As with anything that has to do with supersymmetry the details will be dependent on your exact conventions, but we can obtain the result as follows:
Assume we have two Grassman variables θ1 and θ2. By applying your first formula twice we find
∫dθ1dθ2θ2θ1=1
Now combine these into
θ=θ1+iθ2andˉθ=θ1−iθ2.
We then have
ˉθθ=−2iθ2θ1
and hence
∫dθ1dθ2ˉθθ=−2i
which is exactly your second integral, if we identify the measure
d2θ=dθ1dθ2.
This post imported from StackExchange Physics at 2014-03-09 16:25 (UCT), posted by SE-user Olof