How to solve hard integrals using just u-substitution

[@hannakrupowiecka6311]

Factor the "not" before the integral and you're left with (not ∫fair dx), and the inner integral is fair.

[@herbie_the_hillbillie_goat]

First I chuckled at this. Then I realized that's pretty much what he did. 😂😂

[@LS-Moto]

You know a teacher is a good teacher by being honest and openly saying that they couldn't figure it out, and had to look at the solution.

[@bprpcalculusbasics]

😆

[@bprpcalculusbasics]

Or admits he uses Wolframalpha too hahah

[@sooryanarayana3929]

@@bprpcalculusbasics or the teachers watches bprp

[@gamingresumed1788]

@@sooryanarayana3929 The teacher is bprp

[@tonyhaddad1394]

@@bprpcalculusbasics if we dont be honest with our self , we can t get better , specialy when we talk about the language of the universe (math) !!

[@mrkenzivedran]

For the first integral, you can use the following two subs: the first one is x=sqrt(tan(u)), after which you use sin(u)=z. You will get a result containing sin(arctan(u*u)), and you can always use the basic trigonometric identity to convert that into a more elegant form. If you're interested, I can show you exactly what I did that worked beautifully.

[@somasahu1234]

Yep, show it!

[@fredericchopin6445]

yes, please show it

[@mrkenzivedran]

@@somasahu1234 if you’re interested in the answer, please check your e-mail inbox, I sent it there because youtube doesn’t let me post links in the comments section

[@mrkenzivedran]

@@fredericchopin6445 if you’re interested, please provide me with an e-mail so that I can send you what I’ve done, because youtube doesn’t allow posting links or pictures in the comments section. I think you can do so by typing it out with spaces between each symbol of the e-mail.

[@Nxck2440]

@@fredericchopin6445 Let x^2 = tan u --> 2x dx = sec^2 u du --> dx = (sec^2 u) / (2 tan^(1/2) u) du The integral becomes = 1/2 * integral of (sec^2 u * tan^(-1/2) u) / (tan u * sec^(3/2) u) du = 1/2 * integral of (sec^(1/2) u) / (tan^(3/2) u) du = 1/2 * integral of cos u * sin^(-3/2) u du Let t = sin u --> dt = cos u = 1/2 * integral of t^(-3/2) dt = -1 * t^(-1/2) + C = -1 * sin^(-1/2) u + C = -1 * (sin arctan x^2)^(-1/2) + C Use a triangle to figure this out = -1 * (x^2 / sqrt(1 + x^4))^(-1/2) + C = -sqrt(1 + x^4) / x + C Could also have used the hyperbolic version i.e. let x^2 = sinh u.

[@kobethebeefinmathworld953]

Problem 1: u-sub in a special way Problem 2: u-sub in a more special way

[@bprpcalculusbasics]

Yup just that!

[@davidbrisbane7206]

I imagined pausing the video and solving the problem and then continued watching

[@jehmarxx]

The U world is like school. You have to find a way to get into one and then get out of it afterwards. Lololol.

[@piratekingluffy6430]

For the second integral, u can try by taking u=x^3+1 and factor out x^6+x^3 into (x^3+1)(x^2)(x). I thought it might be complicated but it turned out simple. Then after another substitution i got the same answer.

[@robertveith6383]

That is "you can try ... " You spell out "you," and "u" is a variable.

[@sergeiivanov5739]

But... For some reason, i didn't notice this. Both integrals are solvable. The 1st is by x = 1/t, and the 2nd by t = x^3 + 1 as you have figured out. Sadly, the 2nd integral was solved by multiple substitutions. Namely, i noticed that if q = x^3 + 2, then the whole integral is a sum of integrals (q^(2 + 1/3)dx - 3q^(1 + 1/3)dx + 2q^(1/3)dx). Here we must notice recursion q^(n + 1/3)dx = x(x^3 + 2)/(3n+2) + [(6n+2)/(3n+2)] * q^(n-1+1/3)dx. So, we must solve the last Integral (x^3 + 2)^(1/3)dx and the initial Integral is a sum of three aforementioned integrals. We start by putting p^3 = x^3 + 2, after which we get cycle integral, denoted by I, which is the of p^3 / (p^3 - 2)^(2/3) wrt p. Next, by integration by parts we have an equality I = p(p^3 - 2)^(1/3) - A where A is the integral of (p^3 - 2)^(1/3). The initial integral can be also represented as a sum of A and 2B, viz. I = A + 2B where B is the integral of 1/(p^3 - 2)^(2/3). This is by representing p^3 as p^3 - 2 + 2. Thus, I = p(p^3 - 2)^(1/3) - A = A + 2B. Hence, the problem has been boiled down to determining B. To solve B, i put p^3 - 2 = 1/t and got an Integral which was rationalized by (2t + 1)^2 = u^6, and i had to evaluate (-2du)/((u^3 - 1)^2) which is solved by method of unknown coefficients. Notice that the integral of f(u) /(u^2 + u + 1)^2 is solvable by lowering the power of denominator and then by integration by parts. Here i am lazy cuz the point has been determined. The Integral of this rational function is evaluable. Recursive expression occurs there. So, once we find the answer in terms of u, it can be expressed through x, in terms of x, by plugging inverse substitutions u = (2t + 1)^(1/3) and t = 1/(p^3 - 2). And lastly, p = (x^3 + 2)^(1/3). So, idk

