how to solve these HARD integrals

[@沈博智-x5y]

Paused the video: The way I did the second one (x^6 + x^3)(x^3+2)^(1/3) = x^3(x^3+1)(x^3+2)^(1/3) Integral of x^3(x^3+1)(x^3+2)^(1/3) dx Let u = x^3 + 2 du = 3x^2 dx 1/(3x^2) du = dx u = x^3 + 2 => x^3 = u-2 => x = (u-2)^(1/3) Integral of x^3(x^3+1)(x^3+2)^(1/3) dx = Integral of x^3((u-2)+1)(u)^(1/3) * 1/(3x^2) du = Integral of x(u-1)(u)^(1/3) du = 1/3 Integral of (u-2)^(1/3)*(u-1)*(u)^(1/3) du = 1/3 Integral of u^(1/3)*(u-2)^(1/3)*(u-1) du = 1/3 Integral of (u*(u-2))^(1/3)*(u-1) du = 1/3 Integral of (u^2-2u)^(1/3)*(u-1) du Let w = u^2 - 2u dw = 2u - 2 du = 2(u-1) du 1/3 Integral of (u^2-2u)^(1/3)*(u-1) du = 1/3 *1/2 Integral of (u^2-2u)^(1/3)*2(u-1) du = 1/6 Integral of (w)^(1/3)* dw = 1/6 * w^(1/3 + 3/3) / (4/3) + K = 1/6 * (3/4) * w^(4/3) + K = 3/24 * (u^2-2u)^(4/3) + K = 1/8 * (u(u-2))^(4/3) + K = 1/8 * ((x^3+2)(x^3 + 2 - 2))^(4/3) + K = 1/8 * (x^3(x^3+2))^(4/3) + K = x^4(x^3+2)^(4/3) / 8 + K 0:39 If I let u equal to the inside, it's a no (actually it's a yes technically just look above) The way I did the first one: 1/(x^2(x^4+1)^3/4) = 1/(x^2(x^4(1+1/x^4))^3/4) = 1/(x^2(x^3(1+1/x^4)^3/4)) = 1/(x^5(1+1/x^4)^3/4) Integral of 1/(x^5(1+1/x^4)^3/4)) dx Let u = 1 + 1/x^4 du = -4/x^5 dx Integral of 1/(x^5(1+1/x^4)^3/4)) dx = -1/4 Integral of (-4)/(x^5(1+1/x^4)^3/4)) dx = -1/4 Integral of 1/(u)^3/4 du = -1/4 Integral of u^(-3/4) du = -1/4 (u^(-3/4 + 4/4))/(1/4) + C = -(u^(1/4)) + C = -(1+1/x^4)^(1/4) + C = - ((x^4 + 1)/x^4)^(1/4) + C = - (x^4+1)^(1/4) / x + C The second one took me longer because I failed to see that I could do something with (u-2)^1/3*u^1/3 and combine them.

[@沈博智-x5y]

I prefer definite integrals, because sometimes you don't need to integrate, you just need to look at symmetry to find the area, or look at the graph, or do some clever algebra trick without ever needing to "find an antiderivative".

[@blackpenredpen]

Wow, very nice solution!

[@ernestschoenmakers8181]

It's great to see how many methods there are to solve an integral, it gives you insight in how solutions are related, especially for trig functions.

[@funahead5426]

You think in a complex way, cannot understand your solution but the way taught in this video is very nice and simple.

[@mittunsudhahar634]

Lol I did the same way for both, then I looked at his solution for the second one and realised it was so much simpler.

[@SeeTv.]

I hate doing the arithmetic when evaluating definite integral. I'm definitely an indefinite integral person.

[@blackpenredpen]

Same here lol

[@zbp9541]

@@blackpenredpen same here 😭😃

[@leif1075]

@@blackpenredpen Why and how how wiyld anyone ever think of doing that manipulation in the first integral?? I hope you can please respond..

