What I've been waiting for all my life...

Discussion in 'Space Junk' started by EatMeReturns, Sep 4, 2011.

What I've been waiting for all my life...

Discussion in 'Space Junk' started by EatMeReturns, Sep 4, 2011.

  1. EatMeReturns

    EatMeReturns Happy Mapper Moderator

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    From:
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    indefinite integral(e^(3x)sin(4x)dx)
    integrate by parts:
    u=e^(3x) dv=sin(4x)dx
    du=3e^(3x)dx v=(-1/4)cos(4x)
    indefinite integral = (-1/4)e^(3x)cos(4x) - indefinite integral((-3/4)e^(3x)cos(4x)dx)
    Rearrange fractions:
    indefinite integral = (-1/4)e^(3x)cos(4x) + (3/4) * indefinite integral(e^(3x)cos(4x)dx)
    integrate by parts:
    u=e^(3x) dv=cos(4x)dx
    du=3e^(3x)dx v=(1/4)sin(4x)
    indefinite integral = (-1/4)e^(3x)cos(4x) + (3/4)((1/4)e^3xsin(4x)-((3/4) * indefinite integral(e^(3x)sin(4x)dx))
    Rearrange and multiply out fractions:
    indefinite integral = (-1/4)e^(3x)cos(4x) + (3/16)e^3xsin(4x)-((9/16) * indefinite integral(e^(3x)sin(4x)dx)
    Consolidate similar indefinite integrals:
    (25/16) * indefinite integral = (-1/4)e^(3x)cos(4x) + (3/16)e^3xsin(4x)
    Isolate the indefinite integral:
    indefinite integral = (-4/25)e^(3x)cos(4x)+(3/25)e^(3x)sin(4x)+c
     
  2. ijffdrie

    ijffdrie Lord of Spam

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    have fun, EMR.