Hi, I'm an IB Maths HL student and I've found your videos absolutely brilliant! They cover everything I'm doing in class and are really improving my understanding. I can't thank you enough!
I'm just learning about integrals (and i have a calc final in two days, lol), and i've got to admit that that was the tastiest way to solve the indefinite integral, substituting the trig identity and so forth... Algebraic manipulation as such is truly an art form.
I personally think it is easier to rearrange the denominator to be 1/√1-u^2 trig identity: √(3-2x^2 ) √(3(1- 2⁄3 x^2 ) ) √(3(1-(√2/√3 x)^2 ) ) Let u=(√2/√3) x √3/√2 du = dx after elimination the √3 from top and bottom. integral of 1/(√2(1-u^2)) integral of 1/√1-u^2 is arcsine x so 1/√2 arcsine(u) +C then plug u=(√2/√3) x back in and you get same answer.
OMFG thank you! i was just devastated by my math teacher telling me that we won't learn trig sub this year, and without a teacher it is hard to study stuff you haven't had yet, but now i CAN do it, next year is gonna be a piece of pie!
It seems like there's so much more to this. That's neat that a trig function can be used to find a way to evaluate a non-trig function. Now I want to know why. That seems so elegant... and you show it so well (my instructor rushes through and gives the class about 5 seconds, before most people get a chance to put down their pencils from copying stuff down, where he asks "So is this okay... everyone got it? Good, moving on!") Thanks Sal.
Sal: The first time I saw the video, it clicked. The second time, I followed along with my homework problem and it works beautifully. Now to see if I can do the next one on my own :)
Aabid Panchbhaya m.intmath.com/methods-integration/6-integration-inverse-trigonometric-forms.php Check out the first formula on that page. The integral that Sal is doing fits that form if a is the square root of 3 and u is the square root of 2 times x. If you can't remember that formula during a test, then by all means use a trig sub. I was just saying that trig sub probably wouldn't be the first place I would go here.
@@dorysoldoff1694 Just learning calculus. If I'm not making a mistake, we can just extract sqrt(2) from the radical and then as a constant in front of the integral. Under the radical we get (sqrt(3)/sqrt(2))^2 - x^2. And this gives us the pattern for the arcsin rule.
The way i like to think of it is that it is the integral of 1/sqrt(2)*x dx. The constants are just going to be pulled out front after the integration and looks really ugly when it is inside the integral, so it's much easier to take the constant (1/sqrt(2)) out in the beginning and integrate afterwards. Hope it helps!
Anytime you have a constant, you can simply pull it out in front. This applies to both integrals and derivatives. For example, d/dx 2x^2 is the same as 2 (d/dx x^2), and int(10xdx) = 10 int(xdx). Also try searching for "constant multiple rule integral" for more. Hope this helps!
@espen180 and christopherchubb no, it wouldnt be |cos(θ)| because its an INDEFINITE integral, so u can just drop the abs. value bars. but it it were a DEFINITE integral, then ud have to check the domain of cos(θ) with regard to the upper and lower limits.
I find trigonometric substitution very useful, because, not only does it help solving integrals involving the radicals, but it also helps solving integrals involving inverse trigonometric and inverse hyperbolic functions by derivation. In other words, with this method in mind, you don't have to remember two complicated theorems about which the integration of a function is a family of the two inverse functions is stated, while there are many basic integration rules to remember already. I just love it.
wheever u have a sqrt(a^2-x^2), its always better to use x=asin(theta), cuz when u get dx, u wont be stuck with any negatives (cuz derivative of sinx is POSITIVE cosx) so, in my opinion id use (cos)^2 = 1 - (sinx)^2 instead of the other one, but u can use either way.
Sometimes the passivity of watching a video turns off the brain to making even simple connections that it otherwise would easily make. He is simply doing an exemplary job of Not making assumptions about the audience's knowledge which is important when you are teaching using a forum that does not lend itself to asking questions/clarifications. Plus, you are getting a nice review, recognizing that you are actually simply building on things u already know- a nice thing IMO.
@uccvertigo In this case I don't think it's necessary, because we are already under the square root. So that means that (2/3)x^2 has to be smaller than 1 or else we'd have an imaginary number.
