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Integrate by parts

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data

    integral tan-1x dx
    i am supposed to integrate this by parts

    2. Relevant equations

    3. The attempt at a solution
    integral tan-1x dx = integral cosx/sinx dx
    u=cos x, du=-sin x dx
    v=ln sin x, dv= sin-1x dx
    integral cosx/sinx dx= cosx ln(sinx) - integral[ ln(sinx)(-sinx) dx]
    is this correct so far?
    Last edited: Feb 25, 2010
  2. jcsd
  3. Feb 25, 2010 #2
    Careful with your terminology, the problem is vastly different if its tan^-1(x) (arctangent) or (tan(x))^-1 (cotangent).

    If it's the inverse function, the correct fraction would be:
    [tex]\arctan{x}= \frac{{\arcsin{x}}}{\arccos{x}}[/tex]

    If it's cotangent, I would set the problem up so that you have [tex]\int{\cos{x} \cdot \csc{x}\,dx}[/tex] so that you have the form [tex]\int{U \cdot V}[/tex]
    Last edited: Feb 26, 2010
  4. Feb 25, 2010 #3
    well, it is arctan x so i need to do some recalculating now
    Last edited: Feb 25, 2010
  5. Feb 25, 2010 #4
    i cant seem to get this problem to work, it seems to repeat itself
  6. Feb 26, 2010 #5
    Ah, sorry I just worked it out and it seems that breaking arctan up wasn't the way to go. Instead, do the following:

    [tex]\int{\arctan{x} \cdot 1 \,dx}[/tex]

    This way, you can differentiate arctan and integrate 1 without having to repeat integration by parts as you would with breaking up the arctan. Let me know if you need more help.
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