Help with integrals?

Question

please help me with these.

evaluate the integral

∫ t^3 e^t dt =

∫ e^ -θ cos2θdθ =

thank you

Answer 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.

Answer 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  

Answer 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.

Answer 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

Answer 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  

Answer 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