MIT Integration Bee Mock 2025 Finals Solutions - Part 1 

john integral himself would be proud

I just recognized that problem 1 polynomial was just the 5th Chebyshev Polynomial

This is beautiful. Made my day

For question 3: integration by parts then a trig sub yields -πln(2)/2 + (0 to π/2) ∫ ln( 1 + sin²(x) ) dx I solved the general form of this remaining integral on my video here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-SFILbUBNS60.html

0,5(sin(x/2^(n-2))-sin(x/2^(n-1))) 0,5(sin(x/2-¹)-sinx+sinx-sin(x/2)+..) =0,5sin(2x) -0,25cos(2x)|(0;π/2)=0,5

Great video👍

crazy good

Hey Lucas I love Silver and I loved your vid as well!! Is there any way that you have like a master sheet of the rules you need to know to be very good at integration ie. Catalans constant and its series representation, eulers reflection formula, glassers master theorem etc.? I find it kind of hard as a newbie to advanced integration to keep track of them all? If you don’t have one, would it be possible if you could help me make one? Thank you!!

Can I get a source for the answer key of the exercise??

$(-1;1)xln(1+(cosx)^sinx)dx=$(-1;1)(-xln(1+(cosx)^sinx)+xsinxlncosx)dx =$(-1;1)0,5xsinxlncosxdx=-cos1lnco s1+cos1-sin1+0,5((sin1-1)ln(1-sni1)+(sin1+1)ln(sin1+1))