Calculus Problem

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I need to find the integral of e^(2x)sin(3x)

I used integration by parts and I let u=e^2x and dv=sin(3x)

My final answer was (-3/8)(e^(2x))(cos(3x)) - (1/8)(e^(2x))(sin(3x)) but it's wrong. Please help!!

  • Calculus Problem -

    u = e^(2x)
    du/dx = 2 e^(2x)
    du = 2 e^(2x) dx

    dv = sin(3x) dx
    v = -(1/3) cos(3x)

    ∫e^(2x)sin(3x) dx
    = (e^(2x) )(-1/3)cos(3x) - ∫(-2/3)(e^(2x))(cos(3x) ) dx
    =(e^(2x) )(-1/3)cos(3x) + (2/3)∫( e^(2x) (cos(3x)) dx

    do it again on that last part:
    let u = e^2x
    du/dx = 2e^(2x)

    let dv = cos(3x) dx
    v = (1/3)sin(3x)
    and ∫( e^(2x) (cos(3x)) dx
    = (1/3)e^(2x) (sin(3x)) - ∫(1/3)sin(3x) (2e^(2x)) dx
    = (1/3)e^(2x) (sin(3x)) - (2/3) ∫e^(2x) sin(3x) dx

    but look at that last part, isn't that what we started with on the original left side of our equation

    let ∫e^(2x)sin(3x) dx = A
    so we have

    A = (e^(2x) )(-1/3)cos(3x) + (2/3)[(1/3)e^(2x) (sin(3x)) - (2/3) A ]
    A = (-1/3)(e^(2x)) (cos(3x) + (2/9)e^(2x) sin(3x) - (4/9)A
    times 9
    9A = -3 e^(2x) cos(3x) + 2 e^(2x) sin(3x) - 4A
    13A = -3 e^(2x) cos(3x) + 2 e^(2x) sin(3x)
    A = (-1/13) ( 3 e^(2x)cos(3x) - 2 e^(2x) sin(3x)

    ∫e^(2x)sin(3x) dx = (-1/13) ( 3 e^(2x)cos(3x) - 2 e^(2x) sin(3x) )

    or as Wolfram has it

    (1/13) ( 2 e^(2x) sin(3x) - 3 e^(2x)cos(3x) )

    http://integrals.wolfram.com/index.jsp?expr=e%5E%282x%29sin%283x%29+&random=false

  • Calculus Problem -

    Thank you so much for showing the steps, I really appreciate the help! (:

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