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Integral of sine function

  1. Nov 30, 2014 #1
    Find the value of the integral:
    a) ∫0π(sinx + 2)dx

    Formula I found:
    integral.gif sin x dx = -cos x + C

    My calculation: F(x) = -cosx + 2x
    => (-cosπ + 2π)-(-cos0) = -1 + 2π + 1 = 2π , but the solution should be 2π +2

    b) ∫0sin(x/2)dx

    My calculation: F(x) = -cosx/2
    => -cosπ + cos0 = 0 ; but the solution should be 4

    What did I wrong in those equations? Can anyone help?
     
  2. jcsd
  3. Nov 30, 2014 #2

    SteamKing

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    sin x + 2 ≠ sin x

    You can't pretend the 2 doesn't exist and then ignore it when you integrate.
     
  4. Nov 30, 2014 #3

    Simon Bridge

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    Take care: cos(0)=1, cos(pi)=-1
     
  5. Dec 1, 2014 #4

    pasmith

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    [itex]-\cos\pi = -(-1) = 1[/itex].

    The integral of [itex]\sin(x/2)[/itex] is [itex]-2\cos(x/2) + C[/itex].
     
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