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?

Using function transformations, compress the integral by a factor of $2$ in the $x$-axis, then multiply by $2$ to get:
$$2 \int_{0}^{\pi/2} \frac{\mathrm{d}t}{3+2\cos(2t)} = 2 \int_{0}^{\pi/2} \frac{\mathrm{d}t}{4 \cos^2 t+1} = 2 \int_{0}^{\pi/2} \frac{\mathrm{\sec^2 t\ d}t}{4 + \tan^2 t+1}$$
and substituting $u = \tan t$, $\mathrm{d} u = \sec^2 t \ \mathrm{d}t$:
$$2 \int_{0}^{\infty} \frac{\mathrm{d} u}{(\sqrt5)^2 + u^2} = \lim_{a \to \infty}2 \left[ \frac{1}{\sqrt5} \tan^{-1} \frac{u}{\sqrt5} \right]_0^{a} = 2 \left[ \frac{1}{\sqrt5} \tan^{-1} \frac{\tan t}{\sqrt5} \right]_0^{\pi/2} $$
$$=\frac{2}{\sqrt5} \left( \tan^{-1} ( \tan \pi/2) - \tan^{-1} (\tan 0)\right)= \frac{2}{\sqrt5} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt5} $$
since $\tan^{-1} (\tan x)= x, x \in [-\pi/2, \pi/2]$.
