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Integration techniques

  1. Mar 19, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int sin(x+\frac{\pi}{6})cosxdx[/tex]



    3. The attempt at a solution


    [tex] u= sin(x+\frac{\pi}{6})[/tex]

    [tex]du=cosxdx[/tex]

    [tex]\frac{du}{cos x}=dx[/tex]

    [tex]\int udu=\frac{1}{2}sin(x+\frac{\pi}{6})+c[/tex]

    Guys how come my answer is different from the one at the back of the book. I'm l the one wrong or this time it is the book ?
     
  2. jcsd
  3. Mar 19, 2009 #2

    Dick

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    You are, this time. If u=sin(x+pi/6) then du=cos(x+pi/6)*dx. The simple substitution doesn't work. Try using sin(a+b)=sin(a)*cos(b)+cos(a)*sin(b) on sin(x+pi/6) before integrating.
     
  4. Mar 19, 2009 #3

    lurflurf

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    sin(x+pi/6)cos(x)=(1/2)(sin(2x+pi/6)+sin(pi/6))
     
  5. Mar 19, 2009 #4

    [tex]\int (sinx cos\frac{\pi}{6}+sin\frac{\pi}{6} cos x)cosx [/tex]

    [tex]\int sinx cosx cos\frac{\pi}{6}+\int sin \frac {\pi}{6} cos^2x[/tex]

    [tex]\frac{1}{2}cos\frac{\pi}{6}\int sin(2x)+\frac{1}{2}sin\frac{\pi}{6}\int(1+cos2x)[/tex]

    Is this now correct ?
     
    Last edited: Mar 19, 2009
  6. Mar 19, 2009 #5

    Dick

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    Yes, that looks right.
     
  7. Mar 19, 2009 #6
    Thanks a lot for the help
     
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