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Trig integral question.

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex]sin32x dx

    2. The attempt at a solution

    [tex]\int[/tex]sin32x dx =
    [tex]\int[/tex](1-cos22x)sin2x dx =
    (-1/2)[tex]\int[/tex](1-cos22x)-2sin2x dx =
    (-1/2)[tex]\int[/tex](1-u2)du where u= cos2x
    (-1/2)(u-(u3/3)) =
    (-cos 2x/2)+(cos32x/6) + C

    Book says (cos3x/3)-(cosx/2) + C
    Where did I go wrong?
    Last edited: Apr 7, 2009
  2. jcsd
  3. Apr 7, 2009 #2


    Staff: Mentor

    It looks like you have the right idea, but you should split the integral into two separate integrals after this step:
    [itex]\int (1-cos^22x)sin2x dx [/itex]
  4. Apr 7, 2009 #3
    Yea integrating by parts seems to be the preferred way to attack these types of problems. The chapter was focusing on u-substitution though so I tried going with that. I just can't tell if I made a mistake or if the book did (there's been a couple times in the past) or if the book's answer is just more simplified and I'm not seeing how.
  5. Apr 7, 2009 #4
    Well I actually checked your work instead of assuming you made a mistake, and you made no mistakes (I got the same thing). The way to do this is definitely a u-substitution and you don't want to split it up as mentioned. Your answer is correct.

    To save work, I also computed the derivative of our answer with Maple and it gave me the integrand that we started with. I did the same for the book's "answer", and it did not. So it looks like the book is incorrect on this one.
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