Integrating a Quadratic boi - Papa's Improvised Session #2 [ integral sqrt(ax^2+bx+c) ]
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Let us burn this integral to ashes :v What a bitch, I even got the answer wrong, 'cause I succ 3: We also have to solve the integral of sec(x) and sec^3(x) on our way through. Enjoy this mental breakdown video lel
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Everything is a standard integral if you want it to be.
"I'll take 'Weiner process' for $1,000 Alex."
But... But I wanted to be your next substitution variable :(
Duncan W
Zeta... as beloved as you are, ur kinda hard to write bruh...
@@kylepoe4321 That's the whole point
lol hi zeta
ξ
Now integrate dx to relax a bit.
Lol
@@PapaFlammy69 this time make it, easy integrals made hard
😂😂😂
dx²
I have just found the physical application of this integral due to pure curiosity. The length of projectile motion curve of an object thrown with an initial speed with some angle in 2D under ideal circumstances is exactly found by this quadratic boi integral wow.
@Hans von Zettour Nope it doesn`t. You`re talking about position: vertically and horizontally distances taken at a given time.
The arc length however is different its the length of the trajectory and you need to integrate some ugly guy but characteristically its this quadratic boi. Check this out: brilliant.org/discussions/thread/arc-length-of-projectile-2/
Can someone abbreviate on this?
@@tommyc5341 Hey Tommy, I don't know why the guy deleted the comment and the link is also broken I guess lol. What do you mean by abbreviate exactly
@@barsercan7203 like what would you integrate to find this out? How do you change initial speed and a 2d angle into an integral?
@@tommyc5341 first we calculate three quantities: vertical and horizontal velocities as functions of time, and the flight time. They all depend on the magnitude and the angle of the initial velocity. Then we do the ordinary arc length integration and it's of this form. You can see how to setup the integral here. The guy does numerical integration in the end lol, but there is the analytic solution. th-cam.com/video/Dr3IlxD3HzQ/w-d-xo.html
Has anyone taken the derivative of his final result?
Don't tempt me
Papa flammy had an identity crisis 😂
7:37
u = (2ax + 1)/sqrt(4ac-b^2)
You are reducing to int sqrt(1+x^2) dx
One can recognize that it looks like cos^2(x) = 1-sin^2(x) => cos(x) = sqrt(1-sin^2(x)), but with a plus instead of the minus:
it's the hyperbolic geometry that we need to deal with: cosh(x)^2 - sinh^2(x) = 1
cosh^2(x) = 1 + sinh^2(x)
let sinh(u) = x
cosh(u) du = dx
int cosh(u) sqrt(1+sinh^2(x)) du
int cosh(u) cosh(u) du
int cosh^2(u) du
cosh^2(x) = 1/2 (cosh(2x) + 1)
int cosh = sinh
int 1 = x
Plugin back in, we'll get sinh(arcsinh(x)) = x and arcsinh(x)
arcsinh(x) = ln(sqrt(1+x^2) + x)
And we find exactly your result.
Hyperbolic geometry matters and is faster than integrating sec^3
Jesus why didn't I think of that
@@PapaFlammy69 is there a way to solve this without trig or hyperbolic identities??
Leif You may try the Feynman method with a definite integral, although this is not guaranteed to work. Otherwise, no, this is impossible with trigonometric or hyperbolic substitutions.
You didn't get absolute value of ln so....?
Mathophilia |sqrt(x^2 + 1) + x| = sqrt(x^2 + 1) + x for all x, because x^2 + 1 > x^2 for all x, hence sqrt(x^2 + 1) > |x| for all x.
I enjoy seeing you. Your demonstrations, accompained by your theatricaly crazy personality remember me that most famous mathematicians of History did their better performance when young, like you. But be carefully, many of them ended very crazy, like you seems to be now, exposing the best of your extraordinay work.
Ja ja ja!!!
Hey, at least he doesn't seem to be on the path of becoming famous on his studies on statistical mechanics
What an emotional rollercoaster
Now integrate x^x.
Series of x^x = exp(x*ln(x)) is ((x*lnx)^n)/n!,
Then you integrate (x*lnx)/n! by parts n times
?????
Profit
Yo stfu u annoying af
@@alexchepilko4298 If u do a u substitution you can solve it without that integration by parts, I think you had to put u=lnx or smt but it ended up alright with an infinite series
I smell lambda w function
10:30
Rewrite sec^3(t) as sqrt(1+tan^2(t))sec^2(t), use the substitution zeta = tan(t) which leads to the integral of sqrt(1+zeta^2), which can be solved with the hyperbolic sub zeta = sinh(r) After this, one has to evaluate the integral of cosh^2(r), which is trivial and left as an exercise to the reader. And then undo all the substitutions
Not quite so standard, and completely different from the method you used, which I would say is a better method. And also entirely possible that this is just wrong.
Also, use zeta as a substitution variable when :)
This generalized integrals make me a happy man. Throw your chalk harder Papa Flammy!
This is the first time I've watched a math video like this the entire way through, thank you papa flambo
Tangents are evil. Hyperbolic sines are the new fad.
that's true if a>0 and b^2-4ac
Angelo Maimone oh, thats why you cant get pi this way if a is -1, b is 0 and c is 1
Actually the only needed condition is for ax²+bx+c >= 0, because no matter what you do, the square root is going to be positive, and therefore the integral is going to be positive
No discriminant is greater than 0 not less than
@@purungo wait so for this question only restriction is ax^2+bx+c is greater than 0
Kyle Sheng Yes
This integral means a lot to me. I tried it so many times before i learn Derrivatives. Then once i learned it and how, i must have done it a 100 times.
lol
I can give the most general answer to this. Firstly, ax^2 + bx + c = a(x^2 + bx/a + c/a). Call b/a = p and c/a = q. Hence ax^2 + bx + c = a(x^2 + px + q). This is important, because if a < 0, the answer has a different form as opposed to if a > 0. In particular, there is a domain restriction depending on the sign of a. If a < 0, then the function is integrable iff b^2 - 4ac > 0.
sqrt[a(x^2 + px + q)] = sqrt(a[(x + p/2)^2 - (p^2 - 4q)/4]) = sqrt([a/4][(2x + p)^2 - (p^2 - 4q)]. Depending on the sign of p^2 - 4q, this has three different answers for a > 0.
