Integral of $\sin(e^t)dt$

An integration by parts that gives the first result will also produce the second. Let $u=e^{-t}$, $dv=e^t\sin(e^t)$. Then $du=-e^{-t}dt$, and we can take $v=-\cos(e^t)$. We are left with $$\int_x^{x+1} e^{-t}\cos(e^t)\,dt.$$ Since in our interval we have $e^{-t}\le e^{-x}$, and $|\cos(e^t)|\le 1$, the desired result follows, with $C=1$.

$$\left|\int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2}du\right|=\left|\int_{x}^{x+1} e^{-t} \cos( e^{t})dt\right| \leq \int_{x}^{x+1} e^{-t} dt = -e^t|_{t=x}^{t=x+1} = \left(1-\frac{1}{e}\right)e^{-x}. $$

I think you can see now your constant $C$.