🔥 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 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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POV: you walk into office hours and suddenly everything clicks 🤓 We were deep in double integrals Jacobians, and then I showed them the shortcut I don’t always get to cover in lecture… and yeah, this is why I love office hours. It turns into a whole vibe—mini math lab, real understanding, and honestly some of my favorite moments with students. If you want the full breakdown (and more examples), it’s up on my YouTube → Math with Professor V xoxo, Professor V #mathwithProfessorV #calculushelp #co
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The bell curve: e^(-x²). It has NO elementary antiderivative — you literally cannot integrate it the normal way. Yet the total area under it is exactly √π. The trick: don't solve I. Solve I². I² = (∫e^-x² dx)(∫e^-y² dy) = ∬ e^-(x² y²) dx dy That's a double integral over the whole plane. And x² y² = r². Switch to polar coordinates. In polar form an extra r appears — and THAT r makes it integrable. Angle gives 2π. Radial gives ½. I² = 2π · ½ = π So I = √π. A flat integral, solved by lifting it int
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