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Math Help - Integration problem

  1. #1
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    Integration problem

    Hi Im having a bit of trouble getting started on the following integration problem,

    any hints would be great

    integral of arctanx/1+x^2 dx
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  2. #2
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    Can i split it up and do 1/(1+x^2) dx and arctanx dx? then multiply the results?
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  3. #3
    MHF Contributor Calculus26's Avatar
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    No there is no product rule or quotient rule for integration

    Let u = arctan(x)

    Then du = 1/(1+x^2) dx

    int(arctan(x)/(1+x^2)dx) = int (udu) = u^2/2 +C = [arctan(x)^2]/2+C
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  4. #4
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    Quote Originally Posted by Calculus26 View Post
    No there is no product rule or quotient rule for integration
    Except that the product rule is exactly what you use, and work backwards through, whenever you do integration by parts. E.g., Balloon Calculus Forum - Viewing Thread

    And the quotient rule is just a special case of the product rule. You might integrate the square of cosec x by working back through it. Balloon Calculus: standard integrals and derivatives

    While I'm here, here's (or will be in two tics) a picture for the present (chain rule) problem, just in case it helps...



    Don't integrate - balloontegrate

    Balloon Calculus Forum
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  5. #5
    MHF Contributor Calculus26's Avatar
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    I agree that int by parts is a product rule of sorts for certain types of products there is no product rule that works on every product as with differentiation.

    In the same light partial fraction decompostion is a quotient rule if you will for rational functions but again there is no integration rule for quotients in general as with differentiation
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  6. #6
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    Quote Originally Posted by Calculus26 View Post
    I agree that int by parts is a product rule of sorts for certain types of products there is no product rule that works on every product as with differentiation.
    Point taken. A product is just as likely to be (as in the present problem) the result of a chain-rule differentiation as a product-rule one. There's no version of either kind of rule for integration, though integration is nearly always a matter of working back through those two rules for differentiation.
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  7. #7
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    hmm, still a bit unsure about this, ill try woking through it and post later to let you know how i got on, thanks everyone.
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  8. #8
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    Quote Originally Posted by Calculus26 View Post
    No there is no product rule or quotient rule for integration

    Let u = arctan(x)

    Then du = 1/(1+x^2) dx

    int(arctan(x)/(1+x^2)dx) = int (udu) = u^2/2 +C = [arctan(x)^2]/2+C
    In this substitution why cant i take 1+x^2 as u?
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  9. #9
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    Quote Originally Posted by offahengaway and chips View Post
    In this substitution why cant i take 1+x^2 as u?
    Exactly! That was what Calculus26 was saying. Unfortunately, tom then started talking about "integration by parts", confusing things.
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  10. #10
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    so it doesnt matter what value i take as u? is that right?
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  11. #11
    MHF Contributor Calculus26's Avatar
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    No No No

    Lets look at this again

    when you take u = arctan(x)

    du = 1/(1+x^2)dx which yields (1+x^2)du = dx

    int(arctan(x)/(1+x^2)dx = int (u (1+x^2)/(1+x^2)dx) = int(udu) which can be easily integrated

    If you let u = 1 + x^2 then du = 2x dx

    you get int (arctan(x)/(2xu)du) which is a mess.

    The idea of u substitution is to obtain an integral in terms of u which you can integrate.

    In general there is only one choice maybe two which will accomplish this.

    What substitution to make ? thatg comes from experience , insight, and good ol' trial and error.
    Last edited by Calculus26; July 8th 2009 at 04:32 AM.
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  12. #12
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    Ok so

    int (arctanx/1+x^2) dx

    u = arctanx

    du/dx = 1/1+x^2

    so:

    int(arctanx/1+x^2) dx = this is where i am stuck
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  13. #13
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    Quote Originally Posted by offahengaway and chips View Post
    Ok so

    int (arctanx/1+x^2) dx

    u = arctanx

    du/dx = 1/1+x^2

    so:

    int(arctanx/1+x^2) dx = this is where i am stuck
    \int{\frac{\arctan{x}}{1 + x^2}\,dx} = \int{\arctan{x}\,\frac{1}{1 + x^2}\,dx}.


    Let u = \arctan{x} so that \frac{du}{dx} = \frac{1}{1 + x^2}.

    So \int{\arctan{x}\,\frac{1}{1 + x^2}\,dx} = \int{u\,\frac{du}{dx}\,dx}

     = \int{u\,du}

     = \frac{1}{2}u^2 + C

     = \frac{1}{2}(\arctan{x})^2 + C.
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  14. #14
    MHF Contributor Calculus26's Avatar
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    Ok the method is called substitution because we substitute

    we have int (arcatan(x)/(1+x^2)dx

    We'll go one step at a time

    Since u = arctan(x) substitute arctan(x) with u

    int (arcatan(x)/(1+x^2)dx = int(u/(1+x^2)dx)

    Now dx = (1+x^2)du so replace dx with this result

    int(u/(1+x^2)dx) = int(u/(1+x^2) *(1+x^2)du)

    Note (1+x^2)/(1+x^2) cancels leaving int(udu)
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  15. #15
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    so int udu= int(arctanx)du

    what is du?
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