[@rasheedmohammed2227]

Indefinite are more fundamental than definite

[@elihowitt4107]

Generally I agree but there are some integrals that don't have an indefinite solution (in terms of 'basic' functions), and some really nice definite integral results that are key to math/science, my favorite being the Gaussian integral, super cool!

[@pieterhofstra3172]

Can someone give this man a bigger whiteboard?? 😂😂

[@Craznar]

Or smaller integrals :)

[@alonbechor1842]

That's what has come up to my mind for so many videos !

[@FireRaptor100]

I noticed that he started using that tiny board after the beginning of the pandemic. I don´t know how both things are related, but it is so.

[@AlexandrBorschchev]

he could easily buy a big whiteboard for less than $500, damn

[@anglaismoyen]

​@@AlexandrBorschchevI have a bigger whiteboard than that which was well under $50 new.

[@ujwalsaisimhay]

The second problem was asked in IIT-JEE in the year 1992! Now that form has become a standard form!

[@kalkal8050]

9:26 I thought you are going to write the power on the wall. Lol!

[@sunilparekh4581]

Same lol😂😂

[@mathteacher313]

[😃⅔]²

[@Happy_Abe]

Wow Randy is really getting around, such a fan

[@black-my9qm]

Hey! I've been studying for an exam and I'm not sure what it is about you, but you help more than these other math teachers and people on KZbin. I wanted to comment this just to let you know how much you help me and these other viewers. Hats off to you, I give a sincere thanks. You've earned my subscription.

[@Kris-P_Boi]

For the first question taking the substitution x^2 as tan(u) also solves the integral in a very concise way.

[@kalkal8050]

Dude, you got a cute little 3blue1brown pi there!

[@lukschs1]

es un asco como enseña ése

[@GamerXBoyZGaming]

i'm still waiting the day where he would forgot the +C 😂

[@FireRaptor100]

Almost forgot the +C. I still remember my calculus teacher: If you don't add the +C, you're giving only one solution, and no infinite. How many points should you get because of that?, use cross multiplication to know it. ∞→ 10 points 1→ ?

[@kuronekonova3698]

10/infinity = 0.

[@ronin4923]

@@kuronekonova3698 Tends to 0

[@ArthurShirinka]

0 Lol

[@laman8914]

We see many of these hard equations, but what we are most interested in is how to identify a situation and then create the equation that fits it. So, how to model real life problems into a mathematical form. I think that skill is more rewarding than solving, because much of the solutions are now done virtually.

[@gardenmenuuu]

Congratulations man you are always dedicating ur time for students around the world thanks for making maths a fun thing

[@gsg1666]

I used the trig substitution: x^2=tan(u) for the first integral. You just have to do some cancellations and convert sec and tan into sin and cos.

[@angelmendez-rivera351]

You can do this, but this is far more tedious to do. Though it gives the correct answer.

[@АлексейЛавренович-х7е]

I passed the previous math analysis exam practically just because of you. Thank you a lot

[@shashankkatiha9439]

substituting x^3 + 2 as u and taking out x^3 common from x^6 + x^3 can also work. Then in the resulting expression (u^2 -2u)^1/3(u-1), u^2 -2u can be resubstituted as t.

[@Revolver03]

Me reading the title: Well.....I mean yes...no...maybe

[@kingbeauregard]

I watch these videos and am sometimes frustrated because I can't figure out any strategies I can use. The first problem, though, I think I can pull this strategy out: if you've got a complicated denominator with polynomials, you'll need some powers of "x" on the top so you can make a "du" to go with the "u" on the bottom. So, multiply top and bottom by whatever powers of "x" are required to make it happen, and don't be afraid of negative exponents on top and bottom.

[@rasheedmohammed2227]

I do the same thing!!

[@arpitsharma4377]

@9:30 Dear bprp, I like the way in which u fit everything in such a small board So beautifully...

[@machoslothman]

Can you do a series answering all the questions from the MIT integration bees explaining all the steps? Thank you!!!!

[@vedants.vispute77]

1:20 We love that as WELL!

[@nithin1729s]

I am getting addicted to your videos . Your videos are so interesting.

