Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integral with cosine and exponential - cos (2t) x e^2t

  1. Jun 4, 2007 #1
    1. The problem statement, all variables and given/known data

    can som1 please give me an idea where to start with this integral.

    integral of: cos (2t) x e^2t

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 4, 2007 #2
    well first start with u sub to make it a tad easier
  4. Jun 4, 2007 #3
    yeh i did that, but i was still stuck, i made u= 2t so i had to find the integral of 1/2(cos (u)xe^u.
  5. Jun 4, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper

    Integrate by parts twice. If I=integral(cos(u)*exp(u)) then you will end up with an expression like I=(something)-I. Then just solve for I.
  6. Jun 4, 2007 #5


    User Avatar
    Science Advisor
    Homework Helper

    another alternative is to write cos(2t) as exponentials using Eulers forlmulas.
  7. Jun 4, 2007 #6
    and how would you do that?
  8. Jun 4, 2007 #7


    User Avatar
    Science Advisor
    Homework Helper

  9. Jun 4, 2007 #8


    User Avatar
    Homework Helper

    Whoops, I don't think any Pre-Calculus students do learn Integration.

    This is a good candidate for Integrate By Parts. I'll give you an example similar to your problem. You can read the example, and see if do the problem on your own.

    The formula is:
    [tex]\int u dv = uv - \int v du[/tex]

    [tex]I = \int e ^ x \sin x dx[/tex]
    Let u = ex, dv = sin x dx
    ~~> du = exdx, and v = -cos x, so your integral will become:
    [tex]I = \int e ^ x \sin x dx = - e ^ x \cos x + \int e ^ x \cos x dx[/tex]

    You should note that, if you let u = ex previously, then, this time, you'll also let u = ex, or you'll end up getting something like: I - I = C (where C is a constant)

    Let u = ex, and dv = cos x dx
    ~~~> du = u = ex dx, and v = sin(x)
    We have:

    [tex]I = - e ^ x \cos x + \int e ^ x \cos x dx + C' = -e ^ x \cos x + \left( e ^ x \sin x - \int e ^ x \ sin x dx \right) + C' = -e ^ x \cos x + e ^ x \sin x - I + C'[/tex]

    Isolate I to one sides yields:
    [tex]2I = -e ^ x \cos x + e ^ x \sin x + C'[/tex]
    [tex]\Rightarrow I = \frac{1}{2} \left( -e ^ x \cos x + e ^ x \sin x \right) + C[/tex] (where C, and C' are the Constants of Integrations.)

    Can you go from here? :)
    Last edited: Jun 4, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook