Show that $\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \frac{\pi}{\sqrt{5}}$ I need to proof that 
\begin{align}
\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \frac{\pi}{\sqrt{5}}
\end{align}
is correct. The upper limit $\pi$ seems to cause me some problems. I thought about solving this integral by using the residual theorem: 
I started with $\gamma: [0,\pi] \to \mathbb{C}, t \to e^{2it}$. Since $\cos(t) = \frac{1}{2}\left(e^{it}+e^{-it}\right)$ and $\gamma'(t) = 2ie^{2it}$ we find that 
\begin{align}
\int_{0}^{\pi} \frac{1}{3+2\cos(t)}\mathrm{d}t = \int_{0}^{\pi} \frac{1}{3+\left(e^{it}+e^{-it}\right)} \cdot \frac{2ie^{2it}}{2ie^{2it}}\mathrm{d}t = \int_{0}^{\pi} \frac{1}{3+\left(\sqrt{\gamma}+\frac{1}{\sqrt{\gamma}}\right)} \cdot \frac{-i\gamma'}{2\gamma}\mathrm{d}t 
\end{align}
I did this with the aim to use 
\begin{align}
\int_{\gamma} f(z) \mathrm{d}z = \int_{a}^{b} (f\circ \gamma)(t)\gamma'(t) \mathrm{d}t,
\end{align}
so we find 
\begin{align}
\int_{0}^{\pi} \frac{1}{3+\left(\sqrt{\gamma}+\frac{1}{\sqrt{\gamma}}\right)} \cdot \frac{-i\gamma'}{2\gamma}\mathrm{d}t = \int_{\gamma} \frac{-i}{6z+2z\sqrt{z}+2\sqrt{z}}  \mathrm{d}z 
\end{align}
At this point I don't know how to continue. Can anyone help?

Use $$\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx.~~~(1)$$
$$I=\int_{0}^{\pi} \frac{dt}{3+2 \cos t}~~~~(2)$$
Using (1), we get
$$I=\int_{0}^{\pi} \frac{dt}{3-2 \cos t}~~~~(3)$$
Add (2) and (3), then
$$2I=\int_{0}^{\pi} \frac{6\,dt}{9-4\cos^2 t}=12\int_{0}^{\pi/2} \frac{dt}{9-4\cos^2 t}=12\int_{0}^{\pi/2} \frac{\sec^2 t}{9\tan^2 t+5}=\frac{4}{3} \int_{0}^{\infty}\frac{du}{u^2+(\sqrt{5}/3)^2}$$
$$2I=\frac{4}{3} \frac{3}{\sqrt{5}}\tan^{-1}(3u/\sqrt{5})|_{0}^{\infty}=\frac{2\pi}{\sqrt{5}}\implies I=\frac{\pi}{\sqrt5} $$ In above we have used $\tan t =u$.