[@yahyaazer2523]

Great video as always

[@darcash1738]

On the 2nd one I factored out x^3, u subbed for x^3+1. Had to isolate x, and followed that line down… …1/3(u^5-u^3)^1/3 (during this step, I did have to bring the u into the cube root, multiplying the now u^3 within, which seemed natural to make it less complicated) Factored out u^3 to bring it outside the cube root, resulting in… …1/3 u(u^2-1)^1/3 W subbed for u^2-1 and followed that line to get 1/6(integration(w^1/3))dw Do the fundamental theorem, and sub back in twice, getting 1/8[(x^3+1)^2-1]^4/3. Then, classic square technique, and the +1 and -1 cancel to get 1/8[x^6+2x^3]^4/3 +c Though my way felt worth the effort, seeing your simplistic approach was surprising. Idk how to get that insane intuition 😅

[@Platinum-nv1zq]

Yeah same pro i struggled with the integral on the right and the integral on tge right is a piece of cake

[@Platinum-nv1zq]

The*

[@zbp9541]

How is your comment was 1 month ago when this video was published today? 🤯🤯 either youtube is drunk or my device has been hacked

[@yahyaazer2523]

@@zbp9541 hahaha same here

[@Platinum-nv1zq]

Idk lol

[@Supercatzs]

@@Platinum-nv1zq man's literally watched the video a month before it was viewed

[@electrovector7212]

I just started integrals and I highly recommend everyone to watch bprp! I know it does sound a casual commercial advertisement but it isn't! I mean he is good at it definitely. There are few who can explain such easily like that and fully understands the point. I just started like a week ago and I already know a lot cuz of bprp. Greet love from somewhere else on earth.

[@electrovector7212]

Besides that great mic as always 😂

[@guidichris]

He's good definitely and indefinitely!!

[@sunilparekh4581]

Are you INDIAN🙄🙄

[@sunilparekh4581]

@@guidichris lol😂😂

[@electrovector7212]

@@sunilparekh4581 no.

[@AloncraftMC]

Hello BPRP! I was wondering if you are going to record a video like the "Ultimate Integral Starter", but with derivatives, something like "Ultimate Derivative Starter". I have watched the entire video of Integrals and now I see all techniques cleaner, and whenever I see Integrals, I remember all the wonderful explanations of every one of the tecniques (u sub, Trig sub, By Parts (DI and u-dv Methods), Partial Fractions...). But after learning all of this amaizing content, I say to myself "I wish there was a Ultimate Derivative Starter made by bprp, It would be so nice!! both for new students and for people like me who wants to see Derivatives again nicely, watching a great long video with all the rules and formulas. Hope you read this comment, I enjoy a lot math watching your videos, and thanks to you I got motivation to study Mathematics at University (Currently I haven't finished high-school). Sorry for my bad english, I'm Spanish :P

[@xyzwds5664]

2nd one : First I took x^3 common =X^3(x^3+1)(x^3+2)^1/3 =( x^3+1) = t ( for substitution) Therefore, 3x^2.dx=dt Also, x=(t-1)^1/3 So if we substitute everything then. It becomes = ((t-1)1/3 .t. (t+1)^1/3) /3 (note :- x^3+1=t , So, x^3 + 2 = t+1) Now, =( (t-1)^1/3.(t+1)1/3. t) /3 ---------------------------- Now the terms above the underlines ( or whatever it is) is, (t^2-1)^1/3 So, = ((t^2-1)^1/3. t. dt) /3 Now, can substitute t^2-1 = u So, 2t. dt = du So, after substituting everything = (U^1/3 . du) /6 = (u^4/3 . 3) 24 = (u^ 4/3 )/8 . Now after , Putting value, = (x^4/8 + c) It's different from your answer teacher but I think in indefinite Everyone will get different values so. Here is my answer.