If p^2 - 4q < 0, then sqrt([a/4][(2x + p)^2 - (p^2 - 4q)]) = sqrt([a/4][(2x + p)^2 + |p^2 - 4q|]). Let 2x + p = sqrt|p^2 - 4q|·sinh(u). Hence 2·dx = sqrt|p^2 - 4q|·cosh(u)·du. After substitution, the integrand becomes |p^2 - 4q|/4·sqrt[a·cosh(u)^2]·cosh(u) = |p^2 - 4q|/4·sqrt(a)·cosh(u)^2 = |p^2 - 4q|/4·sqrt(a)·[cosh(2u) + 1]/2 = sqrt(a)·|p^2 - 4q|/8·[cosh(2u) + 1]. Antidifferentiating gives sqrt(a)·|p^2 - 4q|/16·sinh(2u) + sqrt(a)·|p^2 - 4q|/8·u + C = sqrt(a)·|p^2 - 4q|/8·[sinh(u)cosh(u) + u] + C. u = arsinh[(2x + p)/sqrt|p^2 - 4q|], hence sqrt(a)·|p^2 - 4q|/8·[sinh(u)cosh(u) + u] + C = sqrt(a)·|p^2 - 4q|/8·[(2x + p)/sqrt|p^2 - 4q|·sqrt([2x + p]^2/|p^2 - 4q| + 1) + arsinh[(2x + p)/sqrt|p^2 - 4q|]] + C = sqrt(a)/8·[(2x + p)·sqrt([2x + p]^2 + |p^2 - 4q|) + ln([2x + p]/sqrt|p^2 - 4q| + sqrt[(2x + p)^2 + |p^2 - 4q|]/sqrt|p^2 - 4q|])] + C = sqrt(a)/2·(x + p/2)·sqrt(x^2 + px + q) + sqrt(a)·|p^2 - 4q|/8·ln[x + p/2 + sqrt(x^2 + px + q)] - sqrt(a)·|p^2 - 4q|/8·ln[sqrt(|p^2 - 4q|/4)] + C = [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 + sqrt(a)·|b^2 - 4ac|·ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]/(8a^2) - sqrt(a)·|b^2 - 4ac|·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) + C
If p^2 - 4q = 0, then sqrt([a/4][(2x + p)^2 - (p^2 - 4q)]) = sqrt(a)/2·|2x + p| = sqrt(a)·|x + p/2|. Antidifferentiating gives sqrt(a)/2·(x + p/2)|x + p/2| + C = sqrt(a)/2·sgn[x + b/(2a)]·(x + p/2)^2 = sqrt(a)/2·sgn[x + b/(2a)]·(x^2 + bx/a + c/a) + C
If p^2 - 4q > 0, then sqrt([a/4][(2x + p)^2 - (p^2 - 4q)]) = sqrt([a/4][(2x + p)^2 - |p^2 - 4q|]). Let 2x + p = sqrt|p^2 - 4q|·cosh(u). After substitution, the integrand becomes |p^2 - 4q|/4·sqrt[a·sinh(u)^2]·sinh(u) = sqrt(a)·|p^2 - 4q|·sgn(u)/8·[cosh(2u) - 1]. Antidifferentiating gives sqrt(a)·|p^2 - 4q|/8·sgn(u)·[sinh(u)·cosh(u) - u] + C. If (2x + p)/sqrt|p^2 - 4q| = 1 or > 1, then u = sgn(u)·arcosh[(2x + p)/sqrt|p^2 - 4q|]. Therefore, sqrt(a)·|p^2 - 4q|/8·sgn(u)·[sinh(u)·cosh(u) - u] + C = sqrt(a)·|p^2 - 4q|/8·sgn(u)^2·(2x + p)/sqrt|p^2 - 4q|·sqrt[(2x + p)^2/|p^2 - 4q| - 1] - sqrt(a)·|p^2 - 4q|/8·ln([2x + p]/sqrt|p^2 - 4q| + sqrt[(2x + p)^2 - |p^2 - 4q|]/sqrt|p^2 - 4q|) + C = (x + p/2)·sqrt(ax^2 + bx + c)/2 - sqrt(a)·|b^2 - 4ac|·ln[x + p/2 + sqrt(x^2 + px + q)]/(8a^2) + sqrt(a)·|b^2 - 4ac|·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) + C1·sgn[x + p/2] + C2 = [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 - sqrt(a)·|b^2 - 4ac|·ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]/(8a^2) + sqrt(a)·|b^2 - 4ac|·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) + C1·sgn[x + b/(2a)] + C2.
The conjunction of all three cases is that the antiderivative family of sqrt(ax^2 + bx + c), for a > 0, and b and c real, is [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 - sqrt(a)·(b^2 - 4ac)·ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]/(8a^2) + sqrt(a)·(b^2 - 4ac)·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) + C1·[sgn(b^2 - 4ac)^2 + sgn(b^2 - 4ac)]·sgn[x + b/(2a)] + C2. Since sqrt(a)·(b^2 - 4ac)·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) is just a constant, the above answer may be simplified to [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 - sqrt(a)·(b^2 - 4ac)·ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]/(8a^2) + A·[sgn(b^2 - 4ac)^2 + sgn(b^2 - 4ac)]·sgn[x + b/(2a)] + B, where A and B are arbitrary constants of integration.
