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.

Potato's answer is what's going on geometrically. If you want it analytically:$$\sqrt{1+\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2} \geq \sqrt{\left(\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\right)^2}$$
$$ = \bigg|\sin\left(\frac{\pi}{t}\right) -\frac{\pi}{t}\cos\left(\frac{\pi}{t}\right)\bigg|$$
The above expression is the absolute value of the derivative of $t\sin(\pi/t)$. 
So your integral is greater than 
$$\int_{1 \over n + 1}^{1 \over n + {1 \over 2}}|(t\sin({\pi \over t}))'|\,dt + \int_{1 \over n + {1 \over 2}}^{1 \over n}|(t\sin({\pi \over t}))'|\,dt$$
This is at least what you get when you put the absolute values on the outside, or
$$\bigg|\int_{1 \over n + 1}^{1 \over n + {1 \over 2}}(t\sin({\pi \over t}))'\,dt\bigg| + \bigg|\int_{1 \over n + {1 \over 2}}^{1 \over n}(t\sin({\pi \over t}))'\,dt\bigg|$$
Then the fundamental theorem of calculus says this is equal to the following, for $f(t) = t \sin(\pi/t)$:
$$\bigg|f({1 \over n + {1 \over 2}}) - f(0)\bigg| + \bigg|f({1 \over n}) - f({1 \over n + {1 \over 2}})\bigg|$$
$$= \bigg|{1 \over n + {1 \over 2}} - 0\bigg| + \bigg|0 -{1 \over n + {1 \over 2}}\bigg|$$
$$ = {2 \over n + {1 \over 2}}$$
