Find the indefinite integral.∫θ2cos3θdθ\int_{}^{}\theta^2\cos3\theta d\theta

θ 2 3 sin ⁡ 3 θ + 2 θ 9 cos ⁡ 3 θ − 2 27 sin ⁡ 3 θ + C \frac{\theta^2}{3}\sin3\theta+\frac{2\theta}{9}\cos3\theta-\frac{2}{27}\sin3\theta+C

Find the indefinite integral. ∫ θ 2 cos 3 θ d θ \int_{}^{}\theta^2\cos3\theta d\theta

this problem, we are asked to find the indefinite integral of this function right here. So we're integrating theta squared times the cosine of three theta, all with respect to theta. Now what I can see here is that I have the product of two functions. And so if when I have these cases, sometimes what you can do is use a variable substitution to solve the problem. If I tried doing a u substitution right here on the three theta, it turns out that's not actually going to get rid of the theta squared out front. If I do a u substitution right here, well, it's not going to allow me to clear anything with the three theta inside the cosine. So since my variable substitution is not going to work, it's going to be best if I solve this using integration by parts. So So I know integration by parts process looks like this. We have the integral of u d v is equal to u v minus the integral of v d u. Now when using this formula, my first step is to figure out what my u is. U is something that needs to become more simple when we take its derivative. Now looking at this, I can see that we have this theta squared, and we have this cosine. I know that if I'm dealing with trig functions like sine and cosine, I'm just gonna get those going back and forth when I take the derivative. Because I know that taking the derivative of cosine will give me negative sine. Then if I took the derivative of that, we would get negative cosine because we have that negative sine out in front. Then we just keep going back and forth between sine and cosine. I don't think that's gonna be a lot more simple. But what I do know is that if I took the derivative of theta squared, well, that would give me two theta if I was with respect to theta. That is more simple since we're going to theta having an exponent of two to not having an exponent, so instead having an exponent of one, basically. So that would mean that u should be theta squared. And if I take the derivative of u with respect to theta, I'm gonna get just two theta using the power rule, d theta. So this is how I found my u and my du, and so that's what I'm going to use as the first part for integration by parts. Now next I need to figure out what my d v is. Now d v is gonna be whatever can be easily integrated. Now looking at this, I can see that my cosine three theta term I need to deal with here, and I know how to integrate these trig functions, so I'm gonna say cosine three theta, d theta is gonna be my d v. So what I can do is integrate both sides right here, that's going to give me the integral of dv, which is v, and to get v, well we need to integrate the cosine of three theta. How could I go about doing that? Well, let's set up this integral separately down here. So you have the cosine of three theta that we're integrating with respect to theta. Now what I could do is a variable substitution on this three theta. I could say that that's equal to w, for example. I'm not gonna use u just because I we use u up here, so I'm gonna make this w. So we're gonna say that w is equal to three theta. And if we were to take the derivative of w, that'd be the derivative of three theta, which would just be three with respect to theta. So notice here that this is how we found w and d w, so what I can do is rewrite what we have over here. So going back to here, well what I would end up with is this whole thing is the integral of the cosine of w, and that's going to be times d theta. Now what I could do is I could take this d theta and I could multiply it by three to get three d theta, but if I multiply this by three, I also have to divide by three to show that this would just cancel and be whatever we had at the start. But if I do this, notice we can now take this whole thing and replace it with d w. So what I can do is say this whole thing is one third times the integral of the cosine of w with respect to w. So from here I can go ahead and integrate the cosine, which I know is just going to give me sine. So we end with one third the sine of w which in this case we said was three theta, I'll plug that back in, plus the constant of integration c. So this right here is how we can find the integral of the cosine three theta. So we know how to integrate this meaning I can see that if I wanted to find v up here that's just going to be one third the sine of three theta. We don't typically need to worry about the c when we're doing integration by parts, and you're just finding the individual parts like the v and the the du and all of that. So we can say that this is what our v is, and we've now found these variables. So now that we've found all of these variables up here, well now we can apply it to this part of the formula. So let's go ahead and do that. So writing this all out, we have this integral that we're dealing with, and I'm first gonna have u times v. So So that's going to be theta squared, that's gonna be times one third, then we have sine of three theta. Then I'm going to have minus the integral of v du, which is going to be one third sine three theta, is going to be times du, which is two theta d theta. So now we have this all written out. Now what I'm gonna do is try to simplify this. So I'm gonna put this theta squared on top of the three right here, and then we'll have sine three theta, and then we're going to have minus, and I'll pull this one third and this two outside the integral, so that's gonna be two thirds of the integral of the theta times the sine of three theta with respect to theta. Now what can I do from here? Well, if you look at this, this is still not an integral that we can really solve, because I can't really just set this equal to u. Or I should say we can't really solve this in a straightforward way, because I can't do a variable substitution here, and there's not really a straightforward power rule or strategy I know at this point aside from just using the integration by parts formula again. So that's what I'm actually gonna do here. So what we're gonna do is we're gonna go over integration by parts a second time. So we're going to do the integral of u d v is equal to u v minus the integral of v d u. We now have to apply that for just this portion right here. So this is