Vtyjkyuk

I am trying to find $$int_gamma z Im(z^2) dz,$$
where $gamma$ is the unit circle traversed once, anticlockwise.

My attempt:

let $gamma(t)=e^itimplies gamma'(t)=ie^it tin[0,2pi]$.
beginalign
int_gamma z Im(z^2) dx&=int_0^2pi e^itsin(2t) ie^it dt\
&=iint_0^2pi e^2itsin(2t) dt
endalign
Now, I let
beginalign
I&=iint_0^2pi e^2itsin(2t) dt \
&=ileft(left[frace^2it2isin(2t)right]_0^2pi-frac1iint_0^2pi e^2itcos(2t) dtright) \
&=-int_0^2pi e^2itcos(2t) dt \
&=-left(left[frace^2it2icos(2t)right]_0^2pi+frac1iint_0^2pi e^2itsin(2t) dtright) \
&=iint_0^2pi e^2itsin(2t) dt
endalign

Where am I going wrong? Wolfram says the answer is $-pi$.

edit

I can see an alternative approach. We could express the integrand as, $$(cos(2t)+isin(2t))sin(2t)=cos(2t)sin(2t)+isin^2(2t).$$ But I prefer using integration by parts and would like to see the solution achieved via this approach.

$$I=iint_0^2pi e^2itsin(2t) dt=iint_0^2pi e^2itdfrace^2it-e^-2it2i dt=dfrac12int_0^2pi e^4it-1 dt=-pi$$

You have that

beginalign
2I
&=
int_0^2pi, -e^2itBig(cos(2t) -isin(2t)Big) ,dt
\&=
int_0^2pi, -e^2itBig(cos(-2t) +isin(-2t)Big) ,dt
\&=
int_0^2pi, -e^2it,e^-2it ,dt
\&=
int_0^2pi, -1 ,dt = -2pi
endalign