# elementary Inequality.

Prove: $$\Big|\int_1^\sqrt{3} \! \frac{\sin(x)}{e^x(x^2 +1)} \, \mathrm{d} x \Big| \leq \frac{\pi}{12e}$$

Approach: we know:

$$\Big|\int_1^\sqrt{3} \! \frac{\sin(x)}{e^x(x^2 +1)} \, \mathrm{d} x \Big| \leq \int_1^\sqrt{3} \! \frac{|\sin(x)|}{e^x(x^2 +1)} \, \mathrm{d} x \leq \int_1^\sqrt{3} \! \frac{1}{e^x(x^2 +1)} \, \mathrm{d} x$$ Since $|\sin(x)|$ is bounded by 1 for every $x$. Can we keep estimating this integral, or should we just solve the last integral?

thanks.

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## 1 Answer

Use the fact $1<x<\sqrt{3} \implies e<e^x<e^{\sqrt{3}}$ and $|sin(x)|\leq 1$

$$\Big|\int_1^\sqrt{3} \! \frac{\sin(x)}{e^x(x^2 +1)} \, \mathrm{d} x \Big|\leq \int_1^\sqrt{3} \! \frac{1}{e(x^2 +1)} \, \mathrm{d} x \leq \frac{\pi}{12e}$$

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why? can you explain? –  Chasky Oct 16 '12 at 3:13
@LJym89: Can you see it now? –  Mhenni Benghorbal Oct 16 '12 at 3:15
GOT IT! thanks a lot –  Chasky Oct 16 '12 at 3:16
@LJym89:You are welcome. –  Mhenni Benghorbal Oct 16 '12 at 3:16