Complex integral computation Evaluate$$ \oint f\left( z \right )dz$$ 
where C is the unit circle 
$$\left | z \right |=3$$
$$f\left ( z \right )=\frac{1}{z^{3}+2z^{2}}$$
Let
$$z=3e^{it}$$
$$z^{3}=27e^{i3t}$$
$$dz=3ie^{it}$$
$$2z^{2}=18e^{i2t}$$
Substituting everything into the initial equation and reducing, we get $$\int_{0}^{2\pi}\frac{i}{9e^{i2t}-6e^{it}}$$
How should I proceed further?
Is it sensible to use the identity Cos(x)+iSin(x)=1 for the complex exponential?

One can use the Residue Theorem to show that the value of the integral is zero.  If one wishes to proceed directly, then we have
$$\begin{align}
\oint_{|z|=3} \frac{1}{z^3+2z^2}\,dz&=\frac14 \oint_{|z|=3}\left(\frac{1}{z+2}+\frac2{z^2}-\frac{1}{z}\right) \,dz\\\\
&=\frac i4\int_0^{2\pi}\frac{3e^{it}}{3e^{it}+2}\,dt+\frac i2\int_0^{2\pi}\frac{3e^{it}}{9e^{i2t}}\,dt-\frac i4\int_0^{2\pi}\frac{3e^{it}}{3e^{it}}\,dt\\\\
&=-\frac i4\int_0^{2\pi}\frac{1}{3e^{it}+2}\,dt\\\\
&=-\frac i4\int_0^{2\pi}\frac{3\cos t+2}{12\cos t+13}\,dt-\frac 14\int_0^{2\pi}\frac{3\sin t}{12\cos t+13}\,dt\\\\
&=-\frac i2\int_0^{\pi}\frac{3\cos t+2}{12\cos t+13}\,dt-\frac 14\int_{-\pi}^{\pi}\frac{3\sin t}{12\cos t+13}\,dt\\\\
&=-\frac i2\left.\left(\frac{t}{4}-\frac12 \arctan\left(5\cot(t/2)\right)\right)\right|_{0}^{\pi}+\frac {i}{8}\left.\log\left(12\cos t +12\right)\right|_{-\pi}^{\pi}\\\\
&=0
\end{align}$$
