How can I integrate e^-x sinx using parts?

[misogynisticfeminist]

I've got a function [tex] \int e^{-x}sinx dx [/tex]

From what I know, only functions which has one or more products with a finite number of successive differentials can be evaluated using integration by parts. Because for [tex]\int v du [/tex]in our choice of du, we want to cut down on the number of times we have to evaluate it using integration by parts again.

Since both [tex]e^{-x}[/tex] and [tex]sinx[/tex] have infinite nos. of successive differentials, how do i evaluate that?

 

[xanthym]

You need to integrate by parts TWICE, after which the Sin() will recur and can be combined with the original Sin() on the same side of the equation. Divide both sides by 2 to arrive at your answer.

~~

 

[misogynisticfeminist]

hey, i didn't saw that, thanks alot...

 

[Curious3141]

There is a much neater way to integrate the function. Hint : What is the imaginary part of [tex]e^{(-1 + i)x}[/tex] ?

EDIT : Nevermind, here's the whole solution since it's only an extra (but "cool" method). z is the constant of integration, with c being the imaginary part of the constant.

[tex]\int e^{-x}\sin x dx = Im(\int e^{(-1 + i)x} dx) = Im(\frac{1}{-1 + i}e^{(-1 + i)x} + z)[/tex]

which can be further simplified to [tex]-\frac{1}{2}e^{-x}(\sin x + \cos x) + c[/tex]

 

[dextercioby]

I would normally second the complex variable method (since it goes nicely in the theory of Laplace transformations),but the exarcise required part integration,which of course,requires in turn less mathematical knowledge...

Daniel.