What Integration Technique Do I Use? Example 1

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11 жыл бұрын

Thanks to all of you who support me on Patreon. You da real mvps! $1 per month helps!! :) / patrickjmt !! 30 Challenging Integrals with Detailed Solutions: www.teacherspayteachers.com/Pr...
What Integration Technique Do I Use ? Example 1. Determining how to integrate a particular function can be challenging. In this video, I try to show how I approach them.

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@ringo3650 9 жыл бұрын

People ended up here, either proving that their school teachers are being lousy, or they don't pay attention during the lecture. Or perhaps simply that Patrick is cool.

@dextermorgan9609 9 жыл бұрын

. 念來過反會才頭木 All of the above for me my friend.

@hurrdurr25 8 жыл бұрын

+. 念來過反會才頭木 more likely the teachers are lousy. people that don't pay attention probably wouldn't care to learn how to do the work and never make it to these videos.

@thewiseone9798 8 жыл бұрын

+. 念來過反會才頭木 Why do you have Chinese characters in your name?

@patrickjmt 8 жыл бұрын

+. 念來過反會才頭木 i think the correct answer is: all of the above (especially the last part)

@ConnorJW96 8 жыл бұрын

+. 念來過反會才頭木 Integration and differentiation are always good things to practice, as they crop up all the time. I find myself coming back to this video (and others by PatricksJMT's) to refresh my memory. Very useful!

@jackmathieson1903 Жыл бұрын

In my opinion, knowing which method of integration to use is the hardest part of A Level Maths. I've got my Pure exam tomorrow.

@Agreedtodisagree 11 жыл бұрын

I can't emphasize this enough, practice practice and practice. If you think you got it, then choose a hard problem and then try it. And kudos to this guy for having such a love for math.

@mileywaffle 10 жыл бұрын

Thank you SO MUCH. I've been subbed and watching your videos for 2 years now. I really appreciate that you keep it free and accessible.

@douglasleigh2831 3 жыл бұрын

The

@ahmadzaka9885 4 жыл бұрын

Not even kidding, this video probably just saved my life, you have no idea how much this helped me.

@Peter_1986 10 жыл бұрын

Awesome, integration is just what I need to practice for my exam in Linear Algebra & Integral Calculus. Thanks a lot.

@ThePharphis 11 жыл бұрын

After a few years of calc I'm feeling pretty confident about most integrations, but I still think this video is helpful.

@kelbyreyes3599 8 жыл бұрын

why do i always have a feeling that i'm doing everything wrong :(

@youareajackass2005 8 жыл бұрын

+Kelbypursuer ' it not just you

@markoplahuta8284 6 жыл бұрын

because you are lazy and think everything will just magically look solvable xD So you give up as soon as something starts looking more complicated than it was. First thing, you see people like patrick solve what they have either already solved or they have a lot of experience in solving and of course that it is gonna look like it's supposed to be easy. Secondly, most important thing is to know the basics. Yes, you can overcomplicate a problem to an extent at which you should start over or look up the answer, BUT, be sure that you have at least gotten to that point by using correct "mathematics" (aka didn't miss a minus, didn't do something that you can't with square roots etc. Basically don't integrate some function, before you know what that function is, how it behaves, how its plot looks, what can you do with it with some algebra . And do keep going, even when it looks like next step will require too much effort to do. You slowly train that way

@vanessasarahilizarragaoliv6591 4 жыл бұрын

hey nono, don´t feel like that, you must trust yourself, you´re learning so you are fixing your mistakes, it´s ok while you are understanding what you do

@DylanM333 3 жыл бұрын

because calculus is a deep ocean of knowledge and you literally build billions of brain pathways to be successful.

@jacobbchapman5644 2 жыл бұрын

Integrals are hard.

@intelliza7248 9 жыл бұрын

Write the numerator as (1-e^x) +2.e^x and then split it in two parts ,integration of first part is x and to integrate the second part substitute u=(1-e^x)

@212ntruesdale 5 жыл бұрын

BINGO!!!!!!!

