Finding $\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt$ I'd like to ask something about the following integral:
$$
\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt
$$
I rewrote and took another variable. 
$$
-i\int_0^{2\pi}\frac{e^{is}-e^{-is}+8i}{e^{is}+e^{-is} + \frac{10}3} dt
\quad = \quad 
-i\int_{C(0,1)^+}\frac{z-\frac{1}{z}+8i}{z+ \frac1z+ \frac{10}3} \cdot \frac{-i}{z}dt
\quad = \quad
- \int_{C(0,1)^+} \frac{z^2-1+8iz}{z(z^2+1+\frac{10}{3} \cdot z)} dz
$$
The only root of $z(z^2+1+\frac{10}{3} \cdot z)$ inside the unit disk is $0$, so I thought that the integral would equal:
$$
Res_{z=i} \ = \ 2 \pi i \cdot \frac{0-1+8i\cdot 0}{0^2+1+\frac{10}{3}\cdot 0} \quad = \quad 2\pi i
$$
I don't understand why $i$ appears in this value. Can you please tell me what I did wrong?

You can do this way. Clearly
$$
\int_0^{2\pi}\frac{\sin t + 4}{\cos t + \frac{5}3} dt=\int_0^{2\pi}\frac{\sin t}{\cos t + \frac{5}3} dt+4\int_0^{2\pi}\frac{1}{\cos t + \frac{5}3} dt
$$
and 
$$
\int_0^{2\pi}\frac{\sin t}{\cos t + \frac{5}3} dt=-\int_0^{2\pi}\frac{1}{\cos t + \frac{5}3} d(\cos t + \frac{5}3)=0.
$$
You only need to calculate
$$ \int_0^{2\pi}\frac{1}{\cos t + \frac{5}3} dt $$
and I think you can finish it.