If a < 0, then let x = y - b/(2a), hence ax^2 + bx + c = a[y^2 - by/a + b^2/(2a)^2] + b[y - b/(2a)] + c = ay^2 - by + b^2/(4a) + by - b^2/(2a) + c = ay^2 + (4ac - b^2)/(4a) = (b^2 - 4ac)/(4|a|) - |a|y^2. Then let y = sqrt(b^2 - 4ac)/(2|a|)·sin(t), converting the integrand to sqrt(|a|)·(b^2 - 4ac)/(2a)^2·|cos(t)|·cos(t), which is sqrt(|a|)·(b^2 - 4ac)/(2a)^2·cos(t)^2 for -π/2 + 2nπ < t < π/2 + 2nπ, and -sqrt(|a|)·(b^2 - 4ac)/(2a)^2·cos(t)^2 otherwise. The antiderivative is sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·[t + sin(t)cos(t)] + C for -π/2 + 2nπ < t < π/2 + 2nπ, and -sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·[t + sin(t)cos(t)] + C otherwise. cos[arcsin(u)] = sqrt(1 - u^2) if -π/2 + 2nπ < t < π/2 + 2nπ, and -sqrt(1 - u^2) otherwise. t = arcsin[2|a|·y/sqrt(b^2 - 4ac)]. Therefore, the antiderivative is sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·(arcsin[2|a|·y/sqrt(b^2 - 4ac)] + 2|a|·y/sqrt(b^2 - 4ac)·sqrt(1 - [2ay]^2/[b^2 - 4ac])) if -π/2 < arcsin[2|a|·y/sqrt(b^2 - 4ac)] - 2nπ < π/2, and -sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·(arcsin[2|a|·y/sqrt(b^2 - 4ac)] - 2|a|·y/sqrt(b^2 - 4ac)·sqrt(1 - [2ay]^2/[b^2 - 4ac])) otherwise. If n = 0, then -π/2 < arcsin[2|a|·y/sqrt(b^2 - 4ac)] < π/2, which is true by definution, hence the antiderivative is sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·(arcsin[2|a|·y/sqrt(b^2 - 4ac)] + 2|a|·y/sqrt(b^2 - 4ac)·sqrt(1 - [2ay]^2/[b^2 - 4ac])) + C. This can be simplified to sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·arcsin[2|a|·y/sqrt(b^2 - 4ac)] + sqrt(|a|)/2·y·sqrt[(b^2 - 4ac)/(2a)^2 - y^2] + C = sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·arcsin[2|a|·y/sqrt(b^2 - 4ac)] + y/2·sqrt[(b^2 - 4ac)/(4|a|) - |a|y^2] + C = [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 + sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·arcsin[2|a|·y/sqrt(b^2 - 4ac)] + C.
The conjunction of both cases is that the antiderivative of sqrt(ax^2 + bx + c) for a nonzero and a, b, and c real is [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 + sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·([1 - sgn(a)]/2·{arcsin[2|a|·y/sqrt(b^2 - 4ac)]} - [1 + sgn(a)]/2·{ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]}) + A·[1 + sgn(a)]·[sgn(b^2 - 4ac)^2 + sgn(b^2 - 4ac)]·sgn[x + b/(2a)]+ B. This is the most general answer.
7:18 *OOF*
Dude your just like me, loves math and has epic savage comedy and puns
That sec^3(x) integral was something I didn't remembered, but I generalized for any n € N. It sounds hard but at the end the result is just impressive ;)
Why not dive into some integrals in polar coordinates? Love me some Jacobians
What you can actually do is turn it into a differential equation and then sub u for ax^2+bx+c.
I did that and got the following:
Starting with dy/dx = sqrt(u)
, where
u = ax^2 + bx + c, so 0 = ax^2 + bx + (c - u)
x = [-b + sqrt(b^2 - 4a(c - u))] / 2a
du/dx = 2ax + b = sqrt(b^2 - 4a(c - u))
dx/du = 1/sqrt(b^2 - 4a(c - u))
dy/dx * dx/du = sqrt(u)/sqrt(b^2 - 4a(c - u))
dy/du = sqrt(u / (b^2 - 4ac + 4au))
With k = b^2 - 4ac, we have
dy/du = sqrt(u / (k + 4au))
,
which may be harder, but I'm not sure.
Do you want to try to integrate sqrt(u / (k + 4au))?
That's a spicy boi
Ok this is fucking... I don't know how i can explain this in english so i'll say that in italian. MA CHE CAZZO DI ROBA ERA!!!!!!! AVREBBERO FATTO MENO MALE AL BRACCIO 30 SEGHE DI FILAAA!!!!!
antonio bellucci
HAHAHAHAHHA quando non sai come rovinarti la giornata intera e ti metti a fare integrali xD
Ahahah; che bello vedere altri italiani in un canale come questo. Comunque si, per mettersi a fare questo integrale devi essere leggermente masochista...
Pensavo di essere l'unico italiano a guardare questa follia
Why not substitute t=x+b/2a and say (4ac-b^2)/4a^2=k^2 and then integrate sqrt(t^2+k^2) with the normal methods and substitute back. It makes everything more neat.
(Also, technically flammy uses substitution for a constant but mine are a bit better)
Euler substitutions
1. a>0 sqrt(ax^2+bx+c)=u-sqrt(a)x
2. a0 and factor quadratic trinomial
then use substitution sqrt(a(x-x_{1})(x-x_{2}))=(x-x_{1})u
There is another one Euler substitution but these two cover all cases
sqrt(ax^2+bx+c)=xu+sqrt(c) may give an integral which requires less calculation in some cases
With Euler substitutions we can get substitution which reduces trig integrals to integrals of rational functions
for example
sec(x)=u-tan(x) if our integrand is expressed with secants and tangents
cos(x)=(1-sin(x))u if our integrand is expressed with cosines and sines
Another approach is similar to your odd and even decomposition
We can rewrite integral as sum of three integrals
One is integral of rational function ,
second one will be integral of rational function after substitution u=(x+b/(2a))/sqrt(ax^2+bx+c)
third one will be integral of rational function after substitution u=sqrt(ax^2+bx+c)
We can also use partial fraction decomposition of rational factor
and then try to calculate the integral
You assumed that a>0 otherwise answer is not complete
Jacek Soplica I can give the most general answer to this. Firstly, ax^2 + bx + c = a(x^2 + bx/a + c/a). Call b/a = p and c/a = q. Hence ax^2 + bx + c = a(x^2 + px + q). This is important, because if a < 0, the answer has a different form as opposed to if a > 0. In particular, there is a domain restriction depending on the sign of a. If a < 0, then the function is integrable iff b^2 - 4ac > 0.
sqrt[a(x^2 + px + q)] = sqrt(a[(x + p/2)^2 - (p^2 - 4q)/4]) = sqrt([a/4][(2x + p)^2 - (p^2 - 4q)]. Depending on the sign of p^2 - 4q, this has three different answers for a > 0.
