Agreed I believe in using simple examples in the beginning, the idea is to show the concept, before doing a ton on algebra, that distracts from what we are learning. While fighting all the alligators, forgot the purpose was to drain the swamp.
Totally agree. And not just this series of videos. Most of the famous youtube videos about integrals of trigonometric substitution videos explained the steps in a more complicated way instead of just a linear step by step procedure to solve it. They could show all the other steps in the beginning and then later show how to solve it in a more neat way. It took me more time to understand the concept because of all the backtracking.
I'm 14 and I'm probably obsessed with math. I've learned so much through this and textbooks from the library that I probably won't have to *try* in math until college. I'm going into Algebra 1 in 9th grade, but I solved this problem on my own, halfway through the video. Khanacademy is great (so is the library).
"cuz i don't really remember the quotient rule. i've told you in the past that it's somewhat lame..." LOL And yes, I too am addicted to this, partially because of his voice
I just want to say the blackboard tool you are using in this setup looks cool and clean and very easy to follow. Though the chunky messes from the past videos were just as much fun. :-)
hey this is a very good way of doing the integral, but the problem can be solved in less than 2 minutes if you use integration using inverse trig functions, i.e. d/dx 1/a * arctan(u/a) = 1/(u^2+a^2) , so 1/a * arctan(u/a) is the antiderivative
@ 1:13 since this is the arctan formula, i learned that x^2/36 can be rewritten as (x/6)^2, and if u pull the 1/36 out of the integral u are left with 1/36*integral[ dx/(1+(x/6)^2)]. Now with u substitution u can say u=x/6 --> du=(1/6)dx -->6du=dx. substitute back and now u get 1/36* integral [6*(du/1+u^2)]. pulling the six out leaves you with 1/6 and now 1/(1+u^2) is just arctan(u)*1/6 --> which is 1/6 * arctan (x/6) --> (u=x/6)
Actually you can substitute x=6tan(*). Then dx /d(*)=6(sec^2(*))d(*) then dx=6(sec^2(*))d(*) then integral becomes {6sec^2(*)d(*) /36(1/(1+tan^2(*))} This gives integral of {sec^2(*)d(*)/6sec^2(*)} Which gives integral of 1/6{d(*)} This becomes */6 which is the answer. "*" can be substituted by arc tan(x/6).
...........that stuff is beautiful. I would have never seen that, but when it starts coming together at the end and you begin getting cancellations, you just giggle and smile.
Took calc ab in junior year and will take calc bc senior year. Watchin this vid made me realize how much calc I forgot. This vid'll help me refresh before calc b
Here's one thing I've never understood about Trig Substitution: Why are we allowed to set "x" to a Trig Function? Doesn't that restrict the values of x?
I know this is a 7 year old comment but your answer would be that the original function, the one we're integrating has already its domain restricted. (in example, it's a fraction and the denominator cannot be zero or it's a square root and it can't be negative inside the square root) It's handy to know the derivative rules for the inverse trigonometric functions, you'll find their domains are exactly the same as the inverse trig functions. (d/dx)[arc sin f(x)] = f'(x) / sqrt(1-[f(x)]^2), (d/dx)[arc cos f(x)] = -f'(x) / sqrt(1-[f(x)]^2), (d/dx)[arc tan f(x)] = -f'(x) / (1+[f(x)]^2)
This concept of integration is learned with Inverse Trig Functions in Calculus AB/Calculus 1. Trig Substitution is a more difficult concept that is learned in Calculus BC/Calculus 2.
I'm not sure I understand your question, but if you mean manipulating du to fit the given equation, there is a specific rule that your can't do that with variables.
Agreed usually easy to follow, in this case I got lost the moment he went straight to the tangent. I can do the substitutions to remove the radical with no trouble and know which to substitute, in this case I got lost. Hope what I'm saying makes sense.
