Clever substitution integral

[@nathanisbored]

This type of substitution which is used to bypass a trig sub and create a rational integrand is often called a “rationalization sub”, and these substitutions infamously tend to seem pretty unmotivated and require foresight (compared to a trig sub which is more systematic), but as far as I know, any trig sub integrand has a corresponding rationalization sub. You could also multiply it out and do a partial fraction decomposition.

[@rnjsaudwo]

Oh man... It's just... Beautiful !

[@mathwithjanine]

This is awesome! Thanks for sharing this super clever trick with us! :)

[@pablomandelo4311]

Wow! That was a really ellegant method to solve this integral

[@jonasdaverio9369]

That's kind of amazing, I must agree.

[@dr.mohamedfawzy963]

Good job

[@nuranichandra2177]

Really an elegant solution

[@sharpnova2]

it's a nice quick exercise to show how this generalizes to the nth power like you said at the start

[@ludovicmatrand9642]

A simple substitution x=tan(t) works very well and is pretty much faster.Nice vid anyway. Thx for sharing

[@10minutesmathematics]

Thank you sir, it's very helpful

[@Kdd160]

That was quick and clever

[@theimmux3034]

That u-sub is so smart

[@vameza1]

Dear Prof. how are you??? Tell me, the substitution x = tan(theta) isn´t easier?? Thank you and amazing video!

[@soumyaghosh8823]

I think you should try X=tan(thita) So the denamenator becomes sec^4(thita) And your dx= sec^2(thita). Ultimately you will get integral 1/(sec^2(thita)) i.e. cos^2(thita) Now limit of the integral will be 0 to pi/2. Just take 1/2common so (2cos^2(thita))=(1+cos(2thita)) just apply the integral You will finally get 1/2*(pi/2)=(pi/4)

[@manfredwitzany2233]

I have used a shorter way: The function, which should be integrated is even. Therefore f(x)=f(-x) => The integral from - infinity to zero gives the same value as from zero to infinity. Therefore we can extend integration from - infinity to infinity, when multipiying the integral with 1/2. Intrgl(0 inf) dx/(x²+1)² = 1/2 * intgrl(-inf inf) dx/(x²+1)² The path of integration is extended to a half circle with radius going to infinity. The integral of the arc tends to zero, when the radius goes to infinity. The function is analytic, therefore we can aply the residue theorem. The function has poles of second order at x=+i and x=-i As integration is done in the positive region of imaginary part of f(x), only x=+1 is relevant. => 1/2 * intgrl(-inf inf) dx/(x²+1)² = pi i (1/2) Res(i) (1/(1+x²)²) = pi i lim (x->i) d/dx [(x-i)²/(x+i)²(i-i)²] = pi i lim (x->i) d/dx [1/(x+i)²]= pi i lim (x->i) (-2)/(z+i)³ = pi i (-2)/(4 i³) = pi i / (4 i) = pi/4

[@harshchoudhary1606]

Bro substitute x =tan(thetha) its more become easy

[@anarchostalinprimitividiag1030]

@@harshchoudhary1606 did it the same way. this is in no way shorter than his solution though. especially since arguing for the fact that the integral over the half circle tends to zero, requires some more algebra that you havent written out.

[@JSSTyger]

I can tell you that it is less than π/2 right off the bat, but greater than 0. The reasoning is that this function will be closer to the x-axis than 1/(1+x²). The area under 1/(1+x²) is easy to compute because the integral is tan-1(x).

[@EddieEntertainment]

first thought before watching: residue theorem. let's see what you're going for edit: marvellous solution!

[@Differential_Equations]

You can suppose u=tanx also

[@eytansuchard8640]

