00:02
For problem 22, we need to integrate e to the negative y, cosine y, d .y.
00:05
And so we're going to have to do this through integration by parts.
00:08
So let's make u equal to cosine y, cosine y, and du will equal negative sine y, dy.
00:17
And then we're going to make dv equal to e to the negative y, dy, which means when we integrate that, that will just give us negative e to the negative y, and there we go.
00:30
So now we can go ahead and piece this together and we get negative e to the negative y times cosine y minus the integral v times du.
00:41
So negative, we're going to pull out that negative, but then also this is negative.
00:45
So both of those will cancel and we'll still be left with just a negative e to the negative y, sine y, d y.
00:54
And so we now are back to looking like a very similar integral.
00:57
So we're going to have to do this use substitution one more time.
01:00
So let's do a second level of it.
01:03
So u, once again, will equal sine y this time.
01:06
Sign y, du will equal cosine y, d .y.
01:11
We have dv equaling e to the negative y, dy, and v will then equal negative e to the negative y.
01:20
So we can now piece this together for yet a second time.
01:24
So we get negative e to the negative y times cosine of y minus, and then let's distribute this negative.
01:31
So we're going to write all of this inside of it first.
01:33
So u times v, so negative e to negative y, sine y, right? and then minus the integral of v times the u.
01:45
So the negative will come out and make this positive.
01:48
So negative becomes positive, e to the negative y, cosine y, d y...