Excellent explanation. There was a little mistake. For the third problem, the denominator has + between (1-x^2) and sqrt(1-x^2). That being said, you would need change the minus sign in your denominator between u and square root of you, v^2 and v, and finally between square root of (1-x^2) and 1 at the ending result.
12 years and counting but bruh u're still helping a lot of students who are unfortunate on math teachers sskksks hope you're earning what's right for you now.
at the beginning of the day i couldn't be able to explain what U-substitution even had to do with anything but after watching all of patricks U-Substitution videos I've learned more than my teacher or my stupid calc book could ever teach me. Kudos from MA.
It'd be pretty cool if in videos where all you are doing is examples, you showed a preview of each type of problem you will do with links to the time that you do each example. It might not be worth the effort, but just a thought! Thanks for all your videos, they have helped me go from a B math student to 94% in Calc I!
I love your little comments catering to the people LEARNING. One of my biggest problems with some professors is they seem to not address or display any understanding or acknowledging the difficulty something might have for people who are LEARNING. Your comments show you are a very good teacher by acknowledging that "when you don't know how to do something, sometimes it seems like the hardest thing ever".
I never would have thought to do a double substitution. Your work is amazing and I've been watching your Integration videos for a few days now and they're helping me so much! Thank you for these videos.
Thank you soo much for these videos. My teacher shows us the easiest example and then expects us to learn the rest on our own, you pretty much are my Calculus teacher, thanks, you rock!
@AhOhy sorry, i do not do hw questions - it could be stuff for a grade and i am not comfortable answering that. however, i will give you a hint: you picked the wrong u sub.
@cincottallll so teach yourself if you are that interested in it! why wait for someone to 'teach' you? although you need some trig as well before you jump into it.
@vhskeva ha! glad you like the videos so much : ) sometimes i turn on like 15 videos at once and get a surreal patrickjmt effect. my wife thinks i am crazy ( i also do it to annoy her!!)
Once again, thank you for these videos which are concise, explain the theory AND how to do it (the algebra part). My calc I teacher does his best to explain these things but it doesnt make sense. I've been using your videos since last summer's Trig class with him and now for Calc I (my husband used your videos as well when he was in school and is the one who suggested you). What a BLESSING! Your videos are the main reason Im grasping the concepts and how to do the problems. And with finals coming up in 8 days, theyre helping me learn and review everything as well! See ya in Calc II. thanks again!
great video very clear! just to mention that to others who is watching this video, the question involving 2 substitutions does not necesserily need 2 subs, u^2=1-x^2 gives same answer.
the first integral can not be integrated. you would use a series expansion to approximate it to any desired amount of accuracy however. the second integral is just arctan(x). for the third, i assume you mean: 1/[x(4x+1)^1/2] at a glance like you would let u = (4x+1)^1/2 solve for x, find dx, plug it all in and you would have a problem that you would have to use partial fractions on
Thanks a lot:I have an exam tomorrow, I was absent when my class was doing substtitution techniques, I got a handout my teacher made on it so I can study but it was like bullshit...so, there I was looking for help on the internet, and found just the one! It was easy to understand I didnt need to re-watch it many times to grab the idea of it...too bad YOU arent my teacher!!!lol pls keep uploading!! thx.
For the last problem you can just set u to x+4 instead of (x+4)^(1/3) U=x+4 X=u-4 Du=dx Integrate (u-4)*u^(1/3) du To get the same answer... But a little simpler
yes, sorry about that! someone else pointed it out! i add text comments into the video so that people will now be aware of the mistake. Thanks for pointing it out!!!
Rajat Bansal at a community college - I'm transferring to UCSC as a junior, here is a question I pulled from random on an exam: ⌡1/(2-sinx)dx and one I picked just to prove this guy wrong: ⌡(2x+1)/(x^4+x)dx
Patrick we all thank you so much you made our math's struggle an easy task , without you we have not passed our exams thanks we should make a donation to support you for what you do without any financial interests we all love you long live the mathematics and the mathematicians
Hi Patrick, Regarding the third question. Why did you choose such a difficult U substitution? I used U=x+4 and it's much easier to follow. I don't mind the U sub used mainly because it showed me a different way of doing it but I found that more difficult. I'm just saying in case someone was thinking of doing the same I did. Thanks for all the help. You're pretty much a super hero to me!
