Trying to prove $\frac{2}{n+\frac{1}{2}} \leq \int_{1/(n+1)}^{1/n}\sqrt{1+(\sin(\frac{\pi}{t}) -\frac{\pi}{t}\cos(\frac{\pi}{t}))^2}dt$ I posted this incorrectly several hours ago and now I'm back! So this time it's correct.  Im trying to show that for $n\geq 1$:
$$\frac{2}{n+\frac{1}{2}} \leq \int_{1/(n+1)}^{1/n}\sqrt{1+\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2}dt$$
I checked this numerically for several values of $n$ up through $n=500$ and the bounds are extremely tight.  
I've been banging my head against this integral for a while now and I really can see no way to simplify it as is or to shave off a tiny amount to make it more palatable.  Hopefully someone can help me.  Thanks.

Now that I have finished this, I see that it is similar to the approach that Zarrax used, but it looks a bit simpler, so I will post it in addition.
Using the following facts
$$
\sqrt{x^2+1}\ge|x|\tag{1}
$$
$$
|x-y|\ge\mathrm{sgn}(x)(x-y)\tag{2}
$$
with a change of variables $t\mapsto 1/t$, we get
$$
\begin{align}
&\int_{\frac{1}{n+1}}^{\frac{1}{n}}\sqrt{1+\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2}\mathrm{d}t\\
&=\int_n^{n+1}\sqrt{1+(\sin(\pi t)-\pi t\cos(\pi t))^2}\frac{\mathrm{d}t}{t^2}\\
&\ge\int_0^1\left|\frac{\pi \cos(\pi t)}{n+t}-\frac{\sin(\pi t)}{(n+t)^2}\right|\mathrm{d}t\\
&\ge\int_0^1\mathrm{sgn}(\cos(\pi t))\left(\frac{\pi \cos(\pi t)}{n+t}-\frac{\sin(\pi t)}{(n+t)^2}\right)\mathrm{d}t\\
&=\int_0^1\mathrm{sgn}(\cos(\pi t))\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\sin(\pi t)}{n+t}\right)\mathrm{d}t\\
&=\left(\frac{\sin(\pi/2)}{n+\frac12}-\frac{\sin(0)}{n}\right)+\left(\frac{\sin(\pi/2)}{n+\frac12}-\frac{\sin(\pi)}{n+1}\right)\\
&=\frac{2}{n+\frac12}\tag{3}
\end{align}
$$