The timeline of subtitles in Korean language is uncorrected. Check please, Thank you for video, teacher sal 한국어 자막 시간대가 모두 0분 0초로 되어 있어서 자막이 생성되지 않습니다. 수정 부탁드립니다. 감사합니다
damn i think there was a leap in difficulty with these trig subs. Is there a video explaining what i should sub with what? How can i tell when to use sin or cos or tan or sec as a sub?
@daanish24 You are so wrong. It's x^2/36, not x/36, and since there is not an x in the denominator you can not use u sub in the denominator. Also this wasn't really that complex, you're just scared of trig.
yeah JMT videos suck and are slow and boring - Khan videos are much better I just checked out JMT and wow that almost put me to sleep and he picked a super easy example... how does he make this topic more complicated? he just derives simple trig identities you should have learned a year before you got to this point - if that is complicated then you just didn't pay attention in trig class
@sponsoredwalk1 remember this video is to show the trig substitution......SAL is way smarter than u or me.......if he wanted he could have done it in 1 line......we know that : ∫ 1 / (a^2 + x^2)dx = 1/a * arctan(x/a) + c
@sponsoredwalk1 oh!! my bad.......i forgot to tell u that i really liked how u did that......it's my stupid high school where i learned that arctan formula :P.....next i'll try to silmilar problem using ur method :D
if you do the indefinite integral of 1/(36(1+(x/6)^2)) ------------| |------------------ 1/36 * 1/(1+(x/6)^2) = 1/36 * the ind int of 1/(1+(x/6)^2) = 1/36 * arctan(x/6) which is the wrong answer why can't it be done like that? I don't see the logic failing anywhere
***** Algebra 1 in 9th grade? I did that in 7th grade lol and I don't get how you can compare Calculus 2 material to simple algebra. I'm doing this in college right now and it's quite a challenge (we get harder problems than Sal shows).
Sal really went off on a tangent with that derivative.
Ba dum tss
@@Nobody-Nowhere-Nothing Badum bum fsss
When khan starts not making sense to you, that's the moment you know for sure your screwed.
khan is usually good at simplifying math he makes it more complicated than needed on these videos
Agreed I believe in using simple examples in the beginning, the idea is to show the concept, before doing a ton on algebra, that distracts from what we are learning. While fighting all the alligators, forgot the purpose was to drain the swamp.
Totally agree. And not just this series of videos. Most of the famous youtube videos about integrals of trigonometric substitution videos explained the steps in a more complicated way instead of just a linear step by step procedure to solve it. They could show all the other steps in the beginning and then later show how to solve it in a more neat way. It took me more time to understand the concept because of all the backtracking.
Your videos are life savers! I'm taking calculus and a university and learning far more from your videos than from my professor. Thank you so much
I'm 14 and I'm probably obsessed with math. I've learned so much through this and textbooks from the library that I probably won't have to *try* in math until college. I'm going into Algebra 1 in 9th grade, but I solved this problem on my own, halfway through the video. Khanacademy is great (so is the library).
how was college math dude
"cuz i don't really remember the quotient rule. i've told you in the past that it's somewhat lame..."
LOL
And yes, I too am addicted to this, partially because of his voice
I have to say you make this stuff seem like fun, It's almost 1AM and i can't sleep because i want to keep learning.
I just want to say the blackboard tool you are using in this setup looks cool and clean and very easy to follow. Though the chunky messes from the past videos were just as much fun. :-)
hey this is a very good way of doing the integral, but the problem can be solved in less than 2 minutes if you use integration using inverse trig functions, i.e. d/dx 1/a * arctan(u/a) = 1/(u^2+a^2) , so 1/a * arctan(u/a) is the antiderivative
@ 1:13 since this is the arctan formula, i learned that x^2/36 can be rewritten as (x/6)^2, and if u pull the 1/36 out of the integral u are left with
1/36*integral[ dx/(1+(x/6)^2)]. Now with u substitution u can say u=x/6 --> du=(1/6)dx -->6du=dx. substitute back and now u get 1/36* integral [6*(du/1+u^2)]. pulling the six out leaves you with 1/6 and now 1/(1+u^2) is just arctan(u)*1/6 --> which is 1/6 * arctan (x/6) --> (u=x/6)
Thank you so much for making these videos Sal. You may be the only thing keeping me sane in college right now!