I used to think this kind of math was just dumb to do. Then I found Patrick. Anytime I have a question about how to do something I immediately search your website and RUclips Channel. It immediately becomes clearer and a light bulb pops into my head. Now I almost look forward to moving into my higher level math classes at my university as I work towards my Computer Science degree.
You teach way better than my teacher! Finally, something I can understand. Now I can ace the test I have tomorrow. :) Thank you for all your hard work~
omg u saved my life!! my high school calc teacher is terrible and because of that my grade is falling, but with your video you totally saved my ass! thank you!!!
OMG you are actually amazing! you are helping me with so much of my university work! i love you" you are a genius! and such a good teach! you explain things so well! :)
Remember that terms inside square roots can be manipulated algebraically just like a normal fraction. In the same way that (5/x)/(x/5) = (5/x)*1/(x/5) = (5/x)*(5/x) = 25/x^2, sqrt(5/x)/sqrt(x/5) = sqrt(5/x)*sqrt(5/x) = 5/x. Taking the integral is simple from there.
ya, you are wrong : ) you can do a u substitution (let u = v+1) to change it to 1/u ------------- when you integrate that, you get ln(u) which is ln(v+1) in this case
QUESTION around 05:08 i still don't understand why you had to square the square root of U. I'm confused as to why we can't write it as U to the one half and then differentiate. Sorry i haven't done calc in a while and i'm rusty as *$%^.
∫ ex haha! There was a small mistake at the conclusion of the ∫ x/(1-x^2)+sqrt(1-x^2). Patrick writes -ln | sqrt(u) - 1 + C, which causes him to mistake the the closing bracket of the absolute value sign for a # 1, so he gives the final answer (incorrectly) as -ln | sqrt(1-x^2 - 1 | + C, when in fact it should have been -ln | sqrt(1-x^2 | + C. I'm only bringing this up to avoid confusion, I'm so grateful for this video I would never ever post a negative comment. Thank you millions of times!
this helped me in so many ways O: thank youuuuuuu(because what they discussed in school is far away from what is given to us in assignments and exams xD)
Everyone should stop complimenting patrick's teaching style, because then he might go and teach again. This dark prospect must not, cannot, will not happen! What would we do then?!? j/k Thanks for the help. Finally I get u-substitution.
Thank u soooo much for your videos!! You're such a great teacher I always recommend ur videos to my friends 😊 I really find them very useful and they have helped me a lot!! Thank u again!
The way you explain it is so clear and is can be easily understood, and the way you film it particularly helps too. Are you sniffing the marker? lol jk, thank you very much!
Just take u to be 3x+2, this will result in 1/3 du = dx meaning: you will get: 1/3 * 8 outside and the inside of the integral will be 1/u which you can solve into ln(u) So it should be 8/3*ln(3x+2) + C
Hi man GREAT VIDEO!!!!!!! Just one small question. Take the first question for example you took u=e^x so du/dx=e^x, at 1.13 you wrote du=e^xdx. I just wanted to know how you can do that because I thought that du/dx was a symbol not a fraction. (Please get back to me) Once great video!!!!! Really clear and well explained THANKS!!!
Alright. So not once have I looked up anything math related on youtube, but I love it and am going into electrical engineering, and this was under the "Recommended for you" tab. Kinda weird right?
There is something wrong in Final answer in Example 4 but thank you so much sir for this. You really explained very well and I really understand it also.
Simpler solution to the 3rd problem:We shall simplift the denominator: sqrt(1-x^2)[ sqrt(1-x^2)+1].Therefore:int(x dx/( sqrt(1-x^2)[sqrt(1-x^2)+1]Let's substitute u=sqrt(1-x^2)-----> du=-[x/sqrt(1-x^2)]dx.Conseqeuntly we simplified the given problem:-int(dt/(u+1)) Which is the linear case----> -ln|u+1|+c -----> -ln|sqrt(1-x^2)+1|+c
You have made the last example way more complicated than it needs to be. x=u-4 as u=x+4 and du=dx. Make the substitutions and you still get the same answer but with half the working out.
