The second problem is a bit hard for their level. Firstly, you have to substitute t=e^-x to get int_{0}^{1} -t log(t)/(1+t) dt. Then you integrate by parts by taking t/t+1 as the integrating function and -log(t) as the differentiating one. Then you'll end up with \int_{0}^{1} (t-log(1+t))/t dt . Now use the Taylor expansion of log(1+t) at 0 and then interchange summation and integration (by Dominated Convergence Theorem) to get the answer as the infinite series 1/2^2 -1/3^2 +1/4^2 - 1/5^2 +..... which is 1-\eta(2) where \eta is the Dirichlet eta function. This evaluates to 1-pi^2/12 by using the relation zeta(2)=1/2*eta(2) where zeta is the Riemann zeta funciton.
I just can't make out how they couldn't do the Sec^5x integral , I was given that integral by my maths teacher just yesterday (I am also a jee aspirant currently in 12)
@@antarikshsengupta4643 Bro you are doing it in your home without time bound and they are facing it in a competition with huge pressure to win and also there is time bound.
MIT almost seems like a parallel reality. A world where anti-differentiation is a sport and people film it on their phones, its a world I desire to be a part of :)
it is only the test of speed and accuracy of just one section of mathematics chirag might be rusty coz ,I think the result would have been different if the physics and chemistry subjects accuracy also starts to play a big role in final score
![The most beautiful equation in math, explained visually [Euler’s Formula]](/img/n.gif)
well played Chirag better luck next time
30:15 this dude is insanely good
The second problem is a bit hard for their level. Firstly, you have to substitute t=e^-x to get int_{0}^{1} -t log(t)/(1+t) dt.
Then you integrate by parts by taking t/t+1 as the integrating function and -log(t) as the differentiating one. Then you'll end up with \int_{0}^{1} (t-log(1+t))/t dt . Now use the Taylor expansion of log(1+t) at 0 and then interchange summation and integration (by Dominated Convergence Theorem) to get the answer as the infinite series 1/2^2 -1/3^2 +1/4^2 - 1/5^2 +..... which is 1-\eta(2) where \eta is the Dirichlet eta function. This evaluates to 1-pi^2/12 by using the relation zeta(2)=1/2*eta(2) where zeta is the Riemann zeta funciton.
Second guy in red t shirt is CHIRAG FALOR JEE ADVANCED 2020 AIR 1 🎁👏
I just can't make out how they couldn't do the Sec^5x integral , I was given that integral by my maths teacher just yesterday (I am also a jee aspirant currently in 12)
@@antarikshsengupta4643 I think that when you are stand up and you have 4 minutes to solve is much harder
@@antarikshsengupta4643 Bro you are doing it in your home without time bound and they are facing it in a competition with huge pressure to win and also there is time bound.
4 min time for sec5x is enough at this stage
@@hsjkdsgd it's not sec5x it's (secx)^5
MIT almost seems like a parallel reality. A world where anti-differentiation is a sport and people film it on their phones, its a world I desire to be a part of :)
Well, I guess it's now time for the truth - brandon v luke, who wins? let's see if my favorite from last year can beat my new favorite.
both player mathed well, but my favorite from this year beat luke - well done to both
CHIRAG FALOR JEE ADVANCED 2020 AIR 1
it is only the test of speed and accuracy of just one section of mathematics
chirag might be rusty coz ,I think the result would have been different if the physics and chemistry subjects accuracy also starts to play a big role in final score
I love integration!
Me2
Me2
Me2
Me2
3 2 1 İntegrate!
yoooo chirag falorrr
I am here for chirag falor❤
Same but I wonder how I can make these, but Chirag can't, maybe he was too nervous or his skills got deprived after going to MIT
😢😢
Are you also JEE aspirant❤❤
@@SHUJA-UD-DEEN mee
@@SHUJA-UD-DEEN defniitely deprived, prime chirag falor would've won integration bee (if he knew the college level integration)
40:43 that is crazy. Who is going to solve that