Integral.

**Bert** · May 11th 2006, 08:37 AM

Hello,

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

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

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

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

What do i wrong? Greets.

**CaptainBlack** · May 11th 2006, 09:13 AM

Originally Posted by Bert

Hello,

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

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

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

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

What do i wrong? Greets.

You have:

$ \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:

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

Hence:

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

RonL

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