But what if we dont know how to integrate trig functions or we forget some trig identities ? Integrating rational function for some reasons is introduced earlier than trig functions I like the way blackpenredpen calculated this integral
@MG-hi9sh
5 жыл бұрын
@@holyshit922 That is true, but trig identities are a helpful method that you should learn eventually. I really like the trig sub method here. It is very good.
@MG-hi9sh
5 жыл бұрын
@@holyshit922 Also, they aren't too hard to remember. Once you use them regularly, they stick in your head and become second nature.
@Peter_1986
5 жыл бұрын
This guy always has interesting integrals for me to try out.
@PunmasterSTP
Жыл бұрын
I never thought about trying to see how I could tweak an integral and then try to make that tweak work by using integration by parts. Thanks for the tip!
@astha_yadav
2 жыл бұрын
OMG ! i did pause and try u-sub, then partial and then Integration by Parts and got the answer till here 7:00 but gave up on that integral since i had used up my limited intelligence Idk it justs boots my confidence on seeing you reach the same thing 😭✨
@imme3024
Жыл бұрын
It works by trig substitution, I get the same result as the one in the video. By putting x^3 = tan(x), and by simplifying with sec^2, we get the integral of tan^2 / 3(1+tan^2), which is the same as sin^2(theta) / 3 or ( 1 - cos(2*theta) ) / 6. Finally we get theta/6 - sin*cos(theta) / 6. And we get the same result by replacing theta with arctan(x^3), sin(theta) = x^3 / sqrt(1+x^6) and cos(theta) = 1 / sqrt(1+x^6)
@guktefngrshoo7465
4 жыл бұрын
Me the intellectual: U substitute everything under the square to begin with
@ChrisMMaster0
Жыл бұрын
Damn, you really want to do the Quotient Rule 😨
@stevenbateson8217
6 жыл бұрын
I solved it by squaring both the numerator and denominator then making two substations. First I let u equal x cubed. I subsequently let tan theta equal u, giving me an answer initially in terms of theta, which I back substituted to get an answer in terms of u then again to get it in terms of x. The end result gave me exactly the same answer as your use of integration by parts.
@krishnanadityan2017
2 жыл бұрын
Put x^3=u. Then the integrant would be u^2/3.(1+u^3)^2. This can be easily integrated.
@DanDart
8 жыл бұрын
Omg I heard sped up you so much I got used to your voice and then thought whoa he's got a deep voice xD I think we can edit speed if we like on youtube already so you don't need to do that :p
@blackpenredpen
8 жыл бұрын
Just noticed your comment. I just wanted to keep this video shorter
@sriramsankar8958
4 жыл бұрын
Even i wanted to say the same thing, got really confused 😂
@karangujar317
7 жыл бұрын
I love you,Man :)
@blackpenredpen
7 жыл бұрын
Karan Gujar I love you too
@leif1075
4 жыл бұрын
@@blackpenredpen Wait you forgot to put the x^5 in the numerator at 5:29 in the video...
@thenuka9954
4 жыл бұрын
@@leif1075 ummmmm no he didn't, look at the u-sub process again
@sibongilemakhanya5155
4 жыл бұрын
You are my legend sir more people should subscribe to your channel 😊🙌
@alberteinstein3612
3 жыл бұрын
Tysm for teaching us the D I method. I will definitely use it in my Calculus class
@millej38
8 жыл бұрын
I also speed up my videos for the same reason, to save time. Great job once again. Keep the good work coming.
@blackpenredpen
8 жыл бұрын
Thank you. inserting *hand shake*
@jaysonkrishna7740
Жыл бұрын
I got another method... I seriously dont know how it hit me 😂... But here it goes Divide by x^6 inside the square in both the numerator and denominator Then numerator will be (x^-2)^2 which will be just x^-4 Denominator will be (1+x^-6)^2 which can be written as (1 + (x^-3)²)² Now take x^-3 as tan(t) and the question will become very easy...
@a.nyangao.m
7 жыл бұрын
Second row in the DI is with minus... but the second part has already a minus from the integration... so -*- =+ *-= - or what? The same for the diagonal for that matter
@Spectrojamz
4 жыл бұрын
Thank you Sooooo Much🥺🥺🥺. We'll definitely meet each other one day.
@ariadnevieira986
5 жыл бұрын
If i could hug u, i would, u'r saving my grades
@爸爸到底-s9x
7 жыл бұрын
haha, It's the integral from challenge problem 13 of chap 7,you can try some more stuff like this man,it's reality fascinating
@blackpenredpen
7 жыл бұрын
It is!
@Nothing_serious
6 жыл бұрын
I used trig substitution. I made (1+x^6)^2 into (sqrt(1+x^6))^4.
@soutriksarangi5580
6 жыл бұрын
blackpenredpen...please make a video on u-v-w technique of Inequalities
@silentthriller
7 жыл бұрын
Do you have to speed up the time or can you edit it out?
@guillermodelacruz5886
6 жыл бұрын
You are a master!!! Good job
@gopalswamy6849
7 жыл бұрын
Integral of (x^4) (1-x^2)^(3/2) with limits 0 to 1 (A)3π/256 (B)2π/256 (C)3π/265 (D)6π/562 could you please tell me how to simplify this one?
