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Math Help - Is thins transformation valid ?

  1. #1
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    Is thins transformation valid ?

    Okay , i have the Integral of xdx/[(x-2.5)^2-6.25] Substitution |t=x-2.5;dt=dx;x=t+2.5|
    = Integral of tdt/(t^2-6.25) + 2.5 * Integral of dt/(t^2-6.25) ; So i can solve the first one with substitution , and the second part is where my question comes out.
    Can i return the substitution of t back as 2.5 * Integral of dx/[(x-2.5)^2-6.25]

    Seems pretty fine to me, but had to be sure if i`m not making any mistake.
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  2. #2
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    Re: Is thins transformation valid ?

    Writing this "pretty", you want to integrate \int \frac{x dx}{(x- 2.5)^2- 6.25}. (I presume you "completed the square" to get that denominator.)

    Yes, you can make the substitution u= x- 2.5 so that du= dx and x= u+ 2.5. That makes the integral \int \frac{(u+ 2.5)du}{u^2- 6.25}= \int \frac{u du}{u^2- 6.25}- 2.5\int \frac{du}{u^2- 6.25}

    For the second of those integrals you are asking "Can i return the substitution of t back as 2.5 * Integral of dx/[(x-2.5)^2-6.25]?"

    You could but there is no point! It is simpler to do as is:
    2.5 \int \frac{du}{u^2- 6.25}= 2.5\int \frac{du}{(u- 2.5)(u+ 2.5)}
    Use "partial fractions".
    Thanks from MirceM
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  3. #3
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    Re: Is thins transformation valid ?

    Yea i didnt thought of that too.. Actually now i see i wouldn`t make any change (facepalm).. At first i used a trigonometric substitution as in your example u=sqrt(6.25)secZ ; du= sqrt(6.25)secZtanZdZ ; but that drove me to a more complicated integrals + because this is a definite integral i`m solving i forgot that to use the substitution u=sqrt(6.25)secZ , there is condition u>=a which in my case wasn`t. So i returned backwards and the first thing i though was mentioned in my question.

    Anyway thanks again, i feel like i still don`t have the sense of some quick and easy transformations in integrals, so sometimes i must do the hard way first .
    Last edited by MirceM; August 2nd 2014 at 08:33 AM.
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  4. #4
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    Re: Is thins transformation valid ?

    But just saying, I always find things like S dx/(x^2-a^2) very sketchy since you can get coth^-1 or tanh^-1 depending on the domain...
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