The Infinite Cosine Trap (MIT Integration Bee)

Hinzugefügt: 2026-05-15

[00:00:00]Here's a question that stomped many students at one of the most competitive math tournaments in the world. The MIT integration B.

[00:00:10]We need to evaluate this integral X times an infinite product of cosiness integrated from 0 to pi / 4.

[00:00:22]Before we touch a single symbol of algebra here, I want to ask a different question first.

[00:00:29]What does this function actually look like? Let's graph it step by step. The first partial product x * cosine of x / 2. Well, it looks like a gentle wave.

[00:00:45]Now multiply by cosine of x / 4. The curve shifts slightly.

[00:00:53]Now cossine of x / 8. Getting smoother.

[00:00:58]um more familiar somehow.

[00:01:01]Keep going. Cosine of x / 16 over 32.

[00:01:08]Wait, is that just sign? Yes.

[00:01:13]The infinite product of cosiness is drawing the exact graph of sin x.

[00:01:21]That's the mystery we're going to unravel today. And the answer is more beautiful than you'd expect.

[00:01:29]So why does the infinite product of equal sin x /x?

[00:01:38]Let's prove it rigorously.

[00:01:42]First a key principle that guides this whole approach.

[00:01:49]Products are hard and sums are easier.

[00:01:53]The natural log turns products into sums. And that's the entire motivation for what we're about to do. Let u of x be the infinite product. Our integral becomes x * u of x.

[00:02:10]Now take the integral the natural log of both sides.

[00:02:15]And watch the product symbol transforms into a summation.

[00:02:22]We now have log u equals sum of log cosine of x / 2 to the k.

[00:02:29]Now differentiate both sides with respect to x.

[00:02:34]The left side gives u prime over u. On the right chain rule gives us sin /x * 1 / 2 k for each term.

[00:02:49]That's negative tangent of x / 2 the k / 2 the k. So u prime / u equals negative the sum of 1 / 2 the k* tangent of x / 2 to the 2 to the k.

[00:03:08]This still looks complicated but we have a secret weapon.

[00:03:13]So here's the key identity.

[00:03:18]Canent theta minus tangent theta equals to quotangent of 2 theta.

[00:03:26]I want you to hold this in your mind because we're about to use it to collapse an infinite sum down to almost nothing.

[00:03:36]Rearranging negative tangent theta equ= 2 can 2 theta minus canent theta.

[00:03:47]Substitute theta equ= x / 2 to the k into every term.

[00:03:55]Each piece of our infinite sum becomes a difference of two cangent expressions.

[00:04:03]Now write out the first four terms.

[00:04:08]Term one quotent x shown in gold minus 12 cangent x / 2 shown in orange.

[00:04:20]term 2 plus 12 cangent x / 2 orange again minus4 cangent x / 4 purple term free plus 14 cangent x / 4 purple minus 18 can x over 8 blue.

[00:04:53]So, do you see it?

[00:04:56]The orange terms cancel, the purple terms cancel, the blue terms cancel one by one like dominoes falling.

[00:05:11]All that survives is the very first term quotangent x and the limit of the final term as k goes to infinity.

[00:05:26]Now what is that limit 1 / 2 to the k* cent of x / 2 to the k as k goes to infinity.

[00:05:42]Here's the key insight.

[00:05:44]As k grows x / 2 to the k shrinks toward zero.

[00:05:53]And for very small angles can theta is approximately 1 / theta.

[00:06:02]So our expression becomes 1 / 2 k * 2 k /x which is simply 1 /x.

[00:06:13]Therefore, u prime / u equ= cent x - 1 / 8.

[00:06:22]Now integrate both sides.

[00:06:26]The integral of cangent x is log sin x.

[00:06:32]The integral of 1 /x is log x.

[00:06:36]So log u equ= log sin x - log x + a constant.

[00:06:46]So we exponentiate u = c * sin x /x.

[00:06:55]We find z using the behavior near zero.

[00:07:01]So as x approaches zero, every cosine term approaches 1. So u approaches 1.

[00:07:12]And sin x /x also approaches 1 at x= 0.

[00:07:19]Therefore c = 1. And we have proved it.

[00:07:25]The infinite product of cosiness equals sin /x.

[00:07:32]Now we bring every everything together.

[00:07:37]Our original integral x * the infinite product becomes x * sin /x sin x /x.

[00:07:48]Watch the x in the numerator and the x in denominator.

[00:07:54]They cancel. And just like that, our monstrous infinite product integral collapses into the integral of sin x from 0 to pi over 4. All that machinery, all that algebra now reduced to one line.

[00:08:15]The anti-derivative of sin x is negative cosine x. So evaluating from 0 to p<unk> / 4 cosine of p<unk> / 4 = cosine of 0 cosine of<unk> /x is <unk>2 / 2 cosine of 0 is 1. So our final answer 1 - <unk>2 / 2 approximately 0.293 293.

[00:08:41]And there is it. There. There it is.

[00:08:46]The exact area under sin x from 0 to pi /4. And that is the same shaded region we showed at the very beginning. The graph told us the answer before we even started.

[00:09:02]And the algebra, we just use it to confirm it. That's the beauty of math.

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