Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Integration by parts formula. The integral UDV is equal to UV minus the integral VDU. This derives from the product rule. DDX of UV is just the first term times the derivative of the second plus the second term times the derivative of the 1st. So if we thought about just writing it as a differential, we'd have DUV equaling UDV plus VDU. Then if we took the integral of each of those pieces in terms of DX, we'd have the integral D DX of UV DX, which is just UV. By the fundamental theorem of calculus, the integral and the derivative are going to undo each other. The next piece. It's a function U, but it was in terms of DV. So we don't know how to evaluate that. So that really is just the integral UDV plus the next term V In terms of DU. Once again, those are two different functions, so we can't integrate those. Now, if we solve for the integral of U DV, we can see that the integral UDV is really just equal to UV minus the integral V DU. The goal of the integration by parts is to take something that's complex and to split it up into things that are more manageable that we know how to integrate. So where we might not be able to integrate UDV, we might be able to integrate VDU. That's our goal. Generally, you're going to choose DV first to be as much of the integrand as you can easily integrate. Then U is whatever is leftover. Also, in general, choose for you in this order logarithms, then algebra, then trig, then exponents, algebra being polynomials, DV in the opposite order, exponents, trig, algebra, and logs. That's kind of a general rough rule. It won't always work. What we really, really want to do is we want to figure out how much of the integrand we can easily integrate. And usually we can integrate an exponent pretty readily or a trig, or an algebra logarithm. So let's look at an example. If we have the integral of Theta, cosine, Pi, Theta, D, Theta, what we're going to do is we're going to let our DV be whatever we can easily integrate. So if I let DV be this cosine Pi Theta D Theta, then when I take the integral of that I get V equaling 1 / π sine Pi Theta. We're going to leave the plus C off until the very last step. So if that's our DV, we easily integrated that. Then whatever's left is our U. So our U is just Theta in this case. So our du is D Theta. So the rule says it's U * v So Theta times 1 / π sine Pi Theta minus vdu. So 1 / π sine Pi Theta DUD Theta. Well, we can easily integrate this now because we know that the integral of sine is really just negative cosine. Take the derivative of this Pi, Theta and divide. So we get +1 / π ^2 cosine Pi, Theta, and then it's going to be a + C because now we've evaluated all the integrals. So the original integral of Theta, cosine, Pi Theta, D, Theta by the integration of parts, figuring out our DV being the thing that's the most complex that we can evaluate. The integral of everything that was left is our U and then using our formula UV minus the integral vdu evaluate it, and this is what it's equivalent to. Let's look at another one. If we have the integral of lane X + X ^2 DX, we're going to let there's not much there that we can easily integrate. There's only really two things that are being multiplied. So we're going to let our DV be DX. That's the easiest thing there that we can easily and readily integrate. So if DV is our DX, we know v = X. That leaves everything left to be our U. So our U is going to be lane X + X ^2. Well, we do know how to find the derivative of this DU is going to equal. The derivative of a natural log is one over. So X + X ^2 is going to go in the denominator. The derivative of the X + X ^2 by the chain rule is going to be 1 + 2 X and that's going to go in the numerator and that's going to be times the DX. So now when we use our formula, it's going to be U * v. So this XL N X + X ^2 minus the integral of VDU. So here's my X, here's my 1 + 2 X. And the denominator, I'm going to factor out an X so that these two X's can cancel 1 + X DX. So simplifying it, we're going to get XL N X + X ^2 minus. Now we're going to do some fancy footwork here. I have this 2X plus one, but if I had two X + 2, I could factor out A2 and this would cancel. So we're going to add 1. But if we add one, in order to keep the equation balanced, we're going to subtract 1. So we're going to have two X + 2 -, 1, which is really the same thing as 1 + 2 X. Now I'm going to split this up into two separate terms, the two X + 2 being over the denominator and then the -1 being over the denominator. I also see that I have a negative in front, so we have negative factoring out the 2. The X plus ones will cancel. So we end up with negative, the integral 2 DX plus. Here. The negative of the negative made it a +1 / 1 plus XDX. Well, we can evaluate both these integrals. This integral is really just two X and this one is just our natural log of whatever was in the denominator because the derivative of that denominator is just one. So we have plus the natural log of 1 + X. We remember absolute values because we have to have it of a positive natural log, so plus C So this integral of natural log of X + X ^2 DX, which we don't know how to solve, we can do some manipulation and some algebra to come up with a solution where we can actually solve it. The next example, we want to use X ^3 sine X DX so we can easily evaluate sine X DX. If we have sine X DX and we take the integral of it, we get negative cosine X. That leaves whatever was left as X ^3 and we get 3X squared DX. So our U times RV minus our integral of VDU. Well, that simplified it down, but this still is going to be complex and we can't evaluate it. So we're going to do the same thing again. We're going to let our DV be here's a negative and a negative. So those are going to turn positive. So we're going to let our DV be cosine XDX. We're going to let our U be 3 X squared. So when we take the integral, we get sine X and when we take the derivative we get 6 XDX. So taking just the second piece down here, we have EU * v right here, minus the integral of VDUVDU. Well, we're going to evaluate it again. So now when we look at this, we're going to have the integral here. We're going to let our DV be the sine XDX. We're going to let our U be 6X. And I'm just going to leave the negative out in front for a minute. So then the integral here is negative cosine X. The derivative is 6 DX. So we're going to have this negative X ^3 cosine X + 3 X squared sine X -, U * v. So minus a negative makes it a positive 6X cosine X minus because of this minus here the integral of VDU, so another negative cosine XDX. But the formula says it's minus a negative. So it really is going to end up with three negatives there, the minus, the