Battle 1, integral of cos(x^2) vs integral of cos(ln(x)), @1:00 Battle 2, integral of ln(1-x^2) vs integral of ln(1-e^x), @7:55 Battle 3, integral of x^(x/ln(x)) vs integral of x^x, @16:23 Battle 4, integral of x*sqrt(x^3+4) vs integral of x*sqrt(x^4+4), @19:29 Battle 5, integral of x/ln(x) vs integral of ln(x)/x, @32:25 Battle 6, integral of ln(ln(x)) vs integral of sqrt(x*sqrt(x)), @34:00 Battle 7, integral of sqrt(sin(x)) vs integral of sin(sqrt(x)), @36:13 Battle 8, integral of sqrt(tan(x)) vs integral of tan(sqrt(x)), @40:52 Battle 9, integral of tan^-1(x) vs integral of sin^-1(x)/cos^-1(x), @59:13 Battle 10, integral of 1/(1-x^2)^(2/3) vs integral of 1/(1-x^2)^(3/2), @1:04:23 file: docs.wixstatic.com/ugd/287ba5_3f60c34605f1494498f02a83c2e62b29.pdf
SIR THE RESOURCES AND LINKS TO LEARN MATHEMATICS THAT YOU SAID IN YOUR VIDEO WITH fematika ARE STILL NOT UPLOADED IN THE DESCRIPTION OF THE VIDEO , please do upload those links
Another approach to the integral of ln(1-x^2) dx would be to factor the inside and then use the product rule of logarithms to get the integral of ln(1-x) + ln(1+x) dx. It's a bit easier to solve this way.
Just a real minor point of #4: you could also do a hyperbolic trig substitution instead, and you'd get a simple inverse hyperbolic sine term in the final answer instead of the natural logarithm. That natural logarithm is also convertible to the inverse hyperbolic sine.
The way I like to think about the Integral of cos(x^2): with some clever substitutions and Euler's formula it can be shown that it can be written in terms of the integral of e^(x^2) and since that cannot be defined in terms of elementary functions, thus the integral of cos(x^2) cannot be
Chirayu Jain It’s quite hard to prove it mathematically. I think we need to know Galois theory from advanced abstract algebra in order to do so. I actually don’t have experience in it unfortunately.
One-hour long video but u definitely spent a lot more time than that! Your effort should be appreciated! And also the patreon list grows longer everytime 😁👍 PS it's 1am here in HK and yr thumbnail looks cool with some chill 😆
To integrate arcsin(x)/arccos(x) from x = -1 to x = t < 1, let x = cos(θ). Then dx = -sin(θ) dθ. The integrand is now -arcsin(cos(θ))·sin(θ)/θ. The bounds are from θ = π to θ = arccos(t). On the interval (0, π), which is the codomain and range of arccos(t), arcsin(cos(θ)) = π/2 - θ. Therefore, the integrand is -(π/2 - θ)·sin(θ)/θ. Factoring -1 will change the bounds to run from θ = arccos(t) to θ = π, with integrand (π/2 - θ)·sin(θ)/θ. By linearity, this gives the integrals of (π/2)·sin(θ)/θ and -sin(θ). The first integral is equal to (π/2)·(Si(π) - Si(arccos(t))), and the second is equal to cos(π) - cos(arccos(t)) = -(1 + t). Then the total integral is simply equal to [(π/2)·Si(π) - 1] - (t + Si[arccos(t)]). Call (π/2)·Si(π) - 1 = C, so the integral is simply C - t - Si(arccos(t)). Done! For the record, Si(x) is defined as the integral from s = 0 to s = x of sin(s)/s. We can extend the answer to other intervals, but this requires some caution, since arcsin(cos(θ)) = π/2 - θ is no longer true in other intervals.
This is the best video You have made - of those I've seen. I was especially happy to know that ln(ln(x)) is a non-fundamental function. That question has been bothering me for years.
22:21 Euler's substitution sqrt(u^2+4)=t-u would be better idea here Last one third Euler substution (with roots) or integrating by parts also are good option
On the first one, it was obvious, because cos(ln x)=(x^i+x^-i)/2. Power rule, separate real and imaginary coefficients, and put it back to trig functions. Even if you're not going to use complex numbers, you can guess the right integral because cos is like an exponential and goes well with ln and poorly with x^2.
15:05 In the last two terms of that answer (before the +C) it was not necessary to use absolute value around the ln input. Respond to this comment if you can figure out why!
Either way you need to do integration by parts. Personally, I broke up the ln but if makes sense to use IBP with a bit of work extra then go for it. As long as you get an answer and understand the process
GhostyOcean no you don’t need to do integration by parts with the method he stated. After you split the ln you can split the integral and solve them both by u sub
19:57 you can do both u-sub and trig-sub at the same time by letting x^2=2tan(theta) ;) then, xdx is nicely equal to sec^2 and the rest is just the usual
Number 9 is a pretty straightforward battle, once you know the formula for antiderivatives of inverse functions. As long as a function has an elementary antiderivative, its inverse has an antiderivative of the form, xf^-1(x) - F(f^-1(x)). Once you know tan(x) has antiderivative ln|sec(x)| + C, you just plug tan^-1(x) into the formula and do some trig identities on sec(tan^-1(x)) to get the same result.
Would it actually be faster to integrate cos(ln(x)) by using the complex definition of the cosine? You would then need to integrate (x^i+x^-i)/2, which is just a matter of integrating polinomials.
