1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Mark44

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