If p^2 - 4q < 0, then sqrt([a/4][(2x + p)^2 - (p^2 - 4q)]) = sqrt([a/4][(2x + p)^2 + |p^2 - 4q|]). Let 2x + p = sqrt|p^2 - 4q|·sinh(u). Hence 2·dx = sqrt|p^2 - 4q|·cosh(u)·du. After substitution, the integrand becomes |p^2 - 4q|/4·sqrt[a·cosh(u)^2]·cosh(u) = |p^2 - 4q|/4·sqrt(a)·cosh(u)^2 = |p^2 - 4q|/4·sqrt(a)·[cosh(2u) + 1]/2 = sqrt(a)·|p^2 - 4q|/8·[cosh(2u) + 1]. Antidifferentiating gives sqrt(a)·|p^2 - 4q|/16·sinh(2u) + sqrt(a)·|p^2 - 4q|/8·u + C = sqrt(a)·|p^2 - 4q|/8·[sinh(u)cosh(u) + u] + C. u = arsinh[(2x + p)/sqrt|p^2 - 4q|], hence sqrt(a)·|p^2 - 4q|/8·[sinh(u)cosh(u) + u] + C = sqrt(a)·|p^2 - 4q|/8·[(2x + p)/sqrt|p^2 - 4q|·sqrt([2x + p]^2/|p^2 - 4q| + 1) + arsinh[(2x + p)/sqrt|p^2 - 4q|]] + C = sqrt(a)/8·[(2x + p)·sqrt([2x + p]^2 + |p^2 - 4q|) + ln([2x + p]/sqrt|p^2 - 4q| + sqrt[(2x + p)^2 + |p^2 - 4q|]/sqrt|p^2 - 4q|])] + C = sqrt(a)/2·(x + p/2)·sqrt(x^2 + px + q) + sqrt(a)·|p^2 - 4q|/8·ln[x + p/2 + sqrt(x^2 + px + q)] - sqrt(a)·|p^2 - 4q|/8·ln[sqrt(|p^2 - 4q|/4)] + C = [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 + sqrt(a)·|b^2 - 4ac|·ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]/(8a^2) - sqrt(a)·|b^2 - 4ac|·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) + C
If p^2 - 4q = 0, then sqrt([a/4][(2x + p)^2 - (p^2 - 4q)]) = sqrt(a)/2·|2x + p| = sqrt(a)·|x + p/2|. Antidifferentiating gives sqrt(a)/2·(x + p/2)|x + p/2| + C = sqrt(a)/2·sgn[x + b/(2a)]·(x + p/2)^2 = sqrt(a)/2·sgn[x + b/(2a)]·(x^2 + bx/a + c/a) + C
If p^2 - 4q > 0, then sqrt([a/4][(2x + p)^2 - (p^2 - 4q)]) = sqrt([a/4][(2x + p)^2 - |p^2 - 4q|]). Let 2x + p = sqrt|p^2 - 4q|·cosh(u). After substitution, the integrand becomes |p^2 - 4q|/4·sqrt[a·sinh(u)^2]·sinh(u) = sqrt(a)·|p^2 - 4q|·sgn(u)/8·[cosh(2u) - 1]. Antidifferentiating gives sqrt(a)·|p^2 - 4q|/8·sgn(u)·[sinh(u)·cosh(u) - u] + C. If (2x + p)/sqrt|p^2 - 4q| = 1 or > 1, then u = sgn(u)·arcosh[(2x + p)/sqrt|p^2 - 4q|]. Therefore, sqrt(a)·|p^2 - 4q|/8·sgn(u)·[sinh(u)·cosh(u) - u] + C = sqrt(a)·|p^2 - 4q|/8·sgn(u)^2·(2x + p)/sqrt|p^2 - 4q|·sqrt[(2x + p)^2/|p^2 - 4q| - 1] - sqrt(a)·|p^2 - 4q|/8·ln([2x + p]/sqrt|p^2 - 4q| + sqrt[(2x + p)^2 - |p^2 - 4q|]/sqrt|p^2 - 4q|) + C = (x + p/2)·sqrt(ax^2 + bx + c)/2 - sqrt(a)·|b^2 - 4ac|·ln[x + p/2 + sqrt(x^2 + px + q)]/(8a^2) + sqrt(a)·|b^2 - 4ac|·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) + C1·sgn[x + p/2] + C2 = [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 - sqrt(a)·|b^2 - 4ac|·ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]/(8a^2) + sqrt(a)·|b^2 - 4ac|·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) + C1·sgn[x + b/(2a)] + C2.
The conjunction of all three cases is that the antiderivative family of sqrt(ax^2 + bx + c), for a > 0, and b and c real, is [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 - sqrt(a)·(b^2 - 4ac)·ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]/(8a^2) + sqrt(a)·(b^2 - 4ac)·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) + C1·[sgn(b^2 - 4ac)^2 + sgn(b^2 - 4ac)]·sgn[x + b/(2a)] + C2. Since sqrt(a)·(b^2 - 4ac)·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) is just a constant, the above answer may be simplified to [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 - sqrt(a)·(b^2 - 4ac)·ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]/(8a^2) + A·[sgn(b^2 - 4ac)^2 + sgn(b^2 - 4ac)]·sgn[x + b/(2a)] + B, where A and B are arbitrary constants of integration.
