1. ## Integral.

Hello,

Who can help me with this integral? given is $\displaystyle \int e^{-x}sin(x)dx$

I can try it by integrations by parts thus $\displaystyle u=sin(x) \rightarrow \ u'=cos(x) \ \ \ v'=e^{-x} \ \rightarrow \ v=-e^{-x}$

Then $\displaystyle \int e^{-x}sin (x) = \ - sin (x) e^{-x}+\int e^{-x}cos(x)$

$\displaystyle =-sin(x)e^{-x}-cos(x)e^{-x}-\inte^{-x}sin(x)$ now I can put the integral on the right to the left but then I find 0

What do i wrong? Greets.

2. Originally Posted by Bert
Hello,

Who can help me with this integral? given is $\displaystyle \int e^{-x}sin(x)dx$

I can try it by integrations by parts thus $\displaystyle u=sin(x) \rightarrow \ u'=cos(x) \ \ \ v'=e^{-x} \ \rightarrow \ v=-e^{-x}$

Then $\displaystyle \int e^{-x}sin (x) = \ - sin (x) e^{-x}+\int e^{-x}cos(x)$

$\displaystyle =-sin(x)e^{-x}-cos(x)e^{-x}-\inte^{-x}sin(x)$ now I can put the integral on the right to the left but then I find 0

What do i wrong? Greets.
You have:

$\displaystyle \int e^{-x}\sin(x)dx=-\sin(x)e^{-x}-\cos(x)e^{-x}-\int e^{-x}\sin(x) dx$

So taking the last term on the right over to the left:

$\displaystyle 2\int e^{-x}\sin(x)dx=-\sin(x)e^{-x}-\cos(x)e^{-x}$

Hence:

$\displaystyle \int e^{-x}\sin(x)dx=-e^{-x}(\sin(x)+\cos(x))/2$

RonL