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Problem with integration

  1. Aug 10, 2007 #1
    Hi guys. I have a problem with an integration problem I've recently stumbled upon.

    [tex]\int \frac{1}{sin^{-1} x\sqrt{1-x^2}} dx}[/tex]

    This is directly integrable, but if I do it by parts, I get a weird result of [tex]\int \frac{1}{sin^{-1} x\sqrt{1-x^2}} dx} = 1 + \int \frac{1}{sin^{-1} x\sqrt{1-x^2}} dx}[/tex]!! Why is that so?

    EDIT: Why doesn't my LaTeX work?!
     
    Last edited: Aug 10, 2007
  2. jcsd
  3. Aug 10, 2007 #2

    arildno

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    "Hi guys. I have a problem with an integration problem I've recently stumbled upon.

    [tex]\int\frac{1}{sin^{-1} x\sqrt{1-x^2}} dx}[/tex]

    This is directly integrable, but if I do it by parts, I get a weird result of [tex]\int\frac{1}{sin^{-1} x\sqrt{1-x^2}} dx} = 1 + \int\frac{1}{sin^{-1} x\sqrt{1-x^2}} dx}[/tex]!! Why is that so?

    EDIT: Why doesn't my LaTeX work?!"
    Use lower case letters, tex, rather than TeX
     
    Last edited: Aug 10, 2007
  4. Aug 10, 2007 #3

    arildno

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    fffffffffggggggggggggg
     
  5. Aug 10, 2007 #4
    Thanks arildno :)
     
  6. Aug 10, 2007 #5

    CompuChip

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    [tex]\frac{1}{\sin^{-1} x\sqrt{1-x^2}}[/tex]
    is that
    [tex]\frac{1}{\sin^{-1}(x)} \cdot \frac{1}{\sqrt{1-x^2}}[/tex]
    or
    [tex]\frac{1}{\sin^{-1}\left( x\sqrt{1-x^2} \right)}[/tex]
     
    Last edited: Aug 10, 2007
  7. Aug 10, 2007 #6

    [tex]\frac{1}{sin^{-1}(x)} \cdot \frac{1}{\sqrt{1-x^2}}[/tex]

    this.
     
  8. Aug 10, 2007 #7

    arildno

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    Remember that:
    [tex]\frac{d}{dx}\sin^{-1}(x)=\frac{1}{\sqrt{1-x^{2}}}[/tex]
     
  9. Aug 10, 2007 #8
    Yes if you do it by parts, we integrate [tex]\frac{1}{\sqrt {1-x^{2}}}[/tex]

    Then we differentiate [tex]\frac{1}{sin^{-1}x}[/tex]. Then you'll end up with something weird, 1=0?!
     
    Last edited: Aug 10, 2007
  10. Aug 10, 2007 #9
    But why would you try to use integration by parts for this, when this problem practically begs for a different method? Have you tried making a substitution?
     
  11. Aug 10, 2007 #10
    To answer the OP's question, integrals can differ by a constant. You can't just "cancel" your integrals in the equation where you seem to have 1=0. If you differentiate both sides of the equation, everything works out nicely. At least in basic calculus, the indefinite integral is simply defined to be the inverse of the derivative. Because the derivative of a constant is 0, indefinite integrals must be defined only up to a constant.

    Also, if you actually want to evaluate the integral, listen to d_leet.
     
  12. Aug 10, 2007 #11

    CompuChip

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    The same "problem" as in the original post can be much easier seen by integrating 1/x by parts (writing it as [itex]1 \cdot 1/x[/itex]), which will also give you
    [tex]\int \frac{dx}{x} = 1 + \int \frac{dx}{x}[/tex]
    and hence seems to prove that 1 = 0.
     
  13. Aug 11, 2007 #12
    to the OP: if y=arcsin(x) what is y'
     
  14. Aug 12, 2007 #13

    Gib Z

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    He made it clear he knew how to do this with a substitution in his first post!! musicheck's post is exactly where the OP needed to hear.
     
  15. Aug 14, 2007 #14
    Thank you musicheck. And to the other posters: thanks, but I do know that this expression is directly integrable.

    Musicheck, what you mean is that indefinite integrals can differ simply by a constant, since on both sides of the integral we will get arbitary constants, right?

    So what I am assuming here is that integration by parts will not always lead us to an answer (I know sometimes it does lead to a more complicated expression to integrate, but here, it leads to the indefinite integral repeating itself? Why is that so? Is there something to do with the definition of integration by parts itself?
     
  16. Aug 14, 2007 #15
    I think he meant that integral expressions are an equivalence class of functions such that two functions f and g belong to the same equivalence class iff f(x)+c(x)=g(x) for some constant function c(x) (meaning that c'(x)=0).

    For example, the derivative of x+C where C is an arbitrary constant in respect to x is 1. Therefore the antiderivative of 1 is x+C. This could be just x, x+1, or any other x+C since they all have the same derivative.
     
    Last edited: Aug 14, 2007
  17. Aug 15, 2007 #16

    Mute

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    I don't think you got what Musicheck was getting at. What he was saying is that when you integrate your function by parts the first time, you're neglecting the arbitrary constant that comes out of the integration which produces the first term on the right hand side. Using the 1/x example, you don't actually get

    [tex]\int \frac{dx}{x} = 1 + \int \frac{dx}{x}[/tex]

    Rather, what you get is

    [tex]\int \frac{dx}{x} = (1 + C) + \int \frac{dx}{x}[/tex]

    where C is an "arbitrary" constant which is equal to -1, as you can see by cancelling the indefinite integrals on both sides. Plugging C = -1 into that expression, you get

    [tex]\int \frac{dx}{x} = \int \frac{dx}{x}[/tex]

    which is what we expect.

    To reinforce this, consider a definite integration by parts:

    [tex]\int_1^t \frac{dx}{x} = x \frac{1}{x}|_1^t + \int_1^t \frac{dx}{x} = (1-1) + \int_1^t \frac{dx}{x} = \int_1^t \frac{dx}{x}[/tex]

    Integrations by parts reproducing the original integral in the expansion is also fine in general. Consider:

    [tex]\int_{-\infty}^x e^t \cos t dt = e^t \cos t |_{-\infty}^{x} + \int_{-\infty}^x e^t \sin t dt[/tex]

    = [tex]e^x \cos x + \left( e^t \sin t |_{-\infty}^{x} - \int_{-\infty}^x e^t \cos t dt \right)[/tex]

    The original integral is on both sides of the expression, but with opposite signs, so you get

    [tex]\int_{-\infty}^x e^t \cos t dt = \frac{1}{2}\left(e^x \cos x + e^x \sin x \right)[/tex]

    So, integrals being repeated during integrations by parts is fine (even in the indefinite integral case, so long as you are careful with the arbitrary constants).
     
    Last edited: Aug 15, 2007
  18. Aug 20, 2007 #17
    Ah okay I see what you are going at. But arbitrary constants can appear on both the LHS and RHS of the equations, that's what I was trying to mean.
     
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