3.1 Integration by Parts
[STARTS AT 31:53]
Professor: OK, next and last section, this section. Integral from 0 to pi over 2 of x cosine x. I believe we saw something similar to this earlier, so it would be wise to choose x to be our u and cosine to be our dv, since we can easily differentiate and actually get rid of the x there.
So the derivative is 1. Integral there is sine x. So x sine x minus the integral of sine x-- integral of sine x is going to be cosine, negative cosine, so we'll get plus cosine x and x sine x evaluated from 0 to pi over 2, which is-- all right, so if I plug-- let's do x times sine x and cosine x.
And so x sine x of pi over 2 plus cosine of pi over 2 is pi over 2. And then that's minus. I'll plug in 0 for there. Well, that's just going to be 1.
I'll double check that, of course. 0, because we have an x term here. That's going to become 0, and cosine of 0 is 1. And that is, in fact, 1.
So pi over 2 minus 1 is our result there. OK, that's the end of our section on integration by parts.
[ENDS AT 34:03]