## anonymous one year ago Evaluate the integral inside! thanks!!

1. anonymous

|dw:1442195407457:dw|

2. anonymous

would it be like this so far? |dw:1442195517625:dw| ? not too sure about the other parts :/

3. zepdrix

This is how I like to think of it when doing by-parts. $$\large\rm u$$ is the thing that you're trying to "destroy" or breakdown. $$\large\rm dv$$ is usually the part that is cyclical, won't breakdown, like sine/cosine or exponential.

4. zepdrix

So when I look at this problem, I'm thinking hmm $$\large\rm t^2$$ is probably a good $$\large\rm u$$ because we differentiate it and it will breakdown, ya?

5. zepdrix

$$\large\rm t^2\quad\to\quad 2t\quad\to\quad 0$$

6. anonymous

yes :)

7. zepdrix

$\large\rm u=t^2$$\large\rm dv=\sin\beta\theta~d\theta$So what do we get for these $\large\rm du=?$$\large\rm v=?$

8. zepdrix

Oh woops, lol not theta

9. zepdrix

$\large\rm u=t^2$$\large\rm dv=\sin\beta t~d t$

10. anonymous

hahaa :P du = 2t and v = -cosBtdt ^sorry!! using a B instead because i cannot find how to put the symbol in

11. anonymous

oops and now i did it!! v = -cos Bø dt lol

12. zepdrix

Hmm the $$\large\rm v$$ looks a little off.

13. zepdrix

You could do a little u-sub, err we already used the letter u.... you could do a little m-sub to deal with integrating your dv. but that's really tedious, you wanna get comfortable with easy integrals like this.

14. zepdrix

$\large\rm \int\limits \cos(2x)dx=?$Like this one, can you solve without u-sub?

15. anonymous

:/ how would i fix v then? but du = 2t ?

16. zepdrix

yes

17. zepdrix

du = 2t dt

18. anonymous

19. zepdrix

You can use a method which I think is called "advanced guessing",$\large\rm \int\limits\limits \cos(2x)dx=\sin(2x)$I'm thinking that it's going to be something like this, ya?

20. anonymous

yes!!

21. zepdrix

Take a derivative to check and see if you have the correct solution.$\large\rm (\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)$Woops! We got a little bit too much back. See that extra 2? That tells us that:$\large\rm \int\limits 2\cos(2x)dx=\sin(2x)$Which means:$\large\rm \int\limits \cos(2x)dx=\frac{1}{2}\sin(2x)$Chain rule told us to multiply by 2 on the outside. So when we integrate, we need to do the reverse to compensate for the missing 2. We need to divide by 2.

22. zepdrix

So this will come up A LOT, so try to get used to it. When you have a coefficient on the x, you end up dividing by that coefficient when you integrate.

23. anonymous

ohh okay!! and so for our v, how can we apply the same method?

24. zepdrix

$\large\rm \int\limits \sin\beta t~d t=\frac{1}{\beta}(-\cos\beta t)$Going backwards, sine gives us -cosine and we have to divide by the beta coefficient on x.

25. anonymous

ohh okay!! and so we have that as our v? so we do the integral udv = uv - integral v du ?

26. zepdrix

$\large\rm u=t^2\qquad\qquad\qquad\qquad du=2t~d t$$\large\rm dv=\sin(\beta t)d t\qquad\qquad v=-\frac{1}{\beta}\cos(\beta t)$Let's list all of our pieces together before setting up the integral. Unless you're following on paper, then you're prolly ok heh.

27. zepdrix

Yes, integral time \c;/

28. anonymous

ooh okay!! and so we get this?|dw:1442196531964:dw|

29. anonymous

-1/beta cos (beta t)t^2 - not sure how to integrate that :/ but 2t dt = 0?

30. zepdrix

Ok good :) Looks like integration-by-parts is required again. Clean things up before choosing your $$\large\rm u$$ and $$\large\rm dv$$ though:$\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta}\int\limits t \cos(\beta t) d t$

31. zepdrix

And then we need to apply by-parts to this orange portion, ya?$\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta}\color{orangered}{\int\limits\limits t \cos(\beta t) d t}$

32. anonymous

yes:)

33. anonymous

and what happens next? we integrate the red?

34. zepdrix

The orange, yes. We need to establish $$\large\rm u$$ and $$\large\rm dv$$ for the orange part. Just a note: Earlier I had said something about how the t^2 will break down and I posted this:$\large\rm t^2\quad\to\quad 2t\quad\to\quad 0$Lil mistake there, 2t doesn't differentiate to 0, it differentiates to 2. Whatever, not a big deal :) We'll just take our derivative carefully

35. zepdrix

I pulled the 2 outside of the integral, hopefully that isn't confusing. So we still want the u to break down. So we'll choose our parts this way:$\large\rm u=t$$\large\rm dv=\cos(\beta t)d t$

36. zepdrix

Getting your other two pieces give you,$\large\rm du=1dt$$\large\rm v=\frac{1}{\beta}\sin(\beta t)$

37. anonymous

okay!!

38. zepdrix

Any confusion there? :U The problem is getting pretty long, so it's easy to get lost lol

39. zepdrix

So draw your by-parts thing again :D

40. zepdrix

|dw:1442197278644:dw|

41. anonymous

okay :) let's see if i understood correctly lol :P so we have this? |dw:1442197308540:dw|

42. zepdrix

|dw:1442197426918:dw|

43. anonymous

okay! and so now i simplify ?

44. zepdrix

$\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta}\color{orangered}{\int\limits\limits\limits t \cos(\beta t) d t}$ $\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta}\color{orangered}{\left[\frac{1}{\beta}t \sin(\beta t)-\frac{1}{b}\int\limits \sin(\beta t)d t\right]}$Yes, maybe distribute the 2/B.

45. zepdrix

b=Beta, typo :c

46. zepdrix

I can type it out if you want lol It's getting so long >.<

47. anonymous

hehe okay!! thank you!! that would be very helpful :P I'm getting a bit lost in it lol :P

48. zepdrix

$\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)-\frac{2}{\beta^2}\int\limits \sin(\beta t)d t$Something like that. Ok one last piece to integrate! No more by-parts required, yay!

49. anonymous

oooh wow okay!! haha got so messy :P so integrate it to -cos dt ?

50. zepdrix

Yes. And again, since we have a coefficient on the t, we have to divide by that.

51. anonymous

how does that look again?

52. zepdrix

$\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)-\frac{2}{\beta^2}\color{orangered}{\int\limits\limits \sin(\beta t)d t}$ $\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)-\frac{2}{\beta^2}\color{orangered}{\frac{1}{\beta}(-\cos(\beta t))}$

53. zepdrix

$\large\rm =-\frac{1}{\beta}t^2 \cos(\beta t)+\frac{2}{\beta^2}t \sin(\beta t)+\frac{2}{\beta^3}\cos(\beta t)+C$Somethinggggg like that? 0_o Boy this one is a doozy when they throw in the beta and the squared term.

54. anonymous

whoahhh wowzers okay :P so this is done? :O

55. zepdrix

yes, imma check my work real quick though, to make sure i didn't mess up anywhere

56. anonymous

okie :)

57. zepdrix

yayyyy looks like we did it correctly \c:/ I know that one was pretty insane to get through. You maybe would want to think about practicing with something like this:$\large\rm \int\limits x^2\sin(2x)dx$Might just be easier without all those betas floating around.

58. anonymous

ooh okay!! yay!! thanks so much!! :D