Is it just me or when ever kahn explains something in calculus the calculus becomes too easy, either way khan rocks the infinity :D
calculus + nice monitor + 720p video + fullscreen = learning easy on the eyes.
Actually you can substitute x=6tan(*). Then
dx /d(*)=6(sec^2(*))d(*)
then dx=6(sec^2(*))d(*)
then integral becomes {6sec^2(*)d(*) /36(1/(1+tan^2(*))}
This gives integral of {sec^2(*)d(*)/6sec^2(*)}
Which gives integral of 1/6{d(*)}
This becomes */6 which is the answer. "*" can be substituted by arc tan(x/6).
...........that stuff is beautiful. I would have never seen that, but when it starts coming together at the end and you begin getting cancellations, you just giggle and smile.
we can find the answer without substitution using the standard result- integral 1/(a^2+x^2)dx=1/a arctan(x/a)+c.
Thanks so much, tomorrow I'll have an exam, and I didn't know how to solve those problems, now I think I can:)
Took calc ab in junior year and will take calc bc senior year. Watchin this vid made me realize how much calc I forgot. This vid'll help me refresh before calc b
thanks so much. You make it seem very simple and understandable
Here's one thing I've never understood about Trig Substitution:
Why are we allowed to set "x" to a Trig Function? Doesn't that restrict the values of x?
I know this is a 7 year old comment but your answer would be that the original function, the one we're integrating has already its domain restricted. (in example, it's a fraction and the denominator cannot be zero or it's a square root and it can't be negative inside the square root)
It's handy to know the derivative rules for the inverse trigonometric functions, you'll find their domains are exactly the same as the inverse trig functions. (d/dx)[arc sin f(x)] = f'(x) / sqrt(1-[f(x)]^2), (d/dx)[arc cos f(x)] = -f'(x) / sqrt(1-[f(x)]^2), (d/dx)[arc tan f(x)] = -f'(x) / (1+[f(x)]^2)
" You SMARRRTTT matter of fact you a GENIUS"
This concept of integration is learned with Inverse Trig Functions in Calculus AB/Calculus 1. Trig Substitution is a more difficult concept that is learned in Calculus BC/Calculus 2.
TY you explain it in a way I can understand it rather than having memories and regurgitate it
I'm not sure I understand your question, but if you mean manipulating du to fit the given equation, there is a specific rule that your can't do that with variables.
At 2:57 why is it not plus minus if you square root both sides?
Agreed usually easy to follow, in this case I got lost the moment he went straight to the tangent. I can do the substitutions to remove the radical with no trouble and know which to substitute, in this case I got lost. Hope what I'm saying makes sense.
wow, i don't know why, but stuff like this blows my mind
The timeline of subtitles in Korean language is uncorrected. Check please, Thank you for video, teacher sal
한국어 자막 시간대가 모두 0분 0초로 되어 있어서 자막이 생성되지 않습니다. 수정 부탁드립니다. 감사합니다
damn i think there was a leap in difficulty with these trig subs. Is there a video explaining what i should sub with what? How can i tell when to use sin or cos or tan or sec as a sub?
How do you know what to use 1+Tan^2x or 1+Cot^2x?
Shouldn't sqrt(x^2) be equal abs(x), not x?
I kinda missed the sniff at the end from vid 1
is there a video on t substitution? like t=tan(x/2)?
In some cases you can't just know a formula to do it. In my tests they ask for you to solve using trig substitution methods.
why is d theta become theta? its treated like a constant?
same question as Hawkilla how do you know which one to use ???
