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Finding the Indefinite Integral Extension Questions

  1. May 21, 2012 #1
    1. The problem statement, all variables and given/known data
    ∫8x3e-cos(x4+4)sin(x4+4)dx


    2. Relevant equations
    Let u = cos(x4+4)


    3. The attempt at a solution
    I know the answer does not have the sin in it and only the e remains, because when the integral is found e stays unchanged.
    I could find somewhere online to calculate it, but I want to know how to do it, because questions like this will be on the final exam.
     
  2. jcsd
  3. May 21, 2012 #2

    Dick

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    You have the correct u substitution. Now use it. What's du?
     
  4. May 21, 2012 #3

    micromass

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    So, what do you get after that u substitution??
     
  5. May 21, 2012 #4
    So I've found du and put it into the form

    ∫udv = uv - ∫vdu

    so for v I have 8x3

    du = -sin(4x3) dx

    dv = 24x2 dx

    Which gives me

    ∫cos(x4+4)24x2dx = cos(x4+4)8x3 - ∫8x3(-sin4x3) dx

    But I feel like I'm going down the wrong path, especially since the e isn't present
     
  6. May 21, 2012 #5

    Dick

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    You are going down the integration by parts path. It's not the right way to go. Why do you think du=(-sin(4x^3))dx?
     
  7. May 22, 2012 #6
    So using substitution

    du = sin(x4+4)

    because cos converted to sin doesn't change its sign, I thought it did but its for converting from cos to sin.

    t = u

    ∫f'(g(x))g'(x) dx = ∫f'(t) dt/dx = ∫ f'(t)dt = f(t) + C

    = f(g(x)) + C when substituting back in

    f(x) = original equation and
    t = u = cos(x4+4)
     
  8. May 22, 2012 #7

    Dick

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    df(x)=f'(x)*dx. f(x) here is cos(x^4+4). What is df(x)?
     
  9. May 22, 2012 #8
    so f(x) = cos(x4+4)

    then

    f'(x) = sin(x4+4) + 4x3

    which gives

    ∫cos(x4+4) +4x3 *sin(x4+4) + C
     
  10. May 22, 2012 #9

    Dick

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    I'd suggest you review the chain rule. If f(x)=cos(x^4+4) then the derivative f'(x) is not what you wrote. Give it another try. Tell me what the chain rule says.
     
    Last edited: May 22, 2012
  11. May 22, 2012 #10
    Chain Rule states

    dy/dx = dy/du * dx/du

    ∫8x3e-cos(x4+4)sin(x4+4)

    8∫x3∫e-cos(x4+4)∫sin(x4+4)

    8∫x4/4∫e-cos(x4+4)∫cos(x4+4)+4x3
     
  12. May 22, 2012 #11

    Dick

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    Stop worrying about the integral for a while. You are having trouble differentiating f(x)=cos(x^4+4). Try and get that right first.
     
  13. May 22, 2012 #12
    Finding the derivative of cos(x4+4)

    I got -4x3sin(x4+4)

    Do I then follow the formula above with f(x) and t and then solve for the Integral.
     
  14. May 22, 2012 #13

    Dick

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    Ok, you've got the derivative. If you take u=cos(x^4+4), that means du=(-4x^3)sin(x^4+4)dx. Now you should be able to turn the integral into
    [tex]C \int e^{-u}du[/tex]
    where C is a numerical constant. This is u-substitution.
     
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