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Math Help - u-substitution problem

  1. #1
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    u-substitution problem

    Hi,

    I want to use u-substitution in my self-study-case.

    If I have equation: \int (x^2-1)^2\:dx and then I substitute:

    let\:\:u = (x^2-1)

    du \:=\: 2x \:dx

    dx\:=\:\frac{du}{2x}

    And then I plugin the u:
    \int (u)^2\: \frac{du}{2x}\:\:\Rightarrow \:\:\frac{1}{2x}\:\int (u)^2\:du\:\:= \:\:\frac{1}{2x}\:\times\frac{(u)^3}{3}\:

    And finally plugin the original x:
    \frac{1}{2x}\:\times\frac{(x^2-1)^3}{3}\:

    Whats wrong with my substitution. I want to use let\:\:u = (x^2-1) Is this substitution even possible?
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  2. #2
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    I dont know if substitution would work here, but not 100% sure. I know you could easily open up the brackets to find the integral to be: \frac {x^5}{5} - \frac {2x^3}{3} + x
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  3. #3
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    Quote Originally Posted by jhomie View Post
    I dont know if substitution would work here, but not 100% sure. I know you could easily open up the brackets to find the integral to be: \frac {x^5}{5} - \frac {2x^3}{3} + x
    Yes that is true. I still looking solution for the substitution whether it's possible or not.
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  4. #4
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    Quote Originally Posted by tabularasa View Post
    Yes that is true. I still looking solution for the substitution whether it's possible or not.
    No it is not possible. "Substitution" is essentially the reverse of "chain rule" in differentiation:
    If f is a function of u and u is a function of x, then df/dx= (df/du)(du/dx)
    When differentiating, we are free to just go ahead and multiply that "du/dx" in.

    If we have an integral \int f(u) (du/dx)dx, that is the same as [tex]\inf f(u)du[/itex]. Unfortunately, if that "du/dx" is not inside the integral to begin with, we cannot just put it in: we cannot, for example, say "if u= x^2-1 so that du= 2x dx so \int (x^2- 1)^2 dx= \frac{1}{2x}\int (x^2-1)^2 (2xdx)= \frac{1}{2x}\int u^2 du because you cannot cancel an x outside the integral with an x inside the integral.

    You CAN cancel constants inside and outside the integral because [tex]\int c f(x)dx= c\int f(x)dx[/itex] for c a constant. If that problem were \int (3x-1)^2 dx, then we could say "let u= 3x- 1 so du= 3dx. Then \int (3x-1)^2 dx= \frac{1}{3}int (3x-1)^2 (3dx)= \frac{1}{3}\int u^2 du.

    But in general we cannot make the subsitution u= f(x) unless du/dx is alread inside the integral. That's one thing that make integration so much harder than differentiation.
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  5. #5
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    You cannot do that:
    <br />
\int (u)^2\: \frac{du}{2x}\:\:\Rightarrow \:\:\frac{1}{2x}\:\int (u)^2\:du\:\:= \:\:\frac{1}{2x}\:\times\frac{(u)^3}{3}\:<br />

    You should not treat x as a constat, because it depends on u. The only way to do that would be to express x in a terms of u. You can do that solving the equation for x:
    x^2 - (u+1) = 0 since u+1 >0 we get

    x = \pm \sqrt{u+1}

    Hence the integral is
    <br />
\frac{1}{2}\int u^2\: \frac{du}{\pm \sqrt{u+1}}

    Good luck.
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  6. #6
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    Quote Originally Posted by albi View Post
    You cannot do that:
    <br />
\int (u)^2\: \frac{du}{2x}\:\:\Rightarrow \:\:\frac{1}{2x}\:\int (u)^2\:du\:\:= \:\:\frac{1}{2x}\:\times\frac{(u)^3}{3}\:<br />

    You should not treat x as a constat, because it depends on u. The only way to do that would be to express x in a terms of u. You can do that solving the equation for x:
    x^2 - (u+1) = 0 since u+1 >0 we get

    x = \pm \sqrt{u+1}

    Hence the integral is
    <br />
\frac{1}{2}\int u^2\: \frac{du}{\pm \sqrt{u+1}}

    Good luck.

    Thank you HallsofIvy and albi! This will help me a lot...I think . Self-studying contines...
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