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Easy question du= cos(x)dx

  1. Jul 2, 2008 #1
    We know that

    [tex]\cos{x}\sin{x}=\frac{\sin{2x}}{2}[/tex]

    so why isnt

    [tex]\int\cos{(x)}\sin{(x)}dx=\int\frac{\sin{(2x)}}{2}dx[/tex]

    ?
     
  2. jcsd
  3. Jul 2, 2008 #2

    HallsofIvy

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    They are the same. Why do you think they aren't? It's easy to integrate both sides and then show that the integrals are the same.

    For example, on the left side, you can let u= sin(x) so that du= cos(x)dx and the integral becomes [itex]\int u du= (1/2)u^2+ C= (1/2)sin^2(x)+ C[/itex]

    The integral on the right is simply [itex]-(1/4)cos(2x)+ C'[/itex]. Not obvious that those are the same? cos(2x)= cos2(x)- sin2(x)= (1- sin2(x))- sin2(x)= 1- 2sin2(x) so [itex]-(1/4)cos(2x)+ C'= 1+ (1/2)sin^2(x)+ C'[/itex] which is exactly the same as the first with C= 1+ C'.
     
  4. Jul 2, 2008 #3
    Ah okay, I see. Thank you.
     
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