 Welcome back everyone. Before we end our section on integration by parts, there's one last example I wanna demonstrate for you. Let us consider the function, well let's integrate the function sine of the natural log of x. Now I do claim this is gonna be an application of integration by parts, but you have to be cautious here. This notation does not mean sine times natural log of x. This is actually function composition, right? What we have going on here is that we have the sine of u composed with the natural log of x. We put natural log of x inside of sine. And because we have function composition, it turns out that u substitution actually is gonna be the strategy that we wanna take here. Take the inside function u to be the natural log of x. Then we know that du is equal to, well du of take the derivative that we're giving in one over x dx. And that does help us out a lot here. We can write this thing as the integral of sine of u. And then if you solve for dx here, because I mean after all this is an equation, you can solve for dx. If you solve for dx, you get the dx equals x du, which is a little problematic here because the whole point of switching the variables is to switch all the x's to u's. How do you deal with this residual x that's hanging out there? Well, remember u is equal to natural log of x. That's an equation you could solve for x and see that x is equal to e to the u. And you can make that substitution right here. And if you do that, you will end up with the integral of e to u sine of u du. And so with the appropriate u substitution, we now have a factorization that would work great for integration by parts e to the u sine of u du. Now, I do wanna point out that we have actually done this exact example before. This is an example we use to demonstrate the integration by cycles technique. And so you should click the video reference that you can see right now if you wanna see the details of that. But as a reminder, the anti-derivative was one half e to the u times sine of u minus cosine of u plus a constant. And so if we plug back in the fact that u equals the natural log of x, right? We plug those in here and here and in here. Now, of course, e to the u is equal to x right there. So making that simplification, you're gonna end up with x over two times sine of the natural log, the natural log of x minus cosine times the natural log of x plus an arbitrary constant. And this now gives us the anti-derivative that we were seeking. So the reason I present this example right here is that sometimes we have to realize that we might have to combine techniques we've learned. This integration by parts is what we're learning right now. And I would compare this to the situation that I've run a shop class. You have a little kid who's never learned how to use tools before. Well, if you give the kid a hammer, then everything looks like a nail. He just wants to go whack everything with the hammer because that's the only tool he knows how to use. If you give him a screwdriver, everything looks like a screw. He just wants to drive those in there. But sometimes the appropriate project requires both the hammer and the screw to work together. Maybe because there's nails and screws or maybe you can combine the two together. You could use your screwdriver as a chisel and you could hammer it. That's sometimes an effective technique when trying to fix something in wood shop there. So combined use substitution with integration by parts when appropriate. And that'll then conclude our section on 7.1 about integration by parts. Much like this example illustrates, it's not that we won't ever use integration by parts again. It's just we're not gonna focus on it in the future videos, but we will use it on occasion when it's necessary to evaluate an anti-derivative.