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Finding integral sin(x) cos(x) dx

  1. Oct 26, 2005 #1
    I'm having trouble with the following integral:
    [tex]\int {\sin x \cdot \cos x{\rm{ dx}}} [/tex]
    I've tried to use integration by parts but that doesn't seem to get me anywhere...any tips?
     
  2. jcsd
  3. Oct 26, 2005 #2

    Hurkyl

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    Seems to work for me... can you show your work?

    Incidentally, there are two different ways to make this a trivial problem -- I don't really think I can hint at either without giving the answer away, so I'll just say to think about what you know about trig functions and integrals.

    (And both of these ways are almost certainly shown in your book)
     
  4. Oct 26, 2005 #3
    I define the following:
    [tex]
    \[
    \begin{array}{*{20}c}
    {u' = - \sin x} & {v = \cos x} \\
    {u = \cos x} & {v' =- \sin x} \\
    \end{array}
    \]
    [/tex]
    Then I put this into the formula:
    [tex]
    \[
    \begin{array}{l}
    \int {\sin x \cdot \cos xdx{\rm{ }} = uv - \int {uv'dx} } \\
    = \cos x \cdot \cos x - \int {\cos x \cdot ( - \sin x)dx} \\
    = \cos ^2 x - \int {\cos x \cdot ( - \sin x)dx} \\
    \end{array}
    \]
    [/tex]
    This integration at the end is no easier than what I started with...so what's wrong here?
     
    Last edited: Oct 26, 2005
  5. Oct 26, 2005 #4

    Hurkyl

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    Your v' is wrong.
     
  6. Oct 26, 2005 #5
    How about U-substitution? What can you substitute for u? And du? (major hint)
     
  7. Oct 26, 2005 #6
    Yeah ok, so it should be negative, it's fixed, but I'm not seeing that it makes it any easier though...am I even on the right track here?
     
  8. Oct 26, 2005 #7
    It seems you made a mistake in your concept of integration by parts, but for this particular problem, it yielded the same result.

    [tex]\int{udv}=uv-\int{vdu}[/tex]

    It's minus the integral of v times du, not the other way around. But you don't need to do integration by parts here. A simple U-substitution will work.
     
  9. Oct 26, 2005 #8
    Ah, I got it now. Thanks!
     
  10. Oct 26, 2005 #9
    No prob. :)
     
  11. Oct 26, 2005 #10

    Hurkyl

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    Once you get the sign right, you can solve the equation for the integral. (Remember that you can do a lot more to an equation than just simplifying the RHS)
     
  12. Oct 27, 2005 #11
    Try: u = sinx du = cosxdx
    or try the fact that sin2x = 2sinxcosx
     
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