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Calculus 2 Integral Question (difficult)

  1. Sep 24, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\int[/tex]cos3xdx

    2. Relevant equations
    cos3x = cos2x * cosx


    3. The attempt at a solution

    Let u= cos2x
    du=(x/2) + (cos2x/4x)
    v=-sinx
    dv=cos


    uv-[tex]\int[/tex]vdu

    -cos2xsinx - [tex]\int[/tex]-sinx(x/2 + cos2x/4x)dx

    -cos2xsinx -[tex]\int[/tex]xsinx/2 + (sinxcos2x/4x)

    -cos2xsinx - (1/2)xsinx +(sinxcos2x/4x)

    -cos2xsinx -(sinx-xcosx/2) + (sinxcos2x/4x)

    ....


    aaaand that's about where I get lost/stuck. i can't figure out how to finish off the problem. i get to the last line on the far right and get confused. any help would be much appreciated.
     
  2. jcsd
  3. Sep 24, 2008 #2
    Hey,

    Why don't you try and use the trig identity[tex] \cos^2x +\sin^2x = 1 [/tex] and then try substituting [tex] \sin x = u [/tex]

    Hope that helps,

    -Spoon
     
  4. Sep 24, 2008 #3
    hm ill give it a shot
     
  5. Sep 24, 2008 #4
    No worries, glad to have helped.

    - Spoon
     
  6. Sep 24, 2008 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Here's your first error. The derivative of cos2 x, using the chain rule, is (2 cos x)(cos x)'= - 2sin x cos x. I have absolutely no idea how you got your du!

    You started wrong by trying to use integration by parts when it is not necessary. cos3 x dx= (cos2 x) (cos x dx)= (1- sin2 x) (cos x dx). Now use the obvious substitution.
     
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