@CheapPhysics 3 жыл бұрын

Bingoooooo.....

@johnman5391 2 жыл бұрын

Hey, I'm still learning this stuff can you explain that more simply? Is that method simpler?

@patrickjmt 11 жыл бұрын

yes, i will make one eventually ! it would be a great one to make, i agree

@hari8568 7 жыл бұрын

simply divide num and denom by e^x then we get e^-x+1 divide by e^-x -1 then add to num +1-1 and split integral we get integral e^-x /e^-x -1 + integral e^x /1-e^x.solving by substituting for both integrals by t=e^-x -1 andt=1-e^x in the end we get -log(e^-x -1)-log(1-e^x)

@flupypersonyui3488 9 жыл бұрын

That's gold, Jerry! Gold!... I mean Patrick.

@billwindsor4224 7 жыл бұрын

Great explanations here, Patrick. I purchased the worksheet -- it's excellent (and inexpensive); I encourage you to develop more worksheets, and to give them more visibility in your video posts!

@BubbleGumCMM 11 жыл бұрын

i wanted a video for this! nice timing. THANKS PATRICK!

@mohanarora1 9 жыл бұрын

You are amazing dude . Became a die hard fan of your s after this video

@ghostkr3676 8 жыл бұрын

Thanks man..... I always wondered how to start..... and you cleared my doubt

@satrickptar6265 5 жыл бұрын

I'm thanking my advanced algebra teacher in grade 9 ;-; it really helped me in integral calculus in grade 10!

@patrickjmt 11 жыл бұрын

and mainly, this is how i did it :) i do not claim it is the fastest, but i do claim it works.

@br5757 10 жыл бұрын

You are a wonderful teacher!

@patrickjmt 11 жыл бұрын

the second you can do with a u sub, but you still need to do the same type of substitution i did (i think!) for the first one

@japori2000 11 жыл бұрын

Yes yes yes. Just the video I needed!!

@nickthorpe7567 6 жыл бұрын

I split the numerator up into 1-ex+2ex so that I could split the fraction up into int[ 2ex/(1-ex) + (1-ex)/(1-ex) ]dx Then you get left with Int[2ex/(1-ex) + 1]dx Int[2ex/(1-ex)]dx + x Let u=1-ex Let du=-exdx -Int[2/u]du + x 2ln|u| + x 2ln[1-ex] + x + c Little bit more intuitive to me imo!

@oom_boudewijns6920 Жыл бұрын

Yep for me 2

@dreia2405 7 жыл бұрын

you made this very easy integral complicated, just add e^x and subtract e^x from de numerator

@marcusrayrosales1 7 жыл бұрын

This is way easier than shown here! First, multiply top and bottom by e^{-x/2}, so that: (1+e^x)/(1-e^x) becomes: -(e^{x/2}+e^{-x/2})/(e^{x/2}-e^{-x/2}) ==-cosh (x/2)/sinh (x/2) The integral is easily seen to be: -2ln [sinh (x/2)]■ knowledge of hyperbolic sines and cosines are not necessary here; just do the first step of multiplying by e^{-x/2}, then let u be the denominator. All so much easier than this video shows.

@Infoworld798 8 жыл бұрын

Poti you work very good

@hadireg 7 жыл бұрын

great job! thanks!

@asiyaishal 9 жыл бұрын

what if we take e^[x/2] out from both denominator & numerator..Then substitute the whole dnmtr term with 'u'....to get a integral [1/u]du form... #justAsking #student

@executorarktanis2323 4 жыл бұрын

the first substitution was a eye opener

@patrickjmt 11 жыл бұрын

it does not have an elementary antiderivative ; you have to use series to do this

@baburo101 11 жыл бұрын

Finally, Pat's saving some trees!

@ladicius 11 жыл бұрын

as a current student in differential equations, i have one bit of advice for you: try to find the staedtler liquid point 7 pens (that are actually a 0.4mm tip). as a user of the v5's for many semesters, the switch was absolutely fantastic for me. it's worth giving them a test run. thanks for the videos. i like watching them over breakfast some mornings.

