M.I.T. Integration Bee Question

University of Oxford mathematician Dr Tom Crawford solves a tricky integral from the MIT Integration Bee. The solution requires several changes of variables, completing the square and knowledge of hyperbolic functions.

The first step when solving the integral is to rewrite cosh in terms of exponentials. We then make the substitution u = exp(x) and complete the square in the denominator. A second change of variables for cosh(y) is then used to allow us to simplify the integrand using the hyperbolic identity cosh^(y) – sinh^2(y) = 1. Finally, we integrate cosech(y) and rewrite the resulting expression in terms of logs to get the desired result.

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[gurkiratsinghtha]

When you are done with the u sub then you can use partial fractions to solve the rest, I know that u^2+4u+1 is not Factorable over rationals but it is factorable over reals, i.e. u^2+4u+1 = (u+2+√3)(u-√3+2)

[rob]

At 7:40 you could have made the substitution (u + 2)/√3 = w to arrive at the answer quicker using partial fractions.

[giuseppepalazzo]

Didn't know Machine Gun Kelly was that good at math

[polychromaa]

Alternate method is hyperbolic weierstrass substitution; t=tanh(x/2)

[kenroyadams]

Absolutely fantastic! Thank you so much for taking the time and effort to do this integral.

[a_k__]

I really like the style of your teaching. I always feel I learned something when I watch your videos. I think it would be very useful if you can make a video about different part of Math as a subject. Kind of like a roadmap that shows different parts of math and how it is used in different industries. I think your unique style of teaching is great for it

[ghostek]

WTF@!@! that app is insane i've never seen it. i'm about to start integral calculus for software engineering and knowing there are such tools to assist in learning is going to give me peace of mind when i encounter problems that I may struggle on. the break down on the app is so thorough as well i love it! thx for the video

[yugiohsc]

I love watching the integration bee! It’s the one sport i enjoy watching

[franolich]

Simpler method similar to what others have suggested:

J = Integral [ 0 to infinity : dx / (2 + cosh(x)) ]
= Integral [ 0 to infinity : 2e^x dx / (e^2x + 4e^x + 1) ]

The roots of (y^2 + 4y + 1) are y = -2 +/- sqrt(3)
J = Integral [ 0 to infinity : 2e^x dx / (e^x + 2 - sqrt(3)) (e^x + 2 + sqrt(3)) ]

Let u = e^x + 2 => du = e^x dx
J = Integral [ 3 to infinity : 2 du / (u - sqrt(3))(u + sqrt(3)) ]

By partial fractions:
J = (1/sqrt(3)) * Integral [ 3 to infinity : 1/(u - sqrt(3)) - 1/(u + sqrt(3)) du ]
= (1/sqrt(3)) * [ 3 to infinity : ln( (u - sqrt(3)) / (u + sqrt(3) ) ]
= (1/sqrt(3)) * [ ln(1) - ln( (3 - sqrt(3)) / (3 + sqrt(3) ) ]
= (1/sqrt(3)) * ln( (sqrt(3) + 1) / (sqrt(3) - 1) )

[smellund]

I don’t even begin to understand integrals but still watched the whole thing, good video

[ThatWeirdCat]

i didn't like math because i can't get past 70% in every exam i take in my school, but i like this dude. 10/10 🔥

[hugolindholm]

Why not factorize it into (u+2-sqrt(3))*(u+2+sqrt(3))? Then do partial fraction decomposition?

[angelodiavolo]

Yesterday I studied integrals for the first time. Today I watch this video. I have surprised understood more then I thought I would hahah

[iTeerRex]

It’s one of those problems with 2 stars at the end of the chapter, and after banging your head against the wall for some time. You get the answer with a negative sign 😂
Happy new year everybody, and may this omen be lifted soon.

[Carlos-vnec]

this is so buck wild. I forgot how cool doing integrals can be

[rossg]

Beautifully explained, but even as a decent amateur mathematician I would struggle to know where to start. The way in?

[adrianh]

1:51

"is given here by this function"

it's not a function, it's a number

[ranpancake]

yet another banger ! [also, love the shirt :)]

[danielbenton]

Why don't you take the minus sign out and sub in tanh so you don't need to look any integrals up?

[d]

Took over a minute to do it in my head, including some checking. Getting old ...