often the case. We have to do this repeated integration by parts to actually get to our result. So I'll rewrite out everything we have out front here. So we have three theta and then minus the two thirds and then everything inside these brackets. Now Now that's what I'm going to figure out goes inside. What I first need to do is figure out a new u and d v for this portion of the integral right here. Now what I'm going to set equal to u is going to be whatever I can easily take the derivative of. Well, recall that we had theta squared be the derivative before that we said to or had be the thing we said to u before so we could take the derivative. So what I'm gonna do is just set the theta now to u. So I'm going to say that this next u that we're dealing with is equal to just theta. And then what that means is the derivative of u is going to be the derivative of theta with respect to theta, which is just one d theta. So now that I've found u and du, next I can find my d v. My d v is going to be this portion right here that I can integrate, which is going to be the sine of three theta d theta. And then I need to find my v. Now let's see if there's a pattern I can generally notice to figure out what my v is. Well, recall that when we went ahead and did this integral right here, we were integrating the cosine, and I know that the integral of cosine gives me sine, and notice how we took whatever was inside here and we divided it on the outside. So I know that I'm going to take whatever is right here, I'm going to end up dividing it on the outside, so it's going to be one third. What is the integral of sine? Well, I know the integral of sine gives me negative cosine. So really what I'm gonna end up having is negative one third, that's gonna be the cosine of three theta. So that's gonna be the pattern that happens when you do this integral. Now if you weren't feeling very confident in that result, you could just go ahead and do a variable substitution on the three theta right here like we did before, and then go through the same process that we did up here to figure out what that integral is. But just to save time here, I'm just gonna go ahead and cut to this pattern right here. So that's gonna be what I use to solve this problem. So now that we found all these variables here, let's go ahead and use this side of the equation or the formula for integration by parts. So we have u times v, which is going to be theta times this quantity negative one third cosine three theta, so we're going to theta times negative one third cosine three theta, write it like this, and then we're going to have minus the integral of v du, which is going to be negative one third cosine three theta. Let me move this out of the way right here. Then we're going to have then this that's our v, then du, which we have as d theta. So this is what we get. So we've written this all out, so we're gonna have this theta squared over three, this is all this front part, times the sine of three theta minus two thirds. And then in this case, I'll try to simplify this right here too. So we're gonna have this as negative theta over three cosine three theta, and we're gonna have and then I've canceled this negative sign here, so we'll have plus, and then we have pull this one third out, one third times the integral of the cosine of three theta integrated with respect to theta, and this is what we now have to solve. So all we we really need to do now is focus on this integral right here, which is something we actually know how to solve. In fact, we dealt with this exact integral, the cosine of three theta before. Recall that when we integrated this, we got one third sine three theta. So we know that this whole thing is just gonna come out to one third sine three theta. So if I want to rewrite this, well, I can just say that this whole thing is theta squared over three sine of three theta. We're going to have minus two thirds times, and then we have negative theta over three cosine three theta right there. We have plus, and that's gonna be one third, and then it's going to be times, and then we know this whole thing is just gonna be one third sine of three theta, and then this whole thing plus c. Now as one last step to go ahead and solve this problem, I'm gonna take this negative two thirds that I have right here, I'm going to distribute it into the brackets. And I'll also go ahead and combine the one thirds that we see there as well. So doing this, writing this all out together, it's gonna be theta squared over three times the sine of three theta minus, and then in this case, I'll distribute the two thirds into there. That's going to give me plus two theta over nine. Right? So two times theta and then three times three and then the negatives cancel, so that's gonna be plus two theta over nine and then cosine three theta. So that's gonna be that first piece right there, and then we'll go ahead and distribute right here. And so first of all, I'll combine these. I know that's gonna be one ninth, and then one ninth multiplied by two thirds, so that's gonna be two over 27 because three times nine is 27. So we'll have and then in this case, it's gonna be minus two over 27. Yeah. Because three times three is nine, and then times the three is 27. So, yeah, minus two over 27 and that negative sign would multiply by that positive to give you a negative. So we get minus two over 27, and that's gonna be sine of three theta. And we're going to have plus, and then I multiply this negative two thirds by the constant c, which is just gonna give me a bigger constant c because that's all that is gonna happen. It's just gonna be a constant either way, whether it's multiplied by two thirds or not, so we end up with that constant. And then this right here is the final result and the solution to the problem. So this is how you can solve integration by parts when you have to use this equation multiple times. As you can see, the work can get tedious, but the main thing to recognize is that whatever you set to u in the first place is getting more and more simple as you continue to go through this process. Notice we started with the theta squared, then we got down to just the theta, and then we were able to actually find an integral that we knew how to solve. So So as you can see, going through this process does allow us to get to a result and find the solution to our problem. So hope you found this video helpful. Please let me know if you have any questions, and let's move on.

