Let $f(x)$ be defined by ${\displaystyle\int_x^{x+1}\sin(e^t)dt}$
Prove that $\left|e^xf(x)\right|\leq2$
What I tried doing was that substitute $e^{t} = z$ which converts the integrand to $\dfrac{\sin(z)}{z}$ . I thought of applying some inequality on this but I don't know how to proceed now.
Please help me with this problem.
$\displaystyle|f(x)|=\left|\int_x^{x+1}\sin(e^t)dt\right|=\left|\int_{e^x}^{e^{x+1}}\frac{\sin(z)}{z}dz\right|\leq e^{-x}\left|\int_{e^x}^{e^{x+1}}\sin(z)dz\right|$, as $z\geq e^x$.
As $\displaystyle\left|\int_{e^x}^{e^{x+1}}\sin(z)dz\right|=\left|\cos({e^{x+1}})-\cos e^x\right|\leq2$, we are done.