I solved for 2/3x^2 = cos2o, and I get a different answer... mine comes out as -1/sqrt(2)*arccos(sqrt(2)/sqrt(3)*x), which is NOT equal to 1/sqrt(2)*arcsin(sqrt(2)/sqrt(3)*x). Can I get some help?
I do think you have a point on this one; it's a very divisive issue. However, I still think Sal's "slowness" is justified. I'd wager that most people (including myself) who watch Khan Academy are having a lot of trouble with math anyway. Because videos aren't interactive, I'd much rather risk Sal going over a concept I'm already solid with than for him to skip one that I was never able to understand in class.
@sahiti321 imho, thorough, deliberate teaching is far superior, on every level. If you don't like, pause at the end of the video, look at what he drew, and figure it out for yourself.
Man your really awesome thanks alot. I wanted to ask one thing Can't we solve this by reversing the product rule ? 1 * (3 - 2x^2)^-1/2. then doing the way Sal solved for it before ? I didn't try it yet,but I will after watching this video.
That one is easy! You just have to "see" the 1 multiplying d(Theta). Imagine that Theta is any other ordinary variable: lets call it "x". I'll note integral as "S". Then : S dx = S 1*dx . Remember that 1 is a constant. Therefore : S 1*dx = 1*x (+c) . ... B-t-w the let "k" be any arbitrary constant. We could as well say : S k*dx = k*x (+c) . because, in the other direction, the derivative of k*x (+c) is just equal to k. In the same way: the derivative of 1*Theta is just equal to Theta.
woudl be easier if u would 2/3x^2 wrote as sqrt2/3x^2 and then and then just take substitution t= sqrt2/3x^2, and get the solution for arcsinsqrt2/3x^2) + C...
This isn't too relevant but it's (sin theta)^2, not (sin x)^2. Anyway, the function has no real values for |(2/3) * x^2| > 1, or in terms of x, |x| > sqrt(3/2), because you will end up having to find the square root of a negative number. e.g. Let x = 2 then 1 / sqrt(3 - 2 * x^2) = 1 / sqrt(3 - 2 * 4) = 1 / sqrt(-5)
how is this allowed and why particularly sin & cos combination why not (possibly?) sec & tan or cosec & cot combination whose squares when subtracted also similarly equal to 1
How can you just set 2/3*x^2 = sin(theta) when they are completly different things? 2/3*x^3 can be any real value and sin(theta) can only be max 1 and min -1??
SOMEONE HELP, PLEASE! He states that (2/3)*(x^2) = (sinx)^2 . But what if |(2/3)*(x^2)| is bigger than one? On the other hand |(sinx)^2| can never be bigger than one. How/why is that equivalence allowed?
This is where calc and I hate each other. When you start taking the equation and saying "oh this looks like this other equation I know so I'll just put it in this one's place" I start to hate it. I understand there's a science behind it all, but god is it hard to want to learn with stuff like this.
I wish he would name his videos and descriptions better instead of "Trig Substitution 1: we use trig sub to solve an indefinite integral," "Trig Substitution 2: we use trig sub again," etc
You are correct. An indefinite integral can be calculated without use of actual values of X. If you were to take the definite integral of 1/(sqrt(3-2x^2)) from 0 to 2, you would get a non-real answer because 2 is outside the limits of integration. Foresight will let you see that the non-integrated equation has limits, which are reflected in the integrated equation. Just take a look at the graph of 1/(sqrt(3-2x^2)): www.wolframalpha.com/input/?i=%281%2F%28sqrt%283-2x^2%29%29%29 vs. The graph of arcsin(sqrt(2/3)x)/sqrt(2): www.wolframalpha.com/input/?i=integral+%281%2F%28sqrt%283-2x^2%29%29%29 Both have the same limit.
Once again, Sal coming in soooooo clutch!
Hi,
I'm an IB Maths HL student and I've found your videos absolutely brilliant! They cover everything I'm doing in class and are really improving my understanding. I can't thank you enough!
How do you even think of that....
"I'm gonna do a bunch of these examples so You can get familiar with them."
..
....
.......
* *S N I F F L E* *
RezKue! That was a genius touch that sniffle before the video ended lol 🤪
HAHAHHA lol, I watched this from Khan Academy then moved to RUclips just to see a comment about that Sniffle HAHHAAH
I'm just learning about integrals (and i have a calc final in two days, lol), and i've got to admit that that was the tastiest way to solve the indefinite integral, substituting the trig identity and so forth...