@Andr295 11 жыл бұрын

You rock bro

@kkamous7278 7 жыл бұрын

yes I feel happy from your reply . I get lots of help from your integration videos. you're so genuine . tomorrow I have presentation in university and I got lots oh help from your video s.

@and_still. 6 жыл бұрын

I just subscribed you. Thanx from India.

@guitarttimman 5 жыл бұрын

Very good.

@lich2012 5 жыл бұрын

Hi friend. Good video. Could you explain the method for integrals with several sin and cos founctions where x must be changed for tan(t/2) (or something like that, i dont remember clear right now) please?

@soupisfornoobs4081 2 жыл бұрын

I believe that's called weierstrass substitution, you can look it up and get some interesting insights

@oom_boudewijns6920 Жыл бұрын

Its the halfangle tangent substitution. Pretty easy, just check out wikipedia

@JC-cu2ym 6 жыл бұрын

@patrickJMT cant we use Quotient rule to solve this example?? Instead of substitution?

@matblitz4637 7 жыл бұрын

you. can also use substitute x =-t and then do it

@Darwinion 5 жыл бұрын

Can that final answer be simplified to: x - 2ln(1-e^x) + c..... as ln (e^x) = x ???

@jmadluck 4 жыл бұрын

I thought the same but I'm not sure if the absolute value messes with that

@wyboo2019 3 жыл бұрын

im wondering if you could make it a trig integral by letting x = iu, and then e^x = cos(u) + i sin(u), but im not sure too what extent this is mathematically rigorous

@EverydayXperiment 11 жыл бұрын

I wish I was this good!

@hamid-we7vs 2 жыл бұрын

At 4:40 , could you have used the reverse chain rule after doing the u-substitution ?

@Monadoabyss 11 жыл бұрын

A quicker way: (1+e^x)/(1-e^x) is just -coth(x/2) which can be easily integrated as it's in the form f'(x)/f(x).

@HotPepperLala 11 жыл бұрын

Can you do a video on Euler_substitution? And possibly Elliptic Integrals?

@akarshansaraf721 4 жыл бұрын

You could have easily done by dividing numerator and denominator by e^x/2 and then substituting e^x/2+e^-x/2=u

@chenjasper7137 10 жыл бұрын

GOOOOOOOOOOOOD

@patrickjmt 11 жыл бұрын

age and allergies and new microphones all do that

@sandyshores8921 11 ай бұрын

thank you.

@gozao100 11 жыл бұрын

i have a question is it normal a function to have more then one integral form?i mean a did this exercise and the result was x-2log(1-e^x)+c and it's the same result when i derive this or am i doing something wrong?

@cypher4528 6 жыл бұрын

Hey man, the question (1+e^x)/(1-e^x) Is it OK if I possibly solved it by splitting it into 1/(1-e^x)+e^x/(1-e^x) Where I multiply (e^-x) to numerator and denominator of 1/(1-e^x) And gave (1-e^x)=t and e^x dx= -dt to e^x/(1-e^x)?

@damianflett6360 3 жыл бұрын

No. I do not permit it

@fergusfs70 5 жыл бұрын

In your final answer you leave ln(e^x) which can just be written as x

@blackham7 5 жыл бұрын

But it helps us students see what's going on. He might know that but we might miss that.

@edgaracosta9976 9 жыл бұрын

I'd just like to comment that there is a way to at least know how to approach one type of integration problem. If you are multiplying or dividing two functions of a different type, the only approach is integration by parts. For example, finding the integral of (x^2 + 2)(sinx) dx. This type of problem would be impossible to do through substitution because in the most general cases, you need the functions to be of the same type, that way you have a chance of figuring out which one is the derivative of which.

@dannyawesome63 2 жыл бұрын

I took Calculus at UC Davis back in 2015... and man.. I don’t remember any of this LMAOO

@nascar12141 11 жыл бұрын

I wish I had your awesome penmanship

@sophiacarryl 11 жыл бұрын

Do you have a video on the second theorem of calculus?