If a < 0, then let x = y - b/(2a), hence ax^2 + bx + c = a[y^2 - by/a + b^2/(2a)^2] + b[y - b/(2a)] + c = ay^2 - by + b^2/(4a) + by - b^2/(2a) + c = ay^2 + (4ac - b^2)/(4a) = (b^2 - 4ac)/(4|a|) - |a|y^2. Then let y = sqrt(b^2 - 4ac)/(2|a|)·sin(t), converting the integrand to sqrt(|a|)·(b^2 - 4ac)/(2a)^2·|cos(t)|·cos(t), which is sqrt(|a|)·(b^2 - 4ac)/(2a)^2·cos(t)^2 for -π/2 + 2nπ < t < π/2 + 2nπ, and -sqrt(|a|)·(b^2 - 4ac)/(2a)^2·cos(t)^2 otherwise. The antiderivative is sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·[t + sin(t)cos(t)] + C for -π/2 + 2nπ < t < π/2 + 2nπ, and -sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·[t + sin(t)cos(t)] + C otherwise. cos[arcsin(u)] = sqrt(1 - u^2) if -π/2 + 2nπ < t < π/2 + 2nπ, and -sqrt(1 - u^2) otherwise. t = arcsin[2|a|·y/sqrt(b^2 - 4ac)]. Therefore, the antiderivative is sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·(arcsin[2|a|·y/sqrt(b^2 - 4ac)] + 2|a|·y/sqrt(b^2 - 4ac)·sqrt(1 - [2ay]^2/[b^2 - 4ac])) if -π/2 < arcsin[2|a|·y/sqrt(b^2 - 4ac)] - 2nπ < π/2, and -sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·(arcsin[2|a|·y/sqrt(b^2 - 4ac)] - 2|a|·y/sqrt(b^2 - 4ac)·sqrt(1 - [2ay]^2/[b^2 - 4ac])) otherwise. If n = 0, then -π/2 < arcsin[2|a|·y/sqrt(b^2 - 4ac)] < π/2, which is true by definution, hence the antiderivative is sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·(arcsin[2|a|·y/sqrt(b^2 - 4ac)] + 2|a|·y/sqrt(b^2 - 4ac)·sqrt(1 - [2ay]^2/[b^2 - 4ac])) + C. This can be simplified to sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·arcsin[2|a|·y/sqrt(b^2 - 4ac)] + sqrt(|a|)/2·y·sqrt[(b^2 - 4ac)/(2a)^2 - y^2] + C = sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·arcsin[2|a|·y/sqrt(b^2 - 4ac)] + y/2·sqrt[(b^2 - 4ac)/(4|a|) - |a|y^2] + C = [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 + sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·arcsin[2|a|·y/sqrt(b^2 - 4ac)] + C.
The conjunction of both cases is that the antiderivative of sqrt(ax^2 + bx + c) for a nonzero and a, b, and c real is [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 + sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·([1 - sgn(a)]/2·{arcsin[2|a|·y/sqrt(b^2 - 4ac)]} - [1 + sgn(a)]/2·{ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]}) + A·[1 + sgn(a)]·[sgn(b^2 - 4ac)^2 + sgn(b^2 - 4ac)]·sgn[x + b/(2a)]+ B. This is the most general answer.
Thank you for NOT integrating through the Hyperbolic trig methods and proving it can be done the traditional trig way. That was really fun to watch you work this all out!
This is *nothing but* amazing
Nice! Now the same assuming a b and c are functions of x.
Matron Not possible.
@@angelmendez-rivera351 Those are the words of a quitter!
(but you probably right tho...)
The fact that all great math channels have that tutor guy ads makes me mad
0:06 "No, Nein!"
I can't say who is hotter: Papa Flammy or the integral!
Awesome dad!
Silas Rodrigues Papa Flammy duh!
If you complete the square you see it's the equation of a hyperbola. Rewrite parametrically and integrate, then convert back to cartesian
Man, I'd really love seeing more from you than calcululs or ODEs
This is about to be the smartest 100k subscribers i swear...
"that's the wrong identity bitch boi" had me in tears!!
I admire your courage. You have used b²_ 4ac positive and apositive. There are three other cases for the signs of a and b²_4ac
Top 10 most BRUTAL anime battles
OMG, I tryed this yesterday and I got NOTHING😋 You are the GREATEST Integral Killing Boi!
You're funny, do not worry, Papa Flammy :) And it's kinda interesting to see a "let's try"-like solution, not just writing some pre-calculated/scripted/etc show onto the blackboard ... It's much more intuitive to see how some can think to solve a problem, in my opinion. However, please stop abusing my mother to be part of a substitution and even stating that she is too big for the blackboard :-O [ok, just kidding, surely no personal insult happened and misunderstand happened here :) ].
"Ah! That's going to be a fucking mess!"
I started to cry with laughter
That's pretty flammable indeed papa, sick vid!
This was surprisingly entertaining to watch.
2:21
The moment when you see that not only logos, but now numbers are oversimplified.
"that way, because we are fucking could no nein, that's, that's disgusting"
Sometimes complex integrals like this just need a simple tool to be solved... I could do everything up to 12:08, and when I saw that trig identity sec^2 = tan^2 + 1 everything becomes clear to me (including the fact that I fking suck at basics lol)
Papa Flammy, your videos are good. It will become more better with subtitles.
Thanks, at least.
Love your videos. Keep it up!
Divide by a reducing to x^2+dx+e.
This simplifies the algebra with no loss of generality.
Joe Cotter Actually, this is not true. This is because there is a domain restriction dependent on sgn(a), and if a < 0, then sqrt(ax^2 + bx + c) can be antidifferentiated if and only if b^2 - 4ac > 0. Furthermore, the answer is different for a < 0 than for a > 0. In particular, for a < 0, the answer is expressed in terms of arcsines, whereas for a > 0, the answer is expressed in terms of logarithms, or arsinhes.
Alright so I did this in a different way but it turned out extremely ugly, however took less than 5 minutes.
First factor out the (a) and move it outside of integral. You end up with a second degree polynomial with no constants next to x^2 which we can write as a perfect square, then take a u sub for whatever you ended up with inside the squared bracket with your variable x. What you end up with is solvable with a hyperbolic sub: take u = sqrt(whatever it is the constant you have)sinh(theta)
What you have is u = k.sinh(theta)
Where k is some ugly constant you got when you formed the perfect square, and u is your ugly x plus some ugly constant.
Apply the main hyperbolic identity and have fun. Took like 5 minutes and ended up with two terms very similar to what you got (too lazy to simplify, and the logarithmic term obviously will be acrsinh but also too lazy to check.
tl;dr: factor out the a, form a perfect square, take a hyperbolic sub with sinh, done. Hope I made any sense. Xd
Edit: difference between two squares not perfect squares* lmao sorry
Can someone explain me this "papa" thing? (what/who is)
Thanks!
Mr. Maths is daddy because he's our sugar daddy when it comes to our need for a daily dose of memes and maths.
Flammable Maths oh sorry daddy not to have known that! I'll keep it in mind
At around 7:18, why not multiply the 1 by (4ac-b^2)/(4ac-b^2), then factor out the denominator so you can add it to the radical outside the integral. You can probably still use a trig identity by assuming the resulting 4ac-b^2 = 1.
At 11:48 I was so focused. That flamming flying guy scared the s** outa me.
I've always wanted to do this => notification squad!!!!
