Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 :
    sin2(x)cos4(x)dx
    =(-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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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)​
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Reduction formulas for integral of sin and cos
  1. Integral of sin and cos (Replies: 12)

Loading...