Problem

Evaluate the integral. $\int_{0}^{1}\left(x^{2}+…

01:13

Problem 13 Easy Difficulty

Evaluate the integral.
$\int_{0}^{\pi} t \sin 3 t d t$

Answer

$\frac{1}{3} \pi$

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Biocalculus Calculus for the Life Sciences

Chapter 5

Integrals

Section 5

Integration by Parts

Discussion

You must be signed in to discuss.

Top Calculus 2 / BC Educators

Watch More Solved Questions in Chapter 5

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Video Transcript

Hello and welcome. We are looking at chapter five, Section five Problem 13. So we were asked to find a definite integral now ah t sign of three. T d t. So what we want to do is we want to choose Are you to be something that when we take the derivative of it, it will cancel? So we're solving this by integration by part. Um, so you you would be the tee value then Devi would be the rest of So if you equals t and D'You equal, uh, one DT. And likewise, if Devi is signed of three t, DT and VI will be the anti derivative, that's that would be negative. Um co sign of three t over three. If you take the derivative of that, it should come back to sign of three D. So, um, we haven't set up. Now we can apply the, um, immigration by parts. So just reminder the integral of you to u V equals u V minus sustainable of VD. So we have tea. That's are you times V. That would be times negative co sign of three t over three That's could be subtracted by the integral the this again. Times d'you. So? So we can see that. Are you cancelled out? And so this is an integral we can take without too much more trouble. So, um, you could write this as negative. T co signed three t over three, and then we can now take the integral of negative co sign of three t over three. So we could also pull out that negative to make this a positive here. And that would be, Ah, a little bit easier. Maybe so. Ah co sign of three t over three. We need the anti derivative of that. Um and we can't forget that this is from zero to pie. This is from zero to pie, and this will be a zero to pie as well. So I'll, uh, i'll do that substitution at the very end. I can't like doing it that way. Or you could do it for sure, right? As right as you take the interview, eso this would become We're taking the integral of co sign of three t over three. So that would be sign of three t overnight, right? And so now this whole thing is between from zero to pie. Uh, so this would be negative. Pie co Sign of three pi over three, plus sign of three pi over nine. So this whole thing being subtracted by zero co sign of three time zero over three, plus sign of three times zero overnight. Right? We're almost done here. So, uh, co sign of three pie would be the same as co sign of pie, which is negative one. So this would become negative one times negative pie or pie over three. Um, sign of three pies. The same. A sign of pie that's gonna be zero. And then when we subtract zero times, anything is zero show that these were zero, and then sign of zero is also zero Just looking at you have to think back to your unit circle. So our final answer is high over three. All right. And we are done

Get More Help with this Textbook