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Integration by parts.

  1. Dec 17, 2007 #1
    how would one integrate by parts the following:
    [tex]\int sin^2xdx[/tex]

    thanks!
     
  2. jcsd
  3. Dec 17, 2007 #2

    rsm

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    answer

    hi
    use the fact that sin^2 x = (1-cos2x)/2
    from the formula cos2x=1-2sin^2 x

    Tell me how you wrote that equation
     
  4. Dec 18, 2007 #3

    malawi_glenn

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  5. Dec 18, 2007 #4

    HallsofIvy

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    Are you required to use integration by parts? As rsm said, there are simple and standard substitutions for [itex]sin^2(x)[/itex] and [itex]cos^2(x)[/itex].

    If you are required to use integration by parts, then, since integration by parts requires a product, the obvious thing to do it write this as a product:
    [tex]\int sin^2(x) dx= \int (sin(x))(sin(x) dx)[/tex]
    Let u= sin(x) and let dv= sin(x) dx. Then du= cos(x)dx and v= -cos(x)
    [tex]\int sin^2 x dx= -sin(x)cos(x)+ \int cos^2(x) dx[/tex]
    Now do the same thing with that integral. Of course, what happens is you will get back to your original [itex]\int sin^2(x) dx[/itex]- but with a lot of other things. Solve that equation algebraically for [itex]\int sin^2(x)dx[/itex]
     
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