WEBVTT mathematics/calculus-ii/murray
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Hi there, welcome to educator.com. This is a lecture on integration by parts.
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The main equation for integration by parts is right here.
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The integral of U dV is equal to UV minus the integral of V dU.
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Where this comes from is the product rule in reverse.
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The product rule is something you learned in Calculus 1, and is a way to take derivatives of products of functions.
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This is changing around the product rule and using it as an integration formula.
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The point of integration by parts is that you will be given a hard integral to solve.
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What you are going to do, is take the integral that you are given and split it up into two parts, a U part and a dV part.
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Then you will invoke this formula to convert it into UV minus the integral of VdU, and if you do that right, then the second integral that you get will be an easier integral.
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Then, you can finish the problem by doing that easier integral.
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That is the idea of integration of parts, but of course the best way to learn it is to do lots of examples.
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Let us go ahead and do some examples.
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Here is the first example, a very typical integration by parts problem.
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We are trying to integrate X sin(X) dX.
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Remember, the first part is to split this integral up into U and dV and we are going to let U be just X and dV be sin(X) dX.
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You always put the dX with the dV part.
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Then, we are going to figure out dU and V because those are both parts of the formula before.
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dU, if U is X, dU is just dX and V if dV is sin(x), V is the integral of sin(x).
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The integral of sin(X) is negative cos(X), and remember the integration by parts formula with the integral of U dV is equal to uV minus the integral of VdU.
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Now, the integral that we are given, because we have converted it using our substitution, that is now the integral of U dV.
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Using the integration by parts formula, that converts into UV while UV is minus X cos(x), minus the integral of VdU.
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So, minus the integral of V dU, that is minus cosin(x) dX.
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OK, I am going to cancel these two negative signs.
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Now we have minus X cosin(x).
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Now, this new integral you can see is just cosin(x).
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That is a much easier integral to deal with than we started with.
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The integral of cosin(x) is just sin(x) and I am going to add on a constant just because you always have a constant for an indefinite integral.
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Then, we have our answer is negative X cosin(x) plus sin(x) plus a constant.
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Let us try a trickier example.
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X² e^3x dX.
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Again, we are going to divide it up into a U and a dV.
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We will let U equal X² and dV be e^3x dX.
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Again, we have got to figure out dU and V, so dU, if U is X² is 2x dX.
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v is the integral of e^3x dX.
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That is one third e^3x.
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Then, I am going to write down the integration by parts formula again.
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The integral of U dV is UV minus the integral of V dU and we have taken our example integral and we have split it up in to U dV again.
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Invoking the formula again, that is UV, so that is 1/3 x² e^3x minus the integral of VdU.
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VdU is, there is a 1/3 and there is a 2, I will combine those as 2/3, and on the outside is x^3x dx.
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Now, what we have here is another integral.
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It is easier than the first one because it has an x instead of an x squared.
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However it is still not an integral that we can do directly.
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What we have to do is integration by parts again.
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This is a very common issue with integration by parts.
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We are going to do integration by parts again on this new integral.
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I will let U equal X dV equal e^3x dX, and again fill in dU equals dX and V is 1/3 e^3x
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We still have that first term minus 2/3.
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Now we have the integral of U dV, so again using the integration by parts formula, that is UV, the new U and the new V.
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So, 1/3 x e^3x minus the integral of V dU, minus the integral of 1/3 e^3x dX.
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I am just going to focus on the stuff on the right.
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This is minus 2/9 X e^3x and then the 2 minuses give you a plus 2/9.
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Now the integral of e^3x is 1/3 e^3x.
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If we put all of those parts together we get 1/3 x² e^3x.
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Minus 2/9 e^3x plus 2/27 e^3x.
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2/27 X e^3x plus a constant, and that is the answer.
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The moral of that example is that sometimes you have to do integration by parts twice in the same problem.
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First we had to do integration by parts to reduce the original x² down to an easier integral that just had an X in it.
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But that still was not an integral that we could do directly.
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We had to do integration by parts again to reduce the X, well actually to make the x go away, and give us an integral that we could do directly.
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That is a pretty common story with integration by parts, that you have to do it twice.
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I want to teach you a secret short cut to doing integration by parts problems.
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This is just kind of a book-keeping device, but it can help you do some of these problems really quickly.
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It is called tabular integration and I am going to introduce it with an example.
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I am going to redo the same problem that I just did.
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Remember that problem was x² e^3x dx.
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Here is the secret shortcut.
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What you do is write x² e^3x, and you make a little table here.
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On the left hand side you write down derivatives.
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The derivative of x² is 2x, the derivative of that is 2, and then the derivative of a constant is just 0.
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On the right hand side, so those were all derivatives, you take integrals.
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The integral of e^3x is 1/3 e^3x.
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The integral of that is 1/9 e^3x.
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The integral of that is 1/27 e^3x.
