Integral of ln(cos x)

We calculate the definite integral of ln(cos x) over the interval from 0 to pi/2.

[CoderboyPB]

It's amazing, I understand it, but I would never get to these ideas, especially using the trigonometric identities, but that's the reason why I failed my math studies: I was interested but never reached this manditory meta level ...

But I like this videos, because even, If I failed, I never broke up with the mathmatics.

Greetings from Germany, your vids help me to come to these hard corona times in social distance.
Stay healthy :-)

[chae]

Wow...this was really impressive. I've just started Calculus 2 and am learning integration by parts. It's a tricky process that requires so much knowledge in order to do efficiently/effectively. I'm okay on my Pythagorean identities, but I need to get all the double and half angle identities down. Definitely subscribed and will keep watching these cool problems. Thanks Michael!

[ashotdjrbashian]

This integral (with sinx instead of cosx) was calculated by Euler more than 250 years ago. You have used a similar idea, but Euler goes backwards, and that makes calculation so much easier. First, he does the substitution x=2t and makes I=ln2*pi/2+ two integrals of 2ln(sint) and 2ln(cost), both from 0 to pi/4. In the second one he does change t=pi/2-u, which makes that an integral of 2ln(sinu) from pi/4 to pi/2. Combining together gives him 2I and all said. It takes literally two lines!

[wolfmanjacksaid]

Whenever I see a new video like this I say "Ok, great"

[bastiangeissbuhler]

Amazing, im from chile and i was triying to solve that integral for a while and when i give up, Youtube suggest me this video, is like fallen from heaven

[rc]

From the thumbnail I tried doing it in my head by simply integrating the initial term and my answer was negative infinity (obviously), and when I saw the video my jaw literally dropped (again, obviously)
Such an elegant method :D

[andcivitarese]

Integral from 0 to pi of ln(sin x) equals 2 times “I” since sin x is an even function with respect to x=pi/2, and so is ln(sin x)

[erazorheader]

Another way to get rid of ln(z) is to consider the derivative d(z^t)/dt which yield ln(z) at t = 0.
Thus it suffices to calculate int_0^{\pi/2} (cos(x))^t dx which results in beta-function after trivial change of variables. After all it will be possible to represent the answer in terms of digamma function.
But the simplest way here is to use cos(x) = (e^{ix} + e^{-ix})/2,
so ln(cos(x)) = -ln(2) + ix + ln(1 + e^{-2 ix}) = -ln(2) + i x + sum_{n = 1}^\infty (-1)^{n - 1} e^{-2 i n x} / n.
As ln(cos(x)) is a real number, we can throw away the imaginary part (because it is zero):
ln(cos(x)) = -ln(2) + sum_{n = 1}^\infty (-1)^{n - 1} cos(2 n x)/n
As integral from cos(2 n x) within the interval (0, pi/2) yields zero,
we get -ln(2) * pi/2.

[chazzaca]

Great video Michael. If you integrate x/tanx from pi/2 to 0 (using integration by parts) you will use all the properties you've specified here

[bmenrigh]

This is probably the first video where I knew exactly where the method was going. I must have seen a trick like this before. Perhaps in solving certain infinite sums that use the same looping substitution / equality trick.

[BoZhaoengineering]

The dummy variables changes back to x . That helps a lot. And you set the integral equal to I, it helps a lot too.

[vbcool]

Tried using the fact that the definite integral between 0 to pi/2 is same for ln(sin(x)) and ln(cos(x)), then added both leading to the integral being half of the sum of those definite integrals. Then used ln a + ln b = ln ab which yields ln sinx + ln cos x = ln sin 2x - ln 2.

This gives the identity I = I/2 - pi/4 ln 2

[harshpratapsingh]

Or simply use 2 properties of definite integration
But it is good that u showed us each step

[chrisli]

amazing, i believe such beautiful solution comes from countless practices.

[bat]

Can someone tell how he repose u = x at min 3
Initially he posed u = pi/2 - x
Wtf

[hg]

Thanks for the explanation above. I like this sort of maths but I can never achieved the mental acrobatic that you can do but I enjoy stressing my brain in following it. However, using my calculator, it cannot even come out with lncos(x) but using lnsin(x) I got a complex number answer as -1.0888 - 4.4409x10^16i.

[whyyat]

I love Dr Penn's videos, but am I the only one who has to pause the video so my ears can catch their breath?

[dralol]

The real fun is to prove the integral does indeed converge..

[TimeforDROPS]

Why can we do change of variables back to x x = u, when we have ever used the substitution x = pi/2 - u, thanks for your answer

[UnforsakenXII]

I've found a new way to do this integral but it involves gamma functions and exponential integrals. = D