on Complex Integration $\int_{\gamma}\frac{dz}{z^{2}-1}$ I have the following problem that I am working on. 

Let $\gamma:[-\pi,\pi]\to\mathbb{C}$ be defined by $\gamma(t)=2e^{it}$. Find $$\int_{\gamma}\frac{1}{z^{2}-1}\:dz.$$

My understanding of the problem: We note that $\gamma'(t)=2ie^{it}$, so $$\int_{\gamma}\frac{1}{z^{2}-1}=\int_{a}^{b}f(\gamma(t))\:d\gamma(t)=\int_{-\pi}^{\pi}\frac{2ie^{it}}{4e^{2it}-1}\:dt.$$ From here, I'm not really sure how to integrate this function. Would I need to write $e^{it}=\cos(t)+i\sin(t)$ and go from there? Or is this completely the wrong direction? I apologize in advance for any ignorance on my behalf; I'm new to complex analysis. Thanks for any help!
EDIT: We have not covered anything related to the residue theorem yet, so I can't use that here. 
 A: While you can do the integration by doing substitutions (say, by using the tangent half angle substitution,) that's going to be very tedious, and this particular integral can be evaluated more easily if you notice one thing.
Given your reduction to the ointegral:
$$\int_{-\pi}^{\pi} \frac{2ie^{it}\,dt}{4e^{2it}-1},$$
note that the integrand $$f(t)=\frac{2ie^{it}}{4e^{2it}-1}$$
has the property that $f(t+\pi)=-f(t).$ 
So the values of $f(t)$ from $[-\pi,0]$ cancel with the values of $f(t)$ from $[0,\pi]$. Specifically:
$$\begin{align}\int_{-\pi}^{\pi} f(t)\,dt &=
\int_{-\pi}^{0} f(t)\,dt + \int_{0}^{\pi} f(t)\,dt\\
&=\int_{-\pi}^{0}f(t)\,dt + \int_{-\pi}^0 f(t+\pi)\,dt\\
&=\int_{-\pi}^0 \left(f(t)+f(t+\pi)\right)\,dt\\
&=0
\end{align}$$.

More generally, if $\gamma(t)=re^{it}$ traces a circle with center $0$ and $g(z)$ is any "nice" (say continuous) function defined on that circle such that $g(z)=g(-z)$, then $$\int_{\gamma} g(z)\,dz = 0$$

I'll start the "hard" approach of computing the integral.
Using the tangent half-angle substitution, we let $u=\tan \frac{t}{2}$ then $$e^{it}=\cos(t)+i\sin t = \frac{1-u^2+2ui}{1+u^2}=\frac{(1+ui)^2}{1+u^2}$$
So we get:
$$\begin{align}\int \frac{e^{it}\,dt}{4e^{2it}-1} &= 2\int \frac{\frac{(1+ui)^2}{(1+u^2)^2}\,du}{4\frac{(1+ui)^4}{(1+u^2)^2}-1}\\
&=2\int\frac{du}{4(1+ui)^2-(1-ui)^2}\\
&=2\int\frac{du}{(3+ui)(1+3ui)}
\end{align}$$
That middle step is because $1+u^2=(1+ui)(1-ui)$.
Even from this point, there is some work. Partial fractions, then multiply numerators and denominators by conjugates, and some other awful stuff. Blech.
The symmetry that lets us conclude the integral is zero in the first part is still here, but it takes a more complicated argument to see it, because the bounds of our integral have become $(-\infty,\infty)$ ands the substitution is $v=-\frac1u$, which obviously has problems at $u=0.$ Using this substitution, we see that:
$$\frac{du}{(3+ui)(1+3ui)}=\frac{1/v^2\,dv}{(3-v^{-1}i)(1-3v^{-1}i)}$$
And $v^2(3-v^{-1}i)(1-3v^{-1}i)=(3v-i)(v-3i)=-(3vi+1)(iv+3)$. So you get that $$\int_{\epsilon}^{1/\epsilon} \frac{du}{(1+3ui)(3+ui)}=-
\int_{-1/\epsilon}^{-\epsilon} \frac{dv}{(1+3vi)(3+vi)}$$
But to actually compute the left hand side is gonna be gross.
A: Enforcing the change of variables $t= \pi+s$ gives 
$$\begin{align}I&= \int_{-\pi}^{\pi}\frac{2ie^{it}}{4e^{2it}-1}\:dt=\int_{0}^{2\pi}\frac{2ie^{-is}e^{-i\pi}}{4e^{-2is}-1}\:ds\\&=-\int_{0}^{2\pi}\frac{2ie^{-is}}{4e^{-2is}-1}\:ds \\&= -\int_{0}^{\pi}\frac{2ie^{-is}}{4e^{-2is}-1}\:ds-\int_{\pi}^{2\pi}\frac{2ie^{-is}}{4e^{-2is}-1}\:ds \\&\overset{\color{red}{u=s-2\pi}}{=}-\int_{0}^{\pi}\frac{2ie^{-is}}{4e^{-2is}-1}\:ds-\int_{-\pi}^{0}\frac{2ie^{-iu}}{4e^{-2iu}-1}\:du
\\&=-\int_{-\pi}^{\pi}\frac{2ie^{-is}}{4e^{-2is}-1}\:ds \\&=-I \end{align}$$
Hence $$\color{red}{I= \int_{-\pi}^{\pi}\frac{2ie^{it}}{4e^{2it}-1}\:dt=0}$$
