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Integration by Substitution

  1. Jan 28, 2007 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation.

    dy/dx = 4x + 4x/square root of (16-x^2)

    2. Relevant equations

    Substituting using U...

    3. The attempt at a solution

    I'm not sure if that's what I am supposed to do, but I tried using the U substitution...

    4x + 4x/square root of (16-x^2)
    u = 16 - x^2
    du = -2x
    -2du = 4x (I multiplied both sides by 2 and brought the negative over)

    -2 (integration sign [the S]) du + du/square root of u

    Take the anti-derivative...

    -2 x (u)^(-1/2) + C (changed square root into a power)

    -2 x (16-x^2) + C

    Unfortunately, that is not the answer they give me. I'm kinda stuck.. am I going in the right direction?
     
  2. jcsd
  3. Jan 29, 2007 #2
    Everything fell apart right after you made the substitution. Stating that "du = -2x" is incorrect. Rather, du = -2x dx. Then, you find an expression for "dx", sub that in, and integrate with respect to u.

    Edit: It's easier to split the integral into a sum of two integrals, so it's the integral of 4x + the other integral.
     
    Last edited: Jan 29, 2007
  4. Jan 29, 2007 #3
    dy/dx = 4x + 4x/square root of (16-x^2)

    I'd imagine that partial fractions would help here.

    It would be alot easier, I'll have a play with it later. Time for work. Just a thought though.

    [tex]\int 4x+=>[/tex]

    [tex]2x^2+C[/tex]

    [tex]\int \frac {4x}{\sqrt(16-x^2)}=>[/tex]

    Actually from here just substitute du from here.
     
    Last edited: Jan 29, 2007
  5. Jan 29, 2007 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, not partial fractions with that "x" in the numerator.

    Let u= 16- x2. Then du= -2x dx so that -2du= 4xdx.
     
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