Have I calculated this integral correctly? I have this integral to calculate:
$$I=\int_{|z|=2}(e^{\sin z}+\bar z)dz.$$
I do it this way:
$$I=\int_{|z|=2}e^{\sin z}dz+\int_{|z|=2}\bar zdz.$$
The first integral is $0$ because the function is holomorphic everywhere and it is a contour integral. As for the second one, I have
$$\int_{|z|=2}\bar zdz = \int_0^{2\pi}e^{-i\theta}\cdot 2 d\theta=-\int_0^{-2\pi}e^{i\tau}\cdot 2 d\tau=\int_0^{2\pi}e^{i\tau}\cdot 2 d\tau=\int_{|z|=2}zdz=0$$ 
because the function is now holomorphic.  
It seems fishy to me. Is it correct?

Another method for the second integral:
$$
\begin{align*}
\int_{|z|=2}\bar z\,dz &= \int_{|z|=2} \frac{z \bar z}{z} \,dz \\
&= \int_{|z| = 2} \frac{|z|^2}{z}\,dz \\
&= \int_{|z| = 2} \frac{4}{z}\,dz \\
&= 4 \cdot 2\pi i \\
&= 8 \pi i.
\end{align*}
$$
