I am trying to prove that for $n\geq 1$:
$\frac{2}{n+\frac{1}{2}} \leq \int_{\frac{1}{n+1}}^{\frac{1}{n}}\sqrt{1+\sin^2(\frac{\pi}{t}) - \frac{2}{t}\sin(\frac{\pi}{t})\cos(\frac{\pi}{t}) + \frac{1}{t^2}\cos^2(\frac{\pi}{t})}dt$
So far basically what I've done is pulled a $\frac{1}{t^2}$ out side the radical then got rid of the $t^2$ inside the radical and multiplied the $\cos^2(\frac{\pi}{t})$ by a $t^2$ which gives me something under the radical of the form $x^2 + 2xy + y^2$ which I then factor into the form $(x+y)^2$ and then take the square root and cancel the $\frac{t}{t}$ to obtain the integral:
$\int_{\frac{1}{n+1}}^{\frac{1}{n}} |\sin(\frac{\pi}{t}) + \cos(\frac{\pi}{t})|dt$
From here I do the substitution $u = \frac{1}{t}$ and $dt = \frac{-du}{u^2}$ to obtain:
$\int_n^{n+1} \frac{|\sin(\pi u) + \cos(\pi u)|}{u^2}du$
At this point I'm a bit worried I may have reduced the integral too much already. But if I haven't, then I probably need to perform some sort of averaging trick, anyways this is where I'm stuck. Can anyone help me with this? Thanks.
Edit: Ok I just realized I should never have un-factored the square in the title and should have just gotten rid of the $1$ under the radical and then taken the square root. This may make it a lot easier.. Note that $\int \sin(\frac{\pi}{t}) -\frac{1}{t}\cos(\frac{\pi}{t})dt = t\sin(\frac{\pi}{t})$.
Thus if we can figure out for what portion of the integral $\sin(\frac{\pi}{t}) + t\cos(\frac{\pi}{t})$ is positive and for what portion it's negative then we can just split the integral, multiply the latter by $-1$, and then integrate and combine, and that should hopefully give us what we want.