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Calculus II - Trigonometric Integrals

  1. Jul 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Apparently I'm doing something wrong. I'm kind of lost as to what because I looked over my work several times.

    2. Relevant equations

    sin^2 x = ( 1 - cos 2x )/2
    cos^2 x = ( 1 - sin 2x )/2
    integral sin(x) dx = -cos(x)
    integral cos(x) dx = sin(x)
    3. The attempt at a solution

    SEE ATTACHMENT
    Wolfram Alpha says the correct solution is 1/64 (24 x+8 sin(4 x)+sin(8 x))+constant
    Thanks for any help you can provide!
     

    Attached Files:

    Last edited: Jul 30, 2011
  2. jcsd
  3. Jul 30, 2011 #2

    eumyang

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    The bolded is not right. It should be
    [tex]\cos^2 x = \frac{1 + \cos 2x}{2}[/tex]
     
  4. Jul 30, 2011 #3
    Thanks for your response. My book specifically told me those were true and even use the fact that they were to evaluate a integral. For example:
     
    Last edited: Jul 30, 2011
  5. Jul 30, 2011 #4
    this
     

    Attached Files:

  6. Jul 30, 2011 #5
    So i guess there's a plus sign and I forgot to include it becuase i used minus instead

    sin^2 x = ( 1 - cos 2x )/2
    cos^2 x = ( 1 + sin 2x )/2

    cos^2 x = ( 1 + sin 2x )/2 instead of cos^2 x = ( 1 - sin 2x )/2

    That is one mistake in my work... assuming that formula is right
     
  7. Jul 30, 2011 #6
    oh man im an idiot thanks
     
  8. Jul 30, 2011 #7
    I guess I'm still doing something wrong, don't know what though
     

    Attached Files:

  9. Jul 30, 2011 #8
    cos^2(x) = 1/2(1 + cos(2x))
     
  10. Jul 30, 2011 #9
    Ya I know. I still don't see what I did wrong though in post #7
     
  11. Jul 30, 2011 #10

    eumyang

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    It's actually the same as the answer you got from Wolfram. You just need to factor out another 1/8.
    [itex]\frac{1}{8}\left( 3x + \sin 4x + \frac{\sin 8x}{8}\right) + C[/itex]
    [itex]= \frac{1}{8}\left( \frac{24x}{8} + \frac{8\sin 4x}{8} + \frac{\sin 8x}{8}\right) + C[/itex]
    Do you see it now?
     
  12. Jul 30, 2011 #11
    The answers don't seem to match though for some reason and I don't have a clue as to what I'm doing wrong...

    The answer I got was
    1/8 [ 3x + sin(4x) + sin(8x)/8 ]+ c
    wolfram alpha returned
    1/32 (12 x-8 sin(2 x)+sin(4 x))+constant

    if they are equal to each other I should be able to plug in a value for x in both expressions and get the same results... this however doesn't occur

    1/8 [ 3*5 + sin(4*5) + sin(8*5)/8 ]+ c is about 2.000760549 + c

    1/32 (12*5 -8 sin(2*5)+sin(4*5))+constant is about 2.039534817 + c

    It would appear to be that i am off by a little bit... which is strange cause I don't see what I did wrong

    and ya they would appear to be equal to each other but i guess they are not
     
  13. Jul 30, 2011 #12
    anyone?
     
  14. Jul 30, 2011 #13

    Ray Vickson

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    For *indefinite* integrations, the C is an arbitrary constant. There is no reason the two C's should be the same. However, you ought to get the same results for the definite integral from a to b. Have you tried it?

    RGV
     
  15. Jul 30, 2011 #14
    Ya I got different results when I integrated both functions from 2 to 10

    for 1/8 [ 3*x + sin(4*x) + sin(8*x)/8 ]
    I got roughly 18.01464011

    for 1/32 (12*x -8 sin(2*x)+sin(4*x))
    I got roughly 18.13678944
     
  16. Jul 30, 2011 #15
    WA gave me 1/64(24x + 8sin(4x) + sin(8x)) + constant, which is equivalent to the answer you say you got above.
    For the other answer that you say WA gave you, I took its derivative in WA and it gave sin4x, so it looks like you had a mistake when you entered it in.
     
  17. Jul 30, 2011 #16
    huh weird thanks
     
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