I really loved integrals at high school and University but right now, due to engineering degree, i have overspecialised myself in matrixes and recursive algorithms. Due to your videos my passion has sparked once again!! Do you have any theory pdfs? You seem to have a talent in explaining things!!
that can be simplified further. I recognize 3+2sqrt(2) as (1+sqrt(2))^2, so the final answer can be written as (pi^2-2ln^2(1+sqrt(2)))/8 (the 2 from the exponent gets taken out twice because it's a ln^2)
Feels like you jumped a lot of in between steps in the first 4 mins worth of manipulations. A bit too quick to follow vs doing the step by step substitutions methodically. Very cool integral nonetheless!
@@maths_505 alive and well! I'm literally always around, I just don't comment every video. But the like goes up on every one even before I watch it! How about some multi-variable calculus? Solids of revolution, Greene and Stokes' Theorem etc.?
I really loved integrals at high school and University but right now, due to engineering degree, i have overspecialised myself in matrixes and recursive algorithms. Due to your videos my passion has sparked once again!! Do you have any theory pdfs? You seem to have a talent in explaining things!!
I'm gonna start my complex analysis course in September. Everyone will have access to pdfs and homework help for the questions I assign.
@@maths_505🎉🎉🎉🎉lets gooo
these integrals are becoming more insane by the day... I love it!
The titling on this video is excellent. Thank you for the clear labeling.
that can be simplified further. I recognize 3+2sqrt(2) as (1+sqrt(2))^2, so the final answer can be written as (pi^2-2ln^2(1+sqrt(2)))/8
(the 2 from the exponent gets taken out twice because it's a ln^2)
how do you make the step on 4:25. The first integral is logic but why can you convert it to that integral from 0 to 1
I changed the variable to ut
I think it becomes more beautiful if you write this as
I = pi^2/8 - ln^2 (1 + sqrt(2))^2 / 16 = pi^2/8 - ln^2 (1+sqrt(2)) / 4
Where does that sum of fractions @5:36 come from?
It's a partial fraction decomposition
Thank you for your effort.
You should make some videos with chat gpt
Great Integral
can i ask where you find these integrals?
This one's from a subscriber
@@maths_505thanks tryna get better 💪
👏👏👏👏👏👏
3+2sqrt(2) = (1+sqrt(2))^2
so [ln(3+2sqrt(2))]^2 = [2*ln(1+sqrt(2))]^2 = 4*[ln(1+sqrt(2))]^2 = 4*arsinh(1)^2
OMG... Wolfram Alpha computes this result for the indefinite integral:
integral arctan(cot^2(x)) dx
= x tan^(-1)(cot^2(x)) + 1/4 (2 cos^(-1)(-sqrt(2)) tan^(-1)((-1 + sqrt(2)) cot(x + π/4)) - 2 cos^(-1)(sqrt(2)) tan^(-1)((1 + sqrt(2)) cot(x + π/4)) - π tan^(-1)(1 - sqrt(2) tan(x)) - π tan^(-1)(sqrt(2) tan(x) + 1) - (π - 4 x) tan^(-1)((-1 + sqrt(2)) tan(x + π/4)) + (π - 4 x) tan^(-1)((1 + sqrt(2)) tan(x + π/4)) + i (cos^(-1)(sqrt(2)) - 2 tan^(-1)((1 + sqrt(2)) cot(x + π/4))) log((sqrt(2) (1 - i cot(x + π/4)))/(i cot(x + π/4) + sqrt(2) - 1)) - i (2 tan^(-1)((-1 + sqrt(2)) cot(x + π/4)) + cos^(-1)(-sqrt(2))) log((sqrt(2) (i cot(x + π/4) + 1))/(i cot(x + π/4) + sqrt(2) + 1)) + i (2 tan^(-1)((1 + sqrt(2)) cot(x + π/4)) + cos^(-1)(sqrt(2))) log(-(i (-2 + sqrt(2)) (cot(x + π/4) - i))/(i cot(x + π/4) + sqrt(2) - 1)) - i (cos^(-1)(-sqrt(2)) - 2 tan^(-1)((-1 + sqrt(2)) cot(x + π/4))) log((i (2 + sqrt(2)) (cot(x + π/4) + i))/(i cot(x + π/4) + sqrt(2) + 1)) - i (2 tan^(-1)((1 + sqrt(2)) cot(x + π/4)) + 2 tan^(-1)((-1 + sqrt(2)) tan(x + π/4)) + cos^(-1)(sqrt(2))) log(((1/2 + i/2) e^(-i x))/sqrt(sqrt(2) - sin(2 x))) - i (-2 tan^(-1)((1 + sqrt(2)) cot(x + π/4)) - 2 tan^(-1)((-1 + sqrt(2)) tan(x + π/4)) + cos^(-1)(sqrt(2))) log(((1/2 - i/2) e^(i x))/sqrt(sqrt(2) - sin(2 x))) + i (2 tan^(-1)((-1 + sqrt(2)) cot(x + π/4)) + 2 tan^(-1)((1 + sqrt(2)) tan(x + π/4)) + cos^(-1)(-sqrt(2))) log(-((1/2 - i/2) e^(-i x))/sqrt(sin(2 x) + sqrt(2))) + i (-2 tan^(-1)((-1 + sqrt(2)) cot(x + π/4)) - 2 tan^(-1)((1 + sqrt(2)) tan(x + π/4)) + cos^(-1)(-sqrt(2))) log(((1/2 + i/2) e^(i x))/sqrt(sin(2 x) + sqrt(2))) - Li_2(-((-1 + sqrt(2)) (-i cot(x + π/4) + sqrt(2) - 1))/(i cot(x + π/4) + sqrt(2) - 1)) + Li_2(-((1 + sqrt(2)) (-i cot(x + π/4) + sqrt(2) - 1))/(i cot(x + π/4) + sqrt(2) - 1)) + Li_2(((-1 + sqrt(2)) (-i cot(x + π/4) + sqrt(2) + 1))/(i cot(x + π/4) + sqrt(2) + 1)) - Li_2(((1 + sqrt(2)) (-i cot(x + π/4) + sqrt(2) + 1))/(i cot(x + π/4) + sqrt(2) + 1))) + constant
Professor:-the paper comprises of only 1 integral worth 100 pts
Integral :-
I=3(pi/4)^2-[1-1/3(1-1/3+1/5)+1/5(1-1/3+1/5-1/7+1/9)-1/7(1-1/3+1/5-1/7+1/9-1/11+1/13)....-]...questa è la soluzione corretta 1,03...
Feels like you jumped a lot of in between steps in the first 4 mins worth of manipulations. A bit too quick to follow vs doing the step by step substitutions methodically. Very cool integral nonetheless!
HE'S ALIVE!!!
HE'S ALIVVVVVVVVVEEEEEEEEE!!!!
Thankfully it was fast enough to hear from you 😂?
How you been mate?
@@maths_505 alive and well! I'm literally always around, I just don't comment every video. But the like goes up on every one even before I watch it!
How about some multi-variable calculus? Solids of revolution, Greene and Stokes' Theorem etc.?
I like the complex versions of them better. The complex analysis series starts in the first week of September.
And thank you so much G
First
Come back to the dark side pls lord Vader
Vader has been gone for too long indeed....
@@maths_505 I meant the thumbnail lmao. Vader is dead since 80s tho
@@bartekabuz855 oh so I don't have to use the infinity stones to snap him back?
Ah well....I'll save em for another occasion.