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Integral Question

  1. Jun 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Integrate[Cos[(2x+3)^(1/3)]


    2. Relevant equations



    3. The attempt at a solution

    Would I use simple substitution?

    Would it just simply be 1/2*Sin((2x+3)^(1/3))??
     
  2. jcsd
  3. Jun 12, 2009 #2

    Cyosis

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    Homework Helper

    You will have to use a substitution yes. No your answer is wrong. When you're integrating always differentiate your final answer to see if it gives the correct result.

    That said do you have any ideas about the kind of substitution you want to use?
     
  4. Jun 12, 2009 #3
    Your proposed solution completely disregards the 1/3 power.

    I would suggest integration by parts.
     
  5. Jun 12, 2009 #4
    Indeed.

    U=(2x+3)^1/3

    Du= ((2x+3)^(-2/3))/(2/3)

    dv=Cos(X)

    V= Sin(x)

    then use U*du=d*v-integral[du*v] ???
     
  6. Jun 12, 2009 #5

    Cyosis

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    The substitution is correct, but your du is not. The expression should be multiplied by 2/3, not divided.

    So we have the following:

    [tex]
    du=\frac{2}{3} \left(\frac{1}{(2x+3)^{\frac{1}{3}}}\right)^2 dx
    [/tex]

    Now write the bracket expression in terms of u.
     
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