@patrickjmt 11 жыл бұрын

then what

@amak1131 11 жыл бұрын

This is indeed the hard part of Calc 2. Actual problem aside (my Calc 2 professor gave crazy integrals to do), the hard part is which method to use.

@yasploofyh8358 Жыл бұрын

Hows life now? What are you up to?

@212ntruesdale 5 жыл бұрын

You made a challenge out of a very easy integral if you make one very clever substitution. Instead of 1-e^x, make the numerator 1-e^x+2e^x. Then you have two integrals, intdx + 2int[e^x/1-e^x]. Couldn't be simpler!

@shashwatbajpai1350 3 жыл бұрын

You could have multiplied numerator and denominator by e^(-x/2) then you could have directly substituted your denominator as u then your numerator would have been - du and you would have got the answer in 3 steps

@DrHansPeterWurst 11 жыл бұрын

that one is really fun...

@bcarray 9 жыл бұрын

what kind of pen is that? i had a same one and loved it but i can't find one anymore :'(

@pvic6959 8 жыл бұрын

+bcarray www.amazon.com/s/ref=nb_sb_noss?url=search-alias%3Daps&field-keywords=precise+7+pen that might be of help

@kkamous7278 7 жыл бұрын

you are the great man. I love your work. you work for human. I love u so much. alas I meet with you . . I watch your video's. I think you are the best man in world of mathematics. May you live long. my prayer with you.

@patrickjmt 7 жыл бұрын

+Muhammad Kamran Khan thanks for the kind words, I appreciate it :)

@cypher4528 6 жыл бұрын

I only noticed that you are Left Handed after watching 10 of your videos.

@beta5770 11 жыл бұрын

CHALLENGE ACCEPTED

@zhubarb 11 жыл бұрын

Thank you, this is elementary and useful information.

@oom_boudewijns6920 Жыл бұрын

Brother always see if u can write the numerator as the denominator. I immedoately see that u can add 2ex to the nominator to get 1-ex+2ex. Split integral and notice one is x and one is ln(1-ex) cos derivative of ex is ex In the end, a usub is not even needed !

@venkateshmangrulkar4776 6 жыл бұрын

In denominator quadratic is compulsory in partial fraction thats why i don't understand why u solve this by partial fraction

@ghost880106 11 жыл бұрын

I love the pens that you use. Just saying. Haha

@patrickjmt 11 жыл бұрын

nope, never done anything at all with them.

@TheHateraboi 11 жыл бұрын

can you do a video on e^(pi i) = -1??

@OUSSA 2 жыл бұрын

There is a much easier way in to write the top as 1-exp(x)+2exp(x) to simplify the integral and get 1+2(exp(x))/(1-exp(x))

@rsmith5645 7 жыл бұрын

How do I integrate 1/(x^1/3+x^1/4)+ln(1+x^1/6)/(x^1/3+x^1/2)dx?

@damianflett6360 3 жыл бұрын

Do you still need to know

@JS-ns8dr 3 жыл бұрын

@@damianflett6360 LMAOOO

@damianflett6360 3 жыл бұрын

@@JS-ns8dr can’t believe they never got back to me.

@peregrinefal6607 11 жыл бұрын

Why not split the first problem to: 1/(1-e^x) + e^x/(1-e^x) ?

@iBoom3RxX 9 жыл бұрын

At 6:34, where did the denominator of -1 come from?

@LoneWolfKTG 9 жыл бұрын

From the second u-sub that he shows immediately after...

@iBoom3RxX 9 жыл бұрын

Thanks mate!

@LoneWolfKTG 9 жыл бұрын

Ginzaki no prob!

@LoneWolfKTG 9 жыл бұрын

Ginzaki oh and Dark Souls ftw !