Awesome Dr. Peyam. Thank you. I wish you could help with the field strength coefficients of the Geometric Chronon Field Theory when the far observer r -> 0. These are 95/96 for the electron, 4/Pi for the Muon and 1.556198537190343965638770314399... for the Tau lepton. Unlike QM, the theory is based on events and not on particles. Such an idea was first to appear in the work of H. S. Snyder in 1947 but was algebraic rather than geometric. The idea of the Geometric Chronon Field Theory is that misalignment of events means forces and thus matter. The measurement of how much events are misaligned is by using the Reeb vector of a gradient of scalar functions. One of these scalar functions is very similar to the Robert Geroch time function (Geroch splitting theorem). It is a scalar field and not a coordinate of time. In a big bang manifold or any manifold with a Cauchy surface, a Geroch function must exist and is the maximal proper time t(q) between each event q and the Cauchy surface but it is not a coordinate because it is not guaranteed that this time can be measured along a unique curve . In a big bang manifold, the Cauchy surface is shrunken to the big bang limit. The model is of a physically accessible observed space time and of an observer spacetime. Dr. Sam Vaknin's idea from 1982 was that time is fundamental to physics and not matter. This approach is diametrically opposed to the trend to get spacetime as emergent from endless matrices. I think the derivation of the mass ratio between the Muon and the electron will interest you and also the calculation of the inverse Fine Structure Constant will be interesting to you too. 4/Pi does not come from the theory. It is actually a critical value in QFT. It should come from the theory itself. The result that not only ordinary mass generates gravity but also charge does is well understood. Negative charge generates weak anti-gravity and positive charge generates positive gravity. As hot and not sparse galaxies age and lose electrons, the gravity becomes stronger than expected. Also the super massive BH of the galaxy must be greater than expected.

[@nadiyayasmeen3928]

Amazing

[@f3ynman44]

Could you solve this integral with Complex Integration?

[@oscarlin8450]

what a clever sub! I can only think of let x=tanθ to deal with this integral before watching your video!

[@AlfonsoNeilJimenezCasallas]

All that business is about a good application of algebra and calculus, impressive!

[@saptashwabaisya4202]

won't just normal substitution that is x=tan(@) work? anyways thanks for the alternative approach professor peyam!

[@emanuelvendramini2045]

it works, but maybe, not all integrals have an exit via substitution, so, we need to know other methods, i guess.

[@pbj4184]

@@emanuelvendramini2045 Yes this particular sub x=1/u can be used in a lot of places where some conditions are met. Like the bounds being "invertible". Like say the bounds (0,∞). Applying the sub changes the bounds to (∞,0). But since dx=-du/u^2, the minus again flips the bounds to (0,∞). But since we have an extra 1/u^2 now in the integrand, it can be used to massage the integrand into what we want. A similar example is of (1,∞). Applying the sub gives you (1,0) and the minus flips it to (0,1). So if you have the bounds (0,∞), you can break it into (0,1) and (1,∞) and apply this sub on the second integral to get (0,1). Now since both integrals have bounds of (0,1), the integrands can be combined into one integral. Whenever you have good bounds like (0,1) , (1,∞) , (0,∞), you should try this sub if it looks like this sub might help

[@saptashwabaisya4202]

@@emanuelvendramini2045 yes I agree with you it worked here but it not might work elsewhere..

[@saptashwabaisya4202]

@@pbj4184 yes I agree with you

[@saptashwabaisya4202]

@Anubhav Mahapatra yes

[@kshitijpal5534]

This is lengthy. Simply substitute x=tan@ . After substituting it we will get a simple integration of cos^2@

[@drpeyam]

Nah

[@kshitijpal5534]

Sir, it's right. Just try putting x=tan@

[@drpeyam]

Nah, watch the video, you’re missing the point

[@BigLeagueChewie]

I solved it using Feynman’s trick, letting $I(t)=\int_{0}^{\infty} 1/(x^2+t^2) dx=\pi/(2t)$. Differentiating w.r.t. t and then letting t=1 yields the desired result.

[@drpeyam]

I like that!!!

[@elkincampos3804]

Let be q(x)=x^2+1 then q'(x)=2*x. Note that 1/(2*x*(x^2+1))= A/(2*x)+(B*x+C)/(x^2+1). That is 1=A*(x^2+1)+(B*x+C)*(2*x). If x=0 then 1=A.Therefore C=0 and B=-1/2. Then the 1/(x^2+1)^2=1/(q(x))^2=A*q(x)/(q(x)^2+(B*x+C)*q'(x)/(q(x)^2) we can use Parts.

[@muskyoxes]

"that's a lot of squaring" - should have squared the ones too.

[@diegoenrique03]

Nice

[@The1RandomFool]

Next up, try the integral from 0 to infinity of dx / (x^2 + 1)^n where n is a natural number :D It's not that hard with complex analysis.

[@yoav613]

yes with the residue you find it is pi2^(1-2n)(2n-2)!/((n-1)!)^2

[@oyibechibundu628]

Can you help me to integrate arccos(cosx/(1+2cosx)). I have tried every trick. But they do not work .

[@Happy_Abe]

How could you do 2 different U subs, how can U=1/x and U=x

[@shapirogensichwa]

U=x was not a substitution, let me give you an exemple : 1+2+3 is the sum of "k" from 1 to 3 but you can also say that its the sum of "p" from 1 to 3.

