
good work but can you see me integral with suite
how to integrate x^3(√ 4-x^2)dx
Break the x^3 up as x^2*x, and then let u = 4-x^2. That's your hint.
gracias por la explicacion amigo, saludos desde venezuela. El idioma no es barrera para transmitir conocimientos
the thumbnail says dt; you just put a t in the numerator
Ah, that’s a mistake.
The thumbnail said “dt” but is in terms of x
wait but i thought 1-2cos(2x) = (sin^2(x))^2 not sin^2(x)
dx , but not dt in title
You used a blue pen.
NYC qst
The term power reduction formula refers to the formula which you derived in the video
ruclips.net/video/y-9iRPENXoU/видео.html
and using it in this context is confusing
dx not dt
Ojala hablaras español
No es un inglés muy complicado, además, si pones atención a lo que hace en cuanto a la parte matemática, podría hablar en cualquier idioma y no habría problema alguno.
So I'm on top of my game now? I haven't even started my calculus 1 course....
I would integrate this by parts and then substitute for square root
eh, isn't that kinda too much work?
He is so young here lol
Kinez zadao alo care gore oznaciiiiii.
You need to work on your pronunciation
unnecessary comment, clearly English is not his first language, however he is still trying and helping us learn. He is doing something for us (which he doesn't have to do) and you are being offensive.