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Integrating xsinxcosxdx

  1. Feb 11, 2009 #1
    How do you integrate

    xsinxcosxdx
     
  2. jcsd
  3. Feb 11, 2009 #2

    mathman

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    Inegrate by parts: u=x, dv=sinxcosxdx=sinxd(sinx)
    You get x(sinx)2/2 -integral of (1/2)(sinx)2dx

    You should be able to proceed (using double angle formula for cos to get rid of (sinx)2/2).
     
  4. Feb 18, 2009 #3
    Since
    2sin(x)cos(x) = sin(2x)
    you can write the integrand function
    [tex]x/2 \cdot \sin (2x)[/tex]
    you can use first the substitution
    [tex]y=2x[/tex]
    and then use integration by part formula to integrate
    [tex]y/4 \cdot \sin (y)[/tex]
    it is EASY if you choose to derive [tex]y/4[/tex] and integrate [tex]\sin(y)[/tex].
     
  5. Feb 20, 2009 #4
    You can solve any question like this by expressing sin(x), cos(x), etc in terms of their exponential form and multiplying everything out.

    cos(x) = [exp(ix)+exp(-ix)]/2
    sin(x) = [exp(ix)-exp(-ix)]/(2i)
     
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