[@Happy_Abe]

@@shapirogensichwa he seemed to say it as a substitution

[@bouch7643]

I think x = e(-y) can be interesting as well or numerator: 1= (x2+1)-x2

[@drpeyam]

How would the first one help?

[@bouch7643]

@@drpeyam I don't have my pen/paper right now but d (e(y))/d (y) is equal to itself. I thought of e (-y) to get the e (y) appear in the numerator. It will likely work out the same as you did.

[@shadow-ht5gk]

Wouldn’t partial fractions work here as well?

[@shadow-ht5gk]

Ah never mind, you need more than one irreducible factor.

[@LostInThoughtsAllAlone2]

Hello, i don't quite understand a part of the solution and i'd be thankful if someone could explain it. In the start we say that u = 1/x and then we find that integral '' I '' 0-->infinity dx/(x^2+)^2 equals 0--->inifnity u^2/(1+u^2)^2 however after finding this we set u = x which is different than the initial statement that u = 1/x. How does part work make work?

[@drpeyam]

You relabel the variables, that’s all

[@harshchoudhary1606]

Bro put x=tan(thetha) its really cool

[@SHUBHAMKumar-qd1pi]

Put x = tan(theta)

[@jonasdaverio9369]

At first, it looked like a Parceval identity, but it would probably be 5 times as much work

[@aneeshsrinivas9088]

couldn't you just use bprp's differentiation under the integral sign formula to find the antiderivitive?

[@drpeyam]

How?

[@carlosgiovanardi8197]

i tried the integral at 0:32 nice results

[@carlosgiovanardi8197]

verify the audio!

[@kasparovmarconi2608]

Saludos de perú

[@anupamamehra6068]

put x=1/t and then do... it works

[@suryakkn9382]

Hi, Sir. What app(pen&whiteboard) are you using?

[@chrismcclure1968]

What software was being used?

[@guitar_jero]

Ah yes, the old “no quita, ni pone” :)

[@plugandsocket500]

Could you do the same for the case where the denominator is raised to any positive integer? That is: Denominator = (1+x*x)^n Where n is a positive integer

[@DragonriderplaysMC]

Not as far as I can tell since the differential from the sub will always be -u^-2 du which only cancels nicely when 2n-2=2 i.e. n =2

[@shlomi8307]

What software did you use doctor?

[@drpeyam]

Microsoft whiteboard

[@nevenazivic5237]

where i can find that program where u write

[@drpeyam]

Microsoft whiteboard

[@chologonzalo999]

Petam, how can you first say that u=1/x and then say x=u? Is that valid?

[@drpeyam]

Yes

[@thequarrymen58]

@@drpeyam HOWWWWWWWWWWWWWW????????????????????????????

[@aadityabhetuwal5990]

reminds me of 0 to π/2 ln(sin x). dx

[@pbj4184]

Was it the adding both versions to simplify the integral that reminded you?

[@aadityabhetuwal5990]

@@pbj4184 yes

[@nicogehren6566]

Dr. can u help me: integral of cos (cot x -tan x)dx from 0 to pi

[@thetheoreticalnerd7662]

Try Weiertrauss Sub

[@bandamkaromi]

good sugget, I don't think about it!

[@physjim]

cool

[@bouch7643]

Why is the minus reverse the integral infinity to zero TO zero to infinity.

[@deadmayday6702]

int from a_b = - int from b_a

[@bouch7643]

@@deadmayday6702 integrals are surface calculus or simple additions. From a to b should be equal to b to a.

[@bouch7643]

@@deadmayday6702 it seems there is an ongoing debate math.stackexchange.com/questions/232455/is-integration-from-a-to-b-same-or-b-to-a-or-is-negative Are integrals surfaces under the curve or can it be just calculus that can be negative. In this latest case, I am wrong. I just cant find it intuitive to turn a surface into a negative one just because we change the direction of a to b, to b to a.

[@bouch7643]

@@deadmayday6702 I thought of this today although I fell sick in love with a woman right in the middle of the street. I think of integrals as surfaces whom "signe" cannot be reversed from + to -. From a pure analytical overview, it is right.

[@mathematics6199]

Broo this is not good method at all instead you can substitute x as tan theta 2 steps

[@drpeyam]

Completely missing the point

[@meowwwww6350]

First ,Mr peyam !!!

[@tsunningwah3471]

second!!!!!