# Let $\gamma(t)=2\cos t+i\sin t$, for $0\leq t \leq 2\pi$. Evaluate the integrals: (i) $\int_{\gamma}\sin(z^2)$; (ii) $\int_{\gamma}z^{-1}dz$;

Let $\gamma(t)=2\cos t+i\sin t$, for $0\leq t \leq 2\pi$. Evaluate the integrals: (i) $\int_{\gamma}\sin(z^2)$; (ii) $\int_{\gamma}z^{-1}dz$; (iii) $\int_{\gamma}(z^2+2iz)^{-1}dz$.

I am using the Cauchy theorem to solve this problem and I have come to the following:

(i) $f(z)=\sin(z^2)$ is analytic in all $\mathbb{C}$, so, since $\gamma$ is closed content in $\mathbb{C}$, it has to be $\int_{\gamma}\sin(z^2)=0$.

(ii) $f(z)=z^{-1}$ is analytic in all $\mathbb{C}$ minus $z=0$, so $\int_{\gamma}z^{-1}=\int_{0}^{2\pi}\frac{-2\sin t+i\cos t}{2\cos t+i\sin t}dt$, but I do not know how to solve this integral, is there another way to do this?

(iii) $\int_{\gamma}(z^2+2iz)^{-1}dz=\frac{1}{2i}\int_{\gamma}z^{-1}dz-\frac{1}{2i}\int_{\gamma}(z+2i)^{-1}dz=\frac{1}{2i}\int_{\gamma}z^{-1}dz$, but I come to the same problem as in (ii), since I do not know what $\int_{\gamma}z^{-1}dz$ is, could someone help me please? Thank you.

$(ii)$ Using Cauchy Integral formula we have:
$$\int_{\gamma}\frac{1}{z}dz=2\pi i f(0)$$ where $f(z)$ is the constant function $f(z)=1$ which is entire then $f(0)=1$