[@OriginalSuschi]

BPRP: Almost forgets the +C Me: *PANICC*

[@s.s.m1595]

The first problem can be solved like this too x=1/u du/dx=-1/x² Now it becomes -u³du/√(u^4+1)³ [fourth root] u^4+1=z To solve the second one take x³ common from the left lart and x³+2=u³ From this we also get x³+1=u³-1 The method shown in the video is better

[@doctorohmed]

1. Differentiate the integral formula 2. The result formula is the just problem. So reverse thought for 1. formula is necessary but difficult.

[@FredDeliege]

taking the fourth root of something put to the fourth power: should it be absolute value ?

[@BoringExtrovert]

Same thought

[@not.a.casual]

0:01 *"tunanenunanenuna"* 18 year old me: starts singing *Doraemon* theme song...

[@sunilparekh4581]

Hey I learn from Nishant Vora sir

[@sunilparekh4581]

Why you deleted your comment

[@Yubin_Lee_Doramelin]

Exactly. I sometimes imagine him holding a microphone with a Doraemon plush or even a puppet (LOL).

[@bigotes20random86]

I don't get it

[@mncubing8160]

In the first problem, I recognized 1/x^2 right away and found u=1/x could work.

[@matthewbacchus3633]

He's holding that stuffed Pi like a Totem of Undying. 😆

[@yashrawat5071]

4:56 Oh my goodness that is so great Feeling of every person when substitution works out

[@yashrawat5071]

*Spiderman: Every video I go* Indian Jee Aspirants *Spiderman: I see his face*

[@yashrawat5071]

If you can't taking out of the square root doesn't work , force it into the root

[@JEE2024-e5h]

you can just substitute x=1/t in the first question, right? I've got the same answer but doin' like this and I followed the same procedure as u did in the second one tho. Can you please reply and justify my statement sir?

[@salsichalivre5401]

One of the reastest abilities is not be able to solve these integrals but instead to smoothly change between color pens

[@egillandersson1780]

Well done !

[@AK-du5mm]

The first one could have been done easily just by taking x^16/3 and then putting the inside value equals to 't'

[@rushilphogat]

they taught us the first one in school checkout ncert class 12th integrals exercise-7.5 Q16😭

[@willconroy8744]

solved the second integral before watching it in the video: let u = x^3 + 1 . Thus du = 3x^2 dx. The integrated can be rewritten as x^3(x^3+1)(x^3 + 2 )^(1/3). the x^3 at the front becomes 1/3x * 3x^2, which then becomes 1/3 (u-1)^(1/3), and the 3x^2 can be lumped with the dx at the end to equal du. The x^3 + 1 in the middle is just u, and the (x^3+2)^(1/3) becomes (u+1)^(1/3). The integrand, in terms of u, then becomes u*((u-1)(u+1))^(1/3), or u*(u^2-1)^(1/3), which then is just a simple u-sub that yields 1/8 * (u^2 - 1)^(4/3) + c, or 1/8 ((x^3 + 1)^2-1)^(4/3) + c = 1/8 (x^6 + 2x^3)^(4/3) + c

[@nk_goyal]

Put x=1/t and first one gets solved in a moment

[@SeggieSum]

I love these integrals

[@kikisinozic7475]

I’m watching this and I am nowhere near this level of mathematics. I’m picked advanced math but am dropping to standard because it’s too hard

[@leonardoguerra6501]

Oh, my god, this guy is just a Genius

[@_wetmath_]

was that the doraemon theme song at the start? haven't heard that in a long time

[@angpanha3566]

Almost lost my last brain cell from mistaking the 1/8 as 13/8

[@Amaru1111]

i cracked when he said "keep these things in your mind because maybe you will encounter them in your dream and attack you" lmao

[@oismath]

I think some step in the first problem isn't clear. How can you take the power inside the root directly without considering the root itself?

[@shubhammahajan3745]

I did this problem when i was preparing for jee(entrance exam in india for admissions in engineering colleges) This ones considered the easiest type in indefinite integration..

