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Ok but how was it found?

Que cara foda❤

Thank you sooooo much ❤

you're literally a life saver.

If you remember the right triangles and derive the values from there, the substitution makes more sense. You have to draw those triangles anyways to convert thetas back to x's at the end anyways. The conversion triangle can help jog your memory should you forget whether to use tan, sec, sin.....

Another way to find min f(x,y)=(x+2)²+(y-3)²-13 Since the braced always positive then minf(x, y)=-13

I watch every single video of you sir!! Thanks 🙏

EXCELLENT EXPLANATION

I hope you haven't become tired of TH-cam. Your content is great.

Thank you! Very well presented.

This is also equal to b-a/2 as pi/2-0 /2 =pi/4.

Pofessor Black Shirt, you are one of the best math presenter on youtube. I am a retired EE who spends a lot of time on youtube math channels, guys like you are realy saving guys like me from getting dementia. Your explation of the gamma function is very clear that even an old guy like me have no problem understanding your lecture, keep up the good work.

Thank you so much for sharing these tips and tricks

You are AWESOME!!!!

Excellent tutorial. I enjoyed watching and learning from this tutorial. Thanks.

Excellent tutorial

Let's integrate by parts multiple times and then derive formula for product in terms of Gamma function

great intuition, thanks!

Unfortunately it is very tricky. In writing another comment I got confused when r was between 0 and -1. Looking at the general geometric series of a+ar+ar²+ar³+... then odd powers of r would be negative. Another problem arises as n approaches infinity ("at infinity" is nonsense) then 1-r^n approaches in value up to 1. The finite formula approaches in value down to a as n approaches infinity. It is reasonable to assume that as n approaches infinity that 1/n is 0, but assuming is not always reasonable. Nonstandard Analysis says they are the same, I disagree. Infinity is incomplete, inconsistent and imprecise.

As you say in the video, SHUT UP and learn where your formula comes from. The formula for the sum of a finite geometric series is; a*(1-r^n)/(1-r) where n is a natural > 1, r!=1 and an if n=1. This formula was obtained by multiplying the series by r, using standard algebraic manipulations to obtain the result, valid for all r except 9/9, excuse me 1. Ok, what happens as n approaches infinity. n becomes very large and 1-r^n becomes a very large absolute number if r is greater than 1. Not helpful. If r is 1, "an" then becomes "a" times larger than n. a*(1-r^n)/(1-r) r<1 and n approaching infinity. 1-r^n becomes very close to 1 and the sum becomes very close to a/(1-r). The question becomes, is the Archimedean property violated.? Infinity is incomplete, inconsistent and imprecise.

Revising for calculus 2 and your videos have been extremely concise on each topic! Thanks a ton

wonderful explanation, i appreciate the efforts.

You teach very well man, Tnx

This helped a lot

outstanding sir ....hats off

You are doing gods work brother thank you

❤

Thanks from India 🇮🇳🫶

I don't know if you are reading the comments right now, but if you are, I must say that you have a great channel. I am a physics Olympian and sometimes math can be difficult for me. I came across your channel on reddit while looking for a good resource on differential equations, your videos are very explanatory and the integrals you solved were the ones I couldn't solve and they helped me a lot in this regard. You uploaded your last video 1 year ago, it is very sad that a channel like yours has few views. I know it is impertinent to say this, but please continue uploading videos. I am sure you will be trending one day, because you deserve it. Have a nice day

gracias por la magnifica explicación ,saludos desde México

You can do the same thing with hyperbolic functions using e^x=cosh(x)+sinh(x). Instead of matching the real and imaginary components in the equality, you match the even and odd components.

Very very clear! And the right speed: I can follow, and it is not too slow! And you convey your fascination for math! Just great!

Thanks for the new and simple way of finding INTEGRATION BY PARTS .....!!!

Asombroso!

As always, a wonderful video. Fractional derivatives are a fascinating concept that I hadn't thought of before. It was also enormously helpful to link it with the gamma function, which I only learned about yesterday.

It has truly brightened my day to stumble onto this channel. After a long search, I came upon this extremely clear and thorough description of the gamma function. Continue doing what you're doing; you are a great instructor.

Loved and subscribed !!!!

Mr perfect ❤

Thank you, doing my physics degree with many gap years from my A levels. Your channel is very hlpful

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Y only say that tan(x)=sin(x)/cos(x) Then say u=cos(x) So -du=sin(x)dx So the resultant integral was -∫(ln(u)/u)du With the simple substitution v=ln(u) and dv=du/u and working it the result is ln²(u)/2+c=ln²(cos(x))/2+c

Probably is not the best method, but I use x=tan⁴(θ), dx=4tan³(θ)sec²(θ)dθ So I get the integral of tan⁵(θ)dθ And is not bad with the identity of tan²(θ)=sec²(θ)-1 The integral give me tan⁴(θ)-2tan²(θ)+4ln|sec(θ)|+c tan⁴(θ)=x 2tan²(θ)=2√(x) 4ln|sec(θ)|=4ln|√(1+√(x))|=2ln|1+√(x)|=2ln(1+√(x)) So x-2√(x)+2ln(1+√(x))+c

Easier to understand, thank you so much❤

Γ(z)= ∫₀ ᪲ tᶻ⁻¹ e⁻ᵗ dt

The clarity of your whiteboard script is very helpful.

Interesting. I wonder what real applicaions are on the horizon with fractional derivatives.

thank you very much! ive been struggling on the chain rule in my course

tysm i was about to kms over this

this was awesome and very helpful thanks

Thanks from Afghanistan