Evaluate $\displaystyle\int_\gamma\frac{1}{z^2 + 4}\:dz$ along the positively oriented circle $\lvert z - i \rvert = 2$ $\displaystyle\int_\gamma\frac{1}{z^2 + 4}\:dz$  is not defined at $\pm \: 2i$, and since $2i$ is an interior point of the circle $\gamma := \lvert z - i \rvert = 2$, I need to define another path inside $\gamma$ which encloses the singularity $2i$.
So I choose another path $\gamma_2 := \lvert z - 2i \rvert = 1$, and I know from Cauchy's theorem that $\displaystyle\int_{\gamma + \gamma_2}\frac{1}{z^2 + 4}\:dz = 0$. So I can get an answer by evaluating $-\displaystyle\int_{\gamma_2}\frac{1}{z^2 + 4}\:dz$.
This is where it gets tricky for me. I substitute the path $\gamma_2 := e^{i\theta} + 2i$ into the line integral and after a lot of simplification I find that $-\displaystyle\int_{\gamma_2}\frac{1}{z^2 + 4}\:dz = -i\displaystyle\int_{0}^{2\pi}\frac{1}{e^{i\theta} + 4i}\:d\theta$. 
I entered this integral to wolfram alpha and got the correct answer $\frac{\pi}{2}$, however I have no idea as to how to go about evaluating this integral. I would appreciate some hint as to how to begin solving this integral, or if there is a simpler way to reach the answer I would appreciate a push in that direction too. Please not that I haven't covered residues yet.
 A: $$\displaystyle\int_\gamma\frac{1}{z^2 + 4}$$
$$\displaystyle\int_\gamma\dfrac{1}{4i}(\dfrac{1}{z-2i}-\dfrac{1}{z+2i})$$
The equation of circe is $\gamma := \lvert z - i \rvert = 2$.
$z+2i$ is out range so its integration will be zero.
$$\displaystyle\int_\gamma\dfrac{1}{4i}(\dfrac{1}{z-2i})$$
$$=\dfrac{1}{4i}*2 \pi f(2i)$$
$$f(2i)=i$$
$$=\dfrac{\pi}{2}$$
A: I think I have found an answer to my question after the hint to use partial fractions. I will post it below.
Let $$\gamma := \lvert z - i \rvert = 2$$ $$\gamma_2 := \lvert z - 2i \rvert = 1$$.
By Cauchy's theorem and using partial fractions, 
$$\displaystyle\int_{\gamma}\frac{1}{z^2 + 4}\:dz = -\displaystyle\int_{\gamma_2}\frac{1}{z^2 + 4}\:dz = \frac{i}{4}\displaystyle\int_{\gamma_2}\frac{1}{z + 2i}\:dz - \frac{i}{4}\displaystyle\int_{\gamma_2}\frac{1}{z - 2i}\:dz$$.
since the function $\frac{1}{z + 2i}$ is analytic on and inside $\gamma_2$, I have that
$$\frac{i}{4}\displaystyle\int_{\gamma_2}\frac{1}{z + 2i}\:dz = 0$$  
If I substitute $\gamma_2 := e^{i\theta} + 2i$ into the line integral, I get
$$ - \frac{i}{4}\displaystyle\int_{\gamma_2}\frac{1}{z - 2i}\:dz = - \frac{i}{4}\displaystyle\int_{0}^{2\pi}\frac{ie^{i\theta}}{e^{i\theta} + 2i - 2i}\:d\theta = \frac{-i^2}{4} \: 2\pi = \frac{\pi}{2}$$
So $$\displaystyle\int_{\gamma}\frac{1}{z^2 + 4}\:dz = \frac{\pi}{2}$$
