 Hello, this is a video about integrating using tables and in particular reduction formulas. What a reduction formula is is if you look at any integration formula in the table, if you apply integration formula and the formula itself has an integral within it, that's what is called a reduction formula. So we'll have to use formulas from the tables multiple times for these examples. So in our first example, we're going to find the indefinite integral of x cubed sine of x with respect to x. If you look at the form of this integral, you'll notice that it closely resembles from your table formula 54, which is shown on the screen. We're going to identify what our u is, what n is. Alright, so first n is equal to 3, that's the power on the x, u, which is always your quantity in terms of x is just x itself, which means that the derivative of u with respect to x is 1. So that means du is the same thing as dx, the two quantities are equal to each other. So literally I can take x cubed sine of x dx and I can now expand it using the formula. So what I have here is negative u to the n, so that's what we'll start with, negative x to the third power, cosine of u for cosine of x plus n, which is 3, integral of u to the n minus 1, so that's x to the second, and then cosine of u, which is x, and then du, which we already said is dx. So that's currently where I stand. So now we're happy with the minus x cubed cosine x, we will keep that, that will appear in our final answer. It's now time to look at the 3 times the integral of x squared cosine x dx. So we're looking at the 3, integral of x squared cosine x dx, we now need to find out what formula applies to this integral, and that would be formula 55 from your table. You have u to a power, remember u is always the quantity in terms of x, and then you have a cosine, and that's it. So now we'll identify our n, which is 2, we'll identify our u, which is simply just x. The derivative of u with respect to x is 1, so du equals dx. So this will enable me to use the formula. So as mentioned, we're keeping the minus x cubed cosine x, that will not change, that's actually going to be in our final answer as it is. But now I have a plus 3 times whatever I get when I integrate x squared cosine x. Well when I integrate x squared cosine x, you have u to the n, which is x to the second, sine of u, or sine of x, minus n, or 2, integral of u to the n minus 1, or x to the first, sine of x dx. We're almost there. Last thing I'm going to do here is I'm actually going to distribute that 3 into the brackets. So I'll end up with a 3x squared sine x minus 6 times the integral x sine x dx. So now we're focusing on the 6 times the x sine x dx, the integral of that. So we now know two groupings that will be in our final answer. We now have one more integral to evaluate. So now time to look for the x sine x dx among our integration formulas, that would be formula 52. You have a quantity in terms of x to the first power, and then you have sine. So all I have to do literally here is identify what u is going to be, which is x, the derivative of u with respect to x is 1, and yes, du equals dx. So we'll keep our first two groupings as they are. That is final answer material, and then it's time to integrate the x sine x, there's a minus 6 out front. So it looks like the integral is going to be sine x minus x cosine x. So sine x minus x cosine x. So this means that I now have negative x cubed cosine x plus 3x squared sine x, distributing the minus 6 gives you a minus 6 sine x, and then you'll get a plus 6x cosine x. And don't forget, since this is the final answer and it's an indefinite integral, you have that constant of integration that c. So there you have it, that is your final answer. And there are other integration methods that would also allow you to integrate x cubed sine x, but this is just how they use the reduction formulas from the integration table. Let's do another one, where let's integrate the square root of 3 minus 5x over 2x. Before I even begin looking for any sort of formulas, I first want to take that 2 on the bottom and bring it out front of the integral, so it becomes a half. So I'm integrating square root of 3 minus 5x over x, I'm integrating with respect to x. So you can look at formulas, but you'll notice formula number 19 is the winner, as it's shown on the screen. Your a, which is your constant term is 3. Your b, which is the coefficient of the x under the radical is negative 5 plus b plus negative 5, so b is negative 5, then u in this case is x, which means du is equal to dx. So now I can actually apply this integration formula, this is formula number 19. So I have a half, and then whatever the integral is, so it's 2, square root of a plus b u, that's 3 minus 5x plus a, which is 3, integral of 1 over x, square root of a plus b u again, or 3 minus 5x, and then you have your dx, because remember du is dx. So before I go further though, I do want to go ahead and distribute my half into the brackets, so that'll then leave me with a half times 2, which is 1, so you're left with square root of 3 minus 5x plus 3 over 2, integral of 1 over x, square root of 3 minus 5x dx. So I like the square root of 3 minus 5x, but I do have another integral, which I will have to evaluate. So we are now going to look at that integral that we're trying to evaluate, and we look for the format or what formula it'll go with, and it will be formula number 17. So I need to identify my u, my a, and my b. So let's do that first, a is going to be 3, and b is going to be negative 5, and u is going to be x, which means du equals dx. Now 17 does have two branches to the formula, I'm using the branch where a is greater than 0 because a is 3. So I think we're finally ready to apply the integration formula, so I have square root of 3 minus 5x plus 3 halves, open brackets, apply the integration formula, 1 over the square root of a, that's 1 over radical 3, natural log, open absolute value, you have square root of a plus b, which is 3 minus 5x minus square root of 3, which is a, over square root of 3 minus 5x again, a plus b, plus square root of 3, close the absolute value, final answer you can then distribute the 3 halves into the brackets, and you'll literally get square root of 3 minus 5x plus 3 over 2 radical 3, natural log, absolute value, and then write what's inside the natural log, the square root of 3 minus 5x minus square root of 3, and the square root of 3 minus 5x plus square root of 3. Once this is the final answer, don't forget to do that plus c at the end, because it is an indefinite integral. So that is using reduction formulas from the integration formula table. Thank you for watching.