negative, and then the one that came from the V. So then when we distribute there, we get this and now we can evaluate the six cosine XDX. So all of that gives us a final solution. Now, we had to do that integration by parts three different times, and there's actually a shortcut for this method, and it's called tabular integration. So looking at this exact same problem again, we're going to look at tabular integration. We're going to take the integral of X ^3 sine XDX, and we're going to figure out one piece where we're going to take the derivative. Now the big key here is to figure out what derivative will eventually go to 0, if possible. So we always want our F of X to be a polynomial if we can. So X ^3, 3, X squared, 6X6, and 0, then we're going to take whatever was left and we're going to take the subsequent integrals. So if I have sine X, then that's negative cosine X, negative sine X, cosine X, sine X. Now, in order for this to work, we're going to do diagonals here, and we're going to change the sine. And the reason we're changing the sine is if you think about the formula, every other time that we do this, the sign would change. So here's our first substitution. But then if we did the substitution again, we would have it being minus the quantity, let's call it U1V1 minus V1 DU 1. So there's the minus there, which would make that a minus, but then the minus of the negative would make it a plus. So the signs are going to change every other line. X ^3 times Negative cosine X or negative X ^3 cosine X3 X squared. Negative of a negative makes it a + 3 X squared. Sine X6X cosine X was a positive -6 sine X + C if we look back right. Here's our answer. Remember, if we highlight it for just a second and control copy, if we look back and paste it down here, you can see that those are the exact same answers, but one was much quicker to get to by just doing this pattern. Let's look at another example of this. If I had X ^3 e XDX X ^3 3 X squared, 6X60 E to the X is going to always be E to the X. So we're going to do it in a diagonal positive negative, positive negative. So X ^3 e to the X -, 3 X squared E to the X + 6 XE to the X - 6 E to the X + C. We could have done the substitution. It would have taken us three times on this one. Also, you can usually figure out how many times based on the exponent on the polynomial. Looking at another one t ^2 e to the four TDT, so t ^2, then 2T2 and 0 E to the 4T. When I integrate, we're going to take the derivative of the exponent and divide so 1/4 E to the 4T and then the derivative of the exponent again and divide 116th E to the 4T and then once again 164th E to the 4T. So the first time is always positive, then negative, then positive. So one 4th t ^2 e to the 4T, then -1 sixth one eighth two times 116th is 1/8 TE to the 4T, then plus 132nd E to the 4T plus C. You could practice integration by parts doing U and VU and DV here Also, let's look at one that we're going to actually have to do a use substitution before we can even get it to the point where we can do the integration by parts. So when we look at this original problem, it's pretty big and complex. Without knowing quite what to do, I'm going to factor out an X in the bottom. And if I could get it to be like an inverse trig function or have the derivative of the denominator somewhere or something, that would be a good thing. So we're going to pull out the X. Well, here's a one plus something quantity squared. That kind of might be something of interest. So if we let U ^2 be this, we'd have U being 2 LNX. So DU is going to be 2 * 1 over XDX. Well here's my 1 / X and here's my DX. So I could take the two to the other side and have 1/2 DU equaling 1X1 over XDX. So I've taken care of this part and I have 1 + U ^2 down here. I still have this lane X that I've got to get rid of. Well, if we knew that U was equaling 2 LNX, then we could think of LNX being 1/2 U. So I'm going to take out this LNX and I'm going to have another 1/2 times EU. Remember I need to leave the variables inside the integrals, but the constants can go out in front. So now I have 1/4 integral U / 1 + U ^2 DU. So now if we let our DV, if we let V, we're going to do another U sub. Although we're using AV sub. If we let V equal 1 + U ^2, then DV would be two UDU or 1/2 DV would be the UDU. So here's my UDU. So I'm going to put in that place 1/2 DV. So now we have 1/8 of 1 / V DV. Well this is just our natural logarithm integral. So 1/8 natural log of the absolute value of v + C So 1/8 of the natural log. Instead of V, we're going to use 1 + U ^2 + C, and then instead of U, we get 1/8 natural log one plus the quantity 2 LNX squared plus C. So this problem looked like I was going to have to use my integration by parts. But once I started a long the route of substitution, it actually came up with something that I could easily integrate and I didn't have to use integration by parts. So there are going to be some problems that are being put in that you could use integration by parts, but maybe there's an easier way. Maybe you don't need to. Looking at this next one, this one's going to have E to the negative Y and cosine YDY. Neither one of those are ever going to have a derivative that goes to 0. So we're going to let our DV be E to the negative YDY, and we're going to let our U be cosine Y. Now, we could have done this in a lot of different ways. We could have let our DV be cosine YDY. But if we look at it in this fashion for right now, we'd have our DU being negative sine YDY and our V Remember we take the derivative of the exponent and divide. So we'd get negative E to the negative Y. So if we rewrite this as the original problem, the original integrand equaling U * v minus the integral vdu. So minus a negative times a negative is a positive. So there were three U's there. So now we have a new integral right here that we're not sure what to do with. So once again, if I let the DV be E to the negative YDY and EU be sine Y, when I take the du I get cosine YDY, and when I figure out the V, it's negative E to the negative Y. So we're going to have this negative E negative Y cosine Y being the same minus U * v. So minus a negative makes it a positive E to the negative Y sine Y. Then it's going to be a -, a negative VDU. So our VDU here, integral negative E to the negative Y cosine YDY. Now the big piece here is that you actually have the same term now on the left and the right. So we're going to take the term on the right and add it to the term on the left, which makes 2 of them. Well, we were trying to solve for one of them, so now all we have to do is divide by two. If I take each side, divide by two, and then I'm going to add my constant C at the end. Thank you and have a wonderful day.