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

[Chadlee88]

Homework Statement

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

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


thanx

Homework Equations

The Attempt at a Solution

 

[ice109]

well first start with u sub to make it a tad easier

 

[Chadlee88]

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]

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]

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

 

[ice109]

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

and how would you do that?

 

[malawi_glenn]

and how would you do that?

http://upload.wikimedia.org/math/8/d/6/8d662412f1815064827ff58e0280d09c.png


http://upload.wikimedia.org/math/9/8/6/986359d5b505762cd4bb6444081878be.png

http://en.wikipedia.org/wiki/Euler's_formula

 

[VietDao29]

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

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 = e^x, dv = sin x dx
~~> du = e^xdx, 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 = e^x previously, then, this time, you'll also let u = e^x, or you'll end up getting something like: I - I = C (where C is a constant)

Let u = e^x, and dv = cos x dx
~~~> du = u = e^x 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? :)