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Math Help - Calculus - Integration

  1. #16
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    must tan be part of the solution, I haven't learned that yet and it is hard for me to grasp! Will I not get the solution if I do not use arctan or take a trigonometric approach. I have only gone as far as intgrating sin and cos back and forth.

    where does the r come from? for u^r?
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  2. #17
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    YES! Because the antiderivative of \frac{1}{1 + x^2} IS \arctan{x}. There are no two ways about it.
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  3. #18
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    Ok, ok.. I'm seeing this.. thank you all!
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  4. #19
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    Quote Originally Posted by bhuang View Post
    I didn't have a u. I went from 1-1/(1+x^2) and I integrated.
    1 becomes x, when integrated. And 1/(1+x^2) I made (1+x^2)^-1 so I could integrate that into ln(1+x^2).
    To differentiate ln(1+ x^2) with respect to x, you would have to use the chain rule. Its derivative is not just \frac{1}{1+ x^2} but that times the derivative of 1+ x^2: \frac{2x}{1+ x^2}.

    That is why, to integrate \int \frac{2x}{1+ x^2}dx you would make the substitution u= 1+ x^2 so that du= 2xdx and not only replace 1+ x^2 in the integral with u but replace 2x dx with du

    "Substitution" in integration is the reverse of the "chain rule" in differentiation. But, unlike differentiation, instead of "adding something", you are "taking away something" and that "something" already has to be there! You can move constants in and out of the integral but not functions of the variable.

    You could, for example, integrate \int\frac{x}{1+ x^2}dx by writing it as \frac{1}{2}\int\frac{2x}{1+x^2}dx and using the substitution u= 1+ x^2 but you cannot do that with \int \frac{1}{1+ x^2} dx because there is no "x" in the numerator and you cannot put one there.

    You really need to know that \int\frac{1}{1+ x^2}dx= tan^{-1}(x)+ C to do this problem. I suggest you talk to your teacher about why this problem was given if you have not yet studied the derivative of arctangent.
    Last edited by mr fantastic; January 10th 2010 at 12:36 PM. Reason: Added a math tag
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