I usually have difficulty in figuring out whether to do substitution before integration by parts or after, is there any way that can help me understand it better? Also, thanks for all the videos, they are a life saver :) !
Another way to approach this would be to let x = sin v and then you have dx = cos v dv and the integral would become (sin v cos v dv) / ((1 - (sin v)^2) + sort(1 - (sin v)^2) which would lead to the same correct answer.
Well, in the third problem if you substitute sqRt (1-x^2) as "u", you can solve it faster, you won't be needing a second substitution. Integral -u/(u^2+u) = integral - 1/( u+1) = -ln (u+1) + C = -ln (sqrt[ (1-x^2) +1] +C
-Question- In the third problem, first substitution, in the denominator, why did u put u - sqrt (u) If the original equation says (I - x^2) + sqrt (1 - x^2 ) Shoudn't it be u + sqrt (u)
I think that there is a slight arithmetic mistake there, u wrote - instead of plus. I substituted sq rt. (1-x^2) = u I think that solves the problem in an easy way without the second substitution.
Excellent explanation.
There was a little mistake. For the third problem, the denominator has + between (1-x^2) and sqrt(1-x^2). That being said, you would need change the minus sign in your denominator between u and square root of you, v^2 and v, and finally between square root of (1-x^2) and 1 at the ending result.
u made a mistake on 4:37- 4:41. It should be a "+" sign
chan tom thanks, I thought I messed up hahah
Nithesh S. he factored out a negative, thus changing the inside functions polarity
Manny Ramirez Nope
Sure he messed up
Mike Kato The solution should be -ln|sqrt(1-x^2)+1|) + c
Just letting you know that your video is still helping people to get through calculus. Thank you very much.
Can you please replace my calculus professor?
mine too
Calculus professor substitution
Mac Just watch his videos instead man. Bin your professor.
Don't you mean SUBSTITUTE? :)
7 years later, and this video still helps Calculus students like me :)
10 years bro...
11 years man...
17 years man...
@@1_adityasingh :D this was made when I was born, and waited for me to click on it my entire life, I guess
13 years
your v's look like square roots :(
Yaaa
Yaaaa
Yaaaaaa
Yaaaaaaa
Yaaaaaaaa
12 years and counting but bruh u're still helping a lot of students who are unfortunate on math teachers sskksks hope you're earning what's right for you now.
at the beginning of the day i couldn't be able to explain what U-substitution even had to do with anything but after watching all of patricks U-Substitution videos I've learned more than my teacher or my stupid calc book could ever teach me. Kudos from MA.
It'd be pretty cool if in videos where all you are doing is examples, you showed a preview of each type of problem you will do with links to the time that you do each example. It might not be worth the effort, but just a thought! Thanks for all your videos, they have helped me go from a B math student to 94% in Calc I!
+Cody Balos that is a good idea, i will keep this in mind for future videos
Many Thanxx, you're my hero. I don't know why my lecturer couldn't teach it as clear as you.
I love your little comments catering to the people LEARNING. One of my biggest problems with some professors is they seem to not address or display any understanding or acknowledging the difficulty something might have for people who are LEARNING. Your comments show you are a very good teacher by acknowledging that "when you don't know how to do something, sometimes it seems like the hardest thing ever".
I never would have thought to do a double substitution. Your work is amazing and I've been watching your Integration videos for a few days now and they're helping me so much! Thank you for these videos.
Sometimes I feel like you are a magical mystical being that just KNOWS.
Thanks for being awesome!
How could i just found your channel!!?? You teach so good, better than my math teacher
+Eat, Love and Eat Again Welcome to the good life! :-)
Thank you soo much for these videos. My teacher shows us the easiest example and then expects us to learn the rest on our own, you pretty much are my Calculus teacher, thanks, you rock!
@AhOhy sorry, i do not do hw questions - it could be stuff for a grade and i am not comfortable answering that. however, i will give you a hint: you picked the wrong u sub.
