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

## Homework Statement

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

thanx

## The Attempt at a Solution

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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.

Dick
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.

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

another alternative is to write cos(2t) as exponentials using Eulers forlmulas.
and how would you do that?

VietDao29
Homework Helper

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

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