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Indefinite integral of sin^4xdx

  1. Feb 20, 2006 #1
    Hi everyone. I'm having some trouble evaluating the following integral

    [tex]
    \int{sin^4xdx}
    [/tex]

    First let me start off by showing what I did.

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

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

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

    [tex]
    =\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{4}\int{(\frac{1}{2}+\frac{1}{2}cos(4x))dx
    [/tex]

    [tex]
    =\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{8}x+\frac{1}{8}\int{cos(4x)}dx
    [/tex]

    [tex]
    =\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{8}x+\frac{1}{32}sin(4x)+C
    [/tex]


    I'm not sure if I have the right answer or not, so could someone please check and see if I did anything illegal. If so, can you please correct me, or even show me a different method entirely.

    Thanks,
    -Zach
     
    Last edited: Feb 20, 2006
  2. jcsd
  3. Feb 20, 2006 #2

    quasar987

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

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

    no?
     
  4. Feb 20, 2006 #3
    Thank you!!
     
  5. Jun 17, 2008 #4
    why is the second cos squared?


    what is next step after that? im stuck here too...
    a lil hint please
     
    Last edited: Jun 17, 2008
  6. Jun 17, 2008 #5
    expand the expression in the integral, then you get the last term being cos sqaured
     
  7. Jun 17, 2008 #6
    what about after that...
     
  8. Jun 17, 2008 #7
    so you expand to get this

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

    then its an integral of 3 terms, so you integrate the 3 terms separately
    did that help?
     
  9. Jun 17, 2008 #8

    Dick

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    There's a trig formula for cos^2(2x) in terms of cos(4x). Just look at what Captain Zappo tried to do, in spite of fluffing a coefficient.
     
  10. Jun 18, 2008 #9
    ∫sin^4(x)

    [sin^2(x)]^2

    [(1 -cos(2x)/2)]^2

    1/4[1 + cos^2(2x) - 2cos(2x)]

    1/4[1 - 2cos(2x) + (1 + cos(4x)/2 ]

    1/4) - (1/2)cos(2x) + (1/8) + (1/8)cos(4x)

    (3/8) - (1/2)cos(2x) + (1/8)cos(4x)

    so ∫sin^4(x)dx = 3/8∫dx - 1/2∫cos(2x) dx + 1/8∫cos(4x) dx

    ∫sin^4(x)dx = (3/8)x - (1/2)sin(2x)*(1/2) + (1/8)sin(4x)*(1/4) + c

    ∫sin^4(x)dx = (3/8)x - (1/4)sin(2x) + (1/32)sin(4x) + c

    is this right?
     
  11. Jun 18, 2008 #10

    Dick

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    Looks fine to me.
     
  12. Jun 18, 2008 #11
    http://integrals.wolfram.com/index.jsp?expr=(sin(x))^4&random=false

    You should use this (for other questions)
     
  13. Jun 18, 2008 #12
    ∫sin^4(x)dx = (3/8)x - (1/4)sin(2x) + (1/32)sin(4x) + c
    correct
     
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