Amazin bro,but you are likely naughty
hey, a question: if u try for instance to integrate sqrt(x^2+2x+1) when a = 1, b = 2 and c = 1, u will get the result of (x+1) times the sqrt(x^2+2x+1) all that over 2, but...wait a sec...according to your solution we can't calculate the integral because the sqrt(4ac-b^2) is 0! So all of this is going to be nothing but meaningless!
INT Sqrt( (x+p)^2 +q^2 ) can be done by parts using unity as one of the functions and a slight algebraic trick. If you are going to use substitution, hyperbolic functions are better than trig in this particular case.
Tony G I can give the most general answer to this. Firstly, ax^2 + bx + c = a(x^2 + bx/a + c/a). Call b/a = p and c/a = q. Hence ax^2 + bx + c = a(x^2 + px + q). This is important, because if a < 0, the answer has a different form as opposed to if a > 0. In particular, there is a domain restriction depending on the sign of a. If a < 0, then the function is integrable iff b^2 - 4ac > 0.
sqrt[a(x^2 + px + q)] = sqrt(a[(x + p/2)^2 - (p^2 - 4q)/4]) = sqrt([a/4][(2x + p)^2 - (p^2 - 4q)]. Depending on the sign of p^2 - 4q, this has three different answers for a > 0.
If p^2 - 4q < 0, then sqrt([a/4][(2x + p)^2 - (p^2 - 4q)]) = sqrt([a/4][(2x + p)^2 + |p^2 - 4q|]). Let 2x + p = sqrt|p^2 - 4q|·sinh(u). Hence 2·dx = sqrt|p^2 - 4q|·cosh(u)·du. After substitution, the integrand becomes |p^2 - 4q|/4·sqrt[a·cosh(u)^2]·cosh(u) = |p^2 - 4q|/4·sqrt(a)·cosh(u)^2 = |p^2 - 4q|/4·sqrt(a)·[cosh(2u) + 1]/2 = sqrt(a)·|p^2 - 4q|/8·[cosh(2u) + 1]. Antidifferentiating gives sqrt(a)·|p^2 - 4q|/16·sinh(2u) + sqrt(a)·|p^2 - 4q|/8·u + C = sqrt(a)·|p^2 - 4q|/8·[sinh(u)cosh(u) + u] + C. u = arsinh[(2x + p)/sqrt|p^2 - 4q|], hence sqrt(a)·|p^2 - 4q|/8·[sinh(u)cosh(u) + u] + C = sqrt(a)·|p^2 - 4q|/8·[(2x + p)/sqrt|p^2 - 4q|·sqrt([2x + p]^2/|p^2 - 4q| + 1) + arsinh[(2x + p)/sqrt|p^2 - 4q|]] + C = sqrt(a)/8·[(2x + p)·sqrt([2x + p]^2 + |p^2 - 4q|) + ln([2x + p]/sqrt|p^2 - 4q| + sqrt[(2x + p)^2 + |p^2 - 4q|]/sqrt|p^2 - 4q|])] + C = sqrt(a)/2·(x + p/2)·sqrt(x^2 + px + q) + sqrt(a)·|p^2 - 4q|/8·ln[x + p/2 + sqrt(x^2 + px + q)] - sqrt(a)·|p^2 - 4q|/8·ln[sqrt(|p^2 - 4q|/4)] + C = [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 + sqrt(a)·|b^2 - 4ac|·ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]/(8a^2) - sqrt(a)·|b^2 - 4ac|·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) + C
If p^2 - 4q = 0, then sqrt([a/4][(2x + p)^2 - (p^2 - 4q)]) = sqrt(a)/2·|2x + p| = sqrt(a)·|x + p/2|. Antidifferentiating gives sqrt(a)/2·(x + p/2)|x + p/2| + C = sqrt(a)/2·sgn[x + b/(2a)]·(x + p/2)^2 = sqrt(a)/2·sgn[x + b/(2a)]·(x^2 + bx/a + c/a) + C
If p^2 - 4q > 0, then sqrt([a/4][(2x + p)^2 - (p^2 - 4q)]) = sqrt([a/4][(2x + p)^2 - |p^2 - 4q|]). Let 2x + p = sqrt|p^2 - 4q|·cosh(u). After substitution, the integrand becomes |p^2 - 4q|/4·sqrt[a·sinh(u)^2]·sinh(u) = sqrt(a)·|p^2 - 4q|·sgn(u)/8·[cosh(2u) - 1]. Antidifferentiating gives sqrt(a)·|p^2 - 4q|/8·sgn(u)·[sinh(u)·cosh(u) - u] + C. If (2x + p)/sqrt|p^2 - 4q| = 1 or > 1, then u = sgn(u)·arcosh[(2x + p)/sqrt|p^2 - 4q|]. Therefore, sqrt(a)·|p^2 - 4q|/8·sgn(u)·[sinh(u)·cosh(u) - u] + C = sqrt(a)·|p^2 - 4q|/8·sgn(u)^2·(2x + p)/sqrt|p^2 - 4q|·sqrt[(2x + p)^2/|p^2 - 4q| - 1] - sqrt(a)·|p^2 - 4q|/8·ln([2x + p]/sqrt|p^2 - 4q| + sqrt[(2x + p)^2 - |p^2 - 4q|]/sqrt|p^2 - 4q|) + C = (x + p/2)·sqrt(ax^2 + bx + c)/2 - sqrt(a)·|b^2 - 4ac|·ln[x + p/2 + sqrt(x^2 + px + q)]/(8a^2) + sqrt(a)·|b^2 - 4ac|·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) + C1·sgn[x + p/2] + C2 = [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 - sqrt(a)·|b^2 - 4ac|·ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]/(8a^2) + sqrt(a)·|b^2 - 4ac|·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) + C1·sgn[x + b/(2a)] + C2.
The conjunction of all three cases is that the antiderivative family of sqrt(ax^2 + bx + c), for a > 0, and b and c real, is [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 - sqrt(a)·(b^2 - 4ac)·ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]/(8a^2) + sqrt(a)·(b^2 - 4ac)·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) + C1·[sgn(b^2 - 4ac)^2 + sgn(b^2 - 4ac)]·sgn[x + b/(2a)] + C2. Since sqrt(a)·(b^2 - 4ac)·ln[sqrt|b^2 - 4ac|/(2|a|)]/(8a^2) is just a constant, the above answer may be simplified to [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 - sqrt(a)·(b^2 - 4ac)·ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]/(8a^2) + A·[sgn(b^2 - 4ac)^2 + sgn(b^2 - 4ac)]·sgn[x + b/(2a)] + B, where A and B are arbitrary constants of integration.
