Calculus

I need help with this integral.

w= the integral from 0 to 5

24e^-6t cos(2t) dt.

i found the the integration in the integral table.

(e^ax/a^2 + b^2) (a cos bx + b sin bx)

im having trouble finishing the problem from here.

asked by Bobby Smith
  1. Write cos(x) as the real part of
    exp(ix)

    The integral of exp(ax)cos(bx) is then the real part of the integral of
    exp[(a + ib)x]

    The integral is:

    1/(a+ib) exp[(a+ib)x] =

    (a-ib)/(a^2 + b^2) exp(ax)*
    [cos(bx) + i sin(bx)]

    The real part of this is:

    exp(ax)/(a^2 + b^2)
    [a cos(bx) + b sin(bx)]

    posted by Count Iblis

Respond to this Question

First Name

Your Response

Similar Questions

  1. Integral

    That's the same as the integral of sin^2 x dx. Use integration by parts. Let sin x = u and sin x dx = dv v = -cos x du = cos x dx The integral is u v - integral of v du = -sinx cosx + integral of cos^2 dx which can be rewritten
  2. Integration by Parts

    integral from 0 to 2pi of isin(t)e^(it)dt. I know my answer should be -pi. **I pull i out because it is a constant. My work: let u=e^(it) du=ie^(it)dt dv=sin(t) v=-cos(t) i integral sin(t)e^(it)dt= -e^(it)cos(t)+i*integral
  3. Calculus

    Hello, I have some calculus homework that I can't seem to get started..at least not on the right track? I have 3 questions 1. integral of [(p^5)*(lnp)dp] I'm using the uv-integral v du formula So first, I'm finding u and I think
  4. please help me calc. have test tom

    d/dx integral from o to x of function cos(2*pi*x) du is first i do the integral and i find the derivative right. by the fundamental theorem of calculus, if there is an integral from o to x, don't i just plug the x in the function.
  5. Calculus

    I have two questions, because I'm preparing for a math test on monday. 1. Use the fundamental theorem of calculus to find the derivative: (d/dt) the integral over [0, cos t] of (3/5-(u^2))du I have a feeling I will be able to find
  6. Integral Help

    I need to find the integral of (sin x)/ cos^3 x I let u= cos x, then got -du= sin x (Is this right correct?) I then rewrote the integral as the integral of -du/ u^3 and then rewrote that as the integral of - du(u^-3). For this
  7. calc

    how do you start this problem: integral of xe^(-2x) There are two ways: 1) Integration by parts. 2) Differentiation w.r.t. a suitably chosen parameter. Lets do 1) first. This is the "standard method", but it is often more tedious
  8. calc

    d/dx integral from o to x of function cos(2*pi*x) du is first i do the integral and i find the derivative right. by the fundamental theorem of calculus, if there is an integral from o to x, don't i just plug the x in the function.
  9. Calculus problem

    Evaulate: integral 3x (sinx/cos^4x) dx I think it's sec3 x , but that from using a piece of software, so you'll have to verify that. Using uppercase 's' for the integral sign we have S 3sin(x)/cos4dx or S cos-4(x)*3sin(x)dx If you
  10. calc asap!

    can you help me get started on this integral by parts? 4 S sqrt(t) ln(t) dt 1 please help! thanks! Integral t^(1/2)Ln(t)dt = 2/3 t^(3/2)Ln(t)- 2/3 Integral t^(1/2) dt = 2/3 t^(3/2)Ln(t) - 4/9 t^(3/2) Simpler method: Integral

More Similar Questions