Algebraic manipulation as such is truly an art form.
thank God for Khan academy. the fact that these are free is the greatest thing ever
I personally think it is easier to rearrange the denominator to be 1/√1-u^2 trig identity:
√(3-2x^2 )
√(3(1- 2⁄3 x^2 ) )
√(3(1-(√2/√3 x)^2 ) )
Let u=(√2/√3) x
√3/√2 du = dx
after elimination the √3 from top and bottom.
integral of 1/(√2(1-u^2))
integral of 1/√1-u^2 is arcsine x
so
1/√2 arcsine(u) +C then plug u=(√2/√3) x
back in and you get same answer.
OMFG thank you!
i was just devastated by my math teacher telling me that we won't learn trig sub this year, and without a teacher it is hard to study stuff you haven't had yet, but now i CAN do it, next year is gonna be a piece of pie!
ain't no way I'm gonna remember this at my final exam R.I.P to my grades
It seems like there's so much more to this. That's neat that a trig function can be used to find a way to evaluate a non-trig function. Now I want to know why. That seems so elegant... and you show it so well (my instructor rushes through and gives the class about 5 seconds, before most people get a chance to put down their pencils from copying stuff down, where he asks "So is this okay... everyone got it? Good, moving on!") Thanks Sal.
Sal: The first time I saw the video, it clicked. The second time, I followed along with my homework problem and it works beautifully. Now to see if I can do the next one on my own :)
Missed a day of my Calc 2 class because of the flu. You are a life-saver khanacademy
I have spent so much time trying to figure out trig sub from my notes and book... I couldn't quite get it until I watched these videos. Thank you!!
now i can actually spam calculus homework w/o using a calculator!!
mad props to you Sal
wow i would've NEVER come up with that on my own, this is brilliant stuff.
This is awesome I have an assignment due tomorrow and my prof did a terrible job explaining this concept. You made it clear as day. Thanks.
You don't need trig substitution for this; just substitute u=x*sqrt(2/3) and du=sqrt(2/3)*dx and use the antiderivative of du/(sqrt(1-u^2))=arcsin u.
Oh. Finding a HD vid inside the playlist makes me all happy
I think I love you! I think every calculus student and professor loves you! Thanks a lot!
Great video as always! But, I feel like this integral can be evaluated more easily with a simple substitution and arcsine rule.
wait which subsitution would you make and which arcsine rule?
Integral of du/sqrt(a^2-u^2) = arcsin(u/a). Make the substitution a = sqrt(3) , u = sqrt(2)x.
can you please explain that again?
Aabid Panchbhaya m.intmath.com/methods-integration/6-integration-inverse-trigonometric-forms.php
Check out the first formula on that page. The integral that Sal is doing fits that form if a is the square root of 3 and u is the square root of 2 times x. If you can't remember that formula during a test, then by all means use a trig sub. I was just saying that trig sub probably wouldn't be the first place I would go here.
@@dorysoldoff1694 Just learning calculus. If I'm not making a mistake, we can just extract sqrt(2) from the radical and then as a constant in front of the integral. Under the radical we get (sqrt(3)/sqrt(2))^2 - x^2. And this gives us the pattern for the arcsin rule.
To me this is the most beautiful concept of integral calculus that students encounter in first year. Incredible!
This makes so much more sense now. Thank you!
Khan academy is just great.
Thank you for being such a support
The way i like to think of it is that it is the integral of 1/sqrt(2)*x dx. The constants are just going to be pulled out front after the integration and looks really ugly when it is inside the integral, so it's much easier to take the constant (1/sqrt(2)) out in the beginning and integrate afterwards. Hope it helps!
Anytime you have a constant, you can simply pull it out in front. This applies to both integrals and derivatives. For example, d/dx 2x^2 is the same as 2 (d/dx x^2), and int(10xdx) = 10 int(xdx). Also try searching for "constant multiple rule integral" for more. Hope this helps!
With his chocolate velvety voice, khan brings hope to students everywhere!
@espen180 and christopherchubb
no, it wouldnt be |cos(θ)| because its an INDEFINITE integral, so u can just drop the abs. value bars.
but it it were a DEFINITE integral, then ud have to check the domain of cos(θ) with regard to the upper and lower limits.
