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Help with integrals?

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please help me with these.

evaluate the integral

∫ t^3 e^t dt =

∫ e^ -θ cos2θdθ =

thank you

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  1. For your first one, we can save time by developing a reduction formula.  Let I = Integ{t^ne^tdt}.  Use integration by parts with u = t^n and dv = e^tdt, so du = nt^(n - 1)dt and v = e^t.  Now

    I = t^ne^t - n*Integ{t^(n - 1)e^tdt.  Use this formula four times.

    Let J = Integ{e^(-t)cos(2t)ddt}  Use integration by parts with u = cos(2t), dv = e^(-t)dt, so du = -2sin(2t)dt and v = -e^(-t).  We get J = -e^(-t)cos(2t) - 2*Integ{e^(-t)sin(2t)dt}.  Now again use integration by parts with u = sin(2t), dv = e^(-t)dt, du = 2 cos(2t)dt, v = -e^(-t).  Substituting above we have

    J = -e^(-t)cos(2t) - 2[-e^(-t)sin(2t) +2J].  Solve this equation (algebraically) for J.


  2. well using the rule ILATE

    u'll get the answer as  

    1) t^3*e^t - 3*t^2*e^t + 6*t*e^t - 6*e^t

    2) I = ( 2*sin2x*e^(-x)  - cos2x*e^(-x)) / 5  

  3. For the first one, you have to use repeated integration by parts.  I'll get you started.

    let u = t^3  dv = e^t dt

         du = 3t^2 dt   v = e^t

    Then  Int(t^3 e^t dt) = u*v - int(v du) = t^3 e^t - Int(3t^2 e^t dt)

    Now repeat process for the new integral and repat until you have only Int(e^t dt) = e^t.

    FOr teh second one, use the Euler identity for the cosine funciton:

    cos2x = (e^(2ix)+e^(-2ix))/2  Then the integrand becomes:

    1/2*[e^(x*(1+2i))+e^(x*(1-2i))]  you can integrate this striaght off

    Integral = 1/2*{e^(x*(1+2i))/(1+2i) + e^(x*(1-2i))/(1-2i)}

    Now you can do some algebra.  You should get:

    Integral = (e^x)/5*{cos2x -i*e^(2ix)}

    You can write e^(2ix) in terms of sin2x and cos2x - that's an exercise for the reader.

  4. ∫(t^3 e^t dt )

    Use integration by parts.

    Let u = t^3.  dv = e^t dt.

    du = 3t^2 dt.  v = e^t

    t^3 e^t - ∫ ( 3t^2 e^t dt )

    t^3 e^t - 3 ∫( t^2 e^t dt )

    Use integration by parts again.

    Let u = t^2.  dv = e^t dt.

    du = 2t dt.  v = e^t

    t^3 e^t - 3 [ t^2 e^t - ∫ [ 2t e^t dt ] ]

    t^3 e^t - 3t^2 e^t + 3 ∫ ( 2t e^t dt )

    t^3 e^t - 3t^2 e^t + 6 ∫ ( t e^t dt )

    And one last time, use integration by parts.

    Let u = t.  dv = e^t dt.

    du = dt.  v = e^t

    t^3 e^t - 3t^2 e^t + 6 [ t e^t - Integral ( e^t dt ) ]

    t^3 e^t - 3t^2 e^t + 6t e^t - 6 Integral ( e^t dt )

    And the last integral is easy.

    t^3 e^t - 3t^2 e^t + 6t e^t - 6e^t + C


  5. ∫ t^3 e^t dt =

    Integration by parts

    dv=e^t

    v=e^t

    u=t^3

    du=3t^2

    ∫ u dv = uv - ∫ vdu

    e^t t^3 - 3  Ã¢ÂˆÂ« t^2 e^t dt -----------------(1)

    consider  Ã¢ÂˆÂ« t^2 e^t dt

    Integration by parts again

    dv=e^t

    v=e^t

    u=t^2

    du=2t

    ∫ u dv = uv - ∫ vdu

    e^t t^2 - 2  Ã¢ÂˆÂ« e^t t dt ---------(2)

    plug (2) into (1)

    (1) becomes

    e^t t^3 - 3 e^t t^2 -6  Ã¢ÂˆÂ« e^t t dt -------------(3)

    consider  Ã¢ÂˆÂ« e^t t dt

    Integrate by parts again

    dv=e^t

    v=e^t

    u=t

    du=1

    ∫ u dv = uv - ∫ vdu

    e^t t -  Ã¢ÂˆÂ« e^t dt

    = e^t t - e^t --------(4)

    plug (4) into (3)

    e^t t^3 - 3 e^t t^2 -6 e^t t - 6 e^t +C  

  6. ∫ t^3 e^t dt

    let u=t³ ; du = 3t² dt

    dv = e^t dt ; v = e^t

    ∫udv = uv - ∫vdu

    ∫ t^3 e^t dt = t³e^t - 3∫t²e^tdt

    let u=t² ; du = 2t dt

    dv = e^t dt ; v = e^t

    ∫t²e^tdt = t²e^t - 2∫te^t dt

    let u = t; du = dt

    dv = e^t dt ; v = e^t

    ∫te^t dt = te^t - ∫e^t dt = te^t - e^t

    ∫ t^3 e^t dt = t³e^t - 3[t²e^t - 2∫te^t dt]

    ∫ t^3 e^t dt = t³e^t - 3[t²e^t - 2[te^t - e^t]]

    ∫ t^3 e^t dt = t³e^t -3t²e^t + 6te^t - 6e^t

    ∫ t^3 e^t dt = e^t [t³ - 3t² + 6t - 6] + c

    Answer: e^t [t³ - 3t² + 6t - 6] + c

    2. I don't know if you wrote ∫ e^(-θ) cos2θdθ or ∫ e^(-θcos2θ) dθ

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