 This lesson is on integration by parts. It is your second technique of integrating. The first one, of course, being UDU substitution, and you have had plenty of practice with that. This one is very interesting because it's a takeoff from product rule of derivatives. So let's go on and derive our formula. The product rule. Well, if we have two functions, and one function is called U and the other is called V, then the product rule for this is written as the derivative of U times V. Now you're wondering why I'm calling them U and V instead of f of x and g of x, primarily because that's what most books call them, when you are doing your integration by parts and developing that. So remember, U is normally your f of x and V is normally your g of x. We want the derivative of U times V, of course, in x's. Well, that's going to be equal to U times the derivative of V plus V times the derivative of U. Now what happens when we want the anti-derivative or when we want to integrate this? Well, we just put the integral symbols up. So now we have the integral of the derivative. Well, an integral of a derivative, we know, is whatever is inside of that. So that's U times V. And that's equal to the integral of U dV plus the integral of V dU. Switching around a little bit and solving for U dV, we get the integral of U dV is equal to the U times the V minus the integral of V dU. This is our basic formula for integration by parts. The integral of some function U times the derivative of another function is equal to the function U times the anti-derivative of the other function minus the integral of the function V times the derivative of U. So let's go on and do our first example. Integrate xE to the x. Well, we're going to look for a function U and a derivative of dV. What we look for is to make our integral get to a point where we can just integrate it with the formulas we know. If we set up U to be x, then dU is equal to dx. That makes dV equal to e to the x dx. So we have our U, which is our x, our dV e to the x dx, the rest of it. So this makes our V equal to e to the x. And of course there's a plus c involved in all of this, but we are going to ignore the plus c's because there can be many of them until the end and just stick it in. So this integral now becomes U times V, which is x e to the x minus the integral of V dU, which is e to the x dx. And lo and behold, by this miracle of integrating by parts, we have a function that we can easily integrate, which is e to the x. And that becomes x e to the x minus e to the x plus we can add in our constant. Again, going back, we picked U to be x. We picked dV to be e to the x. Well, how do we choose these? Most of the time dU is simpler than U. If we choose x squared to be U, its dU is 2x. Secondly, whatever you select for dV, you should be able to find the antiderivative V. You cannot choose dV to be ln of x right now because you do not know what the antiderivative of ln of x is. And then V should be simpler than dV most of the time. Let's look at a problem that does some of this. Evaluate the integral x ln of x dx. If we pick U to be x and dV to be ln of x, I have just said we don't know the antiderivative of ln of x. So we are forced to pick U to be ln of x. And that's okay. So we say U is equal to ln of x. That makes dU equal to 1 over x dx. Hence dV becomes x and V becomes x squared over 2. When we put these together, we'll have U times V. So this will be equal to x squared over 2 ln of x minus V x squared over 2 times dU, which is 1 over x dx. The integral simplifies to 1 half x dx. That's an easy integral to take. Let's go on. x squared over 2 ln of x minus 1 half x squared over 2. So this is 1 fourth x squared plus our constant. So in this case, we had to choose the ln of x for our U because we could not take its antiderivative. So let's go on. This is where we have dV is equal to dx. So evaluate the integral arc sine x dx. The only choice we have for U is the arc sine of x. That makes dU equal to 1 over the square root of 1 minus x squared dx. That makes dV equal to dx and V is equal to x. Again, we were forced to take U to be our arc sine. Let's see how this works. This becomes U times V, so it's x arc sine of x minus the integral of V dU. So that becomes x over the square root of 1 minus x squared dx. Well, we have to look at this again with a substitution, a U dU substitution. So I can use other variables or I can go on and say, well, I'm going to use a U sub 1 and a V sub 1 for my substitution here. So I think I'll just use the U sub 1 is equal to 1 minus x squared. That makes dU sub 1 is equal to negative 2x. So we have to fill in negative 2 here, which means plus 1 half goes outside. So now we have the integral, 1 half integral. U is the 1 minus x squared. The dU is negative 2x dx, so this is dU over the square root of U, which is 1 half integral of 1 over the square root of U is the square root of U times 2. So these cross out and we get a final answer of x arc sine of x plus the square root of U, which is 1 minus x squared plus a constant. Let's go on. Let's do a problem with more than one iteration. We can see that we have an x squared on this one, because we're going to evaluate the integral x squared sine 3x dx. If we do two iterations of derivatives, that will be out and we will only be left with some sort of trig function, either sine or cosine. So