2 of the coolest integrals on YouTube
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- čas přidán 17. 05. 2024
- 2 ridiculously awesome integrals through one generalized solution development using some of your favorite tricks. Enjoy....
The formulae:
• 2 fascinating integral...
My complex analysis lectures:
• Complex Analysis Lectures
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Ecco il passaggio decisivo...moltiplicare (1+x) sopra e sotto...io non c'ero arrivato ...complimenti
At 3:15 I can see a difference of two integrals, where both integrals do not exist for instance for s=2 (but Re(s=2)
Thank you for solving such integrals. So, for non specialists and undergraduate students, we shall proof each integral or theorem before introducing solutions. This channel is the best, and i want to be the best all the time.
Bro is starting a math cult
I am all here for it
Also I got JEE advanced exam on may 26, hoping it goes well.
Nice I love homework
According to Wolfram these last two simplify to...
√⅓ / (4 cosh(π/3) - 2), and...
-½ π sinh(π/6) sech(π/2)
Can you tell how to integrate root sinx from 0 to π/2 is this integral can be done using calculus 1
cosh^2 n+3sinh^2 n=1+4sinh^2 n
What software are you using for writing this math on?
cult? wasn't this a harem??????
It's everything and more
I played with u=1/x substitution and partial fraction decomposition of the factor (1+x^2)/(x^4+x^2+1)I have got
\int_{0}^{1}\frac{\cos{\ln{x}}}{x^2+x+1}dx + \int_{0}^{1}\frac{\cos{\ln{x}}}{x^2-x+1}dx
Now i would expand numerator and denominator to get
\int_{0}^{1}\frac{(1-x)\cos{\ln{x}}}{1-x^3}dx + \int_{0}^{1}\frac{(1+x)\cos{\ln{x}}}{1+x^3}dx
and now i would try to use geometric series expansion
0:16 "You have not been scammed"
Reality: Here's a starter integral before we go to the main course lunch
Woah
Who else used contour integration and residue theorem. ?
Hi,
"ok, cool" : 3:15 , 4:45 , 5:58 , 8:26 ,
"terribly sorry about that" : 5:29 , 5:45 , 10:35 .
quoted theorem is actually really hard, I tried MMA to understand the details 😢but finally get this in youtube
int x^(m-1)/(x^n+1)
czcams.com/video/OnTn3R9_m1Q/video.htmlsi=sqvK0IGvkC9ED9Uh
things become much much easier in the complex number field❤