@cincottallll so teach yourself if you are that interested in it! why wait for someone to 'teach' you? although you need some trig as well before you jump into it.
@vhskeva ha! glad you like the videos so much : ) sometimes i turn on like 15 videos at once and get a surreal patrickjmt effect. my wife thinks i am crazy ( i also do it to annoy her!!)
Once again, thank you for these videos which are concise, explain the theory AND how to do it (the algebra part). My calc I teacher does his best to explain these things but it doesnt make sense. I've been using your videos since last summer's Trig class with him and now for Calc I (my husband used your videos as well when he was in school and is the one who suggested you). What a BLESSING! Your videos are the main reason Im grasping the concepts and how to do the problems. And with finals coming up in 8 days, theyre helping me learn and review everything as well! See ya in Calc II. thanks again!
great video very clear! just to mention that to others who is watching this video, the question involving 2 substitutions does not necesserily need 2 subs, u^2=1-x^2 gives same answer.
10 years later and I still find this video useful. Thank you Sensei
the first integral can not be integrated.
you would use a series expansion to approximate it to any desired amount of accuracy however.
the second integral is just arctan(x).
for the third, i assume you mean:
1/[x(4x+1)^1/2]
at a glance like you would let u = (4x+1)^1/2
solve for x, find dx, plug it all in and you would have a problem that you would have to use partial fractions on
@sam1209 sometimes it is nice to take the scenic route.
hi patrick, at 4:46 u wrote u - sqrt(u) . it should be +
small error but great video nonetheless. thank you!
he wrote the correction
@bobish101 so for x in your u = 16 + 2x part; then plug that in, divide it out and integrate
Today I am a what I'd like to call a successful engineer.
Thank you Patrick.
We all owe you.
You and Sal really helped to pass all my uni math courses, big thanks to you!
This guy is better
Night before Calc BC test feeling pretty great about it. This video reinforced my confidence. Thnx bro
@bamf703 trying to keep the v's from lookin' like u's
yes! you are absolutely correct! it should be u + u^5! thanks for pointing out the error!
Thanks a lot:I have an exam tomorrow, I was absent when my class was doing substtitution techniques, I got a handout my teacher made on it so I can study but it was like bullshit...so, there I was looking for help on the internet, and found just the one! It was easy to understand I didnt need to re-watch it many times to grab the idea of it...too bad YOU arent my teacher!!!lol pls keep uploading!! thx.
@ladyasphodel i prefer complicating things as much as possible.
@JRaby21 u have to use taylor series. it does not have an elementary antiderivative
For the last problem you can just set u to x+4 instead of (x+4)^(1/3)
U=x+4
X=u-4
Du=dx
Integrate (u-4)*u^(1/3) du
To get the same answer... But a little simpler
you have perfect handwritting , explanation skills as well as pace
Finally understood this before 2 hours from the exam! Subscription well deserved sir
yes, sorry about that! someone else pointed it out! i add text comments into the video so that people will now be aware of the mistake. Thanks for pointing it out!!!
i find the hardest thing to do is double or triple substitution.i lost track of them quite a bit.
I want you pens ... that is why people are good at math, sometimes they want to write with cool pen.
It's. A. Fucking. Sharpie.
yep! you are right
That third example is satanic. Would never get that on a test :/
Average? That's like you have to be able to do it in your head in my course - try ⌡√(tanx)dx
What course are you talking +rusko dudesko
Rajat Bansal calc 2
***** k which university or college
Rajat Bansal at a community college - I'm transferring to UCSC as a junior, here is a question I pulled from random on an exam:
⌡1/(2-sinx)dx
and one I picked just to prove this guy wrong:
⌡(2x+1)/(x^4+x)dx
Patrick we all thank you so much you made our math's struggle an easy task , without you we have not passed our exams
thanks
we should make a donation to support you for what you do without any financial interests
we all love you
long live the mathematics and the mathematicians
@omegaplatypus cause i used to get them cofused with ' u ' so i put a little hook on them to help myself
Hi Patrick, Regarding the third question. Why did you choose such a difficult U substitution? I used U=x+4 and it's much easier to follow. I don't mind the U sub used mainly because it showed me a different way of doing it but I found that more difficult. I'm just saying in case someone was thinking of doing the same I did. Thanks for all the help. You're pretty much a super hero to me!