[@jibiteshsaha4392]

Surely a indefinite Integral person

[@yashrawat5071]

No one : My brain: He is holding square-root of 6[1/1² + 1/2² + 1/3² ....... ]

[@skrill500]

Lol for the second integral I factored out an x^3 first for some reason and split it into x^2*x. I had to do 2 subs but it eventually worked

[@maxsch.6555]

Same

[@leif1075]

Did anyone actually solve the first one the way he did..i dont think amyone would ever think of that..what about partial fractions??

[@bimarshadhikari5662]

Well, the technique used in the first question is taught specifically to Indian students aspiring to crack JEE.

[@thatonekid20111]

@@bimarshadhikari5662 what's JEE?

[@bimarshadhikari5662]

@@thatonekid20111 It's a highly competitive engineering entrance test in India.

[@DavesMathVideos]

Any time you see questions like this, there is often some manipulation which simplifies it greatly. Failing that, Either the answer will be simple, but the technique will be tough Technique will be easy but the answer will be something monstrous.

[@sunilparekh4581]

I don't care how the answer looks but technique should be easy 😁😁

[@BoringExtrovert]

For the first one, when taking out x^3, don’t you need to take the absolute value?

[@abdullahifarah5878]

The algebraic manipulation is insane.

[@xbowschool]

We do such questions in 12th here in India

[@caroid9093]

And also 12th grade kids around the world Not just India...

[@lakshya4876]

@@caroid9093 true idk why these dudes keep saying this in all the comments

[@IIT_786]

Because Indian's are best at maths that's ... universal truth ... A lot of engineers all over the world are INDIAN ... Why because they study maths in more depth and difficulty as compared to the rest world except some like China .... 😑😑😑😑😑 Edit: Maybe u too did these types at 16 age but in Indian coachings these are considered the easiest or moderate ones for a 16 y student but I have many really many foreign freinds like from US, Argentina,UK etc who struggle with these in group discussions🤔🤔🤔

[@eliseuantonio6652]

Shouldn't the fourth root of x⁴ be |x| instead of x? Why didn't you use the absolute value?

[@swadhin155]

The first and foremost assumption we take is that we are solving within the limits of the domain of the function. If it wasnt so then you hadnt been able to solve any logarithmic ques without domain. Think urself, log(x^2) doesnt have same domain as 2logx but arent they same?

[@eliseuantonio6652]

@@swadhin155 yes, but in this case x

[@angelmendez-rivera351]

@@swadhin155 x < 0 is in the domain of the function, so your point is moot. OP is correct, absolute values are needed in this step, and because the domain is disconnected, two constants of integration, one for each connected subset, is needed.

[@franky1168]

when you take out of the 4th root the (x^3)^4 , isn't it |x^3| in case x is a negative number?

[@abdullahifarah5878]

I'm an indefinite person. I don't want to add in my limitations while searching for the area. I'll take the arbitrary constant over anything.

[@adityarupda1795]

Hi bprp I don't know that you remember or not but few months ago during that covid 19 lockdown I have emailed you four integral battle problems if you remember than can you make video on them ?????????????

[@inteculus603]

Hmm. The first one was very easy, but the 2nd one really made me scratch my head.

[@jesse2535]

This isn't really a hard integral exercise it's a niche elementary math applied to solving integrals. It doesn't test my knowledge of integrals, it tests if I remember that niche thing my math teacher told us not to worry about.

[@nithin1729s]

Can i draw an imaginary circle with negative radius on complex plane???

[@zdino9116]

I love being the only archeologist understand and solve integrals in my class. NICE VIDEO, DUDE.

[@rodnarciso3725]

Tais toi personne ne s'en fout là

[@maalikserebryakov]

Its a mostly worthless skill outside of pure mathematics examinations. Computers can easily do this stuff.

[@recursiveloop]

Lovely videos. What's he holding the dolls for though?

[@elishagyamfi7486]

Hello great video, I have a question. In the second question why does the x from the outside get raised to the power of 3 once it enters the cubed root

[@harshrajrana7142]

Can we asume x³+2 as t³

[@jamiethomson6649]

And this is why I ended up with a 12 in senior year math :/

[@allmight801]

Just one question is x cubed an odd or even function? My math teacher wrote it as even but i wonder if she made a mistake

[@blackpenredpen]

Yes it should be odd.

