# Integral.

• May 11th 2006, 09:37 AM
Bert
Integral.
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.
• May 11th 2006, 10:13 AM
CaptainBlack
Quote:

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