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?

If $z = 2e^{i \theta}$, then $$\bar{z} dz = 2e^{-i \theta}2i e^{i \theta} d \theta = 4i d \theta$$
Hence, $$\int_{\vert z \vert = 2} \bar{z} dz = \int_0^{2 \pi} 4i d \theta = 8 \pi i$$
