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Indefinite trigonometric integral with an Nth Root

  1. Feb 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Solve: [itex] \int sin(16x) \sqrt[a]{cos(16x)}\,dx [/itex] Answer should be linear in the constant "a"

    3. The attempt at a solution

    [itex] \int sin(16x) \sqrt[a]{cos(16x)}\,dx[/itex] Set: [itex]u=cos(16x), du=-16sin(16x) du[/itex] [itex] ~~\Rightarrow~~ {-1/16}\int \sqrt[a]{u}\,du = {-1/16}(\frac{1}{1+1/a}u^{1+{1/a}})+C = {-1/16}(\frac{1}{1+1/a}cos^{1+{1/a}}(16x))+C[/itex]

    This is what I did initially, then i realized it wasn't linear in the constant "a".
    I have a feeling i may need to substitute a log function in to make "a" a linear constant.
     
    Last edited: Feb 16, 2014
  2. jcsd
  3. Feb 16, 2014 #2

    Dick

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    What you did is fine, except you mixed up some dx's and du's in the presentation. I have no idea what they mean by "Answer should be linear in the constant a". It's clearly not.
     
  4. Feb 16, 2014 #3
    Thanks for the reply

    The question initially doesn't say "Answer should be linear in the constant a", but my homework website doesn't accept "+C" in this question, and gives me that message as a response when i put in my answer.

    Normally I would see my professor about this, but it is reading week.
     
  5. Feb 16, 2014 #4

    Dick

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    Well, there should be a "+C" in the answer if it's an indefinite integral, and I think you did it correctly. Not sure what to say.
     
  6. Feb 16, 2014 #5
    I can only think of two things. The site might require a more standard form after simplifying, and it's possible you inputted the answer incorrectly. How do you input your answers? Is it text based or is there a GUI for you to use?

    For example, if you input cos^(1/2)(x) into WolframAlpha, it reads it as ##\sqrt{\cos{x}} \cdot x## where you might have wanted it to be intepretted as simply ##\sqrt{\cos{x}}##.

    (I brought up this example because this is how you formatted your answer to the problem)
     
  7. Feb 16, 2014 #6

    SammyS

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    ##\displaystyle \frac{1}{1+1/a} = \frac{a}{a+1} ##

    Maybe the website would like this better.
     
  8. Feb 16, 2014 #7

    Dick

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    Well, that's not "linear in a" either. Who knows what the website would like?
     
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