[@sergeiivanov5739]

So, simple and so difficult at the same time. Probably, we can see from where Chebychev theorem comes. We inspect x^m (a+bx^n) ^ p and play with the integrand. And upon opting for some substitution, this gets evaluated. Interesting but too lazy

[@holyshit922]

Integrals of the function in the form x^m(a+bx^n)^p , m,n,p \in Q There are three cases and there is substitution for each cases case 1: p\in Z u^{s} = x , where s = lcm(denominator(m),denominator(n)) case 2: (m+1)/n \in Z u^{s} = (a+bx^n) , where s = denominator(p) case 3: (m+1)/n + p \in Z u^{s} = (b+a/x^n) , where s = denominator(p) For the second integral is also possible integration by parts

[@MrJoshie333_]

What

[@holyshit922]

@@MrJoshie333_ They dont teach you this substitutions

[@MrJoshie333_]

@@holyshit922 Clearly lol

[@holyshit922]

@@MrJoshie333_ They work, check it Anshuman Agrawal also mentioned this types of integrals but didn't show substitutions for all possible cases

[@MrJoshie333_]

@@holyshit922 ah okay thanks. Just curious, which Calc class does one learn this? I've done Calc 1-3 but haven't seen this kind of advanced substitution

[@anshumanagrawal346]

The first one isn't hard if you're taught that kind of integral, it is of the form x^n(ax^m+b)^p where (n+1)/m + p turns out be an integer so you know that if you factor out an x^m from that and substitute it will become solvable easily

[@replicaacliper]

Nothing is hard if you've seen it before

[@SJrad]

After the integral in the 2nd problem, would it be better to simplify the cube root by separating that binomial to a 4th power into 2 separate binomials with one being to the 3rd power and the other to the first, and then move the one with the third power outside the cube root. And also factor out a x^3 from the binomials and take it outside as well

[@sandeepchauhan7596]

that 1st integral is in our NCERT 12th mathematics book (miscellaneous exercise)

[@yoav613]

I can give you more. Try integral lnx x^(1/6)/(1-x^2) from 0 to inf

[@insayno9959]

(pi^2)(sqrt3-2)

[@yoav613]

@@insayno9959 yes wolfram helps.full sol please😅

[@rogerkearns8094]

@@yoav613 Yes, I have elbows, how did you know?

[@yoav613]

@@rogerkearns8094 there is nice sol using digamma function

[@rogerkearns8094]

@@yoav613 Ok ;)

[@polychromaa]

Where can i find more integrals like these? I often struggle to find gems like these online.

[@winter_c]

Theoretically,learning only integration by substitution can solve these two questions. But I beg only very very few people can actually think out of the box like that.

[@anshumanagrawal346]

I'm so proud I figured out both of them 😊

[@tintinfan007]

ME TOO

[@camelliascholl6564]

me 3 days ago: damn my calc final couldn't have been much worse me right now, 2 seconds into this video: never mind, my calc final could have been loads worse

[@tambuwalmathsclass]

The Father of modern calculus

[@Vladimir_Pavlov]

You should write (∜x^4)^3 = |x|^3. In order to reveal the sign of the absolute value, it is formally necessary to consider two cases: x>0 and x < 0. As a result, the answer received by the author will be multiplied by sgn(x): -sgn(x)*∜(1+x^(-4)) +C.

[@violintegral]

This absolute value issue can be resolved more nicely by simply multiplying the (1+x^-4)^(1/4) by 1 of the form x/x. The x in the numerator can become (x⁴)^(1/4), which multiplies with (1+x^-4)^(1/4) to give (x⁴+1)^(1/4). This gives a simplified antiderivative of (x⁴+1)^(1/4)/x+C, which is equivalent to yours using sgn(x). If you realize that the integrand is an even function before integration, it stands to reason that the antiderivative will be odd (with the appropriate constant of integration, in this case, C=0). Looking at the parity of the integrand compared with that of the antiderivative can be a good way to check that your antiderivative is reasonable.

[@loich.9133]

can you do the integral of 1/(x^5)+1 ?

[@antoine2571]

i'm in high school, i didn't see integrals and i'm loving it

[@tonyhaddad1394]

Deja vu 😁 since a wihle on one of your channel , i learned some deep sens from these intgrale

[@sybot.]