I find trigonometric substitution very useful, because, not only does it help solving integrals involving the radicals, but it also helps solving integrals involving inverse trigonometric and inverse hyperbolic functions by derivation. In other words, with this method in mind, you don't have to remember two complicated theorems about which the integration of a function is a family of the two inverse functions is stated, while there are many basic integration rules to remember already. I just love it.
wheever u have a sqrt(a^2-x^2), its always better to use x=asin(theta), cuz when u get dx, u wont be stuck with any negatives (cuz derivative of sinx is POSITIVE cosx)
so, in my opinion id use (cos)^2 = 1 - (sinx)^2 instead of the other one, but u can use either way.
Sometimes the passivity of watching a video turns off the brain to making even simple connections that it otherwise would easily make. He is simply doing an exemplary job of Not making assumptions about the audience's knowledge which is important when you are teaching using a forum that does not lend itself to asking questions/clarifications. Plus, you are getting a nice review, recognizing that you are actually simply building on things u already know- a nice thing IMO.
she is an indian girl who is quite envy about the achievement of sal, this guy is perfect,concise and to the point.
I watched this video without the audio and all I can say is; Brilliant!!!!!!!!!
best teacher on the internet
This was an amazing video. Really easy to understand. Thanks so much Sal
Thank you so much. I was just about to give up of this, this video just saved my self life :D.
@uccvertigo In this case I don't think it's necessary, because we are already under the square root. So that means that (2/3)x^2 has to be smaller than 1 or else we'd have an imaginary number.
this just blew my mind
I agree with Thevidfather, you really get a good intuition of what our doing.
He makes this so much more complicated that it should. You don't need to factor out the 3. Just use it as your "x" by just putting it in a radical.
This is absolutely beautiful!
Man good job explaining this. Well done!!!
thank you so much, you have cleared up so many questions i have had from class. awesome videos sal
You sir, are a gentleman and a scholar.
making vids even when you're sick! thanks Sal.
I solved for 2/3x^2 = cos2o, and I get a different answer... mine comes out as -1/sqrt(2)*arccos(sqrt(2)/sqrt(3)*x), which is NOT equal to 1/sqrt(2)*arcsin(sqrt(2)/sqrt(3)*x). Can I get some help?
It should be under one of the videos in "Solid Revolutions"
I do think you have a point on this one; it's a very divisive issue.
However, I still think Sal's "slowness" is justified. I'd wager that most people (including myself) who watch Khan Academy are having a lot of trouble with math anyway. Because videos aren't interactive, I'd much rather risk Sal going over a concept I'm already solid with than for him to skip one that I was never able to understand in class.
Thank you thank you thank you thank you thank you!!! Why isn't there a standard format for teaching???? You have helped me soo much! :-)
that was fckin magic
Thanks. When I first read this in my text with the absence of an introduction, I didn't know what was going on.
@sahiti321 imho, thorough, deliberate teaching is far superior, on every level. If you don't like, pause at the end of the video, look at what he drew, and figure it out for yourself.
I love you. You have saved me so many times
thank you so much i have an exam in 2 days and this helps a lot
Way to end the video.
And you are a great teacher.
new captions :O
so happy now i dont have to increase the media player size to b able to read what you are doing.
Man your really awesome thanks alot. I wanted to ask one thing Can't we solve this by reversing the product rule ? 1 * (3 - 2x^2)^-1/2. then doing the way Sal solved for it before ? I didn't try it yet,but I will after watching this video.
That one is easy! You just have to "see" the 1 multiplying d(Theta).
Imagine that Theta is any other ordinary variable: lets call it "x". I'll note integral as "S". Then :
S dx = S 1*dx .
Remember that 1 is a constant. Therefore :
S 1*dx = 1*x (+c) .
...
B-t-w the let "k" be any arbitrary constant. We could as well say :
S k*dx = k*x (+c) .
because, in the other direction, the derivative of k*x (+c) is just equal to k.
In the same way: the derivative of 1*Theta is just equal to Theta.
You can just use t for the expression beneath the root and the write it in a form of an exponent. Well at least in this example you can do that.
Great example! Thanx! 😊
Yes, More Videos Please!!!
