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Help simplifying/checking integral

  1. Aug 8, 2011 #1
    OK, so I'm just trying to refresh my calc knowledge before I go off to school. The problem I'm trying to do is:

    [itex]\int sin(2x)tan^{-1}(sin(x))dx[/itex]

    So what I've done is, first, replace sin(2x), giving:

    [itex]\int 2cos(x)sin(x)tan^{-1}(sin(x))dx[/itex]

    or

    [itex]2\int cos(x)sin(x)tan^{-1}(sin(x))dx[/itex]

    Then, I assume that sin(x) = u (and therefore, du = cos(x)dx). This gives me:

    [itex]2\int utan^{-1}(u)du[/itex]

    Then, I integrate by parts.

    [itex]2\int utan^{-1}(u)du = 2(\frac{1}{2}u^{2}tan^{-1}(u)-\int \frac{u^{2}}{2+2u^{2}})[/itex]

    or

    [itex]2(\frac{1}{2}u^{2}tan^{-1}(u)-(\frac{u^{3}}{6}+\frac{u}{2}))[/itex]

    Substituting back in for u gives me:

    [itex]2(\frac{1}{2}sin^{2}(x)tan^{-1}(sin(x))-(\frac{sin^{3}(x)}{6}+\frac{sin(x)}{2}))[/itex]

    Or, to simplify a little:

    [itex]-\frac{sin^{3}(x)}{3}+tan^{-1}(sin(x))sin^{2}(x)-sin(x)[/itex]

    My question is where do I go from here to simplify, and also if the work I've done up to this point is correct/on the right track (it's been a while since I took calc so go easy hahah). Just so you all know, I don't have an answer key for this problem so I'm working off of Wolfram Alpha's answer which is:

    [itex]\frac{1}{2}((cos(2x)-3)(-tan^{-1}(sin(x)))-2sin(x)) + C [/itex]

    Thanks in advance!
     
  2. jcsd
  3. Aug 8, 2011 #2
    There's an error from the integral on the right side of the top equation going to the line below it. Factor out a 2 from the denominator then try long division to make the integrand into something easier to integrate.
     
  4. Aug 8, 2011 #3
    Ahh I knew it was something really dumb like that! Thanks, but new question now. For the long division, what answer should I be getting? Because trying to work it through I'm not getting the right answer. I get an answer of [itex]1-\frac{1}{u}[/itex] and the remainder but I can't get the remainder (unless its [itex]\frac{1}{u^{3}+u}[/itex]). The math doesn't seem to work when I check this can you help me out? Polynomial long division was always a weakness.
     
  5. Aug 8, 2011 #4
    Whoops never mind I got it ([itex]1-\frac{1}{u^{2}+1}[/itex]) dumb mistake.
     
  6. Aug 8, 2011 #5
    You can read about polynomial long division here
    http://www.sosmath.com/algebra/factor/fac01/fac01.html" [Broken]
    But if I can avoid it, I try not to use it...
    For problems like this, there's a trick so you don't have to use long division:[tex]\frac{u^2}{u^2 + 1} = \frac{u^2 + 1 - 1}{u^2 + 1} = \frac{u^2 + 1}{u^2 + 1} - \frac{1}{u^2 + 1} = 1 - \frac{1}{u^2 + 1}[/tex]
     
    Last edited by a moderator: May 5, 2017
  7. Aug 8, 2011 #6
    OK so now I have:

    [itex]2(\frac{1}{2}u^{2}tan^{-1}(u)-\frac{1}{2}(u-tan^{-1}(u))[/itex]

    Or:

    [itex](u^{2}+1)tan^{-1}(u)-u[/itex]

    From here, I substitute sin(x) back in:

    [itex](sin^{2}(x)+1)tan^{-1}(sin(x))-sin(x)[/itex]

    Can I simplify more or am I still making a mistake? Thanks for all the help.
     
  8. Aug 8, 2011 #7
    It's correct and probably as simplified as you can make it; just remember the + c :wink:
     
  9. Aug 8, 2011 #8
    Hahah yes that darn + C :smile: Thanks for all your help!
     
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