# Simple Integration by Parts Question

1. Jul 11, 2012

### Saterial

1. The problem statement, all variables and given/known data
integral of e^-xsinxdx

2. Relevant equations
uv-/vdu

3. The attempt at a solution
u=e^-x
du = -e^-xdx
v=sinx
dv=cosxdx

e^-xsinx-/(-e^-x)sinx
=e^-x(sinx+cosx)

Wolfram alpha is telling me that the indefinite integral is actually "-1/2e^-x(sinx+cosx)" where did the -1/2 come from? :(

2. Jul 11, 2012

### LCKurtz

dv = sin(x)
v = - cos(x) dx

I don't understand what you did in those last two lines.You should have, after that integration by parts,$$\int e^{-x}\sin x\, dx = -e^{-x}\cos x -\int e^{-x}\cos x\,dx$$
You have to do the second integral on the right and solve for your original integral. Then you will see where the 1/2 comes from.