 Welcome back everyone to our lecture series Math 1220 Calculus II for students at Southern Utah University. As usual, I will be your professor today, Dr. Andrew Misalign. It's good to have you. In lecture 10 here, we're going to talk some more about integration by parts, but we're also going to start section 7.2 from Stuart's textbook about derivative, or sorry, anti-derivatives of trigonometric functions, but that will come later. I want to talk a little bit more about integration parts right now in this video. In particular, I want to show you another variation of the integration by parts technique that turns out to be quite effective in the right situation. Let's find the integral of the natural log of dx. Now, many of us are very, very quick to be like, oh, that's easy, the natural log of x dx, that equals 1 over x plus c. But I want to mention that this is completely wrong, right? The issue here is that we're conflating the derivative of the natural log. So we know the derivative of natural log is 1 over x. And sometimes we conflate that with the anti-derivative. We don't want to confuse those things with each other whatsoever. So how does one find the anti-derivative of the natural log? In fact, up to this point in the series, we actually don't know what the anti-derivative of the natural log is. We know it's derivative, but we don't know it's anti-derivative. And so I claim we're going to use integration by parts here because in this situation, if you take the natural log of dx here, we're going to take the natural log of x as one factor. And as the other factor, we're going to take 1 dx, right? And so as we choose our options u and dv, u has to be chosen so we can take its derivative. And dv has to be chosen so we can take its anti-derivative. We don't know the anti-derivative of the natural log otherwise we would be done, in which case we're kind of forced to put the natural log in as u. And so this right here is what I often like to refer to as a variation of integration by parts, which we call integration by hope, right? So imagine you have Princess Leia there saying, Obi-Wan Kenobi, you're only hope, right? What's the hope, right? We don't know the anti-derivative of the natural log of x, but we do know the derivative of ln of x. And so we have this hope that if I take the derivative of ln x and use integration by parts, I hope this will give me the anti-derivative. And in many situations, we can. And the reason it's going to work here is when the natural log of x, one of the reasons we don't know how to take its anti-derivative is because it's a strength of the general function. But when you take the derivative of the natural log of x, you get the algebraic function 1 over x. And over here, if we take dv to be dx, its anti-derivative will just be an x. And the combination of this x with the algebraic function 1 over x actually leads to the correct anti-derivative. So let's see how that would look. So by the integration by parts form, we're going to get u times v, which is x, the natural log of x. That part's now done. Moving on to the integral, we're going to get v, which is x. Times du, which is dx over x. Which this is actually quite nice here. You'll notice x over x cancels out. And we actually end up with a dx. So we're now looking at the situation x natural log of x minus the integral of dx. Which if we take the anti-derivative of dx, that will just be an x. We already did that one, right? x natural log of x minus x plus a constant. This then gives us the correct anti-derivative of the natural log of x. This is something you can memorize for future reference. But this idea of the hope, right? We hoped that because we know the derivative of the natural log, we could find the anti-derivative using integration by parts. And our hope, even though our hope was beyond hope, it turned out it was right. We used the anti-derivative. We used the derivative of ln x to find its anti-derivative. Now really the reason it worked so successfully is because the derivative of the natural log was algebraic. So it interacted with the x that popped up very nicely. Let me give you another example where integration by hope actually works very nicely. Take the integral from 0 to 1 of arc tangent of x dx. Likewise, we don't know the anti-derivative of arc tangent, but we can use integration by hope to find it. Because we do know the derivative of arc tangent. Well, hopefully we know that one. Maybe I speak too forwardly. But in calculus one, we've seen the derivative of arc tangent. That's going to be 1 over 1 plus x squared dx. So we do know the derivative of arc tangent and it is also an algebraic function. If we take dv to be dx, then its anti-derivative will be x. And is there a nice compatibility between x and 1 over x squared? We'll see that actually, that is the case. So this is a definite integral. So we're going to get arc tangent, sorry, x times arc tangent. Evaluate from 0 to 1. And then we're going to subtract from this the integral from 0 to 1. x dx over 1 plus x squared. Now for the first piece, we can just kind of keep it around for a while and deal with it later. We can plug in the 1 and 0 right now. That's what I'm going to choose to do this time around. And so if you plug in the 1 into x, you'll just get x, or you'll just get a 1, right? You get arc tangent of 1, which is something we can compute. We'll come back to that in a second. Then when you plug in 0, you get 0 times arc tangent of 0. That's just going to be a 0. So if you do that arithmetic, you're going to end up with the arc tangent of 1. You can consult your calculator if you want to, but be aware that arc tangent of 1. We're looking for the angle from which tangent equals 1. That is when tangent equals 1 and sine and cosine are equal. This is going to happen for theta is a 45 degree angle. But we should do this in radians because we're trying to measure area, not an angle. We're going to do, we've got a pi-force right there. Now for the next part, we have to find an antiderivative of x over 1 plus x squared. How does one do that? We've been kind of neglected in the recent videos here, but u substitution actually works out really nicely here. If we take u to be 1 plus x squared, then du will equal a 2x dx, which if I want the 2, I can just put a 1 half to compensate for that. So this right here gives me my du. That works out really great. And since it's a definite integral, let's change the bounds as well. So we have x equals 1 and 0. So u will equal if you plug in 1 into this formula, you get a 2. And if you plug 0 into that formula, you get a 1. And so let's make that change there. So like we saw before, our tangent of 1 is going to be pi-force. So we took care of that part. Then we're going to subtract 1 half the integral from 1 to 2 of du over u, right? And the antiderivative of 1 over u is going to be the natural log. So we get pi-force minus 1 half times, I guess I don't need a parenthesis there, the natural log of u as u ranges from 1 to 2. Plug those in there. Be aware that the natural log of 1 is 0. The natural log of 2, we've got to keep that around. And so getting rid of the natural log of 1, which is 0, we end up with pi-force minus the natural log of 2 over 2, which we can approximate that with our calculators if we so choose, but we'll be content with this exact answer right here. And so we see another example of this integration by hope. We didn't know the antiderivative of our tangent of x in this situation, but because we know it's derivative and we know it's derivative algebraic, we hope that integration by parts will help us calculate this definite integral and that's exactly what happened. So we've now seen three versions of integration parts. There's the traditional one, which is really good when you have some power of x and you have some other function right here. So you can just knock off powers of x over and over and over again. We saw integration by cycles, which if you do integration by parts enough, you'll eventually cycle back to the original integral you can solve for it. And then this last one is hope, right? Where we don't know the antiderivative, we do know the derivative and we can use that and hope that we'll find the antiderivative eventually if we work through the calculation here.