@theBang4thebuck 5 жыл бұрын

I’m in calc 2 and I’m about to throw in the towel

@lezero211 9 жыл бұрын

Hmmm can't seem to understand the part where you used ln to solve for dx

@kyler2532 9 жыл бұрын

well it is the relationship between exponents and logarithms. When you multiply ln to e^x ( ln(e^x) ) you will get x. Also in order to make the equals sign true he multiplied the other side with ln. His way is much easier than trying to solve the integral when both u and du=e^x

@mayahmed5476 7 жыл бұрын

Your the best. I have a final tomorrow and ur video helped me alot

@prateekgurjar1651 7 жыл бұрын

you're*

@yashwanthandaluri8934 6 жыл бұрын

sollu

@wolfgangi 7 жыл бұрын

I fucking hate integration by parts especially my memory is wiped out by my crippling alcoholism lol

@realitycheck816 4 жыл бұрын

Lol you could have just divided the numerator and denominator by e^x. Then it becomes an easy log integration. 3 lines maximum lmao

@cepson 2 жыл бұрын

It's weird to me that this problem can be solved by letting u = 0, when u is really e to the x and can't be 0. Yet it works.

@idiosyncraticidiott5015 11 жыл бұрын

Cuz we all recognize -coth(x/2) when we see it

@chasegregory1328 5 жыл бұрын

For every percent I get over 50% on my exam tomorrow, I'll donate $1.

@blackham7 5 жыл бұрын

Can I have the money

@damianflett6360 3 жыл бұрын

How much did you donate

@alexgabel4379 11 жыл бұрын

please try to solve ∫e^(-x^2) dx. There's a way to solve it using a 2nd dimension (by squaring the whole thing) and then you get: ∫∫e^(-x^2-y^2) dxdy, but then I get stuck...

@murugesankgm1297 5 жыл бұрын

you could do it by rationalising it and resolving

@Grantbraner 3 жыл бұрын

Hi Mr. Woody

@kkamous7278 7 жыл бұрын

thanks to my God that you read my comment. l love you sir so much.

@sabithasasikumar239 11 жыл бұрын

Nooooooooooooo.....I discovered patrickJMT through google ;)

@jacobbchapman5644 2 жыл бұрын

What's the difference between an Integral and an improper Integral?

@OUSSA 2 жыл бұрын

Improper means it has an infinity in one of its boundaries or both

@algebrayd 11 жыл бұрын

are you the same person who does the cengagebrain solution videos??

@JlsUno94 9 жыл бұрын

I love you

@veggiekeller 9 жыл бұрын

For the problem finding the integral of (1 +e^x)/(1-e^x) why couldn't you just break up the fraction at the beginning and integrate 2 fractions (1/1-e^x) + (e^x/1-e^x)

@HamzaDjeloud 5 жыл бұрын

This is a very easy way Thank you

@JanCRefsgaard 11 жыл бұрын

ahhh cool, I didn't know about that... but google did ;)

@cyliachipara5658 8 жыл бұрын

hnmmmn z der anatha video...u los m on th las part

@seroujghazarian6343 5 жыл бұрын

7:47 you can't be serious! You should know that ln(e^x)=x, and e^x is always positive for all real x, so you can't tell me it is in absolute value because abs(e^x)=e^x no matter which value x has.

@fastchill 11 жыл бұрын

an other option you should consider is partial integration.

@patrickjmt 11 жыл бұрын

ah, in that case, i will just delete all my videos ;)

@sineadh1994 10 жыл бұрын

where did the absolute value signs come from?? i understood it until there

@Peter_1986 10 жыл бұрын

The absolute value signs make the natural logarithm more general, so that it is valid for negative value of u as well. If you try comparing ln(u) and ln|u| on a graphing calculator then you can see this more clearly.

@lucasm4299 8 жыл бұрын

When we have 1/x and take the antiderivative we have to take into account the domain of the antiderivative. Derivative Ln(x) - 1/x The ln(x) domain is x>0 The values of 1/x respect the domain Antiderivative 1/x - Ln(x) The domain of 1/x is all numbers except 0, ln(x) can't take negative numbers so we have to find a way to extend its domain. |x| solves that problem. If you were wondering, d/dx ln|x| = 1/x