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Then this is just a clever little trick, but say this time we were doing the problem the previous way.
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You draw these little diagonal lines and then you put little positive and negative signs on the diagonal lines alternating plus, minus, plus.
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Then, what you do is multiply along these diagonal lines.
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Those give you the terms of your answer.
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So, you get x² times 1/3 e^3x.
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Now, the next diagonal line is minus 2x/9 e^3x.
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The next diagonal line is plus 2/27 e^3x.
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You attach a constant at the end and there is your answer
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That is just a clever book-keeping way of suppressing all the grunt work of going through the U and dV stuff.
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It works really fast for certain kinds of problems.
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If you have a polynomial like x², times something like e^3x or cos(x) or sin(x) where it is easy to take integrals, then this tabular integration trick works really nicely.
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I want to do a more complicated example, where we have e^x cosin(x) dX.
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Again, this is one where we can not do the previous shortcut, the tabular integration idea, because we do not have a polynomial.
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If we took derivatives of either one of these, e^x or cosin(x), we would never get down to 0.
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The shortcut is not available here.
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But, again, we are going to start with integration by parts.
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Let U equal e^3x, or sorry, e^x, and dV equal cosin(x) dX, and fill in dU is e^x dX.
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A lot of students leave off the dX when they are doing these problems.
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It is really important to include the dX.
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It makes all the notation work out and helps you to track where you are going late on, so do include the dX.
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If dv is cosin(x) dX, then V is the integral of cosin(x), which is just sin(x).
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This integral, again using our formula UV minus the integral of V dU, is UV.
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So, e^x sin(x) minus the integral of V dU.
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That is e^x sin(x) dX.
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That gives us a new integral e^x sin(x) and we are going to do integration by parts again.
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We are agian going to let U equal e^x.
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This time dv is going to be sin(x) dx, and dU is e^x dX.
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V is the integral of sine which is negative cosin(x).
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Invoking our integration by parts formula again, this is UV.
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Negative e^x cosin(x) minus the integral of V dU, that is negative cosin(x).
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I will combine the negatives and that becomes positive there.
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e^x cosin(x) dx.
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I am just going to get rid of the brackets and bring the negative sign through so we get e^x times sin(x) plus e^x cosin(x).
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Minus the integral of e ^x cosine(x) dx.
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Here, we have got a strange thing happening because what we did was integration by parts twice.
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If you look at this integral that we ended up with, which was supposed to be getting easier, it is exactly the same as the integral that we started with.
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That seems like we are spending a lot of time to just go around in circles, because we have done a lot of work and we have come up with the integral that we started with.
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In fact, we can use this to finish the problem.
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The way we do that is we let i be this integral that we started with.
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That means that i is equal to all these calculations that we did.
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Then this integral that we get at the end is i again.
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So what we can do is we can move i over to the other side of the equation.
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What we get is 2I is equal to e^x sin(x) plus e^x cosin(x).
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That means we can solve for i.
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I is just one half the quantity e^x sin(x) plus e^x cosin(x).
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And then of course we have to add on a constant, as always.
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All of a sudden, even though it looked like we were going around in circles, we have solved our integral.
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That is the answer right there.
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So that is another common pattern that can happen with integration by parts.
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It happens a lot when you have combinations of e^x and cosin(x).
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You have e^ax, e to some number x, and then either sine, or cosine, of some number times x.
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You can do this trick of doing integration by parts twice and then you get an expression that you can solve for your original integral.
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I want to show you a little memory trick that can help a lot with integration by parts.
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The difficult part with integration by parts is that you will be given an integral of a whole lot of stuff.
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What you have to do is decide how to break up that integral into a U part, and a dV part.
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That is what make integration by parts tricky, deciding what to make U, and what to make dV.
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The idea is that after you go through the integration by parts formula, you want to get an integral that is easier than the one you started with, not harder.
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That can often be difficult to predict ahead of time.
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There is this little mnemonic that can help you remember how to split up integrals that way.
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Just remember these five letters. L.I.A.T.E., lee-ah-tay.
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What those stand for is the functions that you should use for U, if you have them.
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Let U be the following functions if you have them.
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First of all L stands for ln(x).
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If you see a ln(x), that is your U.
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I stands for inverse functions.
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If you see an inverse trigonometric function, for example arcsine, arctan, and those are also written as sine inverse and tan inverse.
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If you see those, you know that is going to be your U.
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A stands for algebra.
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If you see something like x or x², that is going to be your U.
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T stands for trigonometry.
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Something like sin(x) or cosin(x).
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E stands for exponential, so e^x.
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Work through these functions in order, and whichever one of these you see first, that is going to be your U.
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That is a very effective way of solving integration by parts problems.
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As you work through your homework and try this out on different problems, keep this in mind and try it out.
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I hope it works out for you.
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So, that is the end of the first lecture from educator.com on integration by parts.