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🔥 Day 133 — 2025 MIT Integration Bee (Qualifying) Q17 Evaluate: ∫ sin(x) * sinh(x) dx Wild mix of circular and hyperbolic functions. The fastest path is a classic double “integration by parts” loop: set up I = ∫ sin(x) sinh(x) dx, integrate by parts twice, and watch I appear on both sides so you can solve for it. 👉 Hints • First parts: u = sin x, dv = sinh x dx. • Second parts (on the new integral): u = cos x, dv = cosh x dx. • You’ll end with 2I = (sin x) cosh x − (cos x) sinh x. Would you ca
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🔥 Day 152 — 2026 MIT Integration Bee (Qualifying) Q18 365 Problems for 365 Days Evaluate: Integral of (sin^2(x)/x^2 − sin(2x)/x) dx This one is a “recognize the derivative” speed problem. The integrand looks messy, but it’s literally the derivative of a super simple expression if you think “quotient rule trig identity.” If you try to integrate directly, it’s pain. If you spot the structure, it’s instant. Hints: \t•\tLook for something like (sin^2 x)/x \t•\tDifferentiate it with the quotient rul
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TikTokbrainrot.mathematics
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🔥 Day 121 — 2025 MIT Integration Bee (Qualifying) Q4 Evaluate: Integral from 1 to e^e of [ log( x^(log(x^x)) ) / x^2 ] dx Looks wild, but the trick is to simplify before integrating. If you collapse the nested powers/logs the integrand turns into something that begs for a one-line u-sub. 👉 Hints • Use: log(a^b) = blog(a) and log(x^x) = xlog(x). • You should get (log x)^2 / x, then let u = log x. Would you spot it under time pressure? ⏱️ 💭 Drop your answer method below 👇 ✅ Full step-by-step s
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🔥 Day 132 — 2025 MIT Integration Bee (Qualifying) Q16 Evaluate: ∫ [ sin(4x) * cos(x) ] / [ cos(2x) * sin(x) ] dx It looks nasty, but this one collapses fast. Expand sin(4x), cancel what you can, and the integrand turns into a simple power of cos. From there it’s a one-liner with the double-angle identity. 👉 Hints • sin(4x) = 2 sin(2x) cos(2x) = 4 sin x cos x cos(2x) • Most things cancel → 4 cos^2 x • Use cos^2 x = (1 cos 2x)/2 Would you spot the cancellations under time pressure? ⏱️ 💭 Comment
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