Find the indefinite integral. ∫ θ 2 cos ⁡ 3 θ d θ \int_{}^{}\theta^2\cos3\theta d\theta

θ 2 3 sin ⁡ 3 θ + 2 θ 9 cos ⁡ 3 θ − 2 27 sin ⁡ 3 θ + C \frac{\theta^2}{3}\sin3\theta+\frac{2\theta}{9}\cos3\theta-\frac{2}{27}\sin3\theta+C

θ 3 3 sin ⁡ 3 θ + 2 θ 9 cos ⁡ 3 θ − 2 27 sin ⁡ 3 θ + C \frac{\theta^3}{3}\sin3\theta+\frac{2\theta}{9}\cos3\theta-\frac{2}{27}\sin3\theta+C

θ 3 3 sin ⁡ 3 θ + 3 cos ⁡ θ + C \frac{\theta^3}{3}\sin3\theta+3\cos\theta+C

θ 2 3 sin ⁡ 3 θ + 3 cos ⁡ θ + C \frac{\theta^2}{3}\sin3\theta+3\cos\theta+C

12. Techniques of Integration

Integration by Parts

Calculus

12. Techniques of Integration

Integration by Parts

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Multiple Choice

Find the indefinite integral.
$\int_{}^{} θ^{2} \cos 3 θ d θ$

$\frac{θ^{3}}{3} \sin 3 θ + \frac{2 θ}{9} \cos 3 θ - \frac{2}{27} \sin 3 θ + C$

$\frac{θ^{2}}{3} \sin 3 θ + \frac{2 θ}{9} \cos 3 θ - \frac{2}{27} \sin 3 θ + C$

$\frac{θ^{3}}{3} \sin 3 θ + 3 \cos θ + C$

$\frac{θ^{2}}{3} \sin 3 θ + 3 \cos θ + C$

Verified step by step guidance

Step 1: Recognize that the integral ∫θ²cos(3θ)dθ requires integration by parts. Recall the formula for integration by parts: ∫u dv = uv - ∫v du. Here, we will choose u = θ² and dv = cos(3θ)dθ.

Step 2: Differentiate u and integrate dv. For u = θ², compute du = 2θ dθ. For dv = cos(3θ)dθ, compute v = (1/3)sin(3θ) (since the integral of cos(3θ) is (1/3)sin(3θ)).

Step 3: Substitute into the integration by parts formula. Using ∫u dv = uv - ∫v du, substitute u = θ², v = (1/3)sin(3θ), and du = 2θ dθ. This gives: ∫θ²cos(3θ)dθ = (θ²)(1/3)sin(3θ) - ∫(1/3)sin(3θ)(2θ)dθ.

Step 4: Simplify the remaining integral. The term ∫(1/3)(2θ)sin(3θ)dθ requires another application of integration by parts. Let u = θ and dv = sin(3θ)dθ. Then, du = dθ and v = -(1/3)cos(3θ) (since the integral of sin(3θ) is -(1/3)cos(3θ)).

Step 5: Substitute the second integration by parts result back into the expression. Combine all terms, simplify, and include the constant of integration C. The final expression will involve terms like (θ²/3)sin(3θ), (2θ/9)cos(3θ), and (-2/27)sin(3θ), plus the constant C.

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