If a < 0, then let x = y - b/(2a), hence ax^2 + bx + c = a[y^2 - by/a + b^2/(2a)^2] + b[y - b/(2a)] + c = ay^2 - by + b^2/(4a) + by - b^2/(2a) + c = ay^2 + (4ac - b^2)/(4a) = (b^2 - 4ac)/(4|a|) - |a|y^2. Then let y = sqrt(b^2 - 4ac)/(2|a|)·sin(t), converting the integrand to sqrt(|a|)·(b^2 - 4ac)/(2a)^2·|cos(t)|·cos(t), which is sqrt(|a|)·(b^2 - 4ac)/(2a)^2·cos(t)^2 for -π/2 + 2nπ < t < π/2 + 2nπ, and -sqrt(|a|)·(b^2 - 4ac)/(2a)^2·cos(t)^2 otherwise. The antiderivative is sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·[t + sin(t)cos(t)] + C for -π/2 + 2nπ < t < π/2 + 2nπ, and -sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·[t + sin(t)cos(t)] + C otherwise. cos[arcsin(u)] = sqrt(1 - u^2) if -π/2 + 2nπ < t < π/2 + 2nπ, and -sqrt(1 - u^2) otherwise. t = arcsin[2|a|·y/sqrt(b^2 - 4ac)]. Therefore, the antiderivative is sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·(arcsin[2|a|·y/sqrt(b^2 - 4ac)] + 2|a|·y/sqrt(b^2 - 4ac)·sqrt(1 - [2ay]^2/[b^2 - 4ac])) if -π/2 < arcsin[2|a|·y/sqrt(b^2 - 4ac)] - 2nπ < π/2, and -sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·(arcsin[2|a|·y/sqrt(b^2 - 4ac)] - 2|a|·y/sqrt(b^2 - 4ac)·sqrt(1 - [2ay]^2/[b^2 - 4ac])) otherwise. If n = 0, then -π/2 < arcsin[2|a|·y/sqrt(b^2 - 4ac)] < π/2, which is true by definution, hence the antiderivative is sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·(arcsin[2|a|·y/sqrt(b^2 - 4ac)] + 2|a|·y/sqrt(b^2 - 4ac)·sqrt(1 - [2ay]^2/[b^2 - 4ac])) + C. This can be simplified to sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·arcsin[2|a|·y/sqrt(b^2 - 4ac)] + sqrt(|a|)/2·y·sqrt[(b^2 - 4ac)/(2a)^2 - y^2] + C = sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·arcsin[2|a|·y/sqrt(b^2 - 4ac)] + y/2·sqrt[(b^2 - 4ac)/(4|a|) - |a|y^2] + C = [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 + sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·arcsin[2|a|·y/sqrt(b^2 - 4ac)] + C.
The conjunction of both cases is that the antiderivative of sqrt(ax^2 + bx + c) for a nonzero and a, b, and c real is [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 + sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·([1 - sgn(a)]/2·{arcsin[2|a|·y/sqrt(b^2 - 4ac)]} - [1 + sgn(a)]/2·{ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]}) + A·[1 + sgn(a)]·[sgn(b^2 - 4ac)^2 + sgn(b^2 - 4ac)]·sgn[x + b/(2a)]+ B. This is the most general answer.
Rooh I need to take a train to Paris, and what I see in my Subscriptions videos ?? A boi that will make me miss the train. But I love quadratic funny integral papa flammable boiii, thx for this video :D
Would it have been easier to rearrange it to x=f(y) and solve xy-INT([f(y)]dy?
good ol' hiperbolas, or, square roots of parabolas
According to Wikipedia's article "List of integrals of irrational functions", the solution should be:
((2ax+b)/4a)Sqrt(ax^2+bx+c) + (4ac-b^2)/(8aSqrt(a)) ln(2sqrt(a)sqrt(ax^2+bx+c)+2ax+b)
By doing Euler's substitution:
Sqrt(ax^2+bx+c) = t +/- xSqrt(a)
I managed to obtain the 2nd term but I was never able to simplify the rest to obtain the first term...
Sensational. What a blockbuster
It was quiet good before trigo after that it was lengthy.U really improvised it .
FANTASTIC!!!
I finally managed to solve this thing using the Euler substitution:
I = Integral sqrt(ax^2+bx+c)dx
Let R = sqrt(ax^2+bx+c) = t-Ax where A = sqrt(a)
ax^2+bx+c = t^2-2Atx+ax^2
bx+c = t^2-2Atx
x(b+2At) = t^2-c
x = (t^2-c)/(2At+b)
dx/dt = (2t(2At+b)-2A(t^2-c))/(2At+b)^2
dx = 2(At^2+bt+Ac)/(2At+b)^2 dt
R = t-Ax
R = t-A(t^2-c)/(2At+b)
R = (t(2At+b)-A(t^2-c))/(2At+b)
R = (At^2+bt+Ac)/(2At+b)
I = Integral Rdx
I = 2 Integral (At^2+bt+Ac)^2/(2At+b)^3 dt
I = 2 Integral (at^4+2Abt^3+(b^2+2ac)t^2+2Abct+ac^2)/(2At+b)^3 dt
I = 1/8a Integral (16a^2t^4+32aAbt^3+(16ab^2+32a^2c)t^2+32aAbct+16a^2c^2)/(8aAt^3+12abt^2+6Ab^2t+b^3) dt
After long division:
I = 1/8a Integral (2At+b+((32a^2c-8ab^2)t^2+(32aAbc-8Ab^3)t+16a^2c^2-b^4)/(2At+b)^3) dt
Using partial fractions technique:
I = 1/8a Integral (2At+b+J/(2At+b)+K/(2At+b)^2+L/(2At+b)^3) dt
where:
J(2At+b)^2+K(2At+b)+L = (32a^2c-8ab^2)t^2+(32aAbc-8Ab^3)t+16a^2c^2-b^4
J(4at^2+4Abt+b^2)+2KAt+Kb+L = (32a^2c-8ab^2)t^2+(32aAbc-8Ab^3)t+16a^2c^2-b^4
For t2 terms:
4aJ = 32a^2c-8ab^2
J = 2(4ac-b^2)
For t terms:
4AbJ+2KA = 32aAbc-8Ab^3
2bJ+K = 16abc-4b^3
2bJ+K = 4b(4ac-b^2)
K = 4b(4ac-b^2)-4b(4ac-b^2)
K = 0
For unit terms:
Jb^2+Kb+L = 16a^2c^2-b^4 = (4ac-b^2)(4ac+b^2)
L = (4ac-b^2)(4ac+b^2)-2b^2(4ac-b^2)
L = (4ac-b^2)^2
Thus:
I = 1/8a Integral (2At+b+2(4ac-b^2)/(2At+b)+(4ac-b^2)^2/(2At+b)^3)dt
I = I1+I2
where:
I1 = 1/8a Integral (2At+b+(4ac-b^2)^2/(2At+b)^3)dt
I1 = 1/32aA 4A(At^2+bt)-(4ac-b^2)^2/(2At+b)^2
I2 = (4ac-b^2)/4a Integral dt/(2At+b)
I2 = (4ac-b^2)/8aA ln|2At+b|
--> I1 = R(x/2+b/4a) (32aA)I1 = 16aARx+8AbR
How long did it take you to write this comment
16:40
We have a fucking mess, that's what we have (:
Awesome Video btw.
