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Integrate (sin x^4)

  1. Jan 2, 2006 #1
    may i know how to integ (sin x^4) ??

    the answer is 1/32(12x - 8sin 2x + sin4x)
     
  2. jcsd
  3. Jan 2, 2006 #2

    VietDao29

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    Again, you can use Power-reduction formulas. Then use some Product-to-sum identities, your goal is convert that sin(x) to the power of 4 into some sine or cosine functions to the power of 1.
    Now let's first split sin4x into (sin2x sin2x). Can you go from here?
     
  4. Jan 2, 2006 #3
    i try to subs using cos2x=1-s(sinx)^2 but can't get
     
  5. Jan 2, 2006 #4

    VietDao29

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    So you have:
    cos(2x) = cos2x - sin2x = 2cos2x - 1 = 1 - 2sin2x.
    From there, rearrange them a bit, you will have:
    [tex]\cos ^ 2 x = \frac{1 + \cos(2x)}{2} \quad \mbox{and} \quad \sin ^ 2 x = \frac{1 - \cos(2x)}{2}[/tex]
    These are call Power-reduction formulas.
    So we will now use [tex]\sin ^ 2 x = \frac{1 - \cos(2x)}{2}[/tex].
    [tex]\int \sin ^ 4 x dx = \int (\sin ^ 2 x) ^ 2 dx = \int \left( \frac{1 - \cos(2x)}{2} \right) ^ 2 dx = \frac{1}{4} \int ( 1 - \cos(2x) ) ^ 2 dx [/tex]
    [tex]= \frac{1}{4} \int ( 1 - 2 \cos(2x) + \cos ^ 2 (2x)) dx[/tex].
    Now again use the Power-reduction formulas for cos2(2x).
    Can you go from here?
     
  6. Jan 2, 2006 #5
    ya,that's where i got stuck because i don't know how to get the sin4x.how to obtain 1/32 sin4x??
     
  7. Jan 2, 2006 #6

    VietDao29

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    Did I tell you to use the Power-reduction formulas for cos2(2x). It's the last line of my above post (namely, the #4 post of this thread).
    Since you have:
    [tex]\cos ^ 2 x = \frac{1 + \cos(2x)}{2}[/tex], so that means:
    [tex]\cos ^ 2 (2x) = \frac{1 + \cos(2 \times (2x))}{2} = \frac{1 + \cos(4x)}{2}[/tex].
    Can you go from here?
     
  8. Jan 2, 2006 #7
    oh........okok i saw it.....thanx very much
     
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