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Homework Help: Trigonometric Integrals

  1. Sep 12, 2010 #1
    I got these two questions wrong and I'm wondering what i did wrong.

    The problem statement, all variables and given/known data[/b]
    1. http://i324.photobucket.com/albums/k327/ProtoGirlEXE/q3.jpg [Broken]

    2. http://i324.photobucket.com/albums/k327/ProtoGirlEXE/q4.jpg [Broken]


    The attempt at a solution[/b]
    1. http://i324.photobucket.com/albums/k327/ProtoGirlEXE/100_0639.jpg [Broken]

    2.http://i324.photobucket.com/albums/k327/ProtoGirlEXE/100_0643.jpg [Broken]
    (I put t instead of x while I was doing the problem on accident)
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 12, 2010 #2

    jgens

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    Your work is really difficult to read from those pages, but from what I can gather, here are some of the mistakes that you made ...

    First Problem: You forgot to distribute the 3 across when you expanded 3cos2(2x). Check your work there. Second, I have no idea how you got your expansion for cos3(2x).

    Second Problem: Review your trig identities :wink:
     
  4. Sep 12, 2010 #3

    jgens

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    First Problem: cos(2x)cos(4x) =/= cos2(8x). I'm not really sure how you got that in the first place.

    Second Problem: Do it again, and this time be very careful about how you distribute things so that you don't end up with extra terms.
     
  5. Sep 12, 2010 #4
    for the second one, is it just (1/3)tan^3 x +c?

    for the first one: there was a -cos^3(2x), so I thought I should split it up like cos^2(2x) and cos (2x) then make cos^2(2x)---> -(1/2)+(1/2)cos(4x)(cos(2x)....
     
  6. Sep 12, 2010 #5

    jgens

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    No. Again, be careful about how you distribute things.

    Sure, you can split it up like that, and I followed your work through there. However, cos(4x)cos(2x) =/= cos2(8x). I don't know what would make you think that that was the case.
     
  7. Sep 12, 2010 #6
    oh yeah duh, so it's just cos(8x)
    then I should get (2/8)x-(3/16)sin(2x)+(3/64)cos(4x)+(1/128)cos(16x)+c, right?
     
  8. Sep 12, 2010 #7

    jgens

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    No. What makes you think that cos(2x)cos(4x) = cos(8x)? They're not equal.
     
  9. Sep 12, 2010 #8
    they're not? so then what can I do, if they won't separate?
     
  10. Sep 12, 2010 #9
    omg (1/3)tan^3 x + x +c is wrong too :(
    this is so frustrating
     
  11. Sep 12, 2010 #10

    jgens

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    They're quite clearly not equal. My question is still what makes you think that they should be in the first place? And I would have tackled this problem using integration by parts. It's really easy to derive a reduction formula and that does pretty much all of the work for you.
     
  12. Sep 12, 2010 #11

    jgens

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    Why don't you show your steps? From the answers that you've been posting, it looks like you've been distributing terms incorrectly.
     
  13. Sep 12, 2010 #12
    how can you apply integration of parts to sin^6 x, its just one term?
     
  14. Sep 12, 2010 #13

    jgens

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    You have to manipulate sin6(x) using trig identities first.

    u =/= x
     
  15. Sep 12, 2010 #14
    now I'm terribly lost
     
  16. Sep 12, 2010 #15

    jgens

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    Hopefully this will help you with the first problem ...

    [tex]\int\sin{(x)}^{2n}dx=\int[1-\cos{(x)}^2]\sin{(x)}^{2(n-1)}dx=\int\sin{(x)}^{2(n-1)}dx-\int\cos{(x)}^2\sin{(x)}^{2(n-1)}dx[/tex]

    [tex]\int\cos{(x)}^2\sin{(x)}^{2(n-1)}dx=\frac{\cos{(x)}\sin{(x)}^{2n-1}}{2n-1}+\frac{1}{2n-1}\int\sin{(x)}^{2n}dx[/tex]

    Putting these two together, we see that

    [tex]\int\sin{(x)}^{2n}dx=\int\sin{(x)}^{2(n-1)}dx-\frac{\cos{(x)}\sin{(x)}^{2n-1}}{2n-1}-\frac{1}{2n-1}\int\sin{(x)}^{2n}dx[/tex]

    [tex]\frac{2n}{2n-1}\int\sin{(x)}^{2n}dx=\int\sin{(x)}^{2(n-1)}dx-\frac{\cos{(x)}\sin{(x)}^{2n-1}}{2n-1}[/tex]

    [tex]\int\sin{(x)}^{2n}dx=\frac{2n-1}{2n}\left[\int\sin{(x)}^{2(n-1)}dx-\frac{\cos{(x)}\sin{(x)}^{2n-1}}{2n-1}\right][/tex]

    Can you see how this applies to the first problem that you're working? What I posted here is a generalization, so you'll need to figure the specifics out.
     
  17. Sep 12, 2010 #16

    jgens

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    For the second problem, I'll start with what you already have ...

    [tex]\int u^2+1du=\frac{u^3}{3}+u+C[/tex]

    Since [itex]u=\tan{x}[/itex], it's now just a matter of replacing each [itex]u[/itex] in the equation above with [itex]\tan{x}[/itex]. The last few times, you've haven't been consistent in this last regard and that's what's causing you so much grief.
     
  18. Sep 13, 2010 #17

    jgens

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    What happened to the factor of cos(2x)? You can't just ignore it, so yes, your answer is wrong. The approach that I outlined in one of my previous posts will let you evaluate this integral with very little work, so why don't you try to apply that?

    I'm not going to resolve this one for you since you're making such an obvious mistake. Look at what you wrote above carefully and tell me if it makes sense. It doesn't pay off to be sloppy with your work.
     
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