isnt saying sqrt(sec^(2)(x)) = sec(x) a terrible idea though because of domain restriction? why not use hyperbolic trig? it would be much better
What do you mean by this statement?
@@cooldawg2009 ah I made this comment quite some time ago, but I think what I meant was sqrt(sec^2(x)) is not always sec(x) right? It's |sec(x)|, so you end up having to take cases on the region of integration; however, sqrt(sinh^2(x)) = sinh(x) is always true since sinh is a strictly positive function on the reals. Maybe you could get a hyperbolic sub to work instead which would avoid this problem but it's such a long time ago I can't remember
Love your content papa
You're an absolute mad man, Papa Flamy! I think I might be in love ;)
Holy shit I was doing the exact thing few hours ago . Great video by the way !
Alright. What I'm thinking: let u = sec, then du = -sin sec^2, then integral sin sec^3 = sec^2, but that's the wrong integral... So sad. If only there were a function which differentiated to itself, oh yeah, e^x ! So let's try the sum of the two integrals sec^3 and (icsc)^3, which is (cos^3 -(isin)^3)/(isincos)^3
Factoring the difference of cubes to get (cos-isin)(cos^2+isincos-sin^2)/(isincos)^3
Decomposing to get (e^ix)(cos^2-sin^2)/(isincos)^3
=(e^ix)cos(2x)8/(isin(2x))^3 = 8i(e^ix)cot(2x)^3, an interesting integral if you could solve it, I have no idea where to go from that.... Not exactly better than sec^3 XD
I forgot another term of 1/(isincos)^2 that I wanted to add near the end, which is also part of the integral... Oops
Brilliant video!!!
This is wonderful.
BEAUTIFUL PAPA!!! ❤
U really make me laugh bro 😂😂😂 Awesome approach 👏👏
Thanks for your videos carrying me in an integral assignment yeet
He's still giving heart and comments
Brother suppose if its a definite integral having proper limits where the quadratic f(x) is positive and quadratic have roots..then how we would approach ?? Please help me out
Ok brother
@@AnshPr2311 😂😅
Isn't it time to explore multiple integrals? Let's find the barycenter of some funky shape!
0:25 You got it papa.
Integral of sin(sin(sin(x))) dx next?
PianoBanana horseshoe integration is one powerful weapon
@17:25, did you say it was the wrong identity bc you forgot a square on the tan?
I go to these videos after fucking up my calculus exams to not feel stupid anymore :)
Can we solve it by finding roots of the equation
8:28 my life
i love his random swearing
time to check by differentiation??
ok not really
great video!!
i love you papa
Plug the values a=1 b=5 c=-6 , it doesn't work you get sqrt of a negative number , it has to be sqrt(abs(4ac - b^2))
But there is also another sqrt of a negative number somewhere else and it will all just cancel out
@@loneranger4282 Right, I used wolfram to calculate the integral and indeed
sqrt(4ac-b^2) cancels out, but still it's not clear in the video he must have pointed it out
Well if you want to go through the complex domain otherwise what if a
10:15 => lol best part of this video
You are funny man xD and I love your content!
Now integrate cbrt(ax^2 + bx + c)
Skylar Deslypere I think this is non-elementary. In particular, the antiderivatives of cbrt(x^2 + 1) and cbrt(x^2 - 1) are nonelementary.
Great videos, know what i'm subscribing.
this is a cool integral my boi
I wish I had you as my math teacher
you should do some more multidimensional integrals and make papa fubini happy. then its also harder to find some help in your beloved tafelwerk 😂😂
You didn't solve the problem, or the problem wasn't full explained. You never said anything about the values a, b, c and their restrictions. I would say that 4ac-b**2 would have to be greater than zer in order to get any real solution, which is a fancy way of saying that the polynomial under the square root needs to have real roots greater than zero. (b**2 - 4ac is the descriminant). I think this works for positive or negative a, b, c if the first equality I said holds and if a>0 of course. Other than that it looks good.
Charlie McGuire you mean b**2 - 4ac has to be greater than zero
Not 4ac - b**2
Charlie McGuire Not at all. If a > 0, then the discriminant can be any value. There is also an answer for a < 0. Covering all cases gives you the answer of [x + b/(2a)]·sqrt(ax^2 + bx + c)/2 + sqrt(|a|)/2·(b^2 - 4ac)/(2a)^2·([1 - sgn(a)]/2·{arcsin[2|a|·y/sqrt(b^2 - 4ac)]} - [1 + sgn(a)]/2·{ln[x + b/(2a) + sqrt(x^2 + bx/a + c/a)]}) + A·[1 + sgn(a)]·[sgn(b^2 - 4ac)^2 + sgn(b^2 - 4ac)]·sgn[x + b/(2a)]+ B. This is the most general answer.
Well done!!! Fkn mess lol 😂😂😂
Now check by differentiation.
What a wonderful shit storm ❤