Happy birthday to me 🥰

@algorithminc.8850

2 жыл бұрын

Definitely keep with age mentally and physically monotonically increasing ... cheers ...

@Ak-ly5qn

2 жыл бұрын

Happy birthday Dr Peyam.

@nicolasdiez4239

2 жыл бұрын

Happy Birthday, pro

@jacobzetterfeldt3652

2 жыл бұрын

Happy birthday 🎂

@anushkrajbordia1873

2 жыл бұрын

Happy Birthday Sir... May you have all the happiness in the world

That's a really cool one I think, I should suggest it to the calculus 1 teacher for whom I'm a teaching assistant.

better if we use identity related to limits of integrals and then integrand will be 1/(1+cos(x)) then use cos(x)=2cos^2(x/2)-1 and finally integrand will be 1/(2cos^(x/2) ) which is derivative of tan(x/2) wrt x. under the limits answer will be 1

Congrats Peyam! Have a really good day! And thx for the beautiful movie!

Alternative suggestion: The whole series can be written as a summation of (-sinx)^n from 0 to n. Switch the integral and sum, you get gamma function.

@gamingresumed1788

2 жыл бұрын

It wouldnt be alternative sigh though

@erikkonstas

2 жыл бұрын

Problem is, it's (-sinx)^n...

@mdrifaturrahman4443

2 жыл бұрын

Oh yeah I made a typo, thanks guys

HAPPY BIRTHDAY VV WHOLESOME OF U TO DO A BLACK PEN RED PEN VIDEO ON UR BDAY. RLY SHOWS HOW GREAT U GUYS R

My man has got a golden rolex on... Love you Peyam

happy birthday ! 🥳

Hello Dear Dr Peyam. At first: *happy birthday 🎉 and Best Wishes* . And second: thank you for such an interesting math problem. I watched it and at the same time I learned and also laughed (for your jokes about party, bprp and ...) Thank you so much (again happy birthday)

loved it...thanks

Happy birthday Dr.Peyam

Here before aash! Happy Bday Dr. P!

fascinating!

Happy birthday dr pyam

Loved it

you can substitute sin(x) with {(2tan(x/2))/(1+tan^2(x/2))} its quite simpler.

great integral great video

haha are you over 30 yet ?? Happy birthday Doc Your use of multiply top n bottom by 1-sine reminds me of the famous secant integral by Isaac Barrow , Newton's professor He multiplied sec by cos top n bottom then cos^2 bottom is 1-sin^2 then partial fractions , result the log ( 1-sin / 1+sin ). So , if Newton discovered calculus how come Barrow 1. used integral calculus before Newton was born 2. Barrow was the first to point out that differentiation and integration are inverses = the Fundamental Theorem of Calculus ? Turns out calculus discovered before Newton by dozens before Newton eg Fermat differentiation , Cavalieri integration both before Newton was born But Newton famous coz stunning application to Solar system Gravity Another beaudy Doc !! eres muy inteligente mi amigo jijijiji

For \int_0^{\pi/2} \frac1{1+sin(x)}dx I have used the substitution t=tan(x/2) and I did not have problem with improper primitive.

Happy Birthday Dr πyumh cake 🎂🎂

8:19 we are allowed to use L'Hospital rule in that case? I know that one of conditions for L'Hospital rule is the denominator function to not be 0 for x\in D\{x0}, where D is the domain definition of numerator and denominator functions, and x0 is the value which are we calculating the limit. In this example, cos(x)=0 for infinitly many points, not just for pi/2.

Best sir

The sum of the series was trivially obvious as long as |sin(x)|

Dr Peyam = The Party!!!

Happy Birthday, 1+sinx= sin(π/2)+sinx=2*sin(x/2 +π/4)*cos(x/2-π/4).After substitution y=x/2+π/4 we have integral 1/((siny)^2) dy where y go π/4 till π/2. We have then -(0-1)=1

My first thought was to use the Wallis identity for powers of sin(x) from 0 to pi/2

what this really means is if you take the 1 out of the original integral, that all of those sine terms perfectly annihilate one another. most notably you can take all of the odd powers and move it to the other side to get: integral 0 to pi/2 of sum sin^(2n) dx = integral 0 to pi/2 of sum sin^(2n+1) dx. the left sum becomes: sum 1/(1-sin^2) - 1 = (some steps later) = tan^2 so we find an interesting expression for the integral of tan^2 in terms of an infinite integral of odd powers of sin(x), though since the first diverges, we know the second therefore also diverges.

BlackPenRedPen Can hold it Hhhhh. The two of you are amazing, God bless you.❤️🌹

the other video black integral red integral was filmed in the same shirt so probably filmed together but that video was released in dec 2020 and its sep 2021 now! what were you doing @Dr Peyam

@drpeyam

2 жыл бұрын

See description

Hey it is 1/(1+sinx) and further it is 1/(sin(x/2)+cos(x/2))^2 and further it can be reduced to 1/ cost^2 which is sect^2 ( by suitable substitution ofc) then it is much simpler right?

@mathematics6199

2 жыл бұрын

Nevermind Actually I didn't see the solution , which is much better

@Vladimir_Pavlov

2 жыл бұрын

dx/(1+sinx)=dx/(cosx/2 +sinx/2)^2 =dx / [(cosx/2)^2*(1+tanx/2)^2] = = 2dy/y^2. Integration limits for the variable y: lower = 1 , upper =2. Is that what you meant?)

Hey Dr peyam can you do some International Math Olympiad problems

@vnever9078

2 жыл бұрын

It would be so helpful

@mathematics6199

2 жыл бұрын

Hey could you please suggest some problems, I will do over my channel

Thanks for presenting a nice integral. Happy birthday to you Dr 3.14159.....m . My best wishes are always with you . Go ahead fearlessly.

F(x)=2/(1+tan(x/2))+C ?

How did he integrate from 0 to π/2 when the infinite sum wouldn't converge if x=π/2?

@iridium8562

2 жыл бұрын

Oh he used the limit at the end, aight

What if you integrate from -π/2 to 0 ?

@adiaphoros6842

2 жыл бұрын

Diverges, because the function has singularities at x = (4n-1)π/2

I like You! Greetings from Poland :D

But sin(π/2) is 1 so why we can do that? And note that the ratio is |r|

@drpeyam

2 жыл бұрын

Improper integral

@ajiwibowo8736

2 жыл бұрын

@@drpeyam oh yes forgot about that bite, thats why you use limit, so sorry Ln(1-Sinx)

I thought you speak austrian german in this video 😹

Professor. you have a beautiful haircut !

But sinx can be equal to 1 in the interval. But for a geometric series it must be less than 1.

@drpeyam

2 жыл бұрын

That’s fine, improper integral

@determinantmatrix9584

2 жыл бұрын

sinx = x

@nikhilnagaria2672

2 жыл бұрын

The point is more proven by the fact that the antiderivative is not defined at x=π/2 which is just sin(x)=1. Hence, you can view the given problem as lim(b->π/2)int from 0 to b ..... and then sin(x)

But it is an improper integral. At pi/2 the series you’re integrating is divergent.

Be my teacher

Another nice integral for your cool integrals playlist🤗

2:29 Can we sue this doctor for malpractice ;)

@drpeyam

2 жыл бұрын

😂😂

you could call l'hopital's rule the "loppy-lu" :)

nice shirt

error spotted

@drpeyam

2 жыл бұрын

Where?

@neilmccafferty7830

2 жыл бұрын

@@drpeyam belated apologies, I got my signs mixed up. second line where the square was taken out of the brackets.

Nice way to copy BPRP 😂😂😂😂

wtf