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Help with an integral

  1. May 16, 2005 #1
    Could anyone help me solve the following integral?

    [tex]\int \sin(2x).\sin(x)dx[/tex]

    Thanks
     
    Last edited: May 16, 2005
  2. jcsd
  3. May 16, 2005 #2

    arildno

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    Use the product/sum formula for trigonometric functions.
     
  4. May 16, 2005 #3
    Do you mean [tex]sin(2x)=2.sinx.cosx[/tex] ? Like this:

    [tex]\int 2sinx.cosx.sinx.dx[/tex]
    [tex]= 2 \int sin^2x.cosx.dx[/tex]
    [tex]= 2 \int (1 - cos^2x)cosx.dx[/tex]
    [tex]= 2 \int cosx dx - 2\int cos^3x.dx[/tex]

    Is that what you mean? I can solve the first integral... But not the second.
     
  5. May 16, 2005 #4

    arildno

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    It wasn't what I meant, but since your own procedure can be used as well, I'll help you out along the track you chose.
    You have established:
    [tex]\int\sin(2x)\sin(x)dx=2\int\sin^{2}x\cos(x)dx[/tex]
    Now, use the substitution
    [tex]u=\sin(x)[/tex]
    Then, we have:
    [tex]\frac{du}{dx}=\cos(x)\to{dx}=\frac{du}{\cos(x)}[/tex]
    Thus, we have gained:
    [tex]\int\sin(2x)\sin(x)dx=2\int{u}^{2}du=\frac{2}{3}u^{3}+C=\frac{2}{3}\sin^{3}x+C[/tex]
     
  6. May 16, 2005 #5
    Thanks! It's actually easier than I thought.
     
  7. May 16, 2005 #6

    arildno

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    If you are interested in another way to do this, we have for any choices a,b the equalities:
    [tex]\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b),\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)[/tex]
    Combining these, we gain:
    [tex]\sin(a)\sin(b)=\frac{1}{2}(\cos(a-b)-\cos(a+b))[/tex]
    That is,
    [tex]\sin(2x)\sin(x)=\frac{1}{2}(\cos(x)-\cos(3x))[/tex]

    This is what I meant with "using the product/sum formula".
     
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