@herbcruz4697
7 жыл бұрын
Gopal Swamy A. I can e-mail you my solution, if you'd like. To begin with, though, start with a trig substitution. Let x=sin(theta). Then dx=cos(theta)*d(theta). Now, change the limits of integration. When x=0, theta=0, and when x=1, theta=(pi/2).
@jokerrecon8420
6 жыл бұрын
Answer is 3pi/256, since the given integral is just 0.5*Beta(5/2,5/2). Now use the relation between Beta and Gamma functions, and a well-known fact that gamma(1/2) = sqrt(pi).
@klausolekristiansen2960
4 жыл бұрын
Combining the DI and du methods.
@nilmadhabghosh3448
8 жыл бұрын
Dear sir I solved this by substituting, as x^6=tan^2(z)
@ben1996123
7 жыл бұрын
or you could do x=tan(u)^(1/3) to get integrate (x^4/(1+x^6))^2 dx = integrate 1/3 sin(u)^2 du = arctan(x^3)/6 - sin(2arctan(x^3))/12
@samegawa_sharkskin
3 жыл бұрын
I LOVE YOUUUUUU
@Kyle_da_athlete
6 жыл бұрын
This can be done by trig sub also . If anyone wants I can make a video
@mihaiciorobitca5287
6 жыл бұрын
TheGeekSoup it is easier ;)
@eshward8448
6 жыл бұрын
Yes, by trigonometric substitution of x^3=tan theta, it becomes easier. It reduces to integration of I/3* sin^2(theta), which gives the same result. Product rule need not then be applied.
@rik69x03
5 жыл бұрын
Yes make a video
@_TharunKumar
3 жыл бұрын
Is it tabular integration ? :thinking_face:
@greatestever6983
6 жыл бұрын
6:25 neg times neg is positive , but you subtract it....if it was neg times positive, then you'll add bc you'll factor out a neg.....[uv-int (v*du)].....however, great video!
@allaboutformulae7282
6 жыл бұрын
hey, I think intergrate of x^4 is (x^5)/5, right?
@cuteshe69
4 жыл бұрын
Ya
@snejpu2508
7 жыл бұрын
black pen red pen blue pen
@mihaiciorobitca5287
6 жыл бұрын
I used trig substitute for x^3=tan(u) and it worked until to make a mistake....,but after revising i get the core t answer YAY
@saxbend
7 жыл бұрын
that first diagonal term appears to have the wrong sign. The integral fraction has -1 as the numerator and it is also on a negative row, so shouldn't the product of that and Xcubed be positive?
@jesusalexisovallesgiuseppe5002
7 жыл бұрын
saxbend no
@mihaiciorobitca5287
6 жыл бұрын
Jesus Alexis Ovalles Giuseppe yes
@JensenPlaysMC
5 жыл бұрын
@@mihaiciorobitca5287 Negative only holds true for the term on the left not the right
@foo4ever1
8 жыл бұрын
Jesus. Well played, sir
@blackpenredpen
8 жыл бұрын
: )
@kostantinosss00
7 жыл бұрын
on D I why you stop on 3x^2 ? becuse i see other video and you say must go to 0 (3x^2) ,6x,x,0
@georgepennington908
6 жыл бұрын
UniverseInfinity there are 3 places you can stop: 1. If you get to 0 2. If the integral repeats 3. If you can do the integral of the product of the row, which is why he stops here
@briandube8366
8 жыл бұрын
Woow nice I was confused a lil
@obinnanwakwue5735
7 жыл бұрын
*Attempts to solve problem with u-substitution* Okay, cool, I got it, let's see what he says. *13 seconds in and he says not to use u-substitution* WHAAAAAAAAAAAAAT?! >_
@mikeljanhajrullaj5784
6 жыл бұрын
Can you resolve the integral of dx/(x + V(x^2 + 1)) ??? ...
@ernestschoenmakers8181
4 жыл бұрын
The trick is just multiply numerator and denominator with its conjugated form. Now the integral becomes a simple one which you can do yourself.
@sonicjia118
7 жыл бұрын
How about use partial fraction? This should be easy to integrate these I think.
@alexlostado2046
7 жыл бұрын
1+x^6 is not factorable, but other denominators such as 1-x^6 should be able to be integrated by partial fractions
@ernestschoenmakers8181
7 жыл бұрын
Wrong 1+x^6 is factorable namely: (1+x^2)(x^2+sqrt(3)*x+1)(x^2-sqrt(3)*x+1) so you can do the integral by partial fractions.
@Convergant
7 жыл бұрын
DI integration by parts, with x^4x/1+x^6 for D and x^4x/1+x^6 dx for I?
@ernestschoenmakers8181
4 жыл бұрын
Do you mean x to the power of 4x?
@astha_yadav
2 жыл бұрын
I thought you had inhaled helium, in the beginning of the video 😂
@Nerdrage2012
8 жыл бұрын
Thank you so much for this video. Can you do an integral, where you are forced to use ever method? I know it could be long. I love your strategy, "wouldn't it be nice if you could change this integral? what would you change so that you can make this question easier"?
