Problem

Evaluate the integral. $ \displaystyle \int \c…

03:29

Problem 8 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int t^2 \sin \beta t dt $

Answer

$$ -\frac{1}{\rho} +^{2} \cos \beta t+\frac{\partial}{\beta^{2}} t \sin \beta^{t}+\frac{2}{\beta^{2}} $$

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Video Transcript

first problem, we're gonna be doing what's known as integration by parts. So we're giving an indefinite integral t squared sine beta T d t. That's what we want to do is performing integration by parts eso The things we have to pick are you? Do you the and D V Um, and at first it can be tricky, but hopefully we'll get the hang of it. So we're gonna pick you to be t squared. Ultimately, we wanted to be in the form of UDV, so it makes sense that this would be are you? And then this is gonna be our DVD, so Devi will put as sign beta t DT. So with that in mind, we know that our d you would obviously be to t d t and then RV would be the, um, anti derivative of this right here. So that's gonna end up giving us, uh, negative one over beta co sign Beta t. So now we want to do this right here. Integration by parts tells us that the integral of you devi is equal to U V minus the integral of V d U. So what that's gonna look like is we're gonna have our integral Being equal to t squared is you. V is a negative one. Over. Beta co sign Beta T minus the integral V. So will copy this again. We're gonna duplicate the and then do you? So do you is gonna be to t d team. So now that we have this, we've at least made it somewhat simpler. But we still have Thio solve a lot more so out here, where we're gonna get is t squared over beta cursing. Beat a tea plus two over beta will factor that out, and we'll also factor off the negative. So now that although we have is t coastline beta T de teeth Now, this is much simpler because if we do the integral of part one, using integration by parts, we now can do an integration by parts the second time letting you equal t and d you equaling DT. So what we end up getting is once we perform that integration by parts, we'll get to t beta squared sine beta t plus two over Beta Cube co sign beta teeth. And keep in mind that this is just a portion of it, because then we have to plug it back in and what we end up getting as a result is going to be. We need to make sure we add see at the end, because this is an indefinite integral, um and it's going in to this portion right here. So our final answer is going to be tu minus beta squared T squared. That's the T squared coming from here. And then that's over being a cubed. So we multiplied the beta cubed here, um, and brought the Beta cube in order to be able to combine the fractions that we have right here. So that's what we did there. And then that's gonna be cosign beta t, which is how we got this portion right here. We're just combining both ideas. And then that's going to be plus to t over beta squared sign beta t that's taking this portion right here and combining it with, um, the to T and the beta. This is just staying as is, whereas this portion right here was a combination of these three. So now that we have all this, we just need to add the plus C because it is an indefinite integral, and we see that this is our final answer for the problem.

Carson M.

California Baptist University

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Problem

Evaluate the integral. $ \displaystyle \int \c…

Problem 8 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int t^2 \sin \beta t dt $

Answer

$$ -\frac{1}{\rho} +^{2} \cos \beta t+\frac{\partial}{\beta^{2}} t \sin \beta^{t}+\frac{2}{\beta^{2}} $$

More Answers

Related Courses

Calculus: Early Transcendentals

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Watch More Solved Questions in Chapter 7

Video Transcript

Calculus: Early Transcendentals

Grace H.

Anna Marie V.

Michael J.

Joseph L.

Integration Review - Intro

Definite vs Indefinite