$\int_C z^2 dz$
$C: y=\sin x$
$x=0$ to $x=3$.
My attempt: I'm not sure if I have this correct - this is a line integral? If so, a line integral is the same concept as a regular integral in the x-y plane: it is the sum of all the (base x height)'s of an infinitesimally thin rectangle from $x_{initial}$ to $x_{final}$. The base in this case is the curve $y=\sin x$ and the height is the function $z^2$.
$$x=t $$ $$y=\sin t$$ $$0\leq t \leq 3$$
This is where I get stuck and start to maybe wonder if I have the type of integral wrong....
The path is defined by $z=t+i \sin{t}$, $dz = (1+i \cos{t}) dt $; thus
$$\begin{align}\int_C dz \, z^2 &= \int_0^3 dt \, (1+i \cos{t}) (t+i \sin{t})^2\\ &= \int_0^3 dt \, (1+i \cos{t}) (t^2-\sin^2{t} +i 2 t \sin{t})\\ &= \int_0^3 dt \, (t^2-\sin^2{t} - 2 t \sin{t} \cos{t} + i (2 t \sin{t} + t^2 \cos{t} - \sin^2{t} \cos{t})) \end{align}$$
You should be able to do out all of those integrals as usual.
ADDENDUM
Of course as pointed out in a comment you can avoid all of this trauma by simply observing the analyticity of the integrand. Thus the integral is simply
$$\frac13 \left ( (3+i \sin{3})^3 - 0^3 \right) = \frac13 (3+i \sin{3})^3$$