[@allmight801]

@@blackpenredpen thank you

[@norgesguardians5682]

Odd. When the power is odd you get minus: (-x)³=-(x)³. Even power is an even function: (-x)²=x². (:

[@allmight801]

@Jessen thank you

[@allmight801]

@@norgesguardians5682 thank you

[@toaj868]

When people make integration problems do they just differentiate a function and then rearrange and rephrase the derivative until it is sufficiently difficult?

[@angelmendez-rivera351]

Basically

[@entitydoesvids]

i want the blue pi plushie so badd

[@catnipppppp]

What happened to big board

[@gauranshmukesh-po6ec]

1st one is a ncert question of class 12 and this is actually easy😂😂😊

[@swadhin155]

Damn they are easy, most of the time in such complex ones, there just lies to factor out or create one exponential term of x that directly converts into a substitutable question. And Im a definite integral person btw, cuz thats all brain and indefinite is just try this try that!

[@kevinhermawan369]

That white board is too small :(

[@satishchaudhary7978]

Shut up

[@vaibhavm2916]

Now don't say even this was a lost video.

[@ilias-4252]

Shouldnt it be absolute value of x on the first question? it s techinically 4rth root of x^12 =sqrt of x^6. Correct me if i m wrong

[@angelmendez-rivera351]

You are correct. In fact, not only should the answer account for this, but it should also account for the disjointed domain of the integrand by using different constants of integration in the different connected sets of the domain.

[@chathuraviduranga7168]

your teaching is superb can you teach 3D vectors

[@sunilparekh4581]

Why you are keeping doraemon theme song in the beginning 😁😁

[@aryagujarathi6308]

I have a question could we let u = x^2 in the first one and let u= x^3 in the second one?

[@dudono1744]

So, when you have a root, just say what you have inside is u and manage to have du showing up

[@angelmendez-rivera351]

If possible, yes, this is the preferrable strategy

[@tomiokascuteness1909]

It's been 3 days I had started learning definite integration so I don't know maybe I am which one but until then proud to be indefinite integral person ..🙃 Btw love your videos they are actually helpful and interesting 🙂

[@sunilparekh4581]

Are you INDIAN

[@sunilparekh4581]

Then start learning from Unacademy jee channel, it is literally very helpful 👍👍

[@sunilparekh4581]

I learn from Nishant Vora sir😁😁

[@nithin1729s]

@Sunil Parekh jaldi se jalsi Unacademy ka promotion kar deta hu , cool lagunga

[@Atchunachu]

These are so easy

[@tradingchris]

Y I can't guaranteed this

[@1024-p4t]

where are the captions?!

[@kamalahamed8586]

Very good

[@pituitlechat3807]

Strange : I write your 2 first expressions (before 3:29) : the first one is even and the second one is odd???? How is possible?

[@tajpa100]

since last tuesday i'm patreon of your channel. however, I couldn't see my name in this video published just 18 hours ago.

[@blackpenredpen]

Hey there, this video was actually made and unlisted published about 4 months ago. I have a different setting now and a much longer beard. When I make new videos, I will be sure to include your name. Thank you for your support.

[@astraestus8828]

I kinda hate Definite Integrals. Hate the extra work lol.

[@27.peterbahaladrielsinaga76]

What an idea mann😢

[@nathanaelmoses7977]

If you fight an integral Dont cry just look at the sky And say Wouldn't it be nice

[@swarajandhale4088]

I love to do both ❤❤❤

[@motivational803]

Bro try black book

[@MindFlaw]

▶All time your problem is highlighted in my life, being a good person I love your teaching and I'm happy to accept you as my teacher. Heartful thanks for hard work. ✔

[@gurusaran7193]

I get these in my highschool test bruh Why does my School do this...

[@fomoCoder69]

In india we call this the kutur putur technique