# Help proving that $\frac{2}{n+\frac{1}{2}} \leq \int_{\frac{1}{n+1}}^{\frac{1}{n}}\sqrt{1+(\sin(\frac{\pi}{t}) -\frac{1}{t}\cos(\frac{\pi}{t}))^2}dt$

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.

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This doesn't seem to be true for $n = 1$, where LHS is $4/3$ and RHS is about $0.7277$. – Cocopuffs Jul 11 '12 at 20:22
Ok actually it is for $n\geq 1$ so I'm not sure what to think then, what did you use to check the integral? – Thoth Jul 11 '12 at 20:28
It doesn't work for $n = 2$ either, unless I made an error entering this in to WolframAlpha. You might want to double check that. – Cocopuffs Jul 11 '12 at 20:31
I'm just getting a series expansion, how do you get wolfram definite integrator to approximate it? – Thoth Jul 11 '12 at 20:35
I entered "integral of sqrt(1 + (sin(pi/t) - 1/t * cos(pi/t))^2 )dt for t=1/3 to 1/2" – Cocopuffs Jul 11 '12 at 20:37