Missed the opportunity to use the opportunity to use the upper boundary as pi M. Another solution: Call I = int_0^pi sin^2(x) dx I = int_0^pi/2 sin^2(x) dx + int_pi/2 ^ pi sin^2(x) dx By symmetry, I = 2 int_0^pi/2 sin^2(x) dx By calling J = int_0^pi/2 sin^2(x) dx cos^2(x+pi/2) = sin^2(x) Then J = int_0^pi/2 cos^2(x) dx sin^2(x) = 1 - cos^2(x) J = int_0^pi/2 1 dx - int_0^pi/2 sin^2(x) = int_0^pi/2 dx - J By adding J in both sides 2J = int_0^pi/2 dx I = 2J = pi/2

some videos on this channel are simple, almost trivial. Others are what I can only assume is senior level undergraduate or above. You never really know what you're going to get, but its always fun to watch.

The difference of difficulty between two videos is insane, but the work is always great ! Keep up :)

There is also a way were you don't have to find an anti-derivative. Let's call the integral I If you let u = pi/2 - t you get I= integral from -pi/2 to pi/2 of cos^2(u) du. Now split the integral in an integral from -pi/2 to 0 and one from 0 to pi/2. Notice that cos^2(u) has an period of length pi ( (cos u+pi)^2 = (-cos u)^2 = cos^2(u) ). So the Integral from -pi/2 to 0 is equal to the integral from pi/2 to pi ( just add the period to the upper and lower bound or let v = pi + u). Now we can combine the two integrals again and get that I = integral from 0 to pi of cos^2(u) du But then 2I = integral of sin^2(t) dt + integral of cos^2(t) dt = integral of sin^2(t) + cos^2(t) dt = integral of 1 (all these integrals go from 0 to pi) so 2I = pi or I =pi/2

Nice that you derived the power reducing formula for (sin X)^2....I like this particular integral, multiple ways to solve, some are less painful than others

I literally was solving a question online and decided to find the recursive formula for sine (because this method was already written) and you put a video about this, lol

Oh! thank you very much ...Now, I can evaluate this integral for Alternating current.

Very happy teacher

Nice Classic !

You could also write sin^2(x) as 1-cos^2(x) and use partial integration

where have you been?? also great video!!

May you do also integral of sin(mx)sin(nx)dx from 0 to pi? With n and m different

Can you find volume in between 2 cylinders using single integrals, senpai!?!?!? :( if so please make video

So the integral from 0 to pi/2 would be pi/4 (using u=pi/2-t, and the fact that sin^2(t)+cos^2(t)=1 (edited pi/2 into pi/4)

First:D

Why not integrate by parts?

do by parts twice and cry.