we will need to take U to be x squared dU to be 2x dx. That makes dV equal to sine 3x dx and V equals negative 1 third cosine 3x plus some constant. So let's put that into the formula. U times V, so we have negative x squared over 3 cosine 3x minus V. Well, we'll have plus 1 third cosine 3x times dU, which is 2x dx. So we'll put a 2 there and an x dx here. So we still have an x sitting in our integral. We need to undo that one. So let's go back and call all these U's sub 1's and dV's sub 1's and V's sub 1's and then go on and do another iteration where we go U sub 2 is equal to x. That makes dU sub 2 is equal to dx. dV sub 2 is equal to cosine 3x dx. V sub 2 is equal to 1 third sine 3x. Putting all this together, we will have negative x squared over 3 cosine 3x plus 2 thirds U times V. U is x and V is 1 third sine 3x. So it's 2 thirds times x over 3 sine 3x minus the integral of V dU, which is 1 third sine 3x dx. Integrating the sine 3x dx, we get negative 1 third cosine 3x. So let's put this into our formula and multiply out so we get negative x squared over 3 cosine 3x plus 2 ninths x sine 3x plus 2 over 27 cosine 3x plus C. Let's go on to another kind of integration by parts, one that is cyclical. These are very interesting because they come to a full circle. In this case, you can choose U to be E to the x and dU to be E to the x dx, or you could do it the other way. It doesn't matter which way you do it because you will come back to the same result either way. So let's continue on this way. If I choose U to be E to the x and dU to be E to the x dx, then I have to make dV equal to sine x dx and V is equal to negative cosine x. So let's put that into our formula, U times V, which is negative E to the x cosine x minus V dU. V is negative cosine x, so that would make that a plus, cosine x, and dU is E to the x dx. Well, we have to do this again because we don't know a formula for E to the x cosine x. So let's call these all ones and set this up again. And usually we'll set it up in the same way. U sub 2 is equal to E to the x dU sub 2 is equal to E to the x dx. That makes dV sub 2 equal to cosine x dx and V sub 2 equal to sine x. So now we have negative E to the x cosine x plus U times V, which is E to the x sine x minus the integral of V dU. Well, this one is E to the x sine x dx. Well, lo and behold, we have E to the x sine x under an integral here. This is what we started off with there. If we could pull this one to the other side, which we can, we will create two integrals, E to the x sine x, and that will be equal to E to the x times sine x minus cosine x plus some constant. And now we can divide both sides by two and get some final answer where the two is thrown into our constant of E to the x times sine x minus cosine x all over two. Cyclical kind. You don't see many of these because they have to involve functions that are cyclical in nature in their derivatives and antiderivatives. E to the x and sine or E to the x and cosine are the only three that seem to crop up all the time. Let's go on. There are some other types of integration that you have been working on in the past but are formally presented in this section in your book. And the first one is the integral of sine squared x dx or integral of cosine squared x dx. And as you may have picked up in previous lessons, you use the half angle formula to develop these. So that is equal to the integral and the half angle formula for sine squared is one half times one minus cosine 2x dx. And the one for the cosine squared is equal to integral one half one plus cosine 2x dx. And just doing antiderivatives is very simple on this. Now we have equals one half x and then the antiderivative of cosine is a positive sign. So this is negative one half sine 2x and of course plus rc. The other one is equal to one half again x, again the same antiderivative but with a plus one half sine 2x plus c. Very interesting ones used a lot and probably will be on an AP exam in BC calculus. Let's go on to another type. This next problem involves a Udu substitution and should have been introduced earlier in your book as it is in most other books. What it is is the integral of y times the square root of 1 plus y dy. Obviously we have to make a substitution of u to be 1 plus y so let's do that. This seems to be the only way to go. That makes du equal to dy. We need to do one more thing which is solve for y so we can substitute for that y there. So y becomes u minus 1. Now making these substitutions we get u minus 1 times the square root of u and since du and dy are the same, du. Multiply out and we get u to the one half times u is equal to u to the three halves minus u to the one half du. Integrating this we get u to the five halves times two-fifths minus u to the three halves times two-thirds plus some constant. Go on and substitute back in our 1 plus y for u and we get two-fifths times 1 plus y to the five halves minus two-thirds 1 plus y to the three halves plus a constant. And if you want to factor out you certainly may and it will leave you 1 plus y to the three halves times two-fifths times 1 plus y minus two-thirds plus a constant. And if you want to go further and put the two-fifths and the two-thirds together that's fine too but I'm going to just leave it at this. That concludes your lesson on integration by parts and of course some other interesting integrations.