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Show that $$\left|\int_{1}^{\sqrt{3}}\frac{e^{-x}\sin{x}}{x^2+1}dx\right|\leq \frac{\pi}{12e}$$
I tried it by first doing: $$\left|\int_{1}^{\sqrt{3}}\frac{e^{-x}\sin{x}}{x^2+1}dx\right|\leq \int_{1}^{\sqrt{3}}\left|\frac{e^{-x}\sin{x}}{x^2+1}\right|dx\leq \int_{1}^{\sqrt{3}}\left|\frac{e^{-x}}{x^2}\right|dx$$ But this bound is bigger than the one required.
Then I tried bounding it by using an upper Darboux sum, but it gets too messy and finding the right partition is a challenge.
Since $|e^{-x}\sin x| \le e^{-1}$ on $[1,\sqrt{3}]$,
$$\left|\int_1^{\sqrt{3}} \frac{e^{-x}\sin x}{x^2 + 1}\, dx\right| \le \int_1^{\sqrt{3}} \frac{e^{-1}}{x^2 + 1}\, dx = e^{-1}(\arctan(\sqrt{3}) - \arctan(1))$$$$ = e^{-1}(\frac{\pi}{3} - \frac{\pi}{4}) = \frac{\pi}{12e}.$$
Hint: $e^{-x}\cdot \sin x \leq e^{-1}\cdot 1 = e^{-1} \Rightarrow I \leq \dfrac{1}{e}\cdot \displaystyle \int_{1}^{\sqrt{3}} \dfrac{1}{x^2+1}dx = .....$