1/2* NOT FAIR^2 + C

[@dominicgrew4000]

I recommend trying this integral! x(sin(x))/(cos^2x)+1 dx. Note cos^2x is cosine squared of x.

[@robertveith6383]

Then you should write cos^2(x). If the denominator is supposed to be cos^2(x) + 1, then you need grouping symbols around it.

[@violintegral]

I'm not 100% sure what function you mean to integrate because of your lack of grouping using brackets or parentheses, but x•sin(x)/(cos²(x)+1) certainly does not have an elementary antiderivative. However, the definite integral of that function from 0 to π can be found without too much trouble by use of symmetry.

[@Alians0108]

xsinx ---------------- cos^2(x)+1

[@laurence090]

what happened to the numerator’s role in the final index/power/exponent law simplification? of the u-sub in the first problem •_O

[@AtariDays80]

Side question: Is u-substitution still valid at x=0? At that point, u-du is dividing by zero.

[@沈博智-x5y]

short answer is: it still seems to be valid, however for the first integral, the domain of the integral did not include x = 0 for the second integral, this may be why people prefer to fudge around with the integrand rather than make dx the subject when they find du/dx so that they "avoid dividing by zero". for example in his substitution, u = x^6 + 2x^3 then: u = x^6 + 2x^3 du/dx = 6x^5 + 6x^2 du = (6x^5+6x^2)dx du = 6(x^5+x^2)dx integral of (x^6+x^3)cbrt((x^3+2)) dx = integral of x(x^5+x^2)cbrt(x^3+2) = 1/6 integral of 6(x^5+x^2)cbrt(x^3(x^3+2)) dx = 1/6 integral of cbrt(x^3(x^3+2))(6)(x^5+x^2)dx = 1/6 integral of cbrt(x^6+2x^3)(6)(x^5+x^2) dx = 1/6 integral of cbrt(u)du = (1/(4/3))ucbrt(u)/6 + c = (1/(4/3))(x^6+2x^3)cbrt(x^6+2x^3)/6 + c = x^3(x^3+2)cbrt(x^3(x^3+2))/8 + c = x^4(x^3+2)cbrt(x^3+2)/8 + c but you still end up with the same answer regardless. it may have to do with limits tending to zero. also, the second integral seems to divide by zero when x = -cbrt(2) as well. as 6(x^5+x^2) = 0 iff x^5+x^2 = 0 iff x^2(x^3+2) = 0 iff x = 0 or x = -cbrt(2) provided x is real.

[@Ashirene22]

Isnt the first chebyshev?

[@sg8119]

It's actually not hard at all use trigonometric substitution by putting in x=rt tan u then just simplify to get du in terms of dx

[@cormalan9894]

The answer to the one in the title is NOTFAIRX by the way

[@eepym1tu]

+ C

[@cormalan9894]

@@eepym1tu how did I forget smh

[@ammyvl1]

shame on you

[@teelo12000]

4:50 trollmath: yeah but you can simplify that further, the fourth root of 1 is 1, and the fourth root of x^-4 is x^-1, so the answer is -(1 + x^-1), = -1 - 1/x

[@dlevi67]

More trollmath at the end: just eliminate a power of 3 from the exponents, since there is a cube root: 1/8 (x^2+ 2x)^4, which trobviously simplifies into 1/8 (x^8 +16x^4) = x^8/8 + 2x^4. Oh, plus C.

[@yoav613]

Integral (not fair)=it is going to be on the test

[@ysmashimaro]

x^4^3^(1/4) should be |x|^3 instead of x^3, right?

[@Vibranium375]

Yes

[@tintinfan007]

NICE POKEBALL DUSTER

[@ussgordoncaptain]

Man I tried to find the u sub for (1+x^4)^(1/4) for the first one

[@pneujai]

the second one is easier than the first one i would say

[@ILoveMaths07]

Beautiful!

[@Kazzit_Chang]

I have reading disability so I use inline substitution all the time.

[@_Cake_progresS_]

Fun fact : i not resolve the first but I resolve the second L.

[@KinderHades]

This video shows how little I actually understand about calculus. Looks like I need some practice

[@gamingmusicandjokesandabit1240]

*haters will say that all integrals are unfair to your free/lesiure time*

[@pradyumangangan4784]

Why is this integral unfair?🤔

[@hyx6817]

thanks

[@hatembahri4314]

integrals🤩🤩

[@chriskim7261]

he is pokemon master

[@Аноним-щ3н]

Omg, why are you so emotional

[@bprpcalculusbasics]

？

[@mrflibble5717]

Nice!

[@buckeye49]

notfairx + C

[@nate0___]

I managed to understand this even though I am 13 lol

[@mrocto329]

nothing special, the real pain is AFTER calculus where you get much more abstract

[@tonyhaddad1394]

Channel(s)*

[@mace6402]

Thats wrong tho

[@pablo2448]

What the heck