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Reduction formulas for integral of sin and cos

  1. Feb 12, 2012 #1
    Integrate sin2(x)cos4(x)dx using reduction formulas?
    My book says integral sin2(x)cos4(x)dx= integral cos4(x)-integral cos6x dx
    Now the reduction formula for n=6
    for integral cos6(x)dx= (1/6)cos5(x)sinx+(5/6) integral cos4(x)dx

    Here is the part I don't get: It then says :
    =(-1/6)cos5(x)sinx+(1/6) integral cos^4(x)dx

    I don't get how the 5/6 becomes 1/6 or why 1/6 becomes -1/6 if that makes any sense? Any help would be great! I've been looking at it for awhile, but Im not seeing it for some reason.
  2. jcsd
  3. Feb 12, 2012 #2


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    hi livvy07! :smile:

    (have an integral: ∫ :wink:)

    i don't understand that :confused:

    ∫ sin2(x)cos4(x) dx

    = ∫ [sin(x)][sin(x)cos4(x)] dx

    then integrate by parts (integrating the second bracket, and differentiating the first bracket)​
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