I did it the same way and got the same answer.
I used to think this kind of math was just dumb to do. Then I found Patrick. Anytime I have a question about how to do something I immediately search your website and RUclips Channel. It immediately becomes clearer and a light bulb pops into my head. Now I almost look forward to moving into my higher level math classes at my university as I work towards my Computer Science degree.
i love u patrick because of u i passed my calculus classes
14 years ago! thank you for posting this!
Thanks Patrick, this video helped a lot. The problem at min 8 was just the one I was stuck at.
@D4ntesword yes, that is wht i would do
You teach way better than my teacher! Finally, something I can understand. Now I can ace the test I have tomorrow. :) Thank you for all your hard work~
@mylesbianify well i assume you meant u-substitution. if he meant trig substitution, that is correct
omg u saved my life!! my high school calc teacher is terrible and because of that my grade is falling, but with your video you totally saved my ass! thank you!!!
OMG you are actually amazing! you are helping me with so much of my university work! i love you" you are a genius! and such a good teach! you explain things so well! :)
Remember that terms inside square roots can be manipulated algebraically just like a normal fraction. In the same way that (5/x)/(x/5) = (5/x)*1/(x/5) = (5/x)*(5/x) = 25/x^2, sqrt(5/x)/sqrt(x/5) = sqrt(5/x)*sqrt(5/x) = 5/x. Taking the integral is simple from there.
ya, you are wrong : )
you can do a u substitution (let u = v+1) to change it to 1/u ------------- when you integrate that, you get ln(u) which is ln(v+1) in this case
QUESTION around 05:08 i still don't understand why you had to square the square root of U. I'm confused as to why we can't write it as U to the one half and then differentiate. Sorry i haven't done calc in a while and i'm rusty as *$%^.
∫ ex
haha! There was a small mistake at the conclusion of the ∫ x/(1-x^2)+sqrt(1-x^2). Patrick writes -ln | sqrt(u) - 1 + C, which causes him to mistake the the closing bracket of the absolute value sign for a # 1, so he gives the final answer (incorrectly) as -ln | sqrt(1-x^2 - 1 | + C, when in fact it should have been -ln | sqrt(1-x^2 | + C. I'm only bringing this up to avoid confusion, I'm so grateful for this video I would never ever post a negative comment. Thank you millions of times!
You are the best mathematician I have ever known!
this helped me in so many ways O: thank youuuuuuu(because what they discussed in school is far away from what is given to us in assignments and exams xD)
@mylesbianify no, you can not. it looks like a trigonometric substitution to me (after completing the square)
Everyone should stop complimenting patrick's teaching style, because then he might go and teach again.
This dark prospect must not, cannot, will not happen! What would we do then?!?
j/k
Thanks for the help. Finally I get u-substitution.
I'm failing my math class because my teacher can't explain how to do anything but you are saving me right now I think I will pass. Haha. thank you.
Thank u soooo much for your videos!! You're such a great teacher I always recommend ur videos to my friends 😊 I really find them very useful and they have helped me a lot!! Thank u again!
no problem!
the best lecturer ever..now I like calcus coz of u.
aw yea, no problem!
2k19 and tomarrow is my calculus exam... THANKS BOI👍
well, you can treat it as a fraction!
plus, we are calculating the differential:
the differential of u = e^x
is du = e^x dx
i see by your wide range of the english vocabulary, that you would do even better in an english exam
The way you explain it is so clear and is can be easily understood, and the way you film it particularly helps too. Are you sniffing the marker? lol jk, thank you very much!
Just take u to be 3x+2, this will result in
1/3 du = dx meaning: you will get:
1/3 * 8 outside and the inside of the integral will be 1/u which you can solve into ln(u)
So it should be 8/3*ln(3x+2) + C
second question: how did we know to use substitution the second time for v?
