For f(x)=int_x^(x+1) sin(e^t)dt. Prove that: e^x|f(x)|<=2

For $f\left(x\right)={\int }_{x}^{x+1}\mathrm{sin}\left({\text{e}}^{t}\right)\text{d}t$
Prove that : ${\text{e}}^{x}|f\left(x\right)|\le 2$
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zupa1z
$f\left(x\right)={\int }_{x}^{x+1}\mathrm{sin}\left({\text{e}}^{t}\right)\text{d}t\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{e}^{x}f\left(x\right)={\int }_{x}^{x+1}{e}^{x}\mathrm{sin}\left({\text{e}}^{t}\right)\text{d}t\le {\int }_{x}^{x+1}{e}^{t}\mathrm{sin}\left({\text{e}}^{t}\right)\text{d}t$
Now substitute $z={e}^{t}$ which gives
${e}^{x}f\left(x\right)\le {\int }_{{e}^{x}}^{{e}^{x+1}}\mathrm{sin}\left(\text{z}\right)\text{d}z=\mathrm{cos}\left({e}^{x}\right)-\mathrm{cos}\left({e}^{x+1}\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{e}^{x}|f\left(x\right)|\le |\mathrm{cos}\left({e}^{x}\right)|+|\mathrm{cos}\left({e}^{x+1}\right)|\le 2$