You probably made that future video already, but it is interesting to point out that the most obvious attempt to antidifferentiate arcsin(x)/arccos(x) with respect to x results in the sine integral: A basic trigonometric identity has arcsin(x)=π/2−arccos(x), from which the integrand becomes ½π/arccos(x)−1; then the substitution x=cos(y) with dx=−sin(y)dy results in the sine integral. That is, ∫arcsin(x)/arccos(x) dx = -x−½π∫sin(y)/y dy = −x−½πSi(arccos(x))+C.
Hi, cos(X square) is a function . Geogebra gives a result, if you integrate ( calculate the area) between 2 points Why we can say that this integral does not have a result.thank you For your reply
Cent Uğurdağ Because the antiderivative of cos(x^2) is *not* the area. The antiderivative of cos(x^2) is simply another function, but the area under the curve is a number. Not remotely the same thing. Any software can calculate any area, but if you ask Geogebra to give you the antiderivative, it *cannot* and *will not* give you an answer, because there is no answer.
Hi BPRP, and thank you for the videos :D I guess this comment will go unnoticed, but if I never ask, I'll never know :) Why are half of these functions impossible to integrate? You just mention as a fact that it's impossible but never why. I'm not great at integration, so I don't understand _why_
integrating arcsinx/arccosx is actually doable;much easier to do than the other ones mentioned as undoable previously. its just a bit of subs and ibp and using the Si function.
11:22-11:25 the integral of the thing you are saying needs partial fractions doesn't, actually, because the answer is clearly inverse hyperbolic tangent (Argthx/Argtanhx)
Serouj Ghazarian Well, that's not correct either, since the domain or arctangent is different from the domain of the function we started with. Strictly speaking, partial fractions are the only correct way to get the most general antiderivative, and this can be proven.
It's non-elementary because if you try to do IBP, you get xln(abs(sec(x)+tan(x)))-integral of ln(abs(sec(x)+tan(x)))dx. Here integral of ln(abs(sec(x)+tan(x))) is non-elementary.
Mohammad Zuhair Khan ln in this situation is not friendlier than ln, since the inside of ln would be a complicated expression. In fact, coth is expressible in terms of ln, so that makes your point moot.
m8 im in high school learning quadratics XD could u do a video where u explain calculus and why it works sorry i just kinda don't get what ur doing and just don't get calculus - but i still sub
Its all about analyzing a graph of the function. Integral is giving u a surface area under a function. Derivative is the gradient of a line tangent to the function
My god, integral of x times the square root of (x^4 + x) is a really complicated integral. It would be even more complicated if one had to integrate sec^3(x) from scratch... 34:26 The integral of the square root of (x times the square root of x)?? The integral of the square root of (x + the square root of x)... 🙂 The integral of √(x + √x) Or the integral of 1/√(x + √x) Or the integral of 1/√(1 + √x)
James Oldfield Obviously, it is sqrt(x·sqrt(x)). Also, the integral of x·sqrt(x^4 + x) is non-elementary, and is also not the integral dealt with in the video, and the one in the video was actually very simple.
@@angelmendez-rivera351 You must be a really smart person to find this kind of thing easy. I'm still at the level of basic integration and differentiation, power rule stuff. Like 1/cube root (9x^4) + 3x^3 + x^2. Really, really basic stuff like that...
@@angelmendez-rivera351 I was being cheeky. I know he said √(x√x). I was thinking it was easy (relatively), and that √(x + √x) would be a harder integral to do...
James Oldfield I wouldn't say I'm smart, just math savvy. Anyway, I only said it's easy because that was one of the easier integrals showed in the video. Most of the other ones were more complicated. And it doesn't have anything on the integral of sqrt[tan(x)], or even worse, the cbrt[tan(x)] integral. The integral of sqrt(x + sqrt(x)) is indeed more complicated than the integral of sqrt(x·sqrt(x)). In fact, the integral is very clever. For example, if y = x + sqrt(x), then dy = [1 + 1/{2·sqrt(x)}]dx. Thus, sqrt(x + sqrt(x)) = sqrt(x + sqrt(x))[1 + 1/{2·sqrt(x)}] - sqrt(x + sqrt(x)/{2·sqrt(x)} = sqrt(x + sqrt(x))[1 + 1/{2·sqrt(x)}] - sqrt(sqrt(x) + 1)/2. Now one can split the integral in two parts using linearity. The integral of sqrt(x + sqrt(x))[1 + 1/{2·sqrt(x)}] can be found using the very simple substitution I already mentioned, and this integral will be equal to (2/3)·sqrt(x + sqrt(x))^3 + C. All the remains is evaluating the integral of sqrt(sqrt(x) + 1). Let z = sqrt(x) + 1, so x = (z - 1)^2, and dx = 2(z - 1)dz. This leaves the integral of 2z^(3/2) - 2z^(1/2) with respect to z. This is just a very basic power rule integral, and it gives the antiderivative (4/5)z^(5/2) - (4/3)z^(3/2) + C. Substitute back to get (4/5)·sqrt(sqrt(x) + 1)^5 - (4/3)·sqrt(sqrt(x) + 1)^3 + C. Altogether, the integral of sqrt(x + sqrt(x)) is nicely equal to (2/3)·sqrt(x + sqrt(x))^3 + (2/3)·sqrt(1 + sqrt(x))^3 - (2/5)·sqrt(1 + sqrt(x))^5 + C.