+Apollo-milkchocolate If you have something that can be replaced and the integral could be reduced to something that will work out itself
1. You're the hero Gotham deserves. You pretty much taught me calculus singlehandedly.
2. WHY DO YOUR V'S LOOK SO MUCH LIKE RADICAL SYMBOLS???
Hi man GREAT VIDEO!!!!!!! Just one small question. Take the first question for example you took u=e^x so du/dx=e^x, at 1.13 you wrote du=e^xdx. I just wanted to know how you can do that because I thought that du/dx was a symbol not a fraction. (Please get back to me) Once great video!!!!! Really clear and well explained THANKS!!!
Alright. So not once have I looked up anything math related on youtube, but I love it and am going into electrical engineering, and this was under the "Recommended for you" tab. Kinda weird right?
whats wrong with natural log? : )
Great explanation...very comprehensive examples...Thanks
There is something wrong in Final answer in Example 4 but thank you so much sir for this. You really explained very well and I really understand it also.
Good examples with better explanation, get it boy......Thanks for the video..
2017 and you rock all the time!
At 7:44, why did you not put u=x+4 just because it was a cube root and not a square root?!! Nice videos, btw. They help me a lot!
A tip:
For last example even if it (x+4) raised to power of 1/3 or 1/2 you can just substitute x+4 as u. And enjoy.
you will get same answer...Go try.
I have a test about this today. This just made me understand everything! Thanks
Simpler solution to the 3rd problem:We shall simplift the denominator: sqrt(1-x^2)[ sqrt(1-x^2)+1].Therefore:int(x dx/( sqrt(1-x^2)[sqrt(1-x^2)+1]Let's substitute u=sqrt(1-x^2)-----> du=-[x/sqrt(1-x^2)]dx.Conseqeuntly we simplified the given problem:-int(dt/(u+1)) Which is the linear case----> -ln|u+1|+c -----> -ln|sqrt(1-x^2)+1|+c
You have made the last example way more complicated than it needs to be. x=u-4 as u=x+4 and du=dx. Make the substitutions and you still get the same answer but with half the working out.
are you sure about that ( the answer should be) = (x+4)^3 /3 - 2(x+4)^2 + c
i do not have calculator to chick that :/
Yeah it works!
@TrAnScEnD3nT derivative of e^x is e^x
It is the chain rule. The derivative of ln sinx is 1/sinx times the derivative of sinx, which is cosx
@patrickJMT solve for x that is
The final expression simplifies further to 3u^4[u^3 / 7 - 1] + c = 3/7(x-3)(x+4)^(4/3) + c
AWESOME VIDEO. You sir just saved me from alot of trouble. 👏🏻
I usually have difficulty in figuring out whether to do substitution before integration by parts or after, is there any way that can help me understand it better? Also, thanks for all the videos, they are a life saver :) !
Another way to approach this would be to let x = sin v and then you have dx = cos v dv and the integral would become
(sin v cos v dv) / ((1 - (sin v)^2) + sort(1 - (sin v)^2) which would lead to the same correct answer.
they do not care about comments on youtube. they care about grades on tests : )
whys that? did i do something stupid in the vid or something? : )
This was very helpful. Thanks Patrick.
Well, in the third problem if you substitute sqRt (1-x^2) as "u", you can solve it faster, you won't be needing a second substitution.
Integral -u/(u^2+u) = integral - 1/( u+1) = -ln (u+1) + C = -ln (sqrt[ (1-x^2) +1] +C
-Question-
In the third problem, first substitution, in the denominator, why did u put
u - sqrt (u)
If the original equation says
(I - x^2) + sqrt (1 - x^2 )
Shoudn't it be
u + sqrt (u)
Thanks
yes you are ri8
i also noticed that
I think that there is a slight arithmetic mistake there, u wrote - instead of plus.
I substituted sq rt. (1-x^2) = u
I think that solves the problem in an easy way without the second substitution.
yeah i saw it also. could you make it clear please....
i did the same thing and it's faster and easier to understand for me
I like to clap every time whenever you solve a problem, by explaining it in a excellent way...
XD is so weird how everything is simpler here, but in college is like odd, thank you very much