Evaluate $\int_\gamma z \ \Im(z^2) \ dz$

Question

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

My attempt:

let $\gamma(t)=e^{it}\implies \gamma'(t)=ie^{it} \ \ \ \ t\in[0,2\pi]$. \begin{align} \int_\gamma z\ \Im(z^2) \ dx&=\int_{0}^{2\pi} e^{it}\sin(2t) \ ie^{it} \ dt\\ &=i\int_{0}^{2\pi} e^{2it}\sin(2t) \ dt\ \end{align} Now, I let \begin{align} I&=i\int_{0}^{2\pi} e^{2it}\sin(2t) \ dt\ \\ &=i\left(\left[\frac{e^{2it}}{2i}\sin(2t)\right]_{0}^{2\pi}-\frac{1}{i}\int_{0}^{2\pi} e^{2it}\cos(2t) \ dt\right) \\ &=-\int_{0}^{2\pi} e^{2it}\cos(2t) \ dt \\ &=-\left(\left[\frac{e^{2it}}{2i}\cos(2t)\right]_{0}^{2\pi}+\frac{1}{i}\int_{0}^{2\pi} e^{2it}\sin(2t) \ dt\right) \\ &=i\int_{0}^{2\pi} e^{2it}\sin(2t) \ dt\ \end{align}

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

edit

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

Answer 1

$$I=i\int_{0}^{2\pi} e^{2it}\sin(2t) \ dt=i\int_{0}^{2\pi} e^{2it}\dfrac{e^{2it}-e^{-2it}}{2i} \ dt=\dfrac12\int_{0}^{2\pi} e^{4it}-1\ dt=-\pi$$

Answer 2

You have that

\begin{align} 2I &= \int_0^{2\pi}\, -e^{2it}\Big(\cos(2t) -i\sin(2t)\Big) \,dt \\&= \int_0^{2\pi}\, -e^{2it}\Big(\cos(-2t) +i\sin(-2t)\Big) \,dt \\&= \int_0^{2\pi}\, -e^{2it}\,e^{-2it} \,dt \\&= \int_0^{2\pi}\, -1 \,dt = -2\pi \end{align}