# Prove: $\frac{2}{3\pi} \leq \int_{2\pi}^{3\pi}\frac{\sin x}{x}\;dx \leq \frac{1}{\pi}$

I'd like your help proving that $$\frac{2}{3\pi} \leq \int_{2\pi}^{3\pi}\frac{\sin x}{x} \; dx \leq \frac{1}{\pi}.$$

I tried to bound both inequalities with Taylor series, other positive functions, but without any success.

Thanks a lot!

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Hint: $\sin x$ is non-negative and $\frac 1{3\pi}\leq \frac 1x\frac 1{2\pi}$. –  Davide Giraudo Feb 11 '12 at 17:09

Note that, on $[2\pi, 3\pi]$, we have $$\frac{\sin x}{3\pi} \leq \frac{\sin x}{x} \leq \frac{\sin x}{2\pi}.$$ Now integrate both sides over $[2\pi, 3\pi]$.
HINT: Because $$\int_{k\pi}^{(k+1)\pi}|\sin(t)|\;\mathrm{d}t=2\tag{1}$$ we have $$\frac{2}{(k+1)\pi}\le\int_{k\pi}^{(k+1)\pi}\left|\frac{\sin(t)}{t}\right|\;\mathrm{d}t\le\frac{2}{k\pi}\tag{2}$$