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

1. Jun 4, 2007

1. The problem statement, all variables and given/known data

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

thanx
2. Relevant equations

3. The attempt at a solution

2. Jun 4, 2007

ice109

3. Jun 4, 2007

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.

4. Jun 4, 2007

Dick

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.

5. Jun 4, 2007

malawi_glenn

another alternative is to write cos(2t) as exponentials using Eulers forlmulas.

6. Jun 4, 2007

ice109

and how would you do that?

7. Jun 4, 2007

8. Jun 4, 2007

VietDao29

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:
$$\int u dv = uv - \int v du$$

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

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:

$$I = - e ^ x \cos x + \int e ^ x \cos x dx + C' = -e ^ x \cos x + \left( e ^ x \sin x - \int e ^ x \ sin x dx \right) + C' = -e ^ x \cos x + e ^ x \sin x - I + C'$$

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

Can you go from here? :)

Last edited: Jun 4, 2007