Can you do this crazy integral?

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[two]

I've noticed myself getting better at integrals just by watching Michael Penn videos. I was able to solve this one is under 5 minutes

[kkanden]

such a fun integral! loved how it combined with the 1/x^2 factor! great video michael keep it up :D

[Monkieteam]

I used what we proved about α and β as inspiration to solve the integral in a pretty nice way :
* Take x = sqrt(1+θ²) + θ
Then similarly to the case with α and β, we have that x - 1/x = 2θ.
* Under this change of variable, α and β clearly become -π/6 and π/6 respectively.
* dx becomes (θ/sqrt(1+θ²) + 1)dθ, however since cos(2θ)*θ/sqrt(1+θ²) is odd, that part vanishes in the integral (the bounds are opposite)
* we are left with integral of cos(2θ)dθ from -π/6 to π/6, which gives sin(π/3).

Nice video as always!

[alexrozenbom]

I would have majored in Math if I had seen Michael Penn's videos growing up in the 80s and 90s.

[HershO.]

Very creative! At first I tried doing this through u-sub but soon realised that it was going to be an extraneously long solution, problem being absence of 1 + 1/x^2 term. Although I think this is the best method, if one were to go through the efforts of doing the u-sub completely, then I suppose IBP would've eventually yielded the solution. Just as a back up brute force plan.

[txikitofandango]

Would be cool to generalize this integral from cosine function to general f(x) and adjust integration bounds accordingly

[paulconway]

I absolutely love the way you present your videos!

[manucitomx]

Wow, wow, wow.
Great, if contrived, problem.
Thank you, professor.

[AntonioRadici]

great video as usual. I'm already a patreon supporter and I'm super happy that we are getting closer to the 1k/mo goal!

[Mystery_Biscuits]

I wasn't really sure where he was going with this but then at 7:49, it all clicked, and it's a very slick solution I might add!

[richardheiville]

the trick works fine too for computing integral of 1/(1+x^4) x=0, infinity

[stalecu]

Michael or his editor should be more careful with the thumbnails, the problem in the video isn't the same as the problem in the thumbnail

[preethamjee]

We can use glasser master theorem if the bounds are between -infinity to +infinity

[CaradhrasAiguo]

I did this with a x = e^t substitution since the (x - 1/x) argument to cos reminded of e^t - e^{-t} which is 2sinh(t), then added and subtracted integral on the same log-transformed bounds of the integral of e^{-t} * cos(e^t - e^{-t}) dt. Then I briefly got stuck but the hint came in helpful with the "extra" definite integration of the e^{-t} * cos(e^t - e^{-t}) dt term which is just a negative copy of the original integral, leaving 2 * I(t) = integral from -ln(beta) to ln(beta) of (e^t + e^{-t}) * cos(e^t - e^{-t}) dt, which is an easy u-sub.

[egillandersson]

I had the "aha!" at 4:13 A very funny integral !

[MichaelGrantPhD]

ok that took a twist I wasn't expecting. fun!

[mathhack]

Waou!1 amazing strategy. Starting my breakfast with this Pb.

[thomasbach-zykn]

If you go back to “x“, should it not be “x^2“ instead of the inverse?

[jaredvv]

This reminds me of calc II test in college. It was the ridiculously ugly integral, just really painful. After all the work and effort it simplified out to be just 1+1 =2. Pretty sure the professor was just having a laugh at our expense.

[jensknudsen]

What a lucky coincidence that all that crazyness boils down to such a neat result in the end ;-)