@GenericCoder its the same if it was integral of (1 dx) that would equal, as per power rule, x. same thing with theta
thank you man. you da real mvp!
You must make a restriction on theta to define the inverse of the trig subst
THANK YOU SO MUCH!! This helps me A LOT :)
WHOA YOU'RE MAGICAL! thank you
When are you allowed to break out factors from integrals like he does at 6:10? What's the rule?
@yws1991 I like how you describe his voice. Nice. It fits!
JUST before the video ends.....SNIFF! funny stuff.
great instruction on the trig sub,
woudl be easier if u would 2/3x^2 wrote as sqrt2/3x^2 and then and then just take substitution t= sqrt2/3x^2, and get the solution for arcsinsqrt2/3x^2) + C...
This isn't too relevant but it's (sin theta)^2, not (sin x)^2.
Anyway, the function has no real values for |(2/3) * x^2| > 1, or in terms of x, |x| > sqrt(3/2), because you will end up having to find the square root of a negative number.
e.g. Let x = 2
then 1 / sqrt(3 - 2 * x^2) = 1 / sqrt(3 - 2 * 4) = 1 / sqrt(-5)
What program is being used here for the math? I would like to start a video series on passing some actuarial tests.
"Would you believe me if i said i'm in love " :)
no
wtf
Finally, a comment I can relate with.
Can't we solve this by reversing the product rule ? 1 * (3 - 2x^2)^-1/2. then doing the way Sal solved for it before ?
how is this allowed and why particularly sin & cos combination why not (possibly?) sec & tan or cosec & cot combination whose squares when subtracted also similarly equal to 1
great video!
Thanks alot I understood it now :).
the derivative of sqrt3/sqrt2 sin theta=sqrt3/sqrt2 --costheta, right? Where is the negative sign?
THANK YOU!
Can we get some more intuition on this process Sal?
Good work
wow i never knew you could do that
I stuck on the same problem! Have you got an answer?
How can you just set 2/3*x^2 = sin(theta) when they are completly different things? 2/3*x^3 can be any real value and sin(theta) can only be max 1 and min -1??
In the beginning, how come you can simply equal 2/3x^2 with (sin theta)^2 to find x? Why would they be equal?
*sniff*
Lmao I accidentally read this just as he sniffed
@natdogrocker
Oh I see thanks alot dude. that makes sense.
SOMEONE HELP, PLEASE!
He states that (2/3)*(x^2) = (sinx)^2 .
But what if |(2/3)*(x^2)| is bigger than one?
On the other hand |(sinx)^2| can never be bigger than one.
How/why is that equivalence allowed?
You are great
nice snort in the end!! lol.., #dang.... I learn this on senior high school...
Won't the square root of cos(θ) squared be the absolute value of cos(θ), or |cos(θ)|?
Sal: "ill do an arbitrary change of colors"
Me: Panik
@E90PAT DAT SNIFF AT THE END MADE NE LOL SO HARD
This is where calc and I hate each other. When you start taking the equation and saying "oh this looks like this other equation I know so I'll just put it in this one's place" I start to hate it. I understand there's a science behind it all, but god is it hard to want to learn with stuff like this.
I wish he would name his videos and descriptions better instead of "Trig Substitution 1: we use trig sub to solve an indefinite integral," "Trig Substitution 2: we use trig sub again," etc
you are a god
Why is the integeral of d(theta) = theta can someone explain this please ?
isnt the square root of cosine squared theta the absolute value of cosine though?
I don't get how sine can be substituted for √3/√2. what chart is he reading it from and if so where can i get it
Excuse me, but what happens if X^2>1 ????? or is it possible only when X^2 is in this interval?
You are correct. An indefinite integral can be calculated without use of actual values of X. If you were to take the definite integral of 1/(sqrt(3-2x^2)) from 0 to 2, you would get a non-real answer because 2 is outside the limits of integration. Foresight will let you see that the non-integrated equation has limits, which are reflected in the integrated equation.
Just take a look at the graph of 1/(sqrt(3-2x^2)): www.wolframalpha.com/input/?i=%281%2F%28sqrt%283-2x^2%29%29%29
vs.
The graph of arcsin(sqrt(2/3)x)/sqrt(2): www.wolframalpha.com/input/?i=integral+%281%2F%28sqrt%283-2x^2%29%29%29
Both have the same limit.