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Integral! Cal. 2!

  1. Sep 12, 2004 #1
    [tex]\int 96cos^4(6x) * dx[/tex]

    ok first i take out the 96 cause it's constant!

    [tex] 96 \int cos^4(6x) * dx[/tex]

    [tex] 96 \int (cos^2(6x)^2) * dx [/tex]

    ok now with that setup, i can know use the half-angel formula!

    [tex] 96 \int (\frac{1+cos(12x)}{2})^2 [/tex]

    squared the problem...

    [tex] 24 \int (1+ 2cos(12x) + (cos12x)^2 [/tex]

    now to use the half-angle idents agian..

    [tex] 24 \int 1 +2cos(12x) + 1/2(1+cos24x) [/tex]

    can someone tell me if im doing this correctly before i integral the problem?
     
    Last edited: Sep 12, 2004
  2. jcsd
  3. Sep 12, 2004 #2

    Pyrrhus

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    Yes use Half Angle Identities.

    [tex] cos^2(6x) = \frac{1+cos(12x)}{2} [/tex]

    [tex]\int 96cos^4(6x) dx[/tex]

    [tex] 96\int \frac{1+cos(12x)}{2} \frac{1+cos(12x)}{2}dx[/tex]

    [tex] 96\int \frac{(1+cos(12x))^2}{4} dx[/tex]


    [tex] 24[\int 1 dx + \int 2cos(12x) dx + \int cos^2(12x) dx][/tex]

    You are doing fine...
     
    Last edited: Sep 12, 2004
  4. Sep 12, 2004 #3
    can i take out the 1/4? so 1/4*96 = 24
     
  5. Sep 12, 2004 #4

    Pyrrhus

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    yes you can
     
  6. Sep 12, 2004 #5
    thank you for checking my answer, ok i will cont. where i left off....


    [tex] 24 \int 1 +2cos(12x) + 1/2(1+cos24x) [/tex]


    [tex] 24 \int 1 +2cos(12x) + 1/2+\frac{1}{2}cos(24x) [/tex]


    [tex]24 \int1dx + 48 \int cos(12x) + 12 \int dx + 12 \int cos(24x)[/tex]

    adding alot of integral signs seems to look messy, but i think you can tell what im doing. ok time to find the anti-dervs.

    [tex]24x + \frac{1}{12}(48)sin(12x) + 12 + 12(\frac{1}{24})sin(24x)[/tex]

    k time to solve...
    [tex]24x+4sin(12x) + 12 + \frac{1}{2}sin(24x)[/tex]

    but it's the wrong answer
     
    Last edited: Sep 12, 2004
  7. Sep 12, 2004 #6
    This bit is wrong. 1/2(1 + cos 24x) = 1/2 + (1/2)cos 24x. (1/2)cos x isn't the same thing as cos (x/2).
     
  8. Sep 12, 2004 #7
    ah i see, dont know what i was thinking. i edited my previous post and fixed the problem, but the answer is still wrong
     
  9. Sep 12, 2004 #8

    Pyrrhus

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    [tex] 24[\int 1 dx + \int 2cos(12x) dx + \int cos^2(12x) dx][/tex]

    [tex] 24[\int 1 dx + \int 2cos(12x) dx + \int \frac{1+cos(24x)}{2} dx][/tex]

    [tex] 24\int 1 dx + 24\int 2cos(12x) dx + 24\int \frac{1+cos(24x)}{2} dx [/tex]

    [tex] 24\int 1 dx + 48\int cos(12x) dx + 12\int 1 dx+ 12\int cos(24x) dx [/tex]

    [tex] 24x + 4sin(12x) + 12x+ \frac{sin(24x)}{2} + C [/tex]

    [tex] 36x + 4sin(12x)+ \frac{sin(24x)}{2} + C[/tex]

    Check my work...
     
  10. Sep 12, 2004 #9
    What in the world are you doing here!? Where did those "+ 24"s come from?

    [edit]Nevermind, I was reading it wrong (since it looks cluttered).[/edit]
     
  11. Sep 12, 2004 #10
    OK. I checked your last calculation. Everything is correct now.
     
  12. Sep 12, 2004 #11

    Pyrrhus

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    Thanks but i meant that to whatuptdoc :smile:
     
  13. Sep 12, 2004 #12
    ah i see what i did wrong, thank you
     
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