integrate e^(-2x) sin 4x dx;x=0 to infinity

Question

zepdrix · Answer

$$\large \int\limits\limits_0^{\infty} e^{-2x}\sin(4x)dx$$

Hmm you'll have to do integration by parts twice on this one.
It doesn't matter which you assign to be your $u$ and $dv$, it will the second time you integrate, but not the first time. :)

anonymous · Answer

Seems you can evaluate the indefinate integral by parts several times. Keep an eye out for the same thing appearing on both sides of the equals sign. Then use that solution to solve the definite integral

zepdrix · Answer

Ignore the limits of integration until the end, that will make it easier to work with.

anonymous · Answer

Thanks, I understand I have to do integration by parts twice on this one but I still don't know how to solve it. The Wolphram Alfa tell me that the answer should be 1/5 but how??

anonymous · Answer

Were you find the indefinate integral?

anonymous · Answer

Well, I feel a bit lost! :(

anonymous · Answer

Okay do integration by parts once. Then you end up with another integral that you cant solve but it has a cosine term. So on that new integral do parts again. Now it has a sine term which is what you began with. Get all the integral of e^(-2x)sin(4x) to one side and you have your solution

anonymous · Answer

Ok, I try! Thanks!

anonymous · Answer

hmm if you have a little time i can solve it for you. :)

anonymous · Answer

there the solution.

anonymous · Answer

This question reminds me the first year of the university :)

anonymous · Answer

Great! Thanks a lot! It is the first year and I find it very hard... :(

anonymous · Answer

your welcome. :) good luck!