@daanish24 You are so wrong. It's x^2/36, not x/36, and since there is not an x in the denominator you can not use u sub in the denominator. Also this wasn't really that complex, you're just scared of trig.
definition of awesome should be changed to 'khanacadamy'
That is crazy!
what's wrong with giving the formulas as well? with a formula it takes about 10 seconds instead of 10 minutes
quotient rule is so much easier tbh when dealing with trig functions
Thank you sir
Why wouldn't I do it with the special function method!
Khan makes this topic much more complicated than it needs to be. If you want better videos on trig integrals, watch PatrickJMT's videos.
yeah JMT videos suck and are slow and boring - Khan videos are much better I just checked out JMT and wow that almost put me to sleep and he picked a super easy example...
how does he make this topic more complicated? he just derives simple trig identities you should have learned a year before you got to this point - if that is complicated then you just didn't pay attention in trig class
If only there were clones of you in all schools around the world
the only thing that confuses me is why he uses theta when he could just know the rule that 1/(a^2 + u^2) = (1/a)arctan(x/a) + C
The only thing I don't understand is how to choose which trig. identity to plug in.
what i do is if its a 1-something^2 i do sin^2 + cos^2 = 1
if its 1 + something^2 i do 1 + tan^2 = sec^2
Yeah i wish that the examples that we had to do were this easy :(
Nice to know how to do it but Id rather save myself eight minutes on the test by memorizing the arctan formula.... =(1/a)arctan(x/a)
i have a test and this comment here has saved me
@sponsoredwalk1 remember this video is to show the trig substitution......SAL is way smarter than u or me.......if he wanted he could have done it in 1 line......we know that :
∫ 1 / (a^2 + x^2)dx = 1/a * arctan(x/a) + c
I've never heard sec called by what I'm guessing is its full name, I just always call it sec :P
i got lost when he derived tan2 theta
lol i was looking for a video explaining how to find the integral of sin^2x dx and i came across this...o.O i must need to go back 20 videos
Yaaay I got it! :D
@sponsoredwalk1 oh!! my bad.......i forgot to tell u that i really liked how u did that......it's my stupid high school where i learned that arctan formula :P.....next i'll try to silmilar problem using ur method :D
I am starting to love math he explained it well there are some inbetween steps we must know i try to to explain those on my channel check it out
if you do
the indefinite integral of 1/(36(1+(x/6)^2))
------------| |------------------ 1/36 * 1/(1+(x/6)^2) =
1/36 * the ind int of 1/(1+(x/6)^2) =
1/36 * arctan(x/6) which is the wrong answer
why can't it be done like that? I don't see the logic failing anywhere
Daski69 I was wondering the exact same thing!
+Daski69 Its because when you substitute x/6=t than dt=dx/6 -> 6dt = dx so you get 6/36 int (dt)/1+t^2 = 1/6 arctg(t)+c = 1/6arctg(x/6) + c
i love you.
@daanish24 that's how I would have done it as well
THANK YOU :((((
VERY VERY MUCH :(( I get it now!
i never knew Bob Ross taught Calculus :D
@sponsoredwalk1 awesome!!....BUT remember evry1 is not as sharp as u :D.....this video is for people who is trying to get into the VOID of CALCULUS
Anyone here in 2018?
tru dat.
coool
Are you a wizard?
couldve just used inverse trig
***** Algebra 1 in 9th grade? I did that in 7th grade lol and I don't get how you can compare Calculus 2 material to simple algebra. I'm doing this in college right now and it's quite a challenge (we get harder problems than Sal shows).
When khan calls the quotient rule lame. ouch!
Kuch samajh nhi aarha
yea im lost
That was fucking amazing lol
ayy cool thanks +1
This is ridiculous! Why are you making this so complicated? Is this the Common Core way of doing it?
"Why is college level math complicated"
Really makes you think doesn't it
thumbs up if you saw the arctan right away
all the noobs can't handle khan