The Golden Ratio and the Natural Log: An “Integral” Connection

Hey there, polymaths!

In today's video video, let's take a look at a definite integral that leads to an absolutely delightful evaluation: the natural log of the golden ratio, phi. It's a journey that starts with the integral of 1/sqrt(x^2+1) from 0 to 1/2.

I begin by substituting x = tan u and dx = sec^2 u du. This allows us to simplify the integrand to sec u du using a pythagorean identity, and to shift the bounds of integration to 0 and the inverse tangent of 1/2. From there, we're simply integrating sec u du, which you can do by multiplying it by sec u + tan u over itself.

With another substitution, this allows us to integrate and evaluate ln | sec u + tan u | from the previous bounds. Our lower bound evaluates to ln 1, which is 0, and therefore which goes away. Our upper bound turns into ln (sec (tan^-1 (1/2) + tan (tan^-1 (1/2))). The tan and inverse tan cancel out leaving 1/2, and we can set up a right triangle to figure out that sec (tan^-1 (1/2)) is the same as √5/2.

The grand finale? This gives us back ln (1/2 + √5/2), which is of course ln φ, the natural log of the golden ratio.

#goldenratio #calculus #integration

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[jadenb]

Hi, it’s your student Jaden from a few years ago, just appreciating how much your channel has grown in just a few years! Keep up the great work! Let’s get 100K :)

[octoni]

I’m in a BC calc class and we have finished everything up to u sub integration by parts partials and diff eqs and slope fields, and although I had to pause it you teach it an such an easy way that you can’t get lost. New sub earned thanks man!

[hampus]

Nice compact math video 😄
I dislike the use of "tangent inverse" though, since tangent doesn't have an inverse. But the original note used arctan at least!

[aaronag]

To me it looks like a kids drawing but to my dad, I remember seeing lots of similar equations on his chalk board at home.
Sadly not even my ADHD could pick up any of it 😮

[Samir-zbxk]

I created another cool integral like this

(0 to π/2) ∫ ln( 1 + 4sin²(x) ) dx

The answer is πln(φ)

[saturnslastring]

Have you heard about terrence howard and his 1x1 = 2 "proof"?