 In this video, we provide the solution to question number 10 for practice exam number one for math 1220, in which case we have to evaluate the indefinite integral e to the two theta times sine of three theta d theta. Now the fact that I see this exponential times a sine function, this makes me think I want to use integration by parts, more specifically integration by cycles, because as you take derivatives of e to the two theta or antiderivatives of e to the two theta, you always give back e to the two theta with a constant multiple that is changing. Same thing with sine, when you take the derivative of sine or the antiderivative of sine, you're always going to give back a constant multiple of cosine, for which if you do that again, you'll get a constant multiple of sine. And so if I do integration by parts enough times, I'm going to cycle back to the original expression. So that's going to give me this integration by cycles, okay? So for my first iteration, it doesn't really matter which one you do, just think, just be consistent about it. I'm going to take u to be e to the two theta. That means du is going to equal two e to the two theta d theta. Then for dv, I'm going to take that to be sine of three theta d theta. And that means v is going to equal one third, well actually negative one third cosine of three theta. So if we apply that, I'm also going to use the simplification that the original integral I'm just going to refer to as capital I, because I'm going to need that later on. So I is equal to, if we apply integration by parts, we're going to get negative one third e to the two theta cosine of three theta. Then we're going to subtract from that the integral of v du. So we get a negative one third cosine of three theta. That's the v times by the du, which is two e to the two theta d theta, like so. We of course can simplify that. We can make that a plus negative there, plus plus. Let me just erase that negative sign, make it better like that. And then there's these constant coefficients in there. You don't necessarily have to do too much at once. Negative one third e to the two theta cosine of three theta. Because there's this one third right here. There's this two. I'm going to bring out the two thirds out in front. And then I'm also going to put the e back in front. And so this is what we predicted. We're going to get something that looks like e to the two theta times cosine of three theta. So I'm going to do integration by parts again. That's how we do it with cycles. You have to do it at least twice. Well I guess maybe you could get away with it once, but not in this situation. But make sure you do integration by parts, you're consistent. If you was your exponential beforehand, you're going to do that again. So two e to the two theta d theta. And then for dv, the trig function is going to serve that role this time, cosine of three theta d theta. And so then v is going to equal one third sine of three theta, like so. So putting those things in there, you're going to get i equals negative one third e to the two theta cosine of three theta. Then for this next one, do remember that the two thirds applies to everything coming up just now. We're going to take u times v, so we're going to get one third e to the two theta sine of three theta. And then we should get an integral that resembles the original one we're looking for. We're going to get a minus integral of one third sine of three theta times cosine times two times e to the two theta d theta, like so. So up to a constant multiple, I want to be aware that that integral right there is in fact the original integral. So let's write that one more time. i equals negative one third e to the two theta cosine of three theta. Now if I distribute this two thirds throughout, you're going to get two thirds times a third. So we're going to get two ninths times e to the two theta sine of three theta. Likewise though, here we have a two times a third, so that's a two thirds. So then when we distribute that, we're going to get negative four ninths. And then once we take the coefficients out, we have an e to the two theta sine of three theta. That's the original integral again. So that's going to give you this i right there. So what I want to do is then add to both sides of the equation this four ninths i. Upon doing so on the other side, we have a nine ninths. You add a four ninths to it. That's going to give us a 13 ninths i equal to then the negative one third e to the two theta cosine of three theta. Then we get this two ninths e to the two theta sine of three theta plus a constant. And so then we're going to times both sides of the equation by the reciprocal of this coefficient. We can times both sides by nine thirteenths. We have to do this on the right hand side as well, nine thirteenths. And so then in the end our integral i is equal to what we distribute the nine thirteenths on the other side. You're going to end up with a negative three thirteenths e to the two theta cosine of three theta. On the next one, the ninths cancel out and you end up with a two thirteenths e to the two theta sine of three theta. And then you do times like arbitrary constant by a specific constant. That's just going to give you back a constant again. So I'm just going to still write that as a plus c. And if you want to switch the order, write the sine before the cosine, you can do so. But that then gives us the anti-derivative we're looking for. And we accomplish this using integration by cycles.