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Math Help - Using substitution to prove two equations

  1. #1
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    Using substitution to prove two equations

    Another question, but this time into this sub-forum. So I'm studying for IB finals and I'm stuck with this question:

    a. Given that a>1, use the substitution u=1/x to show that
    integral(1-->a)[1/(1+x^2)]dx=integral(1/a-->1)[1/(1+u^2)]

    I'm able to simplify the expression to: arctan a - arctan 1 = arctan 1 - arctan a. Should I do it this way or is there some other way to do it and how do I get it to actually equal to each other do I have to use the information that a>1?

    b. Hence show that arctan a + arctan 1/a = pi/2

    So I see that arctan 1 = pi/4 right? So that would be okay (add both sides you get pi/2), but since I have to use the substitution arctan for the RHS will also be arctan a instead of arctan 1/a (it's a definite integral) so is there something I should do?

    Thanks in advance.
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  2. #2
    Super Member Random Variable's Avatar
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     \arctan(a) - \arctan(1) = \displaystyle \int^{a}_{1} \frac{dx}{1+x^{2}} = -\int^{\frac{1}{a}}_{1} \frac{1}{1+ \frac{1}{u^{2}}} \frac{du}{u^{2}}

     \displaystyle = \int_{\frac{1}{a}}^{1} \frac{du}{u^{2}+ 1} = \arctan(1) - \arctan(\frac{1}{a})

    therefore,  \displaystyle \arctan(a) + \arctan(\frac{1}{a}) = 2 \arctan(1) = 2 * \frac{\pi}{4} = \frac{\pi}{2}
    Last edited by Random Variable; March 13th 2011 at 08:26 AM.
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  3. #3
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    Can you still clarify a bit how do I exactly prove for part a., because I don't understand what you did on the first line and it doesn't seem to me to prove anything? Thanks.
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  4. #4
    Super Member Random Variable's Avatar
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    It's just asking you to show that if you make that particular substitution, you get that equivalent integral. Then in part (b) you need to evaluate both integrals and set them equal to each other.
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  5. #5
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    I really seem to be in need of some serious revision, because I don't understand why the integral get's a minus in front of it and then the limits are 1 to 1/a, and I'm not sure I understand why you add du/u^2 at the end (namely the u^2 in the denominator)?
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  6. #6
    Super Member Random Variable's Avatar
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    If you let  \displaystyle u = \frac{1}{x} , then  \displaystyle x= \frac{1}{u} and  \displaystyle dx = - \frac{1}{u^{2}}  \ du .

    The lower limit becomes  u = \frac{1}{1} = 1 and the upper limit becomes  u = \frac{1}{a} .
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  7. #7
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    Quote Originally Posted by Random Variable View Post
     \displaystyle dx = - \frac{1}{u^{2}}  \ du .
    This seems like something that's never been taught to us, or in some respect it makes some kind of sense, but what constitutes the minus? I guess I shouldn't have posted in the university math section, but this is what you get when they give this kind of questions to you in IB finals :P
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  8. #8
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    I can't believe that you have never been taught that \frac{d x^n}{dx}= n x^{n-1} because that is what is being used here. With u= \frac{1}{x}= x^{-1}., then x= u^{-1} so that \frac{dx}{du}= -u^{-2}= -\frac{1}{u^2} and dx= -\frac{1}{u^2}du.
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