When I first did the integral I did it precisely as in the video, but then I realized something: the square root of 2 was arbitrary and you can use any number. So I replaced it with just t and thought, if this is I(t) then I'(t) must always be 0 because I(t) is going to be a constant, but why. Turns out if you take the derivative you get a really nasty expression, but what is beautiful is that it is always odd about the line pi/4 in the interval [0,pi/2]! Why? If you find the derivative, then the integrand will have some tangent functions. Now, looking at the unit circle, sine and cosine essentially switch spots around pi/4 so tangent and cotangent essentially switch spots. Thus we switch the tangents with cotangents and do some multiplying by 1 to get them back to tangents and surprisingly at the end we get the same expression but without the negative sign. Therefore it is odd about pi/4. Therefore I'(t)=0. Using t=0 gives that I(t)=pi/4 trivially from there.

One other way I thought of doing this is using x=arctan(u), which gets us to an intimidating integral, but if we then substitute u=1/w then use the I+I=2I trick then we get to the same answer.

Just thought I would share these other ways :) It would be totally cool to see a video about these two ways, if you wanted to show your viewers that there is more than one way to do this integral.

From your excited-about-integrals viewer, Andrew.

Thank you for talking about this recent putnam integral here, mate. This is quality content from you. I won't have to

pay for essaynow, because I've got the information that I needed from this post. Thank you!