@blackpenredpen
8 жыл бұрын
Hi there. I have lots of videos here: blackpenredpen.com/math/Calculus.html these are based on Stewart, 7th ed, ET. Oh, by the way, did you mean: "forced to use ever(y) method"?? If so, I close sqrt(1+e^x) is pretty close to every method being involved. Ah, as I am typing, I just come up with an idea that I can make a integrals with combined methods videos series. = )
@Nerdrage2012
8 жыл бұрын
yes, I meant "every". In the class I am taking we are required to use Stewart 8th edition; a lot of the problems are different, but mostly the same concepts in each one as 7th edition. I was hoping the combination of all methods would help me learn faster before my class starts. I am studying for Calc 2; class starts in 45 days, What area would you recommend I focus most of my energy?
@blackpenredpen
8 жыл бұрын
Integrals, definitely integrals. I always tell my calc2 students to review sect. 5.5, which is the U-sub section during summer before they start calc2. After you have done that, you can then start ch7, which is the new integral technique section, where most of the calc 2 class starts. Once you feel comfortable with integration techniques, you will feel confident in other topics in calc2.
@Nerdrage2012
8 жыл бұрын
I am learning a lot very fast thanks to your videos. Thank you
@ernestschoenmakers8181
7 жыл бұрын
But can the integral of 1/(1+x^6)^2 be done?
@holyshit922
6 жыл бұрын
Int(1/(1+x^6)^2,x)=Int((1+x^6-x^6)/(1+x^6)^2,x) =Int(1/(1+x^6),x)+Int(-x^6/(1+x^6)^2,x) =Int(1/(1+x^6),x)+1/6Int(x*(-6x^5)/(1+x^6)^2,x) =Int(1/(1+x^6),x)+1/6(x/(1+x^6)-Int(1/(1+x^6),x)) =1/6 x/(1+x^6)+5/6Int(1/(1+x^6),x) I think we need partial fraction decomposition
@SalmanKhan-hp8zs
6 жыл бұрын
where 3x gone
@scientadeus1449
5 жыл бұрын
We literally need to slow down the video to watch it
@rainbowbloom575
4 жыл бұрын
Solved this using trig sub.. tan2(u)=x6
@rainbowbloom575
4 жыл бұрын
I just got sin(2tan-1(x3))
@benemwatsena
5 жыл бұрын
Can you help me integrate x/sqr(1+x^6)
@cuteshe69
4 жыл бұрын
Hey substitute tan u for x^3
@g0dsm4ck100
8 жыл бұрын
Good video, nice job!
@manuelodabashian
7 жыл бұрын
can you do it x^4 (1/(1+(x^3)^2) then square the result?
@lakshaymd
7 жыл бұрын
Manuel Odabashian No Integral of (f(x))²≠(integral of f(x))²
@ernestschoenmakers8181
7 жыл бұрын
Without the square it can be done easily by partial fractions cause 1+x^6 is factorable.
@douro20
6 жыл бұрын
I would had done a little more algebra so I could combine everything over the same denominator.
@mhdiaden3120
3 жыл бұрын
thank you please I need trigononetry identities
@mdzakianwar1852
4 жыл бұрын
Sir u r not suffering from Coronavirus 🤔
@papabeardiscgolf1781
6 жыл бұрын
Is it weird that I’m drunk and watching this?
@fernandoperles727
6 жыл бұрын
Great guy
@wduandy
7 жыл бұрын
Where the DI method comes from?!
@blackpenredpen
7 жыл бұрын
Eduardo integration by parts
@Jonas-pd6cz
7 жыл бұрын
Very nice video ;)
@sibongakonkengobese3216
6 жыл бұрын
ive used trig substitution....
@anjelpatel36
6 жыл бұрын
Why don't you monetize your videos???
@adityadhar1358
3 жыл бұрын
I can hear better now
@ferhatidjillali
5 жыл бұрын
روعة
@Kyle-li8wi
7 жыл бұрын
Awesome!
@baronajosesek
7 жыл бұрын
pls pls do integral of sqrt tgx
@mihaiciorobitca5287
6 жыл бұрын
Jose Barona he did :))) your blessing was heart Next : integral of sqrt(sin x) dx
@eliavonsalis1073
5 жыл бұрын
I used u=x^3
@not_vinkami
6 жыл бұрын
5:19 you forgot something such as +C
@maxstone5108
5 жыл бұрын
Isnt it
@holyshit922
6 жыл бұрын
I would calculate this integral in the same way Integrating rational functions with inverse trig substition is bad idea from teaching point of view because trig integrals usually needs integration of rational functions
@matiassantacruz5487
5 жыл бұрын
Great content as always, but it annoys me that he uses tan^-1 instead of arctan...
@botiromondavlatov3299
Жыл бұрын
👍👍👍👍👍👍👍👏👏
@briandube8366
8 жыл бұрын
Your voice
@blackpenredpen
8 жыл бұрын
Lol. I just wanted to shorten the video.
@briandube8366
8 жыл бұрын
nice, great work though on the video.....cool tricks
Пікірлер: 125