Simple Integration by Parts Question

[Saterial]

Homework Statement

integral of e^-xsinxdx

Homework Equations

uv-/vdu

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? :(

 

[LCKurtz]

Homework Statement

integral of e^-xsinxdx

Homework Equations

uv-/vdu

The Attempt at a Solution

u=e^-x
du = -